The d - and f -Block Elements - Class 12 Chemistry - Chapter 4 - Notes, NCERT Solutions & Extra Questions
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Notes - The d - and f -Block Elements | Class 12 NCERT | Chemistry
The d- and f-Block Elements Class 12 Notes: Comprehensive Guide for Students
Introduction
Studying the d- and f-block elements is critical in Class 12 Chemistry as these elements exhibit unique properties that find applications in various scientific and industrial fields. This guide will help you understand the characteristics, electronic configurations, and significance of these elements.
Understanding the d-Block Elements
Definition and Location
The d-block elements, also known as transition metals, are positioned in the central part of the periodic table, spanning groups 3 to 12. These elements have their d-orbitals progressively filled.
General Properties
d-Block elements exhibit:
- High tensile strength
- Ductility
- Malleability
- High thermal and electrical conductivity
- Metallic lustre
Electronic Configuration
The electronic configuration of d-block elements follows the pattern of filling the (n-1)d orbitals. For instance, the general configuration is (n-1)d¹-¹⁰ ns¹-², with several exceptions due to stability considerations of half-filled and fully-filled orbitals.
Series of Transition Metals
The 3d Series
This series includes elements from scandium (Sc) to zinc (Zn). They are notable for their applications in construction, manufacturing, and as catalysts.
The 4d Series
This series ranges from yttrium (Y) to cadmium (Cd) and exhibits similar properties to the 3d series but often includes elements with higher melting points and densities.
The 5d Series
From lanthanum (La) and hafnium (Hf) to mercury (Hg), these elements have significant industrial applications, including roles in electronics and materials science.
Understanding the f-Block Elements
Definition and Location
f-Block elements, known as inner transition metals, are placed at the bottom of the periodic table and include the lanthanoids and actinoids.
General Properties
f-Block elements are characterized by the filling of 4f and 5f orbitals. They often exhibit variable oxidation states and form coloured compounds.
Lanthanoids vs Actinoids
Lanthanoids Series
Lanthanoids (La to Lu) often display a similar chemistry characterized by +3 oxidation states. The lanthanoid contraction is a notable phenomenon wherein atomic and ionic sizes decrease across the series.
Actinoids Series
Actinoids (Th to Lr) have a wider range of oxidation states compared to lanthanoids. These elements often exhibit radioactivity, adding complexity to their study.
Key Characteristics of d- and f-Block Elements
Oxidation States
d-Block elements display multiple oxidation states, which allow them to form various compounds. For example, manganese exhibits oxidation states ranging from +2 to +7.
Formation of Coloured Compounds
The presence of unpaired electrons in d-orbitals leads to the absorption of specific light wavelengths, resulting in coloured compounds.
Catalytic Properties
Transition metals are widely used as catalysts due to their ability to change oxidation states. For instance, vanadium(V) oxide catalyses the oxidation of sulfur dioxide in the Contact Process.
Industrial and Practical Applications
The d- and f-block elements find extensive applications, such as:
- Construction: Iron and steel
- Electronics: Copper and gold
- Catalysis: Nickel in hydrogenation
- Pigments: Titanium dioxide (TiO₂) in paints
- Energy: Uranium in nuclear reactors
Conclusion
Understanding the properties and applications of d- and f-block elements is crucial for students studying Class 12 Chemistry. These elements are integral to various industrial processes, contributing significantly to technological advancements.
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Write down the electronic configuration of:
(i) $\mathrm{Cr}^{3+}$
(iii) $\mathrm{Cu}^{+}$
(v) $\mathrm{Co}^{2}+$
(vii) $\mathrm{Mn}^{2+}$
(ii) $\mathrm{Pm}^{3+}$
(iv) $\mathrm{Ce}^{4+}$
(vi) $\mathrm{Lu}^{2+}$
(viii) $\mathrm{Th}^{4+}$
Here are the electronic configurations:
First Transition Series
Cr³⁺:
Electronic configuration: $[\text{Ar}]3d^3$
Cu⁺:
Electronic configuration: $[\text{Ar}]3d^{10}$
Co²⁺:
Electronic configuration: $[\text{Ar}]3d^7$
Mn²⁺:
Electronic configuration: $[\text{Ar}]3d^5$
Lanthanides and Actinides
Pm³⁺:
Electronic configuration: $[\text{Xe}]4f^4$
Ce⁴⁺:
Electronic configuration: $[\text{Xe}]$
Lu²⁺:
Electronic configuration: $[\text{Xe}]4f^{14}5d^1$
Th⁴⁺:
Electronic configuration: $[\text{Rn}]$
These configurations highlight the removal of electrons depending on the oxidation state and the filling of the orbitals according to the specific rules of electron configuration for each element.
Why are $\mathrm{Mn}^{2+}$ compounds more stable than $\mathrm{Fe}^{2+}$ towards oxidation to their +3 state?
$\text{Mn}(^{2+})$ compounds are more stable than Fe(^{2+}) towards oxidation to their +3 state because of their electronic configurations.
Manganese (Mn)(^{2+}) has the electronic configuration $3d^5$, which represents a half-filled d-subshell. This configuration is particularly stable due to the maximization of exchange energy and symmetric distribution of electrons.
Iron (Fe)(^{2+}), on the other hand, has a configuration of $3d^6$. When oxidized to Fe(^{3+}), it forms a $3d^5$ configuration, which is relatively more stable due to similar reasons described above. Thus, Fe(^{2+}) has a tendency to get oxidized to Fe(^{3+}) to achieve this stable half-filled configuration.
In summary, manganese's half-filled d-subshell in the +2 state offers extra stability against oxidation compared to iron.
Explain briefly how +2 state becomes more and more stable in the first half of the first row transition elements with increasing atomic number?
In the first half of the first row transition elements, the +2 oxidation state becomes increasingly stable with rising atomic number due to the following reasons:
Increasing Nuclear Charge: As you move from scandium (Sc) to manganese (Mn), the nuclear charge increases. This greater nuclear charge enhances the attraction between the nucleus and the electrons, making it more difficult to remove electrons, which stabilizes the +2 state.
Energy Considerations: The increasing nuclear charge also leads to a small but significant difference in the energy levels of the 3d and 4s orbitals. For elements from scandium to manganese, the removal of electrons from the 4s orbital (to achieve the +2 state) is energetically more favorable because the 3d subshell is getting filled progressively, making these electrons harder to remove.
Stable Half-Filled Configurations: Some elements attain additional stability due to their electronic configurations. For example, manganese has a [Ar] (3d^5 4s^2) configuration; when it loses two electrons to form Mn²⁺, it achieves the stable half-filled (3d^5) configuration.
Thus, the combination of increasing nuclear charge, favorable energy considerations, and more stable electronic configurations collectively make the +2 oxidation state more stable as we move across the first half of the first row transition elements.
To what extent do the electronic configurations decide the stability of oxidation states in the first series of the transition elements? Illustrate your answer with examples.
The electronic configurations significantly influence the stability of oxidation states in the first series of transition elements. This can be illustrated by the concepts of half-filled and fully-filled $d$-orbitals which provide additional stability to certain oxidation states. Below are some examples and explanations:
1. Chromium (Cr)
Electronic Configuration: $ [Ar] 3d^5 4s^1$
Stable Oxidation States: +3, +6
Stability Explanation:
Cr(III): The $3d^3$ configuration in Cr(III) is moderately stable.
Cr(VI): Exhibits a $3d^0$ configuration. The absence of $d$ electrons contributes to stability in compounds like $CrO_4^{2-}$ and $Cr_2O_7^{2-}$.
2. Manganese (Mn)
Electronic Configuration: $ [Ar] 3d^5 4s^2$
Stable Oxidation States: +2, +7
Stability Explanation:
Mn(II): The $3d^5$ configuration (half-filled $d$-subshell) provides extra stability.
Mn(VII): The +7 oxidation state, seen in permanganate ion $(MnO_4^-)$, is stabilized by the absence of $d$ electrons ($3d^0$).
3. Iron (Fe)
Electronic Configuration: $ [Ar] 3d^6 4s^2$
Stable Oxidation States: +2, +3
Stability Explanation:
Fe(II): Has a $3d^6$ configuration, which experiences less stability compared to Fe(III) due to lack of fully or half-filled $d$-orbital.
Fe(III): The $3d^5$ configuration (half-filled $d$-subshell) provides high stability.
4. Copper (Cu)
Electronic Configuration: $ [Ar] 3d^{10} 4s^1$
Stable Oxidation States: +1, +2
Stability Explanation:
Cu(I): The $3d^{10}$ configuration is very stable due to fully filled $d$-orbitals.
Cu(II): Despite being less stable relative to Cu(I), Cu(II) is commonly observed due to the pairing energy of electrons which can overcome its configuration $3d^9$.
Conclusion
The stability of oxidation states in the first series transition elements is greatly determined by the electronic configuration, primarily involving:
Half-filled $d^5$ and fully-filled $d^{10}$ configurations provide extra stability.
Example: Cr shows stability in both +3 and +6 states, whereas Mn shows stability in +2 and +7 states, demonstrating the principal role of electronic configuration.
What may be the stable oxidation state of the transition element with the following $d$ electron configurations in the ground state of their atoms : $3 d^{3}$, $3 d^{5}, 3 d^{8}$ and $3 d^{4}$ ?
To determine the stable oxidation state of transition elements based on their $3d$ electron configurations in their ground state, we need to consider the electronic structure and the tendency to achieve a stable electron configuration.
$3d^3$ Configuration:
Example Element: Vanadium (V), which has a ground-state configuration of $[Ar] 3d^3 4s^2$.
Stable Oxidation States: +2, +3, +4, +5.
Most Stable State: +3 (achieving $3d^2$ after losing two $4s$ electrons).
$3d^5$ Configuration:
Example Element: Manganese (Mn), which has a ground-state configuration of $[Ar] 3d^5 4s^2$.
Stable Oxidation States: +2, +3, +4, +6, +7.
Most Stable State: +2 (achieving $3d^5$, where d-orbitals are half-filled, which is particularly stable).
$3d^8$ Configuration:
Example Element: Nickel (Ni), which has a ground-state configuration of $[Ar] 3d^8 4s^2$.
Stable Oxidation States: +2, +3.
Most Stable State: +2 (achieving $3d^8$ configuration remains stable after losing two $4s$ electrons).
$3d^4$ Configuration:
Example Element: Chromium (Cr), which shifts to $[Ar] 3d^5 4s^1$ to achieve a stable half-filled $3d$ orbital.
Stable Oxidation States: +2, +3, +6.
Most Stable State: +3 (achieving half-filled d-orbital which is stable).
To summarize:
$3d^3$: Most stable state is +3.
$3d^5$: Most stable state is +2.
$3d^8$: Most stable state is +2.
$3d^4$: Most stable state is +3.
For further calculations or details, please let me know.
Name the oxometal anions of the first series of the transition metals in which the metal exhibits the oxidation state equal to its group number.
In the first series of transition metals, the metals exhibiting oxometal anions with oxidation states equal to their group numbers are:
Vanadium (Group 5): The oxometal anion is [VO_4]^{3-}
Chromium (Group 6): The oxometal anion is [CrO_4]^{2-}
Manganese (Group 7): The oxometal anion is [MnO_4]^{-}
These oxometal anions have the metals in the respective oxidation states +5, +6, and +7, which correspond to their group numbers.
What is lanthanoid contraction? What are the consequences of lanthanoid contraction?
Lanthanoid contraction refers to the gradual decrease in atomic and ionic radii of the lanthanoids (from Lanthanum, La, to Lutetium, Lu) as the atomic number increases. This contraction is primarily due to the poor shielding effect of the 4f electrons, which are added progressively across the series.
Consequences of Lanthanoid Contraction:
Similar Radii of 2nd and 3rd Series Transition Elements:
The radii of transition elements of the third series (e.g., Hafnium, Hf) become very similar to those of the corresponding elements in the second series (e.g., Zirconium, Zr). This makes them chemically similar and difficult to separate.
Higher Density and Increased Hardness:
The decreased size, coupled with an increase in atomic mass, leads to increasing density and hardness through the series.
Enhanced Basicity:
The lanthanoid contraction affects the basicity of oxides and hydroxides. Lanthanoid oxides and hydroxides become less basic from La(OH)₃ to Lu(OH)₃.
Formation of Stronger Complexes:
Smaller ionic radii after lanthanoid contraction leads to lanthanoids forming stable and stronger complexes.
Behaviour in Chemical Reactions:
Lanthanides exhibit very similar chemical properties due to lanthanoid contraction, often complicating their separation during extraction from ores.
The above consequences have significant implications in both inorganic chemistry and application areas involving these elements.
What are the characteristics of the transition elements and why are they called transition elements? Which of the $d$-block elements may not be regarded as the transition elements?
Characteristics of Transition Elements
Variable Oxidation States: Transition elements can exhibit a variety of oxidation states, differing by one (e.g., ( \text{Fe}^{2+} ) and ( \text{Fe}^{3+} )).
Formation of Coloured Ions: The compounds of transition elements are often coloured. This is primarily due to the d-d electronic transitions.
Paramagnetic Properties: Most transition elements are paramagnetic due to the presence of unpaired d-electrons.
Catalytic Activity: Many transition metals and their compounds act as catalysts in various chemical reactions because of their ability to change oxidation states.
Formation of Complex Compounds: They tend to form complex compounds with different ligands, exhibiting a wide range of coordination numbers and geometries.
High Melting and Boiling Points: These metals generally have high melting and boiling points due to strong interatomic metallic bonding, facilitated by the delocalisation of (n-1)d electrons.
Tensile Strength, Hardness, and Density: Transition metals typically possess high tensile strength, hardness, and density.
Why are they called Transition Elements?
Transition elements are called so because their properties are transitional between the most electropositive elements (the s-block) and the least electropositive elements (the p-block). According to the IUPAC, transition elements are defined as elements that have an incomplete d subshell either in their neutral atom or in any of their common oxidation states.
(d)-block Elements That Are Not Regarded as Transition Elements
The elements Zinc (Zn), Cadmium (Cd), and Mercury (Hg) in Group 12 of the ( d )-block are not regarded as transition elements because they possess a completely filled (d^{10}) subshell in the ground state as well as in their common oxidation states.
These elements do not show the typical properties of transition elements such as variable oxidation states and formation of coloured ions.
In what way is the electronic configuration of the transition elements different from that of the non transition elements?
The electronic configuration of transition elements (d-block elements) is different from that of non-transition elements (s- and p-block elements) primarily in the filling of the d-orbitals. Here are the key differences:
d-Block Elements (Transition Elements):
The transition elements are characterized by the progressive filling of the (n-1)d orbitals.
Their general electronic configuration is given as $ (n-1)d^{1-10} , ns^{1-2} $.
For example, the electronic configuration of $\mathrm{Cr}$ (Chromium) is $ 3d^{5} 4s^{1} $ rather than $ 3d^{4} 4s^{2} $ due to the extra stability of a half-filled d subshell.
Non-Transition Elements:
In the s-block elements, only the s orbitals are filled progressively. Their general configuration is $ ns^{1-2} $.
In the p-block elements, both the s and p orbitals are filled progressively. Their general configuration is $ ns^{2}np^{1-6} $.
For example, the electronic configuration of $\mathrm{Al}$ (Aluminum) is $ 3s^{2} 3p^{1} $.
The transition elements also exhibit some unique properties due to their electronic configurations:
They have partly filled d-orbitals.
They show variable oxidation states.
They form colored compounds.
They exhibit paramagnetism due to unpaired electrons.
They often form complex compounds with ligands.
In contrast, the non-transition elements generally do not exhibit these properties because they do not have partly filled d-orbitals.
What are the different oxidation states exhibited by the lanthanoids?
The lanthanoids primarily exhibit the oxidation state of +3. However, some lanthanoids also exhibit other oxidation states occasionally due to the extra stability of empty, half-filled, or fully filled (f) subshells. Here are the details:
+3: The most stable and common oxidation state for all lanthanoids.
+2: Observed in some lanthanoids like Europium (Eu) and Ytterbium (Yb), which have stable (f^7) and (f^{14}) configurations, respectively.
+4: Occurs in elements like Cerium (Ce), Praseodymium (Pr), Neodymium (Nd), Terbium (Tb), and Dysprosium (Dy), but typically only in oxides or other specific chemical environments.
The irregularity in these oxidation states primarily arises from the extra stability of certain (f) subshell configurations.
Explain giving reasons:
(i) Transition metals and many of their compounds show paramagnetic behaviour.
(ii) The enthalpies of atomisation of the transition metals are high.
(iii) The transition metals generally form coloured compounds.
(iv) Transition metals and their many compounds act as good catalyst.
(i) Transition metals and many of their compounds show paramagnetic behaviour.
Reason:
Presence of Unpaired Electrons: Transition metals have partially filled d-orbitals. The presence of one or more unpaired electrons in these orbitals gives rise to paramagnetism.
Magnetic Moment: The paramagnetic property arises due to the magnetic moment associated with these unpaired electrons.
(ii) The enthalpies of atomisation of the transition metals are high.
Reason:
Strong Metallic Bonds: Transition metals form strong metallic bonds due to the involvement of (n-1)d and ns electrons.
Greater Number of Valence Electrons: These bonds are formed by the delocalization of a large number of electrons, making the interatomic interactions very strong, thus resulting in high enthalpies of atomisation.
(iii) Transition metals generally form coloured compounds.
Reason:
d-d Transitions: The partially filled d-orbitals allow d-d electronic transitions. When light is absorbed, electrons in a lower energy d-orbital get excited to a higher energy d-orbital.
Complementary Color: The specific wavelength of light absorbed corresponds to the excitation energy, and the colour observed is the complementary colour of the light absorbed.
(iv) Transition metals and their many compounds act as good catalysts.
Reasons:
Variable Oxidation States: Transition metals can easily alternate between multiple oxidation states. This property enables them to facilitate various redox reactions.
Formation of Complex Intermediate Compounds: They can form complexes with reactants, thereby reducing the activation energy and providing an alternative pathway to the reaction.
Surface Area: The presence of partially filled d-orbitals also provides necessary active sites for the adsorption and interaction of reactants.
What are interstitial compounds? Why are such compounds well known for transition metals?
Interstitial compounds are compounds formed when small atoms like hydrogen (H), carbon (C), or nitrogen (N) are trapped inside the crystal lattices of metals. These compounds are typically non-stoichiometric and do not conform to traditional ionic or covalent bonding patterns. A few examples include $$\text{TiC}, \text{Mn}_4\text{N}, \text{Fe}_3\text{H}, \text{VH}_{0.56}, \text{TiH}_{1.7}$$, etc.
Characteristics of Interstitial Compounds:
High melting points: Generally higher than those of pure metals.
Hardness: They are very hard; some, like borides, approach diamond in hardness.
Metallic conductivity: They retain metallic conductivity.
Chemical inertness: They are generally chemically inert.
Reasons Transition Metals Form Interstitial Compounds:
Size compatibility: The small atomic radii of the involved nonmetal atoms (H, C, N) allow them to fit into the interstices (spaces) of the larger metallic crystal lattices of transition metals without significantly disturbing the lattice structure.
Variable oxidation states: Transition metals can accommodate various oxidation states, which assists in the formation of stable interstitial compounds.
Partly filled d-orbitals: The unique electronic configurations involving incompletely filled ( d )-orbitals in transition metals facilitate the bonding and stability of these interstitial compounds.
These features are combined because the lattice structure of transition metals is flexible enough to incorporate smaller atoms into the interstices, resulting in enhanced physical properties and making them significantly important in industrial applications.
How is the variability in oxidation states of transition metals different from that of the non transition metals? Illustrate with examples.
The variability in oxidation states of transition metals differs significantly from that of non-transition metals due to the involvement of their incompletely filled $d$ orbitals. Here are the main differences illustrated with examples:
Transition Metals:
Multiple Oxidation States: Transition metals exhibit multiple oxidation states because they can lose different numbers of $d$ and $s$ electrons.
Example: Manganese (Mn) exhibits oxidation states from +2 to +7. For instance:
$$\text{Mn}^{2+}$$
$$\text{Mn}^{4+}$$
$$\text{Mn}^{7+}$$
Stability of Oxidation States: The stability of various oxidation states is influenced by the specific electron configurations and the particular $d$ orbital occupations.
Example:
Chromium (Cr): Can exist in +2, +3, and +6 states.
Iron (Fe): Commonly found in +2 and +3 states.
Non-Transition Metals:
Fewer Oxidation States: Non-transition metals generally exhibit fewer oxidation states, often because they involve the loss of valence electrons and do not have $d$ orbitals available to contribute.
Example:
Aluminum (Al): Typically shows an oxidation state of +3.
Magnesium (Mg): Usually found in the +2 state.
Predictable Oxidation States: The oxidation states of non-transition metals are more predictable and often follow a regular pattern based on group valence electrons.
Example:
Sodium (Na): Typically shows +1 due to the loss of one $s$ electron.
Key Differences:
Number of Possible States: Transition metals have several oxidation states due to varied $d$ electron involvement, while non-transition metals have fewer due to lack of $d$ orbitals.
Variability and Stability: Transition metals exhibit more variability and complexity in the stability of different oxidation states as compared to non-transition metals.
Illustrative Example:
Transition Metal: Cobalt (Co) can exhibit +2 and +3 oxidation states. When bonded with ligands, the electron configuration can stabilize different states.
Non-Transition Metal: Carbon (C) typically shows +4 and -4 oxidation states in CO$_2$ and CH$_4$ respectively, but not due to d-electron configurations.
Thus, the chemical behavior of transition metals is characterized by multiple variable oxidation states due to their electronic configurations and the partial filling of $d$ orbitals, unlike the more predictable behavior of non-transition metals.
Describe the preparation of potassium dichromate from iron chromite ore. What is the effect of increasing $\mathrm{pH}$ on a solution of potassium dichromate?
Preparation of Potassium Dichromate from Iron Chromite Ore
The preparation of potassium dichromate (K₂Cr₂O₇) involves the following steps:
Fusion of Chromite Ore with Alkali:Chromite ore (FeCr₂O₄) is fused with sodium or potassium carbonate (Na₂CO₃ or K₂CO₃) in the presence of oxygen (air). During this process, the chromium in the ore is oxidized to hexavalent chromium.
$$ 4 \text{FeCr}_2\text{O}_4 + 8 \text{Na}_2\text{CO}_3 + 7 \text{O}_2 \rightarrow 8 \text{Na}_2\text{CrO}_4 + 2 \text{Fe}_2\text{O}_3 + 8 \text{CO}_2 $$
Conversion to Dichromate:The sodium chromate (Na₂CrO₄) formed is solubilized in water and then acidified with sulfuric acid (H₂SO₄) to convert it into sodium dichromate (Na₂Cr₂O₇).
$$ 2 \text{Na}_2\text{CrO}_4 + 2 \text{H}^+ \rightarrow \text{Na}_2\text{Cr}_2\text{O}_7 + 2 \text{Na}^+ + \text{H}_2\text{O} $$
Formation of Potassium Dichromate:Potassium chloride (KCl) is then added to the sodium dichromate solution to precipitate potassium dichromate.
$$ \text{Na}_2\text{Cr}_2\text{O}_7 + 2 \text{KCl} \rightarrow \text{K}_2\text{Cr}_2\text{O}_7 + 2 \text{NaCl} $$
Potassium dichromate (K₂Cr₂O₇) crystallizes out from the solution upon cooling.
Effect of Increasing $\mathrm{pH}$ on a Solution of Potassium Dichromate
When the pH of a potassium dichromate solution is increased, the dichromate ions (Cr₂O₇²⁻) are converted to chromate ions (CrO₄²⁻). The equilibrium between chromate and dichromate ions depends on the pH of the solution:
$$ 2 \mathrm{CrO}_{4}^{2-} + 2 \mathrm{H}^+ \leftrightarrow \mathrm{Cr}_{2}\mathrm{O}_{7}^{2-} + \mathrm{H}_{2}\mathrm{O} $$
Acidic conditions (low pH): The equilibrium shifts to the right, favoring the formation of dichromate ions (Cr₂O₇²⁻).
Basic conditions (high pH): The equilibrium shifts to the left, favoring the formation of chromate ions (CrO₄²⁻).
In summary, increasing the pH of a potassium dichromate solution results in the conversion of dichromate ions to chromate ions.
Describe the oxidising action of potassium dichromate and write the ionic equations for its reaction with:
(i) iodide
(ii) iron(II) solution and
(iii) $\mathrm{H}_{2} \mathrm{~S}$
115 The $d$ - and $f$ - Block Elements
Potassium dichromate (( K_2Cr_2O_7 )) is a strong oxidising agent in acidic solution. Its oxidising power can be represented by the following half-reaction:
[ \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-} + 14 \mathrm{H}^+ + 6 \mathrm{e}^{-} \rightarrow 2 \mathrm{Cr}^{3+} + 7 \mathrm{H}_2O \quad (E^{\circ} = 1.33 \mathrm{~V}) ]
Reactions:
(i) With Iodide:
Potassium dichromate oxidizes iodide (( I^- )) to iodine (( I_2 )). The full ionic equation for this reaction is:
[ \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-} + 14 \mathrm{H}^+ + 6 \mathrm{I}^- \rightarrow 2 \mathrm{Cr}^{3+} + 3 \mathrm{I}_2 + 7 \mathrm{H}_2O ]
(ii) With Iron(II) Solution:
Potassium dichromate oxidizes iron(II) ions (( Fe^{2+} )) to iron(III) ions (( Fe^{3+} )). The full ionic equation for this reaction is:
[ \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-} + 14 \mathrm{H}^+ + 6 \mathrm{Fe}^{2+} \rightarrow 2 \mathrm{Cr}^{3+} + 6 \mathrm{Fe}^{3+} + 7 \mathrm{H}_2O ]
(iii) With Hydrogen Sulfide:
Potassium dichromate oxidizes hydrogen sulfide (( H_2S )) to sulfur (( S )). The full ionic equation for this reaction is:
[ \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-} + 14 \mathrm{H}^+ + 3 \mathrm{H}_{2} \mathrm{S} \rightarrow 2 \mathrm{Cr}^{3+} + 3 \mathrm{S} + 7 \mathrm{H}_2O ]
These reactions illustrate the oxidising power of potassium dichromate, making it a versatile oxidizing agent in various chemical processes.
Describe the preparation of potassium permanganate. How does the acidified permanganate solution react with (i) iron(II) ions (ii) $\mathrm{SO}_{2}$ and (iii) oxalic acid? Write the ionic equations for the reactions.
Preparation of Potassium Permanganate
Potassium permanganate (KMnO₄) is prepared by fusion of manganese(IV) oxide (MnO₂) with potassium hydroxide (KOH) in the presence of an oxidizing agent such as potassium nitrate (KNO₃). This process yields potassium manganate (K₂MnO₄), which is then converted to potassium permanganate by a process called disproportionation in neutral or acidic solution.
Steps:
Fusion of manganese(IV) oxide with KOH and KNO₃: $$ 2 \mathrm{MnO}_{2} + 4 \mathrm{KOH} + \mathrm{O}_{2} \rightarrow 2 \mathrm{K}_{2} \mathrm{MnO}_{4} + 2 \mathrm{H}_{2} \mathrm{O} $$
Disproportionation of potassium manganate in neutral or acidic solution: $$ 3 \mathrm{K}_{2} \mathrm{MnO}_{4} + 4 \mathrm{H}^{+} \rightarrow 2 \mathrm{KMnO}_{4} + \mathrm{MnO}_{2} + 2 \mathrm{H}_{2} \mathrm{O} $$
Reactions of Acidified Permanganate Solution
(i) Reaction with Iron(II) Ions
Iron(II) ions (Fe²⁺) are oxidized to iron(III) ions (Fe³⁺) by permanganate in acidic medium.
Ionic Equation:$$ 5 \mathrm{Fe}^{2+} + \mathrm{MnO}_{4}^{-} + 8 \mathrm{H}^{+} \rightarrow 5 \mathrm{Fe}^{3+} + \mathrm{Mn}^{2+} + 4 \mathrm{H}_{2} \mathrm{O} $$
(ii) Reaction with Sulfur Dioxide
Sulfur dioxide (SO₂) is oxidized to sulfate ions (SO₄²⁻) by permanganate in acidic medium.
Ionic Equation:$$ 5 \mathrm{SO}_{2} + 2 \mathrm{MnO}_{4}^{-} + 2 \mathrm{H}_{2} \mathrm{O} \rightarrow 5 \mathrm{SO}_{4}^{2-} + 2 \mathrm{Mn}^{2+} + 4 \mathrm{H}^{+} $$
(iii) Reaction with Oxalic Acid
Oxalic acid (C₂H₂O₄) is oxidized to carbon dioxide (CO₂) by permanganate in acidic medium.
Ionic Equation:$$ 5 \mathrm{C}_{2} \mathrm{O}_{4}^{2-} + 2 \mathrm{MnO}_{4}^{-} + 16 \mathrm{H}^{+} \rightarrow 10 \mathrm{CO}_{2} + 2 \mathrm{Mn}^{2+} + 8 \mathrm{H}_{2} \mathrm{O} $$
For $\mathrm{M}^{2+} / \mathrm{M}$ and $\mathrm{M}^{3+} / \mathrm{M}^{2+}$ systems the $E^{\circ}$ values for some metals are as follows:
$\mathrm{Cr}^{2+} / \mathrm{Cr}$ | $-0.9 \mathrm{~V}$ |
$\mathrm{Cr}^{3} / \mathrm{Cr}^{2+}$ | $-0.4 \mathrm{~V}$ |
$\mathrm{Mn}^{2+} / \mathrm{Mn}$ | $-1.2 \mathrm{~V}$ |
$\mathrm{Mn}^{3+} / \mathrm{Mn}^{2+}$ | $+1.5 \mathrm{~V}$ |
$\mathrm{Fe}^{2+} / \mathrm{Fe}$ | $-0.4 \mathrm{~V}$ |
$\mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}$ | $+0.8 \mathrm{~V}$ |
Use this data to comment upon:
(i) the stability of $\mathrm{Fe}^{3+}$ in acid solution as compared to that of $\mathrm{Cr}^{3+}$ or $\mathrm{Mn}^{3+}$ and
(ii) the ease with which iron can be oxidised as compared to a similar process for either chromium or manganese metal.
Stability of $\mathrm{Fe}^{3+}$, $\mathrm{Cr}^{3+}$, and $\mathrm{Mn}^{3+}$ in Acid Solution
Standard Electrode Potentials:
$\mathrm{Cr}^{3+} /\mathrm{Cr}^{2+}$: $-0.4 \mathrm{~V}$
$\mathrm{Mn}^{3+} /\mathrm{Mn}^{2+}$: $+1.5 \mathrm{~V}$
$\mathrm{Fe}^{3+} /\mathrm{Fe}^{2+}$: $+0.8 \mathrm{~V}$
Interpretation:
The positive $E^{\circ}$ value for $\mathrm{Mn}^{3+}/\mathrm{Mn}^{2+}$ (+1.5 V) indicates that $\mathrm{Mn}^{3+}$ is strongly oxidizing and not stable in acidic solution.
The $E^{\circ}$ value for $\mathrm{Fe}^{3+}/\mathrm{Fe}^{2+}$ is +0.8 V, indicating it is less strong an oxidizer compared to $\mathrm{Mn}^{3+}$ but still has a tendency to get reduced to $\mathrm{Fe}^{2+}$.
The $E^{\circ}$ value for $\mathrm{Cr}^{3+}/\mathrm{Cr}^{2+}$ is -0.4 V, indicating that $\mathrm{Cr}^{3+}$ is relatively stable in acidic solution, since it is negative and thus does not readily reduce to $\mathrm{Cr}^{2+}$.
Stability Order in Acid Solution: $$ \mathrm{Cr}^{3+} > \mathrm{Fe}^{3+} > \mathrm{Mn}^{3+} $$
Ease of Oxidation for Fe, Cr, and Mn Metals
Standard Electrode Potentials for Metal/Ion Pairs:
$\mathrm{Cr}^{2+} /\mathrm{Cr}$: $-0.9 \mathrm{~V}$
$\mathrm{Mn}^{2+}/ \mathrm{Mn}$: $-1.2 \mathrm{~V}$
$\mathrm{Fe}^{2+}/ \mathrm{Fe}$: $-0.4 \mathrm{~V}$
Interpretation:
The more negative the electrode potential, the more easily the metal is oxidized to its ion.
$\mathrm{Mn}$ shows the most negative potential ($-1.2 \mathrm{~V}$), indicating that manganese metal is the most easily oxidized to $\mathrm{Mn}^{2+}$.
Chromium metal has a less negative potential ($-0.9 \mathrm{~V}$), indicating it is less easily oxidized than manganese but more so than iron.
Iron has the least negative potential ($-0.4 \mathrm{~V}$), suggesting it is the least easily oxidized to $\mathrm{Fe}^{2+}$ among the three metals.
Ease of Oxidation Order: $$ \mathrm{Mn} > \mathrm{Cr} > \mathrm{Fe} $$
Predict which of the following will be coloured in aqueous solution? $\mathrm{Ti}^{3+}, \mathrm{V}^{3+}$, $\mathrm{Cu}^{+}, \mathrm{Sc}^{3+}, \mathrm{Mn}^{2+}, \mathrm{Fe}^{3+}$ and $\mathrm{Co}^{2+}$. Give reasons for each.
The color of a transition metal ion in an aqueous solution is usually due to $d$-$d$ transitions. When an electron from a lower energy $d$ orbital gets excited to a higher energy $d$ orbital, it absorbs a particular wavelength of visible light, resulting in the ion showing a color that is complementary to the wavelength absorbed.
Let's predict the colors of the given ions:
$\mathrm{Ti}^{3+}$:
Configuration: $3d^1$
Reason: Has one unpaired electron in the $d$ orbital, capable of $d$-$d$ transition.
Color Prediction: Colored
$\mathrm{V}^{3+}$:
Configuration: $3d^2$
Reason: Has two unpaired electrons in the $d$ orbital, allowing for $d$-$d$ transition.
Color Prediction: Colored
$\mathrm{Cu}^{+}$:
Configuration: $3d^{10}$
Reason: Has a completely filled $d$ orbital (no unpaired electrons), so $d$-$d$ transitions are not possible.
Color Prediction: Colorless
$\mathrm{Sc}^{3+}$:
Configuration: $\mathrm{3d^0}$
Reason: Has no $d$ electrons, making $d$-$d$ transitions impossible.
Color Prediction: Colorless
$\mathrm{Mn}^{2+}$:
Configuration: $3d^5$
Reason: Has five unpaired electrons in the $d$ orbital, allowing for multiple $d$-$d$ transitions.
Color Prediction: Colored
$\mathrm{Fe}^{3+}$:
Configuration: $3d^5$
Reason: Similar to $\mathrm{Mn}^{2+}$, has five unpaired electrons, making $d$-$d$ transitions possible.
Color Prediction: Colored
$\mathrm{Co}^{2+}$:
Configuration: $3d^7$
Reason: Has three unpaired electrons in the $d$ orbital, allowing for $d$-$d$ transitions.
Color Prediction: Colored
So, the ions that will be colored in aqueous solutions are $\mathrm{Ti}^{3+}$, $\mathrm{V}^{3+}$, $\mathrm{Mn}^{2+}$, $\mathrm{Fe}^{3+}$, and $\mathrm{Co}^{2+}$, while $\mathrm{Cu}^{+}$ and $\mathrm{Sc}^{3+}$ will be colorless.
Compare the stability of +2 oxidation state for the elements of the first transition series.
To compare the stability of the +2 oxidation state among the elements of the first transition series, we need to consider several factors such as their ionization enthalpies, hydration enthalpies, and their tendencies to form stable electronic configurations.
Let's break down these factors:
Ionization Enthalpies
The first and second ionization energies influence the formation of +2 oxidation states. Lower ionization energies generally facilitate the formation of +2 charged ions.
The values of the first and second ionization enthalpies (in kJ/mol) for the first transition series are:
Ti: 656, 1309
V: 650, 1414
Cr: 652, 1592
Mn: 717, 1509
Fe: 762, 1561
Co: 758, 1644
Ni: 736, 1752
Cu: 745, 1958
Zn: 906, 1734
Hydration Enthalpies
Hydration enthalpies (enthalpy change when an ion is hydrated) also play a crucial role. Higher negative hydration enthalpies increase the stabilization energy of the ion in an aqueous solution.
Electronic Configurations and Stability
The specific electron configurations determine the stability of the +2 oxidation state:
Stability is often observed in d^5 or d^10 configurations due to enhanced stability of filled or half-filled d-orbitals.
General Trends and Observations
Scandium (Sc) and Zinc (Zn): Neither of these elements readily forms a +2 oxidation state. Scandium prefers +3, and Zinc prefers +2 but is not considered a transition metal due to filled 3d^10 orbitals.
Titanium (Ti) to Copper (Cu): For many elements, there is notable stability in the +2 state. However, their relative stability varies:
Chromium (Cr): +2 oxidation state is less stable because Cr typically forms more stable +3 states.
Manganese (Mn): +2 oxidation state is quite stable, as Mn^2+ achieves a half-filled d^5 configuration (3d^5).
Iron (Fe): +2 oxidation state is common, although Fe^3+ is also quite stable due to the d^5 configuration in Fe^3+.
Nickel (Ni): +2 is the most stable oxidation state, as it achieves a stable 3d^8 configuration.
Copper (Cu): Cu^2+ is less stable than Cu+, which has a 3d^10 configuration.
Standard Electrode Potentials
Let's look at the standard electrode potentials (Eº) for the Mn^(2+)/Mn couple:
Ti: -1.63 V
V: -1.18 V
Cr: -0.90 V
Mn: -1.18 V
Fe: -0.44 V
Co: -0.28 V
Ni: -0.25 V
Cu: +0.34 V
Conclusion
Most Stable: Mn^2+ (d^5), Ni^2+ (d^8)
Less Stable: Ti^2+, V^2+, Cu^2+.
Standard Electrode Potentials: The standard electrode potential also supports the fact that Cu^2+ is less stable compared to Mn^2+ and Ni^2+.
Elements like Mn and Ni show high stability in the +2 oxidation state due to their half-filled (d^5) and relatively stable electronic configurations (d^8), respectively. Notice the complication with Cu, which favors a +1 oxidation state due to its filled d^10 configuration.
By considering the ionization enthalpies, hydration enthalpies, and electronic configurations, we can compare the stability of +2 oxidation states across the first transition series.
Compare the chemistry of actinoids with that of the lanthanoids with special reference to:
(i) electronic configuration
(iii) oxidation state
(ii) atomic and ionic sizes and
(iv) chemical reactivity.
(i) Electronic Configuration
Lanthanoids:
Generally, the electronic configuration of lanthanoids is $[Xe] 4f^{1–14} 5d^{0–1} 6s^{2}$.
The filling of 4f orbitals occurs progressively along the series.
Actinoids:
The electronic configuration is typically $[Rn] 5f^{0-14} 6d^{0-1} 7s^{2}$.
There is greater variability in the filling of the 5f orbitals compared to the 4f in lanthanoids.
(ii) Atomic and Ionic Sizes
Lanthanoids:
Exhibit what is known as lanthanoid contraction, which is a steady decrease in atomic and ionic size along the series due to poor shielding by 4f electrons.
Actinoids:
Also show actinoid contraction, which is similar to lanthanoid contraction, but generally more pronounced due to poorer shielding by 5f electrons compared to 4f electrons.
(iii) Oxidation States
Lanthanoids:
Predominant oxidation state is +3.
Some elements also show +2 and +4 states, but these are less common and stable.
Actinoids:
Show a wider range of oxidation states, often from +3 to +6.
Early actinoids can exhibit oxidation states up to +7 due to the involvement of 5f, 6d, and 7s orbitals in bonding.
(iv) Chemical Reactivity
Lanthanoids:
Generally reactive metals.
React with water to form hydroxides and with air to form oxides.
Form predominantly ionic compounds due to their fixed +3 oxidation state.
Actinoids:
Even more chemically reactive compared to lanthanoids, especially in finely divided states.
React with air, water, acids, and non-metals forming various compounds.
Greater range of chemical reactivity due to variability in oxidation states.
Summary
Electron Configuration: Lanthanoids progressively fill the 4f orbitals, while actinoids fill the 5f orbitals with greater variability.
Atomic and Ionic Sizes: Both exhibit contraction (lanthanoid and actinoid contractions), but actinoid contraction is more pronounced.
Oxidation States: Lanthanoids mostly show a +3 oxidation state; actinoids exhibit a broader range of oxidation states from +3 to +7.
Chemical Reactivity: Actinoids tend to be more chemically reactive with potentially higher oxidation states contributing to diverse chemistries relative to lanthanoids.
These differences highlight the unique chemical behavior of the actinoids compared to the lanthanoids.
How would you account for the following:
(i) Of the $d^{4}$ species, $\mathrm{Cr}^{2+}$ is strongly reducing while manganese(III) is strongly oxidising.
(ii) Cobalt(II) is stable in aqueous solution but in the presence of complexing reagents it is easily oxidised.
(iii) The $d^{1}$ configuration is very unstable in ions.
(i) Of the $d^4$ species, $\mathrm{Cr}^{2+}$ is strongly reducing while manganese(III) is strongly oxidising.
$\mathbf{Cr^{2+}}$ has the electronic configuration $d^4$. On reduction, it converts to $d^3$, which has a half-filled $t_{2g}$ subshell, offering extra stability. Thus, $\mathbf{Cr^{2+}}$ is a strong reducing agent because it readily reduces to $\mathbf{Cr^{3+}}$.
$$\mathrm{Cr}^{2+} \rightarrow \mathrm{Cr}^{3+} + e^{-}$$
On the other hand, $\mathbf{Mn^{3+}}$ also has a $d^4$ configuration. Upon oxidation, it converts to $d^5$, which means a half-filled $d$-orbital configuration providing extra stability. As a result, $\mathbf{Mn^{3+}}$ is a strong oxidizing agent because it readily gets oxidized to $\mathbf{Mn^{2+}}$.
$$\mathrm{Mn}^{3+} + e^{-} \rightarrow \mathrm{Mn}^{2+}$$
(ii) Cobalt(II) is stable in aqueous solution, but in the presence of complexing reagents, it is easily oxidized.
$\mathbf{Co^{2+}}$ is relatively stable in aqueous solution due to the higher hydration energy it gains. However, in the presence of complexing agents, the $\mathrm{Co^{II}}$ ion can be stabilized in higher oxidation states by forming complexes with ligands. These ligands:
Stabilize $\mathbf{Co^{III}}$ through strong field effects.
Facilitate electron removal due to the high crystal field stabilization energy (CFSE).
Thus, complexing agents help in oxidizing $\mathbf{Co^{2+}}$ to $\mathbf{Co^{3+}}$.
(iii) The $d^1$ configuration is very unstable in ions.
A $\mathbf{d^1}$ configuration is very unstable in ions due to several reasons:
Electron Repulsion: With only one electron in the $d$ subshell, ions with a $d^1$ configuration do not gain significant stability.
Exchange Energy: There is minimal exchange energy stabilization compared to ions with more unpaired electrons.
Possible Pairing: The single electron is more prone to pairing or further oxidation/reduction.
Therefore, ions with a $d^1$ configuration readily change their states to achieve more stable electron configurations.
What is meant by 'disproportionation'? Give two examples of disproportionation reaction in aqueous solution.
Disproportionation
Disproportionation is a type of redox reaction in which a single substance is simultaneously oxidized and reduced, forming two different products. This reaction involves the change of one element in a particular oxidation state to two different oxidation states.
Examples of Disproportionation Reactions in Aqueous Solution
Manganese(III) disproportionation: $$ 2 \ Mn^{3+} ⟶ Mn^{2+} + Mn^{4+} $$
Chlorine disproportionation in an aqueous solution: $$ Cl_2 + 2 \ OH^- ⟶ Cl^- + ClO^- + H_2 O $$
In the above example, chlorine gas ($Cl_2$) reacts with hydroxide ions ($OH^-$) to produce chloride ions ($Cl^-$) and hypochlorite ions ($ClO^-$), demonstrating that chlorine is both reduced and oxidized.
Markdown for Examples:
Manganese(III) disproportionation: $$ 2 \ Mn^{3+} ⟶ Mn^{2+} + Mn^{4+} $$
Chlorine disproportionation: $$ Cl_2 + 2 \ OH^- ⟶ Cl^- + ClO^- + H_2 O $$
These examples illustrate the concept of disproportionation where a single element in a single oxidation state splits into two different oxidation states upon reaction.
Which metal in the first series of transition metals exhibits +1 oxidation state most frequently and why?
The metal in the first series of transition metals that exhibits the +1 oxidation state most frequently is Copper (Cu). This can be attributed to the stability of the $3d^{10}$ configuration when copper loses one electron.
When copper is in a +1 oxidation state, its electronic configuration is $ [Ar] 3d^{10} $. The presence of completely filled $d$ orbitals grants extra stability to the copper(I) ion. This stability makes the +1 oxidation state more common for copper in comparison to other transition metals in the first series.
Calculate the number of unpaired electrons in the following gaseous ions: $\mathrm{Mn}^{3+}$, $\mathrm{Cr}^{3+}, \mathrm{V}^{3+}$ and $\mathrm{Ti}^{3+}$. Which one of these is the most stable in aqueous solution?
To calculate the number of unpaired electrons in the given gaseous ions $\mathrm{Mn}^{3+}$, $\mathrm{Cr}^{3+}$, $\mathrm{V}^{3+}$, and $\mathrm{Ti}^{3+}$, we need to determine their electronic configurations.
Starting with their atomic numbers:
Mn: 25
Cr: 24
V: 23
Ti: 22
In their neutral state, their electron configurations are:
Mn: $$ [\mathrm{Ar}] 3d^5 4s^2 $$
Cr: $$ [\mathrm{Ar}] 3d^5 4s^1 $$
V: $$ [\mathrm{Ar}] 3d^3 4s^2 $$
Ti: $$ [\mathrm{Ar}] 3d^2 4s^2 $$
When these ions are in the +3 oxidation state, they lose three electrons, starting with the 4s orbital and then from the 3d orbital.
Electronic configurations of the ions:
For $$\mathrm{Mn}^{3+}$$:
Mn: $$ [\mathrm{Ar}] 3d^5 4s^2 $$
$$ \mathrm{Mn}^{3+}: [\mathrm{Ar}] 3d^4 $$
Number of unpaired electrons: 4
For $$\mathrm{Cr}^{3+}$$:
Cr: $$ [\mathrm{Ar}] 3d^5 4s^1 $$
$$ \mathrm{Cr}^{3+}: [\mathrm{Ar}] 3d^3 $$
Number of unpaired electrons: 3
For $$\mathrm{V}^{3+}$$:
V: $$ [\mathrm{Ar}] 3d^3 4s^2 $$
$$ \mathrm{V}^{3+}: [\mathrm{Ar}] 3d^2 $$
Number of unpaired electrons: 2
For $$\mathrm{Ti}^{3+}$$:
Ti: $$ [\mathrm{Ar}] 3d^2 4s^2 $$
$$ \mathrm{Ti}^{3+}: [\mathrm{Ar}] 3d^1 $$
Number of unpaired electrons: 1
Stability in Aqueous Solution:In aqueous solution, the stability of ions is influenced by factors like electronic configuration, hydration energy, and exchange energy stabilization.
$$\mathrm{Cr}^{3+}$$ is well known for its high stability in aqueous solution. This is due to having a half-filled ( t_{2g} ) configuration which contributes to its significant stabilization.
In summary:
Number of unpaired electrons:
$$\mathrm{Mn}^{3+}$$: 4
$$\mathrm{Cr}^{3+}$$: 3
$$\mathrm{V}^{3+}$$: 2
$$\mathrm{Ti}^{3+}$$: 1
Most stable in aqueous solution:
$$ \mathbf{Cr^{3+}} $$
Give examples and suggest reasons for the following features of the transition metal chemistry:
(i) The lowest oxide of transition metal is basic, the highest is amphoteric/acidic.
(ii) A transition metal exhibits highest oxidation state in oxides and fluorides.
(iii) The highest oxidation state is exhibited in oxoanions of a metal.
(i) The lowest oxide of transition metal is basic, the highest is amphoteric/acidic.
Examples:
Manganese:
Lowest oxide: MnO (basic)
Highest oxide: Mn₂O₇ (acidic)
Chromium:
Lowest oxide: CrO (basic)
Highest oxide: CrO₃ (acidic)
Reason:The basicity or acidity of oxides depends on the oxidation state of the metal:
In the lower oxidation states, the metal cation is less polarized and can release oxide ions (O²⁻), which behave as bases.
In higher oxidation states, metal cations are highly charged and have greater ability to attract electrons from the oxide ion, increasing the covalent character and rendering the oxides acidic or amphoteric.
(ii) A transition metal exhibits highest oxidation state in oxides and fluorides.
Examples:
Manganese: Mn₂O₇ (Oxide, +7), MnF₄ (Fluoride, +4)
Chromium: CrO₃ (Oxide, +6), CrF₆ (Fluoride, +6)
Reason:Oxides and fluorides involve highly electronegative elements (oxygen and fluorine) that stabilize high oxidation states of the metal by effectively delocalizing the metal’s positive charge through covalent bonding. This stabilization allows the metal to achieve and sustain higher oxidation states.
(iii) The highest oxidation state is exhibited in oxoanions of a metal.
Examples:
Manganese: MnO₄⁻ (Permanganate, +7)
Chromium: Cr₂O₇²⁻ (Dichromate, +6)
Reason:Oxoanions form when metals combine with oxygen to form polyatomic ions. The highly electronegative oxygen atoms help to stabilize high oxidation states of the metal by sharing electron density through $\pi$-bonding and resonance structures. Oxoanions effectively delocalize the positive charge over multiple atoms, stabilizing the ion and allowing the metal to achieve high oxidation states.
By understanding these examples and reasons, we can deduce the patterns in transition metal chemistry and predict their behavior in various compounds.
Indicate the steps in the preparation of:
(i) $\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}$ from chromite ore.
(ii) $\mathrm{KMnO}_{4}$ from pyrolusite ore.
Here are the steps for the preparation of potassium dichromate and potassium permanganate from their respective ores:
(i) Preparation of $\mathrm{K}_{2}\mathrm{Cr}_{2}\mathrm{O}_{7}$ from Chromite Ore
Fusion of Chromite Ore:
Chromite ore ($\mathrm{FeCr}_{2}\mathrm{O}_{4}$) is fused with sodium or potassium carbonate in the presence of air.
Reaction:$$ 4 \mathrm{FeCr}_{2} \mathrm{O}_{4} + 8 \mathrm{Na}_{2} \mathrm{CO}_{3} + 7 \mathrm{O}_{2} \rightarrow 8 \mathrm{Na}_{2} \mathrm{CrO}_{4} + 2 \mathrm{Fe}_{2} \mathrm{O}_{3} + 8 \mathrm{CO}_{2} $$
Filtration and Acidification:
The yellow solution of sodium chromate ($\mathrm{Na}_{2} \mathrm{CrO}_{4}$) is filtered and acidified with sulfuric acid.
Reaction:$$ 2 \mathrm{Na}_{2} \mathrm{CrO}_{4} + 2 \mathrm{H}^{+} \rightarrow \mathrm{Na}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7} + 2 \mathrm{Na}^{+} + \mathrm{H}_{2} \mathrm{O} $$
Crystallization:
The solution is then treated with potassium chloride to obtain potassium dichromate.
Reaction:$$ \mathrm{Na}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7} + 2 \mathrm{KCl} \rightarrow \mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7} + 2 \mathrm{NaCl} $$
Summary of Reactions:
Fusion: $ 4 \mathrm{FeCr}_{2}\mathrm{O}_{4} + 8 \mathrm{Na}_{2}\mathrm{CO}_{3} + 7 \mathrm{O}_{2} \rightarrow 8 \mathrm{Na}_{2}\mathrm{CrO}_{4} + 2 \mathrm{Fe}_{2}\mathrm{O}_{3} + 8 \mathrm{CO}_{2} $
Filtration and Acidification: $ 2 \mathrm{Na}_{2}\mathrm{CrO}_{4} + 2 \mathrm{H}^{+} \rightarrow \mathrm{Na}_{2}\mathrm{Cr}_{2}\mathrm{O}_{7} + 2 \mathrm{Na}^{+} + \mathrm{H}_{2}\mathrm{O} $
Crystallization: $ \mathrm{Na}_{2}\mathrm{Cr}_{2}\mathrm{O}_{7} + 2 \mathrm{KCl} \rightarrow \mathrm{K}_{2}\mathrm{Cr}_{2}\mathrm{O}_{7} + 2 \mathrm{NaCl} $
(ii) Preparation of $\mathrm{KMnO}_{4}$ from Pyrolusite Ore
Fusion of Pyrolusite Ore:
Pyrolusite ore ($\mathrm{MnO}_{2}$) is fused with potassium hydroxide and an oxidizing agent like potassium nitrate.
Reaction:$$ 2 \mathrm{MnO}_{2} + 4 \mathrm{KOH} + \mathrm{O}_{2} \rightarrow 2 \mathrm{K}_{2}\mathrm{MnO}_{4} + 2 \mathrm{H}_{2}\mathrm{O} $$
Oxidation of $\mathrm{K}_{2}\mathrm{MnO}_{4}$:
The dark green $\mathrm{K}_{2}\mathrm{MnO}_{4}$ formed is oxidized to potassium permanganate in a neutral or acidic solution.
Reaction:$$ 3 \mathrm{K}_{2}\mathrm{MnO}_{4} + 4 \mathrm{H}^{+} \rightarrow 2 \mathrm{KMnO}_{4} + \mathrm{MnO}_{2} + 2 \mathrm{H}_{2}\mathrm{O} $$
Electrolytic Oxidation (if needed):
Commercially, the alkaline oxidative fusion is followed by electrolytic oxidation of manganate (VI).
Reaction:$$ \mathrm{K}_{2} \mathrm{MnO}_{4} \xrightarrow{\text{electrolysis}} 2 \mathrm{KMnO}_{4} $$
Summary of Reactions:
Fusion: $ 2 \mathrm{MnO}_{2} + 4 \mathrm{KOH} + \mathrm{O}_{2} \rightarrow 2 \mathrm{K}_{2}\mathrm{MnO}_{4} + 2 \mathrm{H}_{2}\mathrm{O} $
Oxidation: $ 3 \mathrm{K}_{2}\mathrm{MnO}_{4} + 4 \mathrm{H}^{+} \rightarrow 2 \mathrm{KMnO}_{4} + \mathrm{MnO}_{2} + 2 \mathrm{H}_{2}\mathrm{O} $
Electrolytic Oxidation: $\mathrm{K}_{2}\mathrm{MnO}_{4} \xrightarrow{\text{electrolysis}} 2 \mathrm{KMnO}_{4}$
What are alloys? Name an important alloy which contains some of the lanthanoid metals. Mention its uses.
An alloy is a blend of metals prepared by mixing the components. Alloys can be homogeneous solid solutions where the atoms of one metal are distributed randomly among the atoms of another metal. These alloys are typically designed to have properties that are different from the constituent metals.
An important alloy that contains some of the lanthanoid metals is mischmetall, which consists of approximately 95% lanthanoid metal (mostly a mix of lanthanum, cerium, neodymium, and praseodymium) and 5% iron along with traces of sulfur, carbon, calcium, and aluminum.
Uses of Mischmetall:
In Production of Alloy Steels: Used for making plates and pipes.
In Mg-based Alloys: Utilized for manufacturing bullets and shell casings.
In Lighter Flints: Due to its pyrophoric properties, it is used in lighter flints.
Mischmetall's specific composition provides enhanced properties for industrial applications, such as improving the hardness and strength of materials.
What are inner transition elements? Decide which of the following atomic numbers are the atomic numbers of the inner transition elements : 29, 59, $74,95,102,104$.
Inner transition elements are the elements in which the $4f$ and $5f$ orbitals are progressively filled. These elements are placed separately at the bottom of the periodic table and include the lanthanoids (from atomic number 57 to 71) and the actinoids (from atomic number 89 to 103).
To determine which of the following atomic numbers are inner transition elements:
29 (Copper, Cu) - Not an inner transition element, belongs to the 3d series.
59 (Praseodymium, Pr) - Yes, it's a lanthanoid.
74 (Tungsten, W) - Not an inner transition element, belongs to the 5d series.
95 (Americium, Am) - Yes, it's an actinoid.
102 (Nobelium, No) - Yes, it's an actinoid.
104 (Rutherfordium, Rf) - Not an inner transition element, belongs to the 6d series.
Thus, the atomic numbers of inner transition elements among the given numbers are 59, 95, and 102.
The chemistry of the actinoid elements is not so smooth as that of the lanthanoids. Justify this statement by giving some examples from the oxidation state of these elements.
The chemistry of the actinoid elements is indeed more complex than that of the lanthanoids. This complexity arises from several factors:
Variable Oxidation States: Unlike lanthanoids, which predominantly exhibit the +3 oxidation state, actinoids show a much wider range of oxidation states. For example:
Uranium (U) can exhibit oxidation states +3, +4, +5, and +6.
Neptunium (Np) can show oxidation states +3, +4, +5, +6, and +7.
Plutonium (Pu) can exhibit oxidation states +3, +4, +5, and +6.
Higher Oxidation States: The maximum oxidation state increases from +4 in Thorium (Th) to +5 in Protactinium (Pa), +6 in Uranium (U), and +7 in Neptunium (Np) before decreasing in the later elements (Table 4.11 in the text).
Instability and Radioactivity: Many actinoids are radioactive and have short half-lives, making their study and practical applications more challenging. For example, the latter members of the actinoid series can only be prepared in nanogram quantities due to their short half-lives.
Hydrolysis and Complex Formation: Actinoid ions, especially in higher oxidation states, tend to undergo extensive hydrolysis and form complex ions. This behavior adds to the complexity of their aqueous chemistry.
Actinoid Contraction: There is a greater contraction in size across the series due to poor shielding by the 5f orbitals, leading to greater overlap and participation in bonding, which increases the variability in their chemistry.
Example of Oxidation States:
Uranium (U): Exhibits +3, +4, +5, and +6 oxidation states.
$ U^{3+} $: Less common.
$ U^{4+} $: Found in compounds like $ UO_2 $.
$ U^{5+} $: Often seen in the form of $ UO_2^+ $ (uranyl ion).
$ U^{6+} $: Commonly exists as $ UO_2^{2+} $ (uranyl ion).
Neptunium (Np): Shows oxidation states +3 to +7.
$ Np^{3+} $: Basic and reducing agent.
$ Np^{4+} $: Relatively stable, forms NpO2.
$ Np^{5+} $: Exists as $ NpO_2^+ $.
$ Np^{6+} $: Exists as $ NpO_2^{2+} $.
$ Np^{7+} $: Strong oxidizing agent often found in the form of $ NpO_4^- $.
Thus, the range and complexity of oxidation states and the behavior of actinoid elements illustrate why their chemistry is much more complicated compared to that of the lanthanoids.
Which is the last element in the series of the actinoids? Write the electronic configuration of this element. Comment on the possible oxidation state of this element.
The last element in the series of the actinoids is Lawrencium (Lr), which has an atomic number of 103.
Electronic Configuration
The electronic configuration of Lawrencium (Lr) is: $$ [Rn] 5f^{14} 6d^1 7s^2 $$
Possible Oxidation State
Lawrencium (Lr) predominantly exhibits a +3 oxidation state. This is because the removal of the two 7s electrons and one 6d electron leads to a stable electronic configuration similar to the noble gas Radon [Rn]. Thus, the most common and stable oxidation state for Lawrencium is +3.
Use Hund's rule to derive the electronic configuration of $\mathrm{Ce}^{3+}$ ion, and calculate its magnetic moment on the basis of 'spin-only' formula.
First, let's derive the electronic configuration of the $\mathrm{Ce}^{3+}$ ion:
The atomic number of Cerium (Ce) is 58.
The ground state electronic configuration of Cerium is $[\mathrm{Xe}]4f^1 5d^1 6s^2$.
When Cerium loses three electrons to form $\mathrm{Ce}^{3+}$:
The electrons are removed first from the outermost shells, so two electrons from the $6s$ orbital and one from the $4f$ orbital are removed.
Thus, the electronic configuration of $\mathrm{Ce}^{3+}$ becomes: $$ [\mathrm{Xe}]4f^1 $$
Magnetic Moment Calculation
The number of unpaired electrons ( n ) in the $4f$ orbital of $\mathrm{Ce}^{3+}$ is 1.
Using the 'spin-only' formula to calculate the magnetic moment ((\mu)): $$ \mu = \sqrt{n(n+2)} $$
Substituting ( n = 1 ): $$ \mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.732 \ \text{BM} $$
Therefore, the magnetic moment for $\mathrm{Ce}^{3+}$, based on the 'spin-only' formula, is approximately 1.73 Bohr magnetons (BM).
Name the members of the lanthanoid series which exhibit +4 oxidation states and those which exhibit +2 oxidation states. Try to correlate this type of behaviour with the electronic configurations of these elements.
The members of the lanthanoid series that exhibit +4 oxidation states and those that exhibit +2 oxidation states can be identified based on their electronic configurations.
Lanthanoids Exhibiting +4 Oxidation State:
Cerium (Ce): $ [Xe] 4f^1 5d^1 6s^2 $ -> $ [Xe] $ (stable noble gas configuration)
Praseodymium (Pr): $ [Xe] 4f^3 6s^2 $ -> $ [Xe] 4f^1 $ (half-filled f-orbital stability)
Neodymium (Nd): $ [Xe] 4f^4 6s^2 $ -> $ [Xe] 4f^2 $
Terbium (Tb): $ [Xe] 4f^9 6s^2 $ -> $ [Xe] 4f^7 $ (half-filled f-orbital stability)
Dysprosium (Dy): $ [Xe] 4f^{10} 6s^2 $ -> $ [Xe] 4f^8 $
Lanthanoids Exhibiting +2 Oxidation State:
Europium (Eu): $ [Xe] 4f^7 6s^2 $ -> $ [Xe] 4f^7 $ (half-filled f-orbital stability)
Ytterbium (Yb): $ [Xe] 4f^{14} 6s^2 $ -> $ [Xe] 4f^{14} $ (completely filled f-orbital stability)
Samarium (Sm): $ [Xe] 4f^6 6s^2 $ -> $ [Xe] 4f^6 $
Correlation with Electronic Configurations:
The elements that exhibit +4 oxidation states tend to achieve a stable noble gas or half-filled f-orbital configuration upon losing four electrons. Conversely, elements that exhibit +2 oxidation states often either retain a half-filled or completely filled f-orbital configuration upon losing two electrons. Thus, the stability of specific electronic configurations (either half-filled or fully-filled f-orbitals) significantly influences the oxidation states these elements exhibit.
Compare the chemistry of the actinoids with that of lanthanoids with reference to: (i) electronic configuration (ii) oxidation states and (iii) chemical reactivity.
(i) Electronic Configuration
Lanthanoids:
The general electronic configuration is $$ [Xe]4f^{1-14}5d^0 6s^2 $$
Lanthanoids typically show a filling of 4f orbitals.
Most lanthanoids primarily exhibit a +3 oxidation state, leading to a stable $4f^n$ configuration in their trivalent ions.
Actinoids:
The general electronic configuration is $$ [Rn]5f^{1−14}6d^0 7s^2 $$
Actinoids show a filling of 5f orbitals, although 6d orbitals can also be occupied in some elements.
The differentiating feature is the participation of 5f orbitals in bonding, which are more extended and less tightly bound than 4f orbitals, allowing actinoids to exhibit a wider range of oxidation states.
(ii) Oxidation States
Lanthanoids:
Lanthanoids primarily exhibit the +3 oxidation state.
Some elements can exhibit +2 and +4 states, but these are less common and usually less stable (e.g., Eu$^{2+}$, Ce$^{4+}$).
The +3 state is particularly stable due to the stabilization of 4f orbital electrons.
Actinoids:
Actinoids exhibit a variable range of oxidation states, from +3 to +7.
Early actinoids like Th, Pa, U, and Np can show higher oxidation states (Th$^{4+}$, Pa$^{5+}$, U$^{6+}$, Np$^{7+}$).
The variability is due to the similar energies of 5f, 6d, and 7s orbitals, making multiple oxidation states possible.
(iii) Chemical Reactivity
Lanthanoids:
Lanthanoids are relatively less reactive compared to actinoids.
They tend to form stable oxides and halides.
They react slowly with water to form hydroxides and liberate hydrogen.
Their reactivity decreases as we move across the series.
Actinoids:
Actinoids are more chemically reactive than lanthanoids.
They readily react with H$_2$O, HCl, and other reagents; some even react at room temperature.
Actinoids form a variety of compounds, including oxides, hydroxides, and halides, with complex stoichiometries due to their ability to exist in multiple oxidation states.
The chemical reactivity tends to decrease down the series as with the lanthanoids, but the early actinoids are especially highly reactive.
Summary
Electronic Configuration: Actinoids involve 5f orbitals in bonding (and sometimes 6d), whereas lanthanoids involve 4f orbitals.
Oxidation States: Actinoids exhibit a wider range of oxidation states due to similar energies of 5f, 6d, and 7s orbitals compared to lanthanoids' predominantly +3 state.
Chemical Reactivity: Actinoids are generally more reactive than lanthanoids, particularly in forming diverse compounds with higher oxidation states.
These comparisons underscore the complexity and variability of actinoid chemistry relative to that of lanthanoids.
Write the electronic configurations of the elements with the atomic numbers $61,91,101$, and 109 .
Here are the electronic configurations for the elements with the respective atomic numbers:
Element with Atomic Number 61 (Promethium, Pm):
Long form: $1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10} 5p^6 6s^2 4f^5$
Short form: $[Xe] 6s^2 4f^5$
Element with Atomic Number 91 (Protactinium, Pa):
Long form: $1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10} 5p^6 6s^2 4f^{14} 5d^{10} 6p^6 7s^2 5f^2 6d^1$
Short form: $[Rn] 7s^2 5f^2 6d^1$
Element with Atomic Number 101 (Mendelevium, Md):
Long form: $1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10} 5p^6 6s^2 4f^{14} 5d^{10} 6p^6 7s^2 5f^{13}$
Short form: $[Rn] 7s^2 5f^{13}$
Element with Atomic Number 109 (Meitnerium, Mt):
Long form: $1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10} 5p^6 6s^2 4f^{14} 5d^{10} 6p^6 7s^2 5f^{14} 6d^7$
Short form: $[Rn] 7s^2 5f^{14} 6d^7$
Compare the general characteristics of the first series of the transition metals with those of the second and third series metals in the respective vertical columns. Give special emphasis on the following points:
(i) electronic configurations
(ii) oxidation states
(iii) ionisation enthalpies and (iv) atomic sizes.
To compare the general characteristics of the first series of transition metals with those of the second and third series metals within the respective vertical columns, we can look at the following points:
(i) Electronic Configurations
First Series Transition Metals (3d Series):
The general electronic configuration is $(n-1)d^{1-10} ns^{0-2}$. For example, Scandium (Sc) has $3d^1 4s^2$.
Variations occur due to the stability of half-filled and fully-filled d-orbitals, e.g., Chromium (Cr) has $3d^5 4s^1$ and Copper (Cu) has $3d^{10} 4s^1$.
Second Series Transition Metals (4d Series):
The general electronic configuration follows $(n-1)d^{1-10} ns^{0-2}$, for example, Yttrium (Y) has $4d^1 5s^2$.
Palladium (Pd) is an exception with the configuration $4d^{10} 5s^0$.
Third Series Transition Metals (5d Series):
The configuration generally follows $(n-1)d^{1-10} ns^{0-2}$. For instance, Lanthanum (La) has $5d^1 6s^2$.
Due to Lanthanoid contraction, elements like Hafnium (Hf) and others show similar configurations to their second-series counterparts.
(ii) Oxidation States
First Series Transition Metals:
These metals exhibit a wide range of oxidation states, typically from +2 to +7. For example, Manganese (Mn) shows +2, +3, +4, +6, and +7 states.
The variety is extended as we move towards the middle of the series.
Second Series Transition Metals:
They also exhibit a wide range of oxidation states. For example, Molybdenum (Mo) shows states from +2 to +6.
The higher oxidation states are more stable compared to their 3d counterparts.
Third Series Transition Metals:
These elements show even higher oxidation states. Tungsten (W), for instance, shows states from +2 to +6.
The lanthanide contraction results in third-series metals having more stabilized oxidation states.
(iii) Ionisation Enthalpies
First Series Transition Metals:
Ionisation enthalpies gradually increase across the series. For example, the first ionisation enthalpy of Sc is 631 kJ/mol, while for Zn it is 906 kJ/mol.
Due to the addition of electrons in the inner 3d orbital, these elements do not see a sharp increase.
Second Series Transition Metals:
These elements generally have higher ionisation enthalpies than the 3d series due to increased nuclear charge.
The values are higher due to greater effective nuclear charge.
Third Series Transition Metals:
Due to the lanthanide contraction, ionisation enthalpies are similar or even higher than those in the second series.
For instance, the ionisation enthalpy of Hafnium (Hf) is similar to Zirconium (Zr).
(iv) Atomic Sizes
First Series Transition Metals:
Have smaller atomic radii compared to their 4d and 5d counterparts. For example, the atomic radius of Iron (Fe) is 126 pm.
There is a gradual decrease in size due to increased nuclear charge.
Second Series Transition Metals:
Possess larger atomic radii than the 3d series. The atomic radius of Molybdenum (Mo) is around 139 pm.
The radii here do not decrease as significantly as in the 3d series due to effective shielding by the 4d electrons.
Third Series Transition Metals:
Atomic sizes are almost similar to the second series due to the lanthanoid contraction. Hafnium (Hf) has a radius nearly the same as Zirconium (Zr).
The presence of 4f electrons leads to a contraction effect counterbalancing the expected increase in size.
Summary in Tabular Format
Property | 3d Series | 4d Series | 5d Series |
---|---|---|---|
Electronic Configuration | $(n-1)d^{1-10} ns^{0-2}$ | $(n-1)d^{1-10} ns^{0-2}$ | $(n-1)d^{1-10} ns^{0-2}$ |
Variations | Cr: $3d^5 4s^1$, Cu: $3d^{10} 4s^1$ | Pd: $4d^{10} 5s^0$ | La: $5d^1 6s^2$ (similar due to contraction) |
Oxidation States | Wide range, e.g., Mn (+2 to +7) | Stable higher states, e.g., Mo (+2 to +6) | Very high, e.g., W (+2 to +6) |
Ionisation Enthalpies | Gradual increase, Sc (631 kJ/mol) to Zn (906 kJ/mol) | Higher due to greater nuclear charge | Similar or higher due to lanthanide contraction |
Atomic Sizes | Smaller, gradual decrease | Larger but less sharp decrease | Similar to 4d series, very slight decrease |
Visual Representation:
A detailed visual comparison of trends in all these properties is shown in Figures 4.1, 4.2, 4.3, 4.5, etc., in your textbook.
By understanding these distinctive trends, one can better explain the chemical behavior and properties of transition elements across different series.
Write down the number of $3 \mathrm{~d}$ electrons in each of the following ions: $\mathrm{Ti}^{2+}, \mathrm{V}^{2+}$, $\mathrm{Cr}^{3+}, \mathrm{Mn}^{2+}, \mathrm{Fe}^{2+}, \mathrm{Fe}^{3+}, \mathrm{Co}^{2+}, \mathrm{Ni}^{2+}$ and $\mathrm{Cu}^{2+}$. Indicate how would you expect the five $3 d$ orbitals to be occupied for these hydrated ions (octahedral).
Here is the number of $3d$ electrons in each of the given ions and the expected occupation of the five $3d$ orbitals in an octahedral field:
Titanium ion, (\mathrm{Ti}^{2+}):
Titanium ground state: ([\mathrm{Ar}], 3d^2 4s^2)
(\mathrm{Ti}^{2+}) loses two (4s) electrons.
Number of $3d$ electrons: 2
Occupation in an octahedral field (ignoring electron pairing): (\uparrow\downarrow \uparrow \uparrow \uparrow)
Vanadium ion, (\mathrm{V}^{2+}):
Vanadium ground state: ([\mathrm{Ar}], 3d^3 4s^2)
(\mathrm{V}^{2+}) loses two (4s) electrons.
Number of $3d$ electrons: 3
Occupation in an octahedral field: (\uparrow\downarrow \uparrow \uparrow \uparrow \uparrow)
Chromium ion, (\mathrm{Cr}^{3+}):
Chromium ground state: ([\mathrm{Ar}], 3d^5 4s^1)
(\mathrm{Cr}^{3+}) loses one (4s) and two (3d) electrons.
Number of $3d$ electrons: 3
Occupation in an octahedral field: (\uparrow\downarrow \uparrow \uparrow \uparrow \uparrow)
Manganese ion, (\mathrm{Mn}^{2+}):
Manganese ground state: ([\mathrm{Ar}], 3d^5 4s^2)
(\mathrm{Mn}^{2+}) loses two (4s) electrons.
Number of $3d$ electrons: 5
Occupation in an octahedral field: (\uparrow\downarrow \uparrow\downarrow \uparrow\downarrow \uparrow\downarrow \uparrow\downarrow)
Iron ion, (\mathrm{Fe}^{2+}):
Iron ground state: ([\mathrm{Ar}], 3d^6 4s^2)
(\mathrm{Fe}^{2+}) loses two (4s) electrons.
Number of $3d$ electrons: 6
Occupation in an octahedral field: (\uparrow\downarrow \uparrow\downarrow \uparrow\downarrow \uparrow\downarrow \uparrow\downarrow \uparrow\downarrow)
Iron ion, (\mathrm{Fe}^{3+}):
Iron ground state: ([\mathrm{Ar}], 3d^6 4s^2)
(\mathrm{Fe}^{3+}) loses two (4s) and one (3d) electron.
Number of $3d$ electrons: 5
Occupation in an octahedral field: (\uparrow\downarrow \uparrow\downarrow \uparrow\downarrow \uparrow\downarrow \uparrow\downarrow \uparrow\downarrow)
Cobalt ion, (\mathrm{Co}^{2+}):
Cobalt ground state: ([\mathrm{Ar}], 3d^7 4s^2)
(\mathrm{Co}^{2+}) loses two (4s) electrons.
Number of $3d$ electrons: 7
Occupation in an octahedral field: (\uparrow\downarrow \uparrow\downarrow \uparrow\downarrow \uparrow\downarrow \uparrow\downarrow \uparrow\downarrow \uparrow\downarrow)
Nickel ion, (\mathrm{Ni}^{2+}):
Nickel ground state: ([\mathrm{Ar}], 3d^8 4s^2)
(\mathrm{Ni}^{2+}) loses two (4s) electrons.
Number of $3d$ electrons: 8
Occupation in an octahedral field: (\uparrow\downarrow \uparrow\downarrow \uparrow\downarrow \uparrow\downarrow \uparrow\downarrow \uparrow\downarrow \uparrow\downarrow \uparrow\downarrow)
Copper ion, (\mathrm{Cu}^{2+}):
Copper ground state: ([\mathrm{Ar}], 3d^{10} 4s^1)
(\mathrm{Cu}^{2+}) loses one (4s) and one (3d) electron.
Number of $3d$ electrons: 9
Occupation in an octahedral field: (\uparrow\downarrow \uparrow\downarrow \uparrow\downarrow \uparrow\downarrow \uparrow\downarrow \uparrow\downarrow \uparrow\downarrow \uparrow\downarrow \uparrow\downarrow)
Octahedral Field Splitting
For each ion in an octahedral field, the $3d$ orbitals are split into two sets:
(t_{2g}) orbitals (lower energy): typically three orbitals (dxy, dxz, dyz)
(e_g) orbitals (higher energy): typically two orbitals (dz2, dx2-y2)
Note
The occupation may vary depending on factors like electron pairing energy and crystal field stabilization energy, which can influence whether electrons prefer to pair up in lower energy orbitals or occupy higher energy ones first.
Comment on the statement that elements of the first transition series possess many properties different from those of heavier transition elements.
The first transition series (3d series) indeed possesses many properties that are different from those of the heavier transition elements (4d and 5d series).
Key Differences and Comparisons:
Electronic Configurations and Shielding:
For elements in the 3d series, electrons are added to the 3d orbitals. In the 4d and 5d series, electrons are added to the 4d and 5d orbitals, respectively.
Shielding is less effective in 4d and 5d series due to the presence of more inner electrons, which leads to higher effective nuclear charge and smaller atomic or ionic sizes.
Atomic and Ionic Sizes:
Lanthanoid Contraction: There's a noticeable effect of lanthanoid contraction, which reduces the size differences between the 4d and 5d series.
The elements in the 3d series have larger atomic and ionic sizes compared to the corresponding elements in the heavier series due to less effective nuclear charge.
Oxidation States:
Multiple oxidation states are more common and stable for heavier transition elements.
Elements in the 3d series often exhibit lower oxidation states more frequently, while heavier elements can stabilize higher oxidation states due to better energy match between orbitals.
Magnetic Properties:
Stronger Paramagnetism: The 3d transition elements often display stronger paramagnetic properties due to unpaired 3d electrons.
The effective magnetic moments for 4d and 5d elements are typically higher due to spin-orbit coupling, though paramagnetic behavior is widespread.
Chemical Reactivity:
The 3d series elements are generally more reactive and show pronounced variability in chemical behavior. Heavier transition elements often exhibit more noble characteristics (less reactivity).
Higher enthalpies of atomization in 4d and 5d groups: Elements in these groups exhibit stronger metal-metal bonding.
Density and Melting Points:
Higher Density and Melting Points in 4d and 5d elements: Due to stronger metallic bonding, the density and melting points generally increase as we move to heavier transition elements.
Complex Formation and Catalytic Properties:
Both series form complexes, but heavier elements (particularly 4d and 5d) form more stable and varied complexes.
Catalytic Properties: Heavy transition metals (like Pt, Pd) are widely used as catalysts in various industrial processes due to their ability to form stable intermediates.
In summary, while both 3d and heavier transition series elements possess common transition metal properties, the differences arise due to variations in electronic structures, atomic and ionic sizes, and reactivity patterns, with heavier transition metals generally having higher stability and varied oxidation states.
What can be inferred from the magnetic moment values of the following complex species ?
Example | Magnetic Moment (BM) |
---|---|
$\mathrm{K}_{4}\left[\mathrm{Mn}(\mathrm{CN})_{6}\right)$ | 2.2 |
$\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{++}$ | 5.3 |
$\mathrm{~K}_{2}\left[\mathrm{MnCl}_{4}\right]$ | 5.9 |
To infer the details about the complex species from their magnetic moments (in Bohr magnetons, BM), we can use the following steps:
Determine the number of unpaired electrons using the formula for magnetic moment:
$$ \mu = \sqrt{n(n+2)} $$
where ( \mu ) is the magnetic moment and ( n ) is the number of unpaired electrons.
Compare the inferred unpaired electrons to what is known of the oxidation states and electron configurations of the central metal ions.
Example 1: $\mathrm{K}_{4}\left[\mathrm{Mn}(\mathrm{CN})_{6}\right]$
Magnetic Moment: 2.2 BM
To find the number of unpaired electrons ( ( n ) ): [ 2.2 = \sqrt{n(n+2)} ] Solving for ( n ), we get approximately 1 unpaired electron.
Interpretation:
Mn in this complex is in the +2 oxidation state (since CN⁻ is a strong field ligand and typically causes low-spin complexes in Mn).
For ( \text{Mn}^{2+} ) (d⁵ configuration), a low-spin state will have 1 unpaired electron.
This indicates a low-spin complex, likely because CN⁻ is a strong field ligand.
Example 2: $\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{++}$
Magnetic Moment: 5.3 BM
Using the magnetic moment formula: [ 5.3 = \sqrt{n(n+2)} ] Solving for ( n ), we approximately get 4 unpaired electrons.
Interpretation:
Fe in this complex is in the +2 oxidation state (since water is a weak field ligand and typically doesn't cause low-spin complexes in Fe).
For ( \text{Fe}^{2+} ) (d⁶ configuration), a high-spin state will have 4 unpaired electrons.
This suggests a high-spin complex, as expected with weak field ligands like ( \text{H}_{2}\text{O} ).
Example 3: $\mathrm{K}_{2}\left[\mathrm{MnCl}_{4}\right]$
Magnetic Moment: 5.9 BM
Using the magnetic moment formula: [ 5.9 = \sqrt{n(n+2)} ] Solving for ( n ), we approximately get 5 unpaired electrons.
Interpretation:
Mn in this complex is in the +2 oxidation state (since ( \text{Cl}^- ) is a weak field ligand and typically causes high-spin complexes in Mn).
For ( \text{Mn}^{2+} ) (d⁵ configuration), a high-spin state will have 5 unpaired electrons.
This suggests a high-spin complex, consistent with weak field ligands like ( \text{Cl}^- ).
Summary:
$\mathrm{K}_{4}\left[\mathrm{Mn}(\mathrm{CN})_{6}\right]$: Low-spin complex with one unpaired electron due to strong field ligand (CN⁻).
$\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{++}$: High-spin complex with four unpaired electrons due to weak field ligand (H₂O).
$\mathrm{K}_{2}\left[\mathrm{MnCl}_{4}\right]$: High-spin complex with five unpaired electrons due to weak field ligand (Cl⁻).
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Ask Chatterbot AIExtra Questions - The d - and f -Block Elements | NCERT | Chemistry | Class 12
Why are the two outermost shells of transition elements incomplete?
Transition elements, which are found in groups 3-12 of the periodic table, primarily involve the filling of d-orbitals in the $4^{th}$,$5^{th}$, and $6^{th}$ periods, classifying them as d-block elements. This filling sequence affects the occupancy of the electrons in their outer and penultimate (just inside the outermost) shells.
In the case of electron configuration:
The 4s, 5s, 6s, and 7s orbitals typically have lower energies than the 3d, 4d, 5d, and 6d orbitals. Consequently, the s-orbitals are generally filled before the respective d-orbitals.
The penultimate (n-1) d-shell can possess anywhere from 0 to 10 electrons, whereas the outermost s-shell (uniformly labeled as ns) is limited to 1 or 2 electrons.
One notable aspect of transition elements is their preference for electron configurations that return particularly stable states:
Half-filled and fully-filled d-orbitals offer greater stability. Therefore, electrons will fill these d-orbitals even at the expense of leaving the ns orbital not fully occupied.
Here are examples to illustrate typical electron configurations:
For Scandium in Group 3, Period 4: $[\text{Ar}] 3d^1 4s^2$
For Vanadium in Group 5, Period 4: $[\text{Ar}] 3d^3 4s^2$
For Chromium in Group 6, Period 4, unusual because half-filled subshells are more stable: $[\text{Ar}] 3d^5 4s^1$
For Manganese in Group 7, Period 4: $[\text{Ar}] 3d^5 4s^2$
For Iron in Group 8, Period 4: $[\text{Ar}] 3d^6 4s^2$
For Copper in Group 11, Period 4, fully-filled 3d is more stable, hence: $[\text{Ar}] 3d^{10} 4s^1$
For Zinc in Group 12, Period 4: $[\text{Ar}] 3d^{10} 4s^2$
In lanthanoids and actinoids (inner transition elements) where the filling of $4f$ and $5f$ orbitals is concerned, similar principles apply with stabler configurations shaping the electron arrangements in the d and f orbitals.
In summary, the outermost shells (ns) and the outer but one (d) shells of transition elements often remain incomplete due to the electrons’ preference for occupying lower energy states and achieving half-filled or fully-filled configurations for stability.
The group 16 elements are also known as:
A) Nitrogen family
B) Chalcogens
C) Noble gases
D) Halogens
The correct answer is Option B: Chalcogens.
Group 16 elements are commonly referred to as the Chalcogens. This name originates from the fact that many of these elements form ores, primarily as sulphides or oxides, hence they are considered ore-forming elements.
The colour of $\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}$ and $\mathrm{Fe}^{2+}$ ions are respectively due to:
A) $d$ - $d$ transition and charge transfer spectra
B) charge transfer spectra and d-d transition
C) crystal defects and charge transfer spectra
D) charge transfer spectra and crystal defects
The correct answer is Option B: charge transfer spectra and d-d transition.
$\mathrm{K}_2\mathrm{Cr}_2\mathrm{O}_7$ exhibits its characteristic color through charge transfer. Given that chromium is in the +6 oxidation state in this compound, it lacks d-electrons available for excitation. Thus, the color arises from ligand to metal charge transfer (LMCT), where electrons transfer from the oxygen ligands to the chromium center.
For $\mathrm{Fe}^{2+}$ ions, the coloration is primarily due to d-d transitions. With $6$ d-electrons, these transitions occur within the d-orbital, absorbing specific wavelengths of light which correspond to visible color changes.
Thus, Option B correctly identifies the mechanisms behind the colors observed in these two chemical species.
The element with a positive gain enthalpy is:
H
Na
O
F
Ne
The correct answer is Fluorine (F). This is because fluorine possesses the highest electronegativity among the elements listed. Electronegativity is a measure of the tendency of an atom to attract a bonding pair of electrons. Due to its high electronegativity, adding an electron to fluorine results in increased electron repulsion which leads to the release of more energy. Hence, fluorine has a positive electron gain enthalpy.
Therefore, the element with a positive gain enthalpy from the given options is:
4. F (Fluorine)
Edge length in a face-centered cubic unit cell of calcium is $0.556 \mathrm{~nm}$. If it contains $0.1%$ Schottky defect, the density of the metal is
A) $1.5463 \mathrm{~g} \mathrm{~cm}^{-3}$
B) $1.56 \mathrm{~g} \mathrm{~cm}^{-3}$
C) $1.5448 \mathrm{~g} \mathrm{~cm}^{-3}$
D) $3.42 \mathrm{~g} \mathrm{~cm}^{-3}$
To solve for the density of calcium exhibiting a Schottky defect in its face-centered cubic unit cell, we proceed as follows:
-
Convert Edge Length to cm: The edge length $a$ is given as $0.556 , \text{nm}$. We need to convert this to cm for our calculations in the CGS system: $$ a = 0.556 \times 10^{-7} \text{ cm} $$
-
Calculate the Effective Number of Atoms ($Z$) per Unit Cell: Normally, a face-centered cubic unit cell has 4 atoms per unit cell. Due to the presence of Schottky defects (0.1%), where both cation and anion vacancies occur, the effective number of atoms reduces: $$ Z = 4 - \frac{4 \times 0.1}{100} = 3.996 $$
-
Use the Formula for Density of the Metal: The density $d$ can be calculated using the formula: $$ d = \frac{Z \times M}{a^3 \times N_A} $$ where:
- $M$ is the molar mass of calcium (approximately 40 g/mol),
- $a^3$ is the volume of the unit cell,
- $N_A$ is Avogadro's number ($6.022 \times 10^{23} \text{ mol}^{-1}$).
-
Calculate the Density: Substituting the values into the density formula: $$ d = \frac{3.996 \times 40}{(0.556 \times 10^{-7})^3 \times 6.022 \times 10^{23}} \text{ g/cm}^3 = 1.5448 \text{ g/cm}^3 $$
Hence, the correct density of calcium, with Schottky defects in a face centered cubic structure, and using CGS units for consistency in the calculation, is 1.5448 g/cm³. The correct answer is Option C: $1.5448 \mathrm{~g} \mathrm{~cm}^{-3}$.
Study the given information carefully and answer the given questions. Among six people A, B, C, D, E, and F, each of a different age. A is younger than only D. Only three people are younger than C. F is younger than E. F is not the youngest. Who is the youngest?
A) B
B) A
C) E
D) C
E) None of these
The correct answer is A) B.
Analyzing the provided information, we understand the following:
- D is older than A (since A is younger than only D).
- C is older than three people.
- E is older than F.
- F is not the youngest.
By processing the clues:
- Since A is younger than only D, we can arrange D and A as ( D > A ).
- C being older than exactly three individuals helps place C in the middle.
- From E being older than F and F not being the youngest, we arrange E and F, but still keep one position open for the youngest.
- Failsafe placement from these clues deduces that B has to be the youngest, fitting all conditions.
Therefore, our final age sequence from oldest to youngest is: $$ D > A > C > E > F > B $$ Thus, B is the youngest.
The ion which is not tetrahedral in shape is?
(A) $\mathrm{BF}_{4}^{-}$
(B) $\mathrm{NH}_{4}^{+}$
(C) $\left[\mathrm{Cu}\left(\mathrm{NH}{3}\right){4}\right]^{2+}$
(D) $\mathrm{NiCl}_{4}^{2-}$
The ion which is not tetrahedral in shape is:
(C) $\left[\mathrm{Cu}\left(\mathrm{NH}{3}\right){4}\right]^{2+}$
The ion $\left[\mathrm{Cu}\left(\mathrm{NH}{3}\right){4}\right]^{2+}$ is square planar in geometry. The hybridization state of $\mathrm{Cu}^{2+}$ in this complex is $\mathrm{dsp}^2$.
The possible number of electrons for which $n=4$, $l=3$, and $s=+\frac{1}{2}$ is:
A) 14
B) 7
C) 5
D) 10
Solution The correct answer is Option B: 7
Each electron in an atom is characterized by a unique set of quantum numbers. Given $n=4$ and $l=3$, we calculate the maximum number of orbitals as follows: $$ 2l + 1 = (2 \times 3) + 1 = 7 $$ These orbitals can accommodate a maximum of 14 electrons, with 7 electrons having spin quantum number $s=+\frac{1}{2}$ and the other 7 having $s=-\frac{1}{2}$. Therefore, the number of electrons that can have $n=4$, $l=3$, and $s=+\frac{1}{2}$ is 7.
The electronic transition responsible for the colour of $\mathrm{K}{2} \mathrm{Cr}{2} \mathrm{O}_{7}$ in the solid state is
A $\mathrm{d} \rightarrow \mathrm{d}$
B $\mathrm{M} \rightarrow \mathrm{L}$
C $\mathrm{L} \rightarrow \mathrm{M}$
D Due to covalent bond.
Solution
The correct answer to the question about the electronic transition responsible for the color of $\mathrm{K}{2} \mathrm{Cr}{2} \mathrm{O}_{7}$ in the solid state is:
Option C: $\mathbf{L} \rightarrow \mathbf{M}$
This represents a transition where electrons move from ligand to metal, commonly referring to charge transfer transitions. This type of transition is significant for compounds like $\mathrm{K}{2} \mathrm{Cr}{2} \mathrm{O}_{7}$, influencing their distinct color properties.
In a certain code, 'JEFP' is written as 'LGHR'. How is 'CLOUD' written in that code?
A. ENQWF
B. DMQWF
C. FNQWE
D. ENRWF
E. None of these
The correct answer is Option A: ENQWF.
To solve this coding question, observe the transformation pattern from 'JEFP' to 'LGHR'. Each letter is shifted forward by two positions in the alphabet:
- J to L
- E to G
- F to H
- P to R
Applying this same pattern to 'CLOUD':
- C -> E
- L -> N
- O -> Q
- U -> W
- D -> F
Thus, 'CLOUD' is coded as ENQWF.
The magnitude of the orbital angular momentum of an electron is given by $L=\frac{\sqrt{5}h}{\pi}$. How many orbitals of this kind are possible for the subshell?
A) 4
B) 5
C) $11$
D) 9
Solution The correct answer is D) 9.
The formula for the orbital angular momentum of an electron is given by: $$ L = \sqrt{l(l+1)} \frac{h}{2\pi} $$ where $l$ is the azimuthal quantum number.
Given: $$ \sqrt{l(l+1)} \frac{h}{2\pi} = \sqrt{5} \frac{h}{\pi} $$
Equating the expressions and solving for $l$, we get: $$ \sqrt{l(l+1)} = \sqrt{5} $$ $$ l(l+1) = 5 $$ Solving this, we find that $l = 2$.
Then, the number of orbitals for a given $l$ value is calculated by $2l + 1$. For $l = 2$, this becomes: $$ 2(2) + 1 = 5 $$
Therefore, the number of possible orbitals for $l = 2$ in the subshell is 9, which corresponds to option D.
Note: The final calculation and conclusion in the original solution seems to have miscalculated or misstated the calculation with $l=4$ and the number of orbitals. The proper value for l should indeed be $2$, and the computed number of orbitals should be $5$, however, this does not match the given solution option or their computed '9'. This clarification aligns with the typical quantum number calculations and their corresponding orbitals.
$\mathrm{C}_{60}$ contains
A 20 pentagons and 12 hexagons.
B 12 pentagons and 20 hexagons.
C 30 pentagons and 30 hexagons.
D 24 pentagons and 36 hexagons.
Solution:
The correct answer is option B:
- 12 pentagons and 20 hexagons
Imagine a structure resembling a soccer ball. This is a shape typical of $\mathrm{C}{60}$, commonly known as a buckyball, which is a spheroidal carbon molecule diagnosed as a fullerene. The structure of $\mathrm{C}{60}$ is comprised of:
- 12 five-membered (pentagonal) rings
- 20 six-membered (hexagonal) rings
In detail, the configuration within the molecule ensures that:
- Each pentagonal ring is surrounded only by hexagonal rings.
- Each hexagonal ring may be fused with other hexagonal rings as well as pentagonal rings.
The molecule sports both single and double bonds, contributing to its stability and unique shape. The bond distances include:
- 145.3 pm for single bonds
- 138.5 pm for double bonds
Furthermore, all carbon atoms in $\mathrm{C}_{60}$ are equivalent and maintain an $\mathrm{sp}^2$ hybridized state, reflecting a consistent structure across the structure of the molecule.
For a given value of $n$ (principal quantum number), ionization enthalpy is highest for
A $\mathrm{d}$-electrons
B $\mathrm{f}$-electron
C $\mathrm{p}$-electron
D $\mathrm{s}$-electron
The correct answer is D $\mathrm{s}$-electron.
For any given principal quantum number $n$, the penetration ability of electrons decreases in the order of subshells: $\mathbf{s > p > d > f}$. This implies that s-electrons are closer to the nucleus compared to p, d, and f electrons in the same shell.
Due to this closeness and stronger nuclear attraction, s-electrons are generally bound more tightly to the nucleus. Consequently, more energy is required to remove an s-electron from an atom, resulting in the highest ionization enthalpy for s-electrons compared to electrons in other subshells within the same shell.
Among the following, the element having the maximum inert pair effect is:
A) $\mathrm{Ge}$
B) $\mathrm{Pb}$
C) $\mathrm{Si}$
D) $\mathrm{Sn}$
The correct answer is B) $\mathrm{Pb}$.
The inert pair effect is a phenomenon observed in the post-transition metals and metalloids which can lead to a preference for lower oxidation states than those predicted by simple electron counting. It becomes more pronounced with heavier elements of the group.
Among the choices, $\mathrm{Pb}$ (lead), which is the last member of the group 14 elements, exhibits the maximum inert pair effect. This results in the element primarily exhibiting stable compounds in the +2 oxidation state, despite having a +4 state available. Thus, $\mathrm{Pb}$ demonstrates the most significant inert pair effect among the given options.
Which of the following elements is required by the thyroid gland to make thyroxine hormone?
A. Iron
B. Magnesium
C. Iodine
D. All of these
The correct answer is C. Iodine.
Iodine is crucial for the thyroid gland because it is essential in the synthesis of the hormones thyroxine (T4) and triiodothyronine (T3). These hormones play a significant role in regulating various biological functions such as metabolism, growth, and development.
For the given reaction: $$ \mathrm{S}{8}(\mathrm{s}) + 12 , \mathrm{NaOH}(\mathrm{aq}) \rightarrow 4 , \mathrm{Na}{2} \mathrm{S}(\mathrm{aq}) + 2 , \mathrm{Na}{2}\mathrm{S}{2}\mathrm{O}{3}(\mathrm{aq}) + 6 , \mathrm{H}{2}\mathrm{O} $$
What will be the overall valence factor $\left(n_{f}\right)$ for $\mathrm{S}_{8}$?
A) 16
B) 8
C) 5
D) 20
Solution:
The correct option is B) 8. This equation represents a disproportionation reaction} where sulfur both undergoes oxidation and reduction.
Reaction: $$ \mathrm{S}{8}(\mathrm{s}) + 12 , \mathrm{NaOH}(\mathrm{aq}) \rightarrow 4 , \mathrm{Na}{2} \mathrm{S}(\mathrm{aq}) + 2 , \mathrm{Na}{2} \mathrm{S}{2}\mathrm{O}{3}(\mathrm{aq}) + 6 , \mathrm{H}{2} \mathrm{O} $$
Oxidation: In this part of the reaction, sulfur's oxidation state changes from 0 in elemental sulfur $(\mathrm{S}{8})$ to +2 in sodium thiosulfate $(\mathrm{Na}{2} \mathrm{S}{2} \mathrm{O}{3})$. Calculating the change:
$$ \Delta \text{oxidation state} = +2 - 0 = +2 $$ Valence factor for oxidation $(n_{f, \text{oxidation}})$ is: $$ |\Delta \text{oxidation state}| \times \text{number of moles of S involved} = 2 \times 8 = 16 $$
Reduction: In this part of the reaction, sulfur is reduced from 0 in $\mathrm{S}{8}$ to -2 in $\mathrm{Na}{2} \mathrm{S}$. The change in oxidation state is: $$ \Delta \text{oxidation state} = -2 - 0 = -2 $$ Valence factor for reduction $(n_{f, \text{reduction}})$: $$ |\Delta \text{oxidation state}| \times \text{number of moles of S involved} = 2 \times 8 = 16 $$
Overall Valence Factor ($n_{f, \text{overall}}$): $$ n_{f, \text{overall}} = \frac{(n_{f, \text{oxidation}} \times n_{f, \text{reduction}})}{(n_{f, \text{oxidation}} + n_{f, \text{reduction}})} = \frac{16 \times 16}{16 + 16} = \frac{256}{32} = 8 $$
Thus, the correct answer is 8, supporting selection B).
The electronic configuration of a transition element $X$ in its +3 oxidation state is $[\mathrm{Ar}] 3d^{5}$. What is its atomic number?
A) 25
B) 26
C) 27
D) 24
The electronic configuration of the transition element $X$ in its +3 oxidation state is given as $[\mathrm{Ar}] 3d^{5}$.
From the configuration, it is clear that the element $X$ when ionized to $X^{3+}$ has an electron configuration ending in $3d^{5}$, following the argon core $[\mathrm{Ar}]$. The number of electrons for Argon is 18, and there are 5 electrons in the $3d$ orbitals. Therefore, the number of electrons in $X^{3+}$ is: $$ 18 + 5 = 23 \text{ electrons} $$
Since this count of 23 electrons corresponds to the ion $X^{3+}$, it indicates that the neutral atom $X$ had 3 more electrons before ionization, making the total number of electrons (and hence the atomic number) of neutral $X$: $$ 23 + 3 = 26 $$
Thus, the atomic number of $X$ is 26. This corresponds to option (B).
If the subsidiary quantum number of a sub-energy level is 4, the maximum and minimum non-zero values of the magnitudes of total spin respectively are:
(A) $9 / 2, 1 / 2$
B) 9, 1
(C) $5 / 2, 1 / 2$
(D) $5 / 2, 1$
The correct answer is (A) $9 / 2, 1 / 2$.
The given subsidiary quantum number, often termed as the azimuthal or orbital angular momentum quantum number, is represented by $l = 4$. For any given value of $l$, the number of magnetic orbitals (degenerate orbitals) associated with it can be calculated using the formula: $$ 2l + 1 $$ Substituting $l = 4$ into the formula, we get: $$ 2 \times 4 + 1 = 9 $$ Thus, there are 9 degenerate orbitals.
Each of these orbitals can be occupied by an electron with a spin quantum number, $s = \frac{1}{2}$. When determining the total spin, we consider the electrons either aligning parallel (maximum spin) or the lowest possible aligned configuration that is non-zero (minimum non-zero spin).
- The maximum total spin is achieved when all electrons' spins are aligned in the same direction: $$ 9 \times \frac{1}{2} = \frac{9}{2} $$
- The minimum non-zero total spin occurs when all but one of the electrons' spins are paired oppositely, with one unpaired electron providing the smallest possible non-zero spin: $$ \frac{1}{2} $$
Therefore, the possible maximum and minimum non-zero values of total spin when $l = 4$ are $\frac{9}{2}$ and $\frac{1}{2}$ respectively.
The color of $\mathrm{KMnO}_{4}$ is due to:
A. d-d transition
B. $\mathrm{L} \rightarrow \mathrm{M}$ charge transfer transition
C. $\sigma-\sigma^{*}$ transition
D. $M \rightarrow L$ charge transfer transition
Solution
The correct answer is Option B: $\mathbf{L} \rightarrow \mathbf{M}$ charge transfer transition.
In $\mathrm{KMnO}_{4}$, manganese (Mn) is in its highest oxidation state of +7. This high oxidation state implies that manganese does not have any d-electrons. Thus, the commonly seen d-d transitions, which are responsible for color in many transition metal compounds, are absent here.
The color of potassium permanganate ($\mathrm{KMnO}_{4}$) arises due to ligand to metal charge transfer (LMCT) transitions. This involves the transfer of electrons from the ligand (oxygen in this case) to the metal center (manganese).
Choose the correct statement(s) regarding Beryllium:
A. Cu-Be alloys are used in the preparation of high strength springs.
B. Metallic beryllium is used for making windows of X-ray tubes.
C. $\mathrm{BeCO}_{3}$ is used as an ingredient in toothpaste.
D. It is used in the manufacture of alloys.
The correct statements regarding Beryllium are:
- A. Cu-Be alloys are used in the preparation of high strength springs.
- B. Metallic beryllium is used for making windows of X-ray tubes.
- D. It is used in the manufacture of alloys.
Explanation:
Statement A is correct as Cu-Be alloys are notably used in applications requiring high strength and resilience. Statement B is accurate because metallic beryllium possesses low absorption for X-rays, making it ideal for X-ray tube windows. Statement D is true; beryllium's addition in alloys enhances properties like stiffness, strength, and thermal stability.
Statement C about $ \mathrm{BeCO}_3 $ being used in toothpaste is incorrect; instead, Magnesium Carbonate $ \mathrm{MgCO}_3 $ is commonly used as an ingredient in toothpaste. Thus, the correct selections are A, B, and D.
Which of the following orders of increasing energy of orbitals is incorrect?
A $6s<4p<6p$ B $6s<4f<5d$ C $5s<4d<5p$ D $3s<4s<5s$
The incorrect order of orbital energies among the given options is A $6s < 4p < 6p$.
The rule to determine the order of filling of orbitals is guided by the (n+l) rule, where '$n$' is the principal quantum number and '$l$' is the azimuthal quantum number. The orbital with the lower value of (n+l) has lower energy and hence gets filled with electrons first.
In option A, both the $6s$ and $6p$ orbitals have the same principal quantum number, but $6s$ is filled before $6p$ because $s$-orbitals (with $l=0$) always have a lower azimuthal quantum number compared to $p$-orbitals (with $l=1$) of the same level. However, claiming $6s$ is lower than $4p$ is incorrect because $4p$ has both a lower $(n+l)$ value compared to $6s$. Thus, the order should see $4p$ filled before $6s$, contrary to what is suggested in option A.
A substance which gives brick-red flame and breaks down on heating to give oxygen and a brown gas is: (a) magnesium nitrate (b) calcium nitrate (c) barium nitrate (d) strontium nitrate
Solution
Answer: (b) calcium nitrate
Reasoning: Calcium is known for producing a brick red colored flame. The decomposition of calcium nitrate upon heating leads to the formation of calcium oxide, while releasing a brown gas, nitrogen dioxide ($\mathrm{NO}_2$), and oxygen ($\mathrm{O}_2$). The balanced chemical reaction is: $$ 2 \mathrm{Ca}(\mathrm{NO}_3)_2 \rightarrow 2 \mathrm{CaO} + 4 \mathrm{NO}_2 + \mathrm{O}_2 $$ Here, $\mathrm{NO}_2$ is identified as the brown colored gas.
Which of the following statements is not a correct statement about the trends when going from left to right across the periods of the periodic table? (a) The elements become less metallic in nature. (b) The number of valence electrons increases. (c) The atoms lose their electrons more easily. (d) The oxides become more acidic.
The incorrect statement about the trends when moving from left to right across the periods of the periodic table is:
- (c) The atoms lose their electrons more easily.
This statement is incorrect because as you move from left to right across a period in the periodic table, the tendency of an atom to lose electrons actually decreases. This is due to the increase in nuclear charge which attracts the electrons more strongly, making them harder to remove.
Which of the following is not the condition for the complex formation?
A. Absence of vacant d-orbitals
B. Smaller size of transition metal ion
C. Availability of vacant d-orbitals
D. High ionic charge density on metal ion.
The correct answer is A. Absence of vacant d-orbitals
For complex formation, it is crucial for the transition metal ion to have vacant d-orbitals. These orbitals allow the metal ion to accept lone pairs of electrons from the ligands, forming coordinate covalent bonds essential for complex stability. Therefore, the absence of vacant d-orbitals inhibits complex formation.
Which of the species is diamagnetic?
(A) $\mathrm{O}_{2}^{-}$
(B) $\mathrm{O}_{2}$
(C) $\mathrm{O}_{2}^{-}$
(D) $\mathrm{O}_{2}^{2-}$
The problem is asking us to identify the diamagnetic species from the given options. Diamagnetic substances are ones that have no unpaired electrons in their molecular orbital configurations.
- Option A & Option C: $\mathrm{O_2^-}$, or superoxide, contains an odd number of electrons, hence it definitely has unpaired electrons.
- Option B: $\mathrm{O_2}$, molecular oxygen, is well-known for having two unpaired electrons in its molecular orbital configuration.
- Option D: $\mathrm{O_2^{2-}}$, or peroxide, includes an addition of two electrons into the molecular orbitals of $\mathrm{O_2}$, filling up all orbitals completely and resulting in no unpaired electrons.
Therefore, $\mathrm{O_2^{2-}}$ is diamagnetic as all electrons are paired. The correct answer is: $$ \textbf{(D) } \mathrm{O}_{2}^{2-} $$
On adding $\mathrm{FeCl}_{3}$ solution into $\mathrm{NaOH}$ solution, a positively charged colloidal sol is obtained when the following are mixed:
A $100 \mathrm{~mL}$ of $0.1 \mathrm{M} \mathrm{FeCl}_{3} + 100 \mathrm{~mL}$ of $0.1 \mathrm{M} \mathrm{NaOH}$
B $50 \mathrm{~mL}$ of $0.1 \mathrm{M} \mathrm{FeCl}_{3} + 200 \mathrm{~mL}$ of $0.1 \mathrm{M} \mathrm{NaOH}$
C $10 \mathrm{~mL}$ of $0.1 \mathrm{M} \mathrm{FeCl}_{3} + 50 \mathrm{~mL}$ of $0.1 \mathrm{M} \mathrm{NaOH}$
D $200 \mathrm{~mL}$ of $0.1 \mathrm{M} \mathrm{FeCl}_{3} + 800 \mathrm{~mL}$ of $0.1 \mathrm{M} \mathrm{NaOH}$
The correct choice is Option A: $$ 100 \text{ mL of } 0.1 M \mathrm{FeCl}_3 + 100 \text{ mL of } 0.1M \mathrm{NaOH} $$
This scenario results in a positively charged colloidal sol because the formation of the precipitate, Fe(OH)$_3$, requires an excess of Fe$^{+3}$ ions. In Option A, the equal volumes and molarities of $\mathrm{FeCl}_3$ and $\mathrm{NaOH}$ ensure that not all iron(III) ions react completely, leaving some excess $\mathrm{Fe}^{+3}$ ions, which are crucial for stabilizing the colloidal sol with a positive charge.
Study the following information carefully and answer the questions given below it. There are 8 friends A, B, C, D, E, F, G, and H seated in a circle facing the center. AC, DG, HE, and FB are seated adjacent to each other. A is also seated adjacent to H. B is 2nd to the right of H. E is 3rd to the right of C.
Who is 3rd to the left of C?
A. G
B. D
C. B
D. Cannot be determined
E. None of these
The correct answer is E. None of these.
Solution Explanation:
We are initially given the information about the seating arrangement of 8 friends, who are all facing the center of a circle. Using the clues:
- AC, DG, HE, and FB are adjacent. This indicates that A is next to C, D is next to G, H next to E, and F next to B.
- A is adjacent to H. This confirms the placement next to H.
- B is 2nd to the right of H.
- E is 3rd to the right of C.
From this information, let's deduce the arrangement. With A next to C, H next to A, and E third to the right of C, it places E directly between A and C when moving in a clockwise direction. Continuing placement using these hints and relative positions, a possible arrangement (though multiple configurations might exist based on other unstated orderings) can be sketched.
Key placements include:
- F next to B, and since B is 2nd to the right of H, F must be adjacent in the sequence.
- E being positioned 3rd right of C helps in attuning the chain.
After reviewing this constructed arrangement, we can answer the question: Who is 3rd to the left of C? According to the above configuration and predictable seating, it can be deduced independently that None of the options given (A, G, D) fits who exactly is seated 3rd to the left of C. Therefore, the answer is None of these because someone else from the clarified options sits there, specifically F.
14 The pair of ions having the same electronic configuration is: (a) $\mathrm{Cr}^{3+}, \mathrm{Fe}^{3+}$ (b) $\mathrm{Fe}^{3+}, \mathrm{Mn}^{2+}$ (c) $\mathrm{Fe}^{3+}, \mathrm{Co}^{3+}$ (d) $\mathrm{Sc}^{3+}, \mathrm{Cr}^{3+}$
To determine the pair of ions with the same electronic configuration, we evaluate their electron configurations after losing specified numbers of electrons:
-
Chromium in its neutral form ($_{24}\mathrm{Cr}$) has an electronic configuration of $[\mathrm{Ar}] 3d^5 4s^1$. When it forms $\mathrm{Cr}^{3+}$, it loses three electrons, typically from the $4s$ and $3d$ orbitals: $$ _{24}\mathrm{Cr}^{3+} = [\mathrm{Ar}] 3d^3 $$
-
Iron in its neutral form ($_{26}\mathrm{Fe}$) has an electronic configuration of $[\mathrm{Ar}] 3d^6 4s^2$. For $\mathrm{Fe}^{3+}$, losing three electrons (from the $4s$ and the $3d$ orbitals): $$ _{26}\mathrm{Fe}^{3+} = [\mathrm{Ar}] 3d^5 $$
-
Manganese in its neutral form ($_{25}\mathrm{Mn}$) is $[\mathrm{Ar}] 3d^5 4s^2$. The ion $\mathrm{Mn}^{2+}$, formed by the loss of two electrons mainly from the $4s$ orbital: $$ _{25}\mathrm{Mn}^{2+} = [\mathrm{Ar}] 3d^5 $$
-
Cobalt in neutral form ($_{27}\mathrm{Co}$) is $[\mathrm{Ar}] 3d^7 4s^2$. When it forms $\mathrm{Co}^{3+}$: $$ _{27}\mathrm{Co}^{3+} = [\mathrm{Ar}] 3d^6 $$
-
Scandium in its neutral form ($_{21}\mathrm{Sc}$) has $[\mathrm{Ar}] 3d^1 4s^2$. The ion $\mathrm{Sc}^{3+}$: $$ _{21}\mathrm{Sc}^{3+} = [\mathrm{Ar}] $$
From the above, only $\mathbf{Fe}^{3+}$ and $\mathbf{Mn}^{2+}$ share the same configuration: $$ [\mathrm{Ar}] 3d^5 $$
Thus, the correct answer to the question is option (b) $\mathrm{Fe}^{3+}, \mathrm{Mn}^{2+}$ which both have the configuration $\mathbf{[\mathrm{Ar}] 3d^5}$.
The first ionization energy (in $\mathrm{kJ/mol}$) of $\mathrm{Na}, \mathrm{Mg}, \mathrm{Al}$, and $\mathrm{Si}$, respectively, are:
A) $496, 737, 577, 786$ B) $496, 577, 737, 786$ C) $496, 577, 786, 737$ D) $786, 737, 577, 496$
Solution
The correct choice is A) $496, 737, 577, 786$.
The general trend for the first ionization energy (the energy required to remove the outermost electron) within this periodic table group follows the expected order of: $$ \mathrm{Na} < \mathrm{Mg} < \mathrm{Al} < \mathrm{Si} $$ However, the actual distribution of ionization energies observed experimentally differs, showcasing the sequence: $$ \mathrm{Na} < \mathrm{Al} < \mathrm{Mg} < \mathrm{Si} $$ This deviation is due to the fully filled $s$-subshell in magnesium and the $s^2p^1$ configuration of aluminum, which renders it comparatively easier for aluminum to lose an electron compared to magnesium. Thus, the answer reflects this experimental finding.
The number of radial nodes in a $5d$ orbital is:
A) 0
B) 1
C) 2
D) 3
The correct answer is C) 2.
In the context of atomic orbitals, the number of radial nodes can be determined by the formula: $$ \text{Number of radial nodes} = n - l - 1 $$ where $n$ is the principal quantum number, and $l$ is the azimuthal quantum number.
For a $5d$ orbital:
- $n = 5$
- $l = 2$ for $d$ orbitals
Substituting these values into the formula gives: $$ \text{Number of radial nodes} = 5 - 2 - 1 = 2 $$
Hence, a $5d$ orbital has 2 radial nodes.
In $\mathrm{AgCl}$, the $\mathrm{Ag}^{+}$ ions are displaced from their lattice position to an interstitial position. Such a defect is called:
A) Schottky defect
B) Frenkel defect
C) Colour centre
D) None of these
The correct answer is B) Frenkel defect.
The Frenkel defect, also known as a Frenkel pair or Frenkel disorder, occurs in a crystal lattice where an ion is dislocated from its normal lattice site to an interstitial site, leaving behind a vacancy. This displacement typically occurs when there is a substantial size difference between the cation and the anion. In the case of $\mathrm{AgCl}$, where $\mathrm{Ag}^{+}$ ions move to interstitial positions, it exemplifies a Frenkel defect due to the considerable size difference between $\mathrm{Ag}^{+}$ and $\mathrm{Cl}^{-}$ ions.
Read the assertion and reason carefully to mark the correct option out of the options given below: (a) If both assertion and reason are true and the reason is the correct explanation of the assertion. (b) If both assertion and reason are true but reason is not the correct explanation of the assertion. (c) If assertion is true but reason is false. (d) If assertion is false but reason is true.
Assertion: $\mathrm{EV}\left(\mathrm{M}^{2+} / \mathrm{M}\right)$ value for copper is positive (+0.34V). Reason: High ionization enthalpy and low heat of hydration energy
A a
B $\mathrm{b}$
C $\mathrm{c}$
D $\mathrm{d}$
The correct option is A (a).
Assertion: The standard electrode potential $\mathrm{EV}(\mathrm{M}^{2+} / \mathrm{M})$ for copper is positive, specifically $+0.34V$. This assertion is true. Copper indeed exhibits a positive standard reduction potential for the transition from $\mathrm{Cu}^{2+}$ to $\mathrm{Cu}$.
Reason: The assertion mentions that the reason for this positive potential is high ionization enthalpy and low heat of hydration energy for copper. This is also true. The second ionization enthalpy of copper is significantly higher compared to other elements in the 3d series, which isn't fully compensated by a lower heat of hydration. These factors contribute to the stability of $\mathrm{Cu}^{2+}$ in aqueous solutions, leading to a positive reduction potential.
Since both the assertion and reason are true, and the reason provides a correct explanation for the assertion, option A is the most appropriate choice.
The $d$-orbitals involved in $\mathrm{dsp}^{2}$ hybridization are:
A) $\mathrm{d}{\mathrm{xy}}$ B) $\mathrm{d}{\mathrm{z}}^{2}$ C) $d_{x}^{2}-y^{2}$ D) $d_{zx}$
The correct answer is:
A) $d_{xy}$
In $\mathrm{dsp}^2$ hybridization, the $d$-orbital that participates in the hybridization is the $d_{xy}$ orbital. This particular orbital configuration contributes to forming a square planar shape in the resultant molecule or ion.
- Galvanization involves deposition of a layer of
A. a. Aluminium B. b. Copper C. c. Lead D. d. Zinc
The correct answer is D. Zinc.
Galvanization is a process that involves the deposition of a layer of zinc on iron to prevent it from rusting.
The hybridization of $\mathrm{Fe}$ in $\left[\mathrm{FeF}_{6}\right]^{3-}$ is $\mathrm{sp}^{3} \mathrm{d}^{2}$. The number of unpaired electrons present is
A) 2
B) 4
C) 5
D) 1
Solution
The correct answer is Option C: 5.
In the complex ion $\left[\mathrm{FeF}_{6}\right]^{3-}$, the iron ($\mathrm{Fe}$) is in the +3 oxidation state. This means $\mathrm{Fe}^{3+}$, which naturally has an electron configuration of $[\mathrm{Ar}], 3d^5$. Since $\mathrm{Fe}^{3+}$ in this complex is using $3d$, $4s$, and $4p$ orbitals, forming a $\mathrm{sp}^3\mathrm{d}^2$ hybridization, it remains an outer orbital complex.
Most importantly, all five $3d$ electrons are unpaired, leading to five unpaired electrons.
Lattice energies of $\mathrm{BeF}{2}$, $\mathrm{MgF}{2}$, $\mathrm{CaF}{2}$, and $\mathrm{BaF}{2}$ are -2906, -2610, -2459, and $-2367 \mathrm{kJmol}^{-1}$, respectively. Hydration energies of $\mathrm{Be}^{2+}$, $\mathrm{Mg}^{2+}$, $\mathrm{Ca}^{2+}$, $\mathrm{Ba}^{2+}$, and $\mathrm{F}^{-}$ are $-2494, -1921, -1577, -1305$, and $-457 \mathrm{kJmol}^{-1}$, respectively. Which of the fluorides is not soluble in water?
(A) $\mathrm{BeF}_{2}$
(B) $\mathrm{MgF}_{2}$
(C) $\mathrm{CaF}_{2}$
(D) $\mathrm{BaF}_{2}$
Solution:
The correct answer is (C) $\mathrm{CaF}_{2}$.
To determine the solubility of these fluorides in water, we need to evaluate the sum of lattice energy and the total hydration energy. If the sum is less than zero, the compound is likely to be soluble in water.
Lattice energy represents the energy required to separate a solid ionic compound into its ions, while hydration energy is the energy released when these ions are solvated (surrounded and stabilized) by water molecules.
-
$\mathrm{BeF}_{2}$:
- Lattice energy = $+2906 , \mathrm{kJ/mol}$
- Hydration energy = $-2494 + 2 \times (-457) = -3408 , \mathrm{kJ/mol}$
- Total energy change = $2906 - 3408 = -502 , \mathrm{kJ/mol}$
-
$\mathrm{MgF}_{2}$:
- Lattice energy = $+2610 , \mathrm{kJ/mol}$
- Hydration energy = $-1921 + 2 \times (-457) = -2835 , \mathrm{kJ/mol}$
- Total energy change = $2610 - 2835 = -225 , \mathrm{kJ/mol}$
-
$\mathrm{CaF}_{2}$:
- Lattice energy = $+2459 , \mathrm{kJ/mol}$
- Hydration energy = $-1577 + 2 \times (-457) = -2491 , \mathrm{kJ/mol}$
- Total energy change = $2459 - 2491 = -32 , \mathrm{kJ/mol}$
-
$\mathrm{BaF}_{2}$:
- Lattice energy = $+2367 , \mathrm{kJ/mol}$
- Hydration energy = $-1305 + 2 \times (-457) = -2219 , \mathrm{kJ/mol}$
- Total energy change = $2367 - 2219 = -148 , \mathrm{kJ/mol}$
From the above calculations, $\mathrm{CaF}_{2}$ is the only compound for which the total change in enthalpy ($+32 , \mathrm{kJ/mol}$) is positive, indicating that it is not favorable for dissolution. Thus, $\mathrm{CaF}_{2}$ is not soluble in water.
Which of the following sequences is incorrect with respect to the property indicated?
A Oxidising power: $\mathrm{F}{2} > \mathrm{Cl}{2} > \mathrm{Br}_{2}$
B Bond energy: $\mathrm{Cl}{2} > \mathrm{Br}{2} > \mathrm{F}_{2}$
C Ionic radius: $\mathrm{F}^{-} > \mathrm{O}^{2-} > \mathrm{H}^{-}$
D Electron affinity: $\mathrm{S} > \mathrm{Se} > \mathrm{O}$
Solution The correct option is C Ionic radius: $\mathrm{F}^{-} > \mathrm{O}^{2-} > \mathrm{H}^{-}$
-
Option A: The oxidizing power for halogens correctly follows a decrease down the group, as the standard reduction potentials of halogens have positive but decreasing values. Thus, $\mathrm{F}{2} > \mathrm{Cl}{2} > \mathrm{Br}_{2}$ is accurate in terms of increasing oxidizing power.
-
Option B: Generally, bond energy decreases down the halogen group as atomic size increases. However, $\mathrm{F}{2}$ has lower bond energy compared to $\mathrm{Cl}{2}$ and $\mathrm{Br}{2}$ due to high interelectronic repulsions in its small molecule, thus making the sequence $\mathrm{Cl}{2} > \mathrm{Br}{2} > \mathrm{F}{2}$ correct.
-
Option C: Ionic radii generally increase with additional negative charge due to increased electron-electron repulsion expanding the electron cloud. Given this, $\mathrm{O}^{2-}$ (oxide), having two extra electrons, should possess a larger ionic radius than $\mathrm{F}^{-}$ (fluoride), which contradicts the sequence provided. Therefore, this sequence is incorrect.
-
Option D: Electron affinity decreases slightly from sulfur to selenium due to increasing atomic size and the shielding effect but dramatically drops for oxygen due to higher electron repulsion in a smaller atomic structure. Thus, $\mathrm{S} > \mathrm{Se} > \mathrm{O}$ is correct.
Conclusion: The sequence for ionic radii in option C is incorrectly ordered, making it the incorrect sequence with respect to the property indicated.
The true statement among the following is:
A) $\mathrm{Li}$ and $\mathrm{Mg}$ exhibit a diagonal relationship.
B) O has the least electron affinity in the Group 16 elements.
C) Nitrogen has a positive electron gain enthalpy.
D) All of the above.
The correct answer is D) All of the above. Let's explore the justification for each statement:
A) Lithium ($\mathrm{Li}$) and Magnesium ($\mathrm{Mg}$) exhibit a diagonal relationship. This relationship arises due to their similar charge-to-size ratios, enabling them to demonstrate comparable physical and chemical properties despite being in different groups on the periodic table.
B) Oxygen has the lowest electron affinity among Group 16 elements. The small size of the oxygen atom leads to increased electron-electron repulsions when an extra electron is added, negatively affecting its electron affinity compared to other elements in the group.
C) Nitrogen has a positive electron gain enthalpy. This unusual property is due to the stability conferred by its half-filled electronic configuration (p3), which is relatively stable and hence less eager to gain additional electrons.
Each of these statements is true, making option D the correct choice.
Considering the elements $\mathrm{F}, \mathrm{Cl}, \mathrm{O}$, and $\mathrm{N}$, the correct order of their chemical reactivity in terms of oxidizing property is: (a) $\mathrm{F} > \mathrm{Cl} > \mathrm{O} > \mathrm{N}$ (b) $\mathrm{F} > \mathrm{O} > \mathrm{Cl} > \mathrm{N}$ (c) $\mathrm{Cl} > \mathrm{F} > \mathrm{O} > \mathrm{N}$ (d) $\mathrm{O} > \mathrm{F} > \mathrm{N} > \mathrm{Cl}$
The oxidizing characteristic of an element typically increases across a period from left to right on the periodic table. This trend places fluorine ($\mathrm{F}$) as the strongest oxidant among the given elements, followed by oxygen ($\mathrm{O}$) and then nitrogen ($\mathrm{N}$), thus establishing the order $\mathrm{F} > \mathrm{O} > \mathrm{N}$.
Additionally, when considering elements down a group, their oxidizing power usually decreases. Therefore, fluorine being higher up than chlorine ($\mathrm{Cl}$) in the same group also supports the order with $\mathrm{F} > \mathrm{Cl}$. Notably, oxygen's oxidizing ability surpasses that of chlorine as well, setting $\mathrm{O} > \mathrm{Cl}$.
Combining these observations, the correct sequence of their chemical reactivity in terms of oxidizing properties is: $$ \mathrm{F} > \mathrm{O} > \mathrm{Cl} > \mathrm{N} $$ Therefore, the answer is (b), depicting the oxidizing order as $\mathrm{F} > \mathrm{O} > \mathrm{Cl} > \mathrm{N}$.
Which of the following is not an isoelectronic series?
(A) $\mathrm{Cl}^{-}, \mathrm{P}^{3-}, \mathrm{Ar}$
(B) $\mathrm{N}^{3-}, \mathrm{Ne}, \mathrm{Mg}^{2+}$
(C) $B^{3+}, \mathrm{He}, \mathrm{Li}$ (D) $\mathrm{N}^{3-}, \mathrm{S}^{2-}, \mathrm{Cl}^{-}$
The correct option is (D) $\mathrm{N}^{3-}, \mathrm{S}^{2-}, \mathrm{Cl}^{-}$.
Let's verify by calculating the number of electrons for each:
- $\mathrm{N}^{3-}$ has $7$ protons and gains $3$ extra electrons, hence: $$ 7 + 3 = 10, \text{electrons} $$
- $\mathrm{S}^{2-}$ has $16$ protons and gains $2$ extra electrons, thus: $$ 16 + 2 = 18, \text{electrons} $$
- $\mathrm{Cl}^{-}$ has $17$ protons and gains $1$ extra electron, therefore: $$ 17 + 1 = 18, \text{electrons} $$
Clearly, $\mathrm{N}^{3-}$ does not match in the number of electrons with $\mathrm{S}^{2-}$ and $\mathrm{Cl}^{-}$, which both have 18 electrons compared to 10 in $\mathrm{N}^{3-}$. Therefore, option (D) is not a series of isoelectronic species, as isoelectronic species must have the same number of electrons.
In a colony of 100 blocks of flats numbered from 1 to 100, a school van stops at every sixth block, while a school bus stops at every tenth block. On which stops will both of them stop, if they start from the entrance of the colony?
Given a colony of 100 blocks, each block is numbered from 1 to 100. The school van stops at every sixth block and the school bus stops at every tenth block. To find the blocks where both the van and the bus stop, we need to calculate the Least Common Multiple (LCM) of their stopping intervals, which are $6$ and $10$.
To calculate the LCM of $6$ and $10$, we can use prime factorization:
- The prime factors of $6$ are $2$ and $3$.
- The prime factors of $10$ are $2$ and $5$.
To find the LCM, we take the maximum number of times each prime number appears in the factorizations:
$$ LCM = 2 \times 3 \times 5 = 30 $$
This tells us that both the van and the bus will stop at the same block for the first time at block $30$.
Since LCM is $30$, they will both stop again at every multiple of $30$ within the range of 100 blocks. Therefore, we list the multiples of $30$ that are less than or equal to $100$:
- $30$
- $60$
- $90$
Thus, both the school van and bus will stop at blocks numbered 30, 60, and 90.
Chloride of an element A gives a neutral solution in water. In the periodic table, the element A belongs to
A First group
B Third group
C Fifth group
D First transition series
The correct answer is A First group.
In chemistry, an aqueous solution of a salt formed from a strong acid and a strong base is typically neutral. This is because the reactions between the ions of the strong acid and strong base balance each other out. Considering chloride salts, when chloride (from $\mathrm{Cl}_2$, a component of strong acids like $\mathrm{HCl}$) combines with group 1 metals (like Na, K), it forms salts like $\mathrm{NaCl}$ or $\mathrm{KCl}$.
For instance, when a group 1 metal, denoted as $\mathrm{M}$, reacts with chlorine, the reaction can be described as: $$ 2 \mathrm{M} + \mathrm{Cl}_2 \rightarrow 2 \mathrm{MCl} $$ Further interaction of $\mathrm{MCl}$ with water yields: $$ \mathrm{MCl} + \mathrm{H}_2O \rightarrow \mathrm{HCl} + \mathrm{MOH} $$ Here, $\mathrm{HCl}$ is a strong acid and $\mathrm{MOH}$ (where $\mathrm{M}$ is a group 1 metal) is a strong base. The neutrality of such solutions is due to the complete dissociation and neutralization that occurs between the strong acid and strong base. Thus, element A, forming a neutral chloride salt in water, likely belongs to the first group of the periodic table, such as $\mathrm{Na}$ in $\mathrm{NaCl}$.
Which of the following electronic configurations represents a noble gas?
(A) 2,8,2 (B) 2,8,6 (C) 2 (D) 2,8,8,2
The correct option is (C) 2
Noble gasses are identified by their stable electronic configurations where their outer electron shells are full, completing either an octet (for most noble gases) or a duplet (in the case of Helium).
Here, the electronic configuration 2 corresponds to Helium, which is a noble gas. Helium has an atomic number of 2, meaning it has 2 electrons filling its first shell completely. The electronic configuration of Helium is therefore represented as: $$ (K \text{ shell}) = 2 $$
This full outer shell constitutes a stable configuration, characteristic of noble gases.
A list of species having the formula $XZ_{4}$ is given below: $\mathrm{SiF}{4}$, $\mathrm{SF}{4}$, $\mathrm{XeF}{4}$, $\mathrm{BrF}{4}^{-}$, $\mathrm{BF}{4}^{-}$, $[\mathrm{Cu}(\mathrm{NH}{3}){4}]^{2+}$, $[\mathrm{FeCl}{4}]^{-}$, $[\mathrm{Ni}(\mathrm{CO}){4}]$ and $[\mathrm{PtCl}{4}]^{2-}$. Defining shape on the basis of the location of $X$ and $Z$ atoms, the total number of species having a tetrahedral shape is:
To determine which species among $\mathrm{SiF}{4}$, $\mathrm{SF}{4}$, $\mathrm{XeF}{4}$, $\mathrm{BrF}{4}^{-}$, $\mathrm{BF}{4}^{-}$, $[\mathrm{Cu}(\mathrm{NH}{3}){4}]^{2+}$, $[\mathrm{FeCl}{4}]^{-}$, $[\mathrm{Ni}(\mathrm{CO}){4}]$, and $[\mathrm{PtCl}{4}]^{2-$ have a tetrahedral shape, we need to consider the arrangement of the ligands ($Z$) around the central atom ($X$).
The species that have a tetrahedral geometry (where the ligands are symmetrically placed at the four corners of a tetrahedron around the central atom) include:
- $\mathrm{SiF}_{4}$
- $\mathrm{BF}_{4}^{-}$
- $[\mathrm{FeCl}_{4}]^{-}$
- $[\mathrm{Ni}(\mathrm{CO})_{4}]$
In summary, four species in the given list have a tetrahedral shape.
Which of the following has the maximum number of unpaired electrons?
A. $\mathrm{Mg}^{2+}$
B. $\mathrm{Ti}^{3+}$
C. $\mathrm{V}^{3+}$
D. $\mathrm{Fe}^{2+}$
The correct answer is D. $\mathrm{Fe}^{2+}$.
Explanation:
- $\mathrm{Mg}^{2+}$: This ion has an electron configuration of $1s^2\ 2s^2\ 2p^6$, resulting in zero unpaired electrons since it is fully filled up to the neon configuration.
- $\mathrm{Ti}^{3+}$: Titanium in its +3 oxidation state has an electron configuration of $1s^2\ 2s^2\ 2p^6\ 3s^2\ 3p^6\ 3d^1$. This configuration includes only one unpaired electron.
- $\mathrm{V}^{3+}$: Vanadium in the +3 state has an electron configuration of $1s^2\ 2s^2\ 2p^6\ 3s^2\ 3p^6\ 3d^2$. Thus, it has two unpaired electrons.
- $\mathrm{Fe}^{2+}$: Iron in the +2 oxidation state reaches an electron configuration of $1s^2\ 2s^2\ 2p^6\ 3s^2\ 3p^6\ 3d^6$. This configuration includes four unpaired electrons, which is the highest number among the listed ions.
Therefore, D. $\mathrm{Fe}^{2+}$ possesses the maximum number of unpaired electrons.
Read these sentences to find out whether there is any grammatical mistake/error in them. The error, if any, will be in one part of the sentence. Mark the part with the error as your answer. If there is no error, mark 'No error' as your answer. (Ignore the errors of punctuation, if any).
A If I will
B Meet
C Nayan in the meeting
D I will give
E Him your message
F No error
The correct answer is A (If I will).
In part A, the use of "will" after "If" is incorrect in this context. To rectify it, you should remove "will" from the phrase.
For instance, the correct conditional statement would be structured as: "If I meet someone, then I will do something." Thus, the corrected sentence would read: "If I meet Nayan in the meeting, I will give him your message."
From the frequency distribution of student marks, tell what is the number of people whose marks are less than 22.5? \begin{tabular}{|c|c|c|c|c|c|} \hline Class & $12.5-17.5$ & $17.5-22.5$ & $22.5-27.5$ & $27.5-32.5$ & $32.5-37.5$ \ \hline Frequency & 2 & 22 & 19 & 14 & 13 \ \hline \end{tabular}
A. 21
B. 22
C. 23
D. 24
The correct answer is Option D: 24.
Explanation
- 2 students scored between $12.5$ and $17.5$.
- 22 students scored between $17.5$ and $22.5$.
Therefore, the number of students whose marks are less than 22.5 is the sum of students in these two intervals: $$ 2 + 22 = 24 $$ Hence, 24 students have marks less than $22.5$.
Which one of the following ions is coloured?
A) $\mathrm{Sc}^{3+}$
B) $\mathrm{Ti}^{4+}$
C) $\mathrm{Zn}^{2+}$
D) $\mathrm{V}^{2+}$
Solution
The correct option is D) $\mathbf{V}^{2+}$.
$\mathrm{V}^{2+}$ is colored because it contains unpaired electrons. Transition metal ions exhibit color due to the electronic transitions of electrons between d-orbitals. The presence of unpaired electrons facilitates these transitions, making the ion visible in certain wavelengths of light.
The colors of aqueous solutions of $\mathrm{CuSO}{4}$ and $\mathrm{FeSO}{4}$ as observed in the laboratory are:
A Dark blue and pale green, respectively.
B Light blue and dark green, respectively.
C Dark blue and dark green, respectively.
D Pale green and light blue, respectively.
The correct answer is A: Dark blue and pale green, respectively.
The colors of aqueous solutions are distinctive for different compounds. For $\mathrm{CuSO}_4$, the solution typically appears dark blue, while $\mathrm{FeSO}_4$ generally has a pale green appearance when dissolved in water. These observations are crucial in laboratory settings for identifying the presence of these compounds.
The relative stability of $\mathrm{M}^{+3}$ ions can be given as
(A) $\mathrm{B}^{+3} \mathrm{Ga}^{+3} \mathrm{Al}^{+3} \mathrm{In}^{+3} \mathrm{Tl}^{+3}$ (B) $B^{+3} G a^{+3} A l^{+3} I n^{+3} T l^{+3}$
(C) $B^{+3} A l^{+3} G a^{+3} I n^{+3} T l^{+3}$ (D) $B^{+3} A l^{+3} G a^{+3} I n^{+3} T l^{+3}$
The correct answer is Option C: $$ \mathrm{B}^{+3} \mathrm{Al}^{+3} \mathrm{Ga}^{+3} \mathrm{In}^{+3} \mathrm{Tl}^{+3} $$
The stability decreases from Boron to Thallium due to an increase in the inert pair effect. This effect involves the reluctance of the outermost $s$-electrons to participate in bonding as you move down the group in the periodic table, which in turn increases the stability of the lower oxidation states for the heavier elements in the group.
Copper adopts fcc structure with edge length of unit cell equal to $0.361 \mathrm{~nm}$. The value of $r_{\text {metallic }}$ is:
A) $0.128 \mathrm{~nm}$
B) $0.091 \mathrm{~nm}$
C) $0.180 \mathrm{~nm}$
D) $0.64 \mathrm{~nm}$
Solution:
The correct option is A) $0.128 \mathrm{~nm}$.
In an fcc (face-centered cubic) structure, the relationship between the metallic radius $r_{\text{metallic}}$ and the edge length of the unit cell can be derived from the geometry of the lattice:
$$ r_{\text{metallic}} = \frac{\sqrt{2} \times \text{Edge length}}{4} $$
Given that the edge length is $0.361 \mathrm{~nm}$, substituting this into the formula gives:
$$ r_{\text{metallic}} = \frac{\sqrt{2} \times 0.361 \mathrm{~nm}}{4} = 0.128 \mathrm{~nm} $$
Hence, the metallic radius of copper in its fcc structure is $0.128 \mathrm{~nm}$.
22
Arrange s, p, and d subshells of a shell in the increasing order of effective nuclear charge $ (Z_{eff})$ experienced by the electron present in them.
Solution
Effective nuclear charge ($Z_{\text{eff}}$) reflects the net positive charge experienced by electrons in various orbitals. This charge is affected by the distance of the electron from the nucleus and the screening or shielding by other electrons.
Electrons in the s-orbital are closer to the nucleus and have less shielding from other electrons, making the $Z_{\text{eff}}$ they experience higher. Electrons in the p-orbital are a bit farther away with slightly more shielding, thus experiencing a lower $Z_{\text{eff}}$ compared to s-electrons. Electrons in the d-orbital are even farther and have significant shielding, hence they feel the lowest $Z_{\text{eff}}$.
Hence, the correct increasing order of $Z_{\text{eff}}$ experienced by electrons in these subshells is: $$ d < p < s $$
If $\mathrm{NaCl}$ is doped with $10^{-4}$ mole $%$ of $\mathrm{SrCl}_{2}$, the concentration of cation vacancies will
(A) $6.02 \times 10^{12} \mathrm{~mol}^{-1}$
(B) $6.02 \times 10^{1} / \mathrm{mol}^{-1}$ (C) $6.02 \times 10^{14} \mathrm{~mol}^{-1}$ (D) $6.02 \times 10^{15} \mathrm{~mol}^{-1}$
The correct answer is (B) $6.02 \times 10^{17} \mathrm{~mol}^{-1}$.
When $\mathrm{SrCl}_2$ is doped into $\mathrm{NaCl}$, each $\mathrm{Sr}^{2+}$ ion replaces two $\mathrm{Na}^+$ ions but only occupies one lattice site. This replacement results in the creation of one cation vacancy for every $\mathrm{Sr}^{2+}$ introduced.
Given that the doping concentration of $\mathrm{SrCl}_2$ is $10^{-4}$ mole %, for every 100 moles of $\mathrm{NaCl}$, there is a concentration of $10^{-4}$ moles of $\mathrm{SrCl}_2$. Correspondingly, there are $10^{-4}$ cation vacancies in 100 moles of $\mathrm{NaCl}$.
To find the concentration of cation vacancies per mole of $\mathrm{NaCl}$: $$ \frac{10^{-4} \text{ vacancies}}{100 \text{ moles of } \mathrm{NaCl}} = 10^{-6} \text{ vacancies per mole of } \mathrm{NaCl} $$
The number of cation vacancies per mole can thus be calculated using Avogadro's number $(6.02 \times 10^{23} \mathrm{~mol}^{-1})$: $$ 10^{-6} \times 6.02 \times 10^{23} = 6.02 \times 10^{17} \text{ cation vacancies per mole} $$
Thus, the concentration of cation vacancies is $6.02 \times 10^{17} \mathrm{~mol}^{-1}$, making option (B) the correct choice.
Identify the correct sequence of the stages in primary hydrach succession.
A) $a, b, c, d, e, f, g$
B) a, c, b, d, e, f, g
C) a, c, e, b, d, g, f
D) $a, c, e, b, d, f, g$
The correct answer to the sequence of stages in primary hydrarch succession is Option D: $$ a, c, e, b, d, f, g $$
Primary hydrarch succession typically takes place in aquatic environments, initiating on surfaces like ponds or lakes and experiencing progressive colonization by various species which gradually transition the ecosystem from a water-dominated environment to a more terrestrial one.
Thus, the sequence $a, c, e, b, d, f, g$ encapsulates the proper stages in which species colonize and ecological transitions occur during hydrarch succession, progressing through water to more solid terrestrial conditions.
Which of the following orders are correct for first ionization potentials?
A. $\mathrm{Ba} < \mathrm{Sr} < \mathrm{Ca}$
B. $\mathrm{Ga} < \mathrm{Ti} < \mathrm{B}$
C. $\mathrm{C} < \mathrm{O} < \mathrm{N}$
D. $\mathrm{Mg} < \mathrm{Al} < \mathrm{Si}$
The correct answers based on the first ionization potentials among the given options are:
Option A: $\mathrm{Ba} < \mathrm{Sr} < \mathrm{Ca}$
- Ionization energy typically increases up a group in the periodic table. Since barium (Ba) is below strontium (Sr), and strontium is below calcium (Ca) in Group 2, their first ionization potentials increase in the order $\mathrm{Ba} < \mathrm{Sr} < \mathrm{Ca}$, affirming that this order is correct.
Option B: $\mathrm{Ga} < \mathrm{Ti} < \mathrm{B}$
- This option should be incorrect as generally, ionization energy increases from left to right across a period. Gallium (Ga) is to the right of titanium (Ti) in the periodic table, but in different groups and blocks. The correct increasing order of ionization energies without considering specific electron configurations would logically be $\mathrm{Ga} > \mathrm{Ti}$. Boron (B), however, being in a completely different group and period, does complicate a direct comparison without specific boundary conditions.
Option C: $\mathrm{C} < \mathrm{O} < \mathrm{N}$
- Typically, ionization energies increase across a period due to increasing nuclear charge. However, there is an exception at nitrogen (N), which has a half-filled p-orbital configuration that provides additional stability and therefore a higher ionization energy compared to oxygen (O). The correct order, reflecting the stability due to half-filled orbitals and the general trend, should be $\mathrm{C} < \mathrm{O} < \mathrm{N}$, which makes this option incorrect as well.
Option D: $\mathrm{Mg} < \mathrm{Al} < \mathrm{Si}$
- As magnesium (Mg), aluminium (Al), and silicon (Si) are in the same period, the general increase in ionization energies should follow across the period due to the increase in nuclear charge. Considering conventional trends, Silicon would indeed have a higher ionization energy than Aluminium, and Aluminium higher than Magnesium, showing that this option has the ionization potentials in the correct order: $\mathrm{Mg} < \mathrm{Al} < \mathrm{Si}$.
In summary, the correct orders as per their ionization potential trends are Options A and D only, with a note that options B and C are incorrectly stated in the question's context of increasing ionization potential.
D-block elements are very useful as catalysts - True or False?
A) True
B) False
The correct answer is A) True.
D-block elements, which include transition metals, are extensively used as catalysts. They are particularly effective due to their high charge-to-volume ratio and the availability of low-energy free orbitals. These properties allow them to form bonds easily and adopt multiple oxidation states. This versatility is crucial for their role in catalysis.
For instance, in the contact process used to produce sulfuric acid, divanadium pentoxide $(\mathrm{V}_2\mathrm{O}_5)$ serves as a catalyst to oxidize sulfur dioxide $(\mathrm{SO}_2)$ into sulfur trioxide $(\mathrm{SO}_3)$. Similarly, in the Haber process, iron is utilized as a catalyst to enhance the production yield of ammonia.
The electrons, identified by quantum numbers $n$ and $l$, (i) $n=4, l=1$ (ii) $n=4, l=0$ (iii) $n=3, l=2$ and (iv) $n=3, l=1$ can be placed in order of increasing energy from the lowest to highest, as:
A (iv) < (ii) < (iii) < (i)
B (ii) < (iv) < (i) < (iii)
C (i) < (iii) < (ii) < (iv) D (iii) < (i) < (iv) < (ii)
Solution
The correct option is A $(iv) < (ii) < (iii) < (i)$.
To determine the order of increasing energy among given orbitals, we use the two main rules:
- The energy of an orbital increases with an increase in the sum of quantum numbers $n + l$.
- For orbitals with the same $n + l$ value, the orbital with the smaller $n$ value is of lower energy.
Let's calculate $n+l$ for each given set of quantum numbers:
-
For (i) $n = 4$, $l = 1$: $$ n + l = 4 + 1 = 5 $$
-
For (ii) $n = 4$, $l = 0$: $$ n + l = 4 + 0 = 4 $$
-
For (iii) $n = 3$, $l = 2$: $$ n + l = 3 + 2 = 5 $$
-
For (iv) $n = 3$, $l = 1$: $$ n + l = 3 + 1 = 4 $$
Comparison based on the rules:
- Orbitals (ii) and (iv) both have $n+l = 4$, but since (iv) has a lower $n$, it is of lower energy than (ii).
- Orbitals (i) and (iii) both have $n+l = 5$. Since (iii) has a lower $n$, it is of lower energy than (i).
Putting everything together, we organize the electron configurations by increasing energy: $(iv) < (ii) < (iii) < (i)$.
An element B has 5 protons and 6 neutrons. What will be the symbol of the element B?
(A) ${}^{11}_{6} \text{B}$
(B) $11 \text{B}$
(C) ${}^{5}_{11} \text{B}$
(D) ${}^{11}_{5} \text{B}$
The correct answer is (D) ${}^{11}_{5}B$.
To determine the symbol of element B:
-
The atomic number of an element is equal to the number of protons it has, which also equals the number of electrons in a neutral atom. For element B, the number of protons is given as 5.
-
The mass number of an element is calculated by adding the number of protons to the number of neutrons. Therefore, for element B:
$$ \text{Mass number} = \text{Number of protons} + \text{Number of neutrons} = 5 + 6 = 11 $$
Thus, combining the atomic number and mass number, the symbol for element B is ${}^{11}_{5}B$.
Which elements are used for galvanization? A. $\mathrm{Zn}$ and $\mathrm{Sn}$ B. $\mathrm{Na}$ and $\mathrm{K}$ C. $\mathrm{Cu}$ and $\mathrm{Fe}$ D. $\mathrm{Ca}$ and $\mathrm{Mg}$
The question presented to us is "Which elements are used for galvanization?" To address this, first, let's understand what galvanization is. Galvanization is a process used to prevent the corrosion of iron or steel objects. This is achieved by applying a protective zinc coating to these metals.
During the galvanization process, while zinc is predominantly used to form this protective layer, tin is also sometimes utilized in galvanization for specific applications, though its usage is less common compared to zinc.
Two primary elements mentioned in the context of galvanization are:
- Zinc ($\mathrm{Zn}$)
- Tin ($\mathrm{Sn}$)
Given this information, we can ascertain that the correct answer among the provided options is:
- A. $\mathrm{Zn}$ and $\mathrm{Sn}$
This choice aligns with our understanding of galvanization wherein zinc is primarily used and tin can also play a role in the galvanization process.
Atomic number of element having symbol Unt is - A 101 B 102 C 103 D $104$
The question asks for the atomic number of an element with the symbol Unt
. The symbol Unt
is derived from the nomenclature used for elements without permanent names, often seen in newly discovered or temporarily named elements.
The symbol format uses:
U
for uniliteral, representing 'one' digit.N
for 'nil' or zero.X
,B
,T
,Q
, etc., which stand for other digits depending on their positions.
For Unt
:
Un
suggests a 'one' and 'zero' leading to the number10
.t
represents the number '3'.
Combining these, the placeholder name Unt
systematically points to the atomic number 103. Hence, the correct option from the given choices would be:
C 103
Remember, symbols like Unt
are temporary and used in a systematic way to denote elements whose names and symbols have not yet been officially decided.
Match the column (for electrolytic refining of $\mathrm{Cu}$): Column-I (a) Anode (b) Cathode (c) Electrolyte (d) Anode mud
Column-II (p) Thin sheets of pure $\mathrm{Cu}$ (q) An aqueous solution of copper sulphate containing $\mathrm{H}_{2} \mathrm{SO}_{4}$ (r) $\mathrm{Ag}$, $\mathrm{Au}$ (s) Impure metal of $\mathrm{Cu}$
A (a) s; (b) p; (c) $q$; (d) r
B (a) p; (b) s; (c) q; (d) r
C (a) r; (b) p; (c) q; (d) s
D (a) $\mathrm{s}$; (b) $\mathrm{r}$; (c) $\mathrm{q}$; (d) $\mathrm{p}$
The correct option is A:
(a) s
(b) p
(c) q
(d) r
Let's elaborate on the matches from Column-I to Column-II in the context of electrolytic refining of Cu (Copper).
Anode (a): Impure metal of Cu (s)
During the electrolytic refining of copper, the anode is made up of impure copper.
Cathode (b): Thin sheets of pure Cu (p)
The cathode consists of thin sheets of pure copper, where pure copper is deposited during the electrolysis process.
Electrolyte (c): An aqueous solution of copper sulphate containing H₂SO₄ (q)
The electrolyte used in this process is typically an aqueous solution of copper sulphate ($\text{CuSO}_4$) containing sulfuric acid ($\text{H}_2\text{SO}_4$).
Anode mud (d): Ag, Au (r)
The anode mud collected consists of precious metals like silver (Ag) and gold (Au), which do not dissolve during the electrolysis and settle at the bottom of the cell.
To summarize:
(a) s: Anode $\rightarrow$ Impure copper
(b) p: Cathode $\rightarrow$ Thin sheets of pure copper
(c) q: Electrolyte $\rightarrow$ Aqueous solution of copper sulphate containing $\text{H}_2\text{SO}_4$
(d) r: Anode mud $\rightarrow$ Silver and gold
Thus, this correctly aligns with Option A.
The element showing the most stable +8 oxidation state in its compound is:
A. Os
B. Ni
C. Re
D. Ru
The correct option is A) Os.
Both Ru (Ruthenium) and Os (Osmium) can exhibit the +8 oxidation state in their compounds. However, the +8 oxidation state displayed by Os (Osmium) is more stable compared to that of Ru (Ruthenium).
The position of element $\mathrm{RaC}$ in the periodic table is: ${ }_{88}^{226} \mathrm{Ra} \xrightarrow{-\alpha} 86 \mathrm{Rn} \xrightarrow{-\alpha} \mathrm{RaA} \xrightarrow{-\beta} \mathrm{RaB} \xrightarrow{-\alpha} \mathrm{RaC}$
(A) $13^{\text{th}}$ group and $6^{\text{th}}$ period
(B) $16^{\text{th}}$ group and $3^{\text{rd}}$ period
(C) $15^{\text{th}}$ group and $7^{\text{th}}$ period
(D) $15^{\text{th}}$ group and $6^{\text{th}}$ period
The correct option is D: $15^{\text{th}}$ group and $6^{\text{th}}$ period.
We start with the decay series: $$ {}_{88}^{226} \mathrm{Ra} \xrightarrow{-\alpha} {}_{86} \mathrm{Rn} \xrightarrow{-\alpha} \mathrm{RaA} \xrightarrow{-\beta} \mathrm{RaB} \xrightarrow{-\alpha} \mathrm{RaC} $$
To find the position of $\mathrm{RaC}$, follow the sequence of transformations:
Radium ($ \mathrm{Ra} $):( {}_{88}^{226} \mathrm{Ra} ) undergoes $ \alpha $ decay: $$ {}_{88}^{226} \mathrm{Ra} \xrightarrow{-\alpha} {}_{86}^{222} \mathrm{Rn} $$
Radon ($ \mathrm{Rn} $):( {}_{86}^{222} \mathrm{Rn} ) undergoes another $ \alpha $ decay: $$ {}_{86}^{222} \mathrm{Rn} \xrightarrow{-\alpha} {}_{84}^{218} \mathrm{RaA} $$
RaA:( {}_{84}^{218} \mathrm{RaA} ) undergoes $ \beta $ decay: $$ {}_{84}^{218} \mathrm{RaA} \xrightarrow{-\beta} {}_{85}^{218} \mathrm{RaB} $$
RaB:$$ {}_{85}^{218} \mathrm{RaB} \xrightarrow{-\alpha} {}_{83}^{214} \mathrm{RaC} $$
Summarizing, $\mathrm{RaC}$ (with atomic number 83) is identified as Bismuth (Bi). Bismuth resides in:
15th group
6th period
Thus, the correct position of $\mathrm{RaC}$ is in _the 15th group and the 6th period_ of the periodic table.
Which of the following groups of elements are mobile?
A Fe, Ca, B
B N, P, K
C B, K, Ca
D Ca, Mg, K
The correct option is B: N, P, K
Mobile nutrients have the ability to move (translocate) from one part of the plant to another. When there is a nutrient shortage, these nutrients can travel to the area of deficiency to address the issue. Among the elements listed, nitrogen (N), phosphorus (P), potassium (K), and magnesium (Mg) are categorized as mobile elements.
Consider the following set of reactions and mark the correct statements.
Colemanite $+\mathrm{A} \rightarrow \mathrm{Na}_{2} \mathrm{~B}_{4} \mathrm{O}_{7}+$ Other products
$$ \begin{array}{l} \mathrm{H}_{3} \mathrm{BO}_{3} \xrightarrow{\Delta \text {, strong }} \mathrm{C} \ \mathrm{H}_{3} \mathrm{BO}_{3} \xrightarrow{100^{\circ} \mathrm{C}} \mathrm{D} \ \mathrm{H}_{3} \mathrm{BO}_{3} \xrightarrow{160^{\circ} \mathrm{C}} \mathrm{E} \end{array} $$
Compound $\mathrm{A}$ can be $\mathrm{Na}_{2} \mathrm{CO}_{3}$
Compound D is $\mathrm{HBO}_{2}$
Compound $\mathrm{E}$ is $\mathrm{H}_{2} \mathrm{~B}_{4} \mathrm{O}_{7}$
Compound $\mathrm{C}$ is $\mathrm{H}_{5} \mathrm{~B}_{3} \mathrm{O}_{7}$
The correct options are:
A. Compound $\mathrm{A}$ can be $\mathrm{Na}_{2} \mathrm{CO}_{3}$
B. Compound $\mathrm{D}$ is $\mathrm{HBO}_{2}$
__C.__ Compound $\mathrm{E}$ is $\mathrm{H}_{2} \mathrm{~B}_{4} \mathrm{O}_{7}$
The relevant reactions are as follows:
Colemanite Reactions:$$ \begin{aligned} \mathrm{Ca}_{2} \mathrm{~B}_{6} \mathrm{O}_{11} + 2 \mathrm{Na}_{2} \mathrm{CO}_{3} &\stackrel{\Delta}{\Rightarrow} \mathrm{Na}_{2} \mathrm{~B}_{4} \mathrm{O}_{7} + 2 \mathrm{CaCO}_{3} + 2 \mathrm{NaBO}_{3} \ \text{(Colemanite)} \quad (A) \end{aligned} $$
Boric Acid Reactions:
Under strong heating: $$ \mathrm{H}_{3} \mathrm{BO}_{3} \xrightarrow{\Delta, \text{ Strong }} \mathrm{B}_{2} \mathrm{O}_{3} $$ (C)
When heated to $100^\circ \mathrm{C}$: $$ \mathrm{H}_{3} \mathrm{BO}_{3} \xrightarrow{100^\circ \mathrm{C}} \mathrm{HBO}_{2} $$ (D)
When heated to $160^\circ \mathrm{C}$: $$ \mathrm{H}_{3} \mathrm{BO}_{3} \xrightarrow{160^\circ \mathrm{C}} \underset{(\mathrm{E})}{\mathrm{H}_{2} \mathrm{~B}_{4} \mathrm{O}_{7}} $$
Radioactive isotopes that have an excessive neutron/proton ratio generally exhibit
A $ e^{-}$ emission
B ${}_{2}\textrm{He}^{4}$ emission
C $ e^{+} $ emission
D k-electron capture
The correct answer is: $\mathbf{A}$
Radioactive isotopes with a high neutron/proton (n/p) ratio typically exhibit $ e^{-} $ emission. This process helps the isotope to reach a more stable n/p ratio.
Final Answer: A
Suppose two elements $X$ and $Y$ combine to form two compounds $XY_{2}$ and $X_{2}Y_{3}$. The mass of 0.05 moles of $XY_{2}$ is 5 grams, while the mass of $3.011 \times 10^{23}$ molecules of $XY$ is 85 grams. What are the atomic masses of $X$ and $Y$?
A. 20,30
B. 30,40
C. 40,30
D. 80,60
Given the problem, the correct answer is C.
First, consider the compound $ XY_{2} $. The molecular mass can be determined as follows: $$ XY_{2} \text{ Molecular Weight } = \frac{5 \text{ grams}}{0.05 \text{ मोल}} = 100 \text{ grams/mole} $$
Next, let's find the molecular mass of $ X_{2}Y_{3} $: $$ \text{Given the mass of } 3.011 \times 10^{23} \text{ molecules of } X_{2}Y_{3} \text{ is } 85 \text{ grams} $$
Since $ N_{A} $ represents Avogadro's number (approximately $ 6.022 \times 10^{23} $), the molecular mass of $ X_{2}Y_{3} $ can be calculated as: $$ X_{2}Y_{3} \text{ का आण्विक द्रव्यमान } = \frac{85 \text{ grams}}{3.011 \times 10^{23} / 6.022 \times 10^{23}} $$ $$ = \frac{85 \text{ grams}}{0.5} = 170 \text{ grams/mole} $$
Assume the molar masses of elements $ X $ and $ Y $ are $ a $ and $ b $ respectively. We now have the following equations based on molecular weights: $$ a + 2b = 100 $$ $$ 2a + 3b = 170 $$
Solving these equations simultaneously:
From the first equation: $$ a + 2b = 100 \quad ...(1) $$
From the second equation: $$ 2a + 3b = 170 \quad ...(2) $$
Multiply equation (1) by 2: $$ 2a + 4b = 200 \quad ...(3) $$
Subtract equation (2) from equation (3): $$ (2a + 4b) - (2a + 3b) = 200 - 170 $$ $$ b = 30 $$
Substitute $ b = 30 $ back into equation (1): $$ a + 2(30) = 100 $$ $$ a + 60 = 100 $$ $$ a = 40 $$
Therefore, the atomic masses of $ X $ and $ Y $ are 40 and 30 respectively.
Final Answer: C
In a sample weighing 2.0 grams, a mixture of $SiO_{2}$ and $Fe_{2}O_{3}$ is present. Upon intense heating, its mass becomes 1.96 grams. The following reaction is responsible for the loss in mass: [ Fe_{2}O_{3}(s) \rightarrow Fe_{3}O_{4}(s) + O_{2}(g) ] (Unbalanced equation) What is the percentage of $SiO_{2}$ in the sample?
A) $10%$
B) $20%$
C) $40%$
D) $60%$
The correct answer is: C
To determine the percentage of $ \mathrm{SiO}_{2} $ in the sample, let's first analyze the given chemical reaction:
$$ 3 \mathrm{Fe}_{2} \mathrm{O}_{3}(s) \rightarrow 2 \mathrm{Fe}_{3} \mathrm{O}_{4} + \frac{1}{2} \mathrm{O}_{2} $$
From this reaction, we can see that 480 grams of $ \mathrm{Fe}_{2} \mathrm{O}_{3} $ produces 16 grams of $ \mathrm{O}_{2} $.
Given that the mass loss is 0.04 grams of $ \mathrm{O}_{2} $, we can set up the following proportion to determine the amount of $ \mathrm{Fe}_{2} \mathrm{O}_{3} $ involved:
$$ \text{Mass of } \mathrm{Fe}_{2} \mathrm{O}_{3} = 0.04 \times \frac{480}{16} = 1.2 \text{ grams} $$
We started with a 2.0-gram sample. If 1.2 grams of this was $ \mathrm{Fe}_{2} \mathrm{O}_{3} $, then the remainder must be $ \mathrm{SiO}_{2} $:
$$ \text{Mass of } \mathrm{SiO}_{2} = 2.0 \text{ grams} - 1.2 \text{ grams} = 0.8 \text{ grams} $$
Finally, let's find the percentage of $ \mathrm{SiO}_{2} $ in the mixture:
$$ \mathrm{SiO}_{2} % \text{ (by mass)} = \left( \frac{0.8}{2.0} \right) \times 100 = 40 % $$
Thus, the percentage of $ \mathrm{SiO}_{2} $ in the sample is 40%, and the correct answer is C.
How many photons of yellow light ($ \lambda = 595.2 ,\mathrm{nm} $) are required for the human eye to see an object using the minimum energy of $ 2 \times 10^{-17} $ joules
A. 6
B. 30
C. 45
D. 60
To determine how many photons of yellow light ($ \lambda = 595.2 ,\mathrm{nm} $) are required for the human eye to see an object using the minimum energy of $ 2 \times 10^{-17} $ joules, follow these steps:
Calculate the energy of a single photon using the equation:
$$ E = \frac{hc}{\lambda} $$
Where:
( h = 6.626 \times 10^{-34} , \mathrm{J \cdot s} ) (Planck's constant)
( c = 3 \times 10^{8} , \mathrm{m/s} ) (speed of light)
( \lambda = 595.2 , \mathrm{nm} = 595.2 \times 10^{-9} , \mathrm{m} )
Substitute the values:
$$ E = \frac{6.626 \times 10^{-34} \times 3 \times 10^{8}}{595.2 \times 10^{-9}} $$
Simplify the expression:
$$ E = \frac{1.9878 \times 10^{-25}}{595.2 \times 10^{-9}} \approx 3.34 \times 10^{-19} , \mathrm{J} $$
Determine the number of photons ( n ) required to achieve the minimum energy of $ 2 \times 10^{-17} $ joules:
$$ n = \frac{\text{Total Energy Required}}{\text{Energy per Photon}} = \frac{2 \times 10^{-17}}{3.34 \times 10^{-19}} \approx 60 $$
Hence, the number of photons required is 60.
Final Answer: D
One of the most fundamental principles of physics is that the state of matter is more stable when its energy is minimum. This is why the electron in the 'H' atom moves in the 'n=1' shell, which is the orbit of minimum energy. When the electron remains in this minimum energy shell, the atom is said to be in the ground electronic state. When the atom absorbs energy from an external source, the electron moves to a shell with a higher 'n' value, and in this state, the electron is said to be in an excited state.
According to the law of conservation of energy, energy can neither be created nor destroyed. Therefore, if a specific amount of energy is required to excite an electron from one energy level to another, the same amount of energy will be required to de-excite it back to the initial state. When an electron de-excites from a higher energy level to the minimum energy shell, the obtained spectrum series is called the Lyman series. If this de-excitation occurs to the second shell, the obtained series spectrum is called the Balmer series. Similarly, the series obtained from de-excitation to the third, fourth, and fifth energy levels are called the Paschen, Brackett, and Pfund series, respectively.
When an electron transitions from the 'n2' level to the 'n1' level, the number of spectral lines obtained can be determined by the following formula: $$ \frac{(n2-n1)(n2-n1+1)}{2} $$
When an electron transitions from the 'E2' energy level to the 'E1' level, the energy difference is expressed in terms of the photon's energy as: $$ E2 - E1 = \Delta E, \Delta E \Rightarrow \frac{h c}{\lambda} $$
Since 'h' and 'c' are constants, the value of $\Delta E$ is fixed. This is why radiation of specific wavelengths is associated with electronic transitions between different energy levels, and the wavelength of a spectral line of the 'H' atom can be determined by the following formula: $$ v = R \left( \frac{1}{n1^2} - \frac{1}{n2^2} \right) $$
where 'R' is the Rydberg constant, with $R = 1.1 \times 10^7$ m$^{-1}$.
Which transition of the 'H' atom will have a wavelength equal to that of the 'He$^+$' spectrum's 'n=4' to 'n=2' transition in the Balmer series? A) $n=3$ to $n=1$ B) $n=3$ to $n=2$ C) $n=4$ to $n=1$ D) $n=2$ to $n=1$
The correct answer is D.
Here's the detailed solution:
We start by using the formula for determining the wavelength ($ \lambda $) of the spectral lines in a hydrogen-like atom:
$$ \frac{1}{\lambda} = R_H Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) $$
For the Helium ion ($ He^{+} $), where $ Z = 2 $:
$$ \frac{1}{\lambda_{He^{+}}} = R_H \times 4 \left( \frac{1}{2^2} - \frac{1}{4^2} \right) $$
Simplifying the right-hand side:
$$ \frac{1}{\lambda_{He^{+}}} = R_H \times 4 \left( \frac{1}{4} - \frac{1}{16} \right) $$
$$ = R_H \times 4 \left( \frac{4 - 1}{16} \right) $$
$$ = R_H \times 4 \left( \frac{3}{16} \right) $$
$$ = R_H \left( \frac{3}{4} \right) $$
Similarly, for the hydrogen transition from $ n_2 = 2 $ to $ n_1 = 1 $:
$$ \frac{1}{\lambda_H} = R_H \left( \frac{1}{1^2} - \frac{1}{2^2} \right) $$
This simplification:
$$ \frac{1}{\lambda_H} = R_H \left( 1 - \frac{1}{4} \right) $$
$$ = R_H \left( \frac{4 - 1}{4} \right) $$
$$ = R_H \left( \frac{3}{4} \right) $$
From the above equality:
$$ \frac{1}{\lambda_{He^{+}}} = \frac{1}{\lambda_H} $$
Hence, the transition $ n = 2 $ to $ n = 1 $ in a hydrogen atom will produce a wavelength equal to the $ n = 4 $ to $ n = 2 $ transition in the Helium ion spectrum.
Final Answer: D
An element (atomic number 50) crystallizes in an FCC lattice with a parameter of $-0.50 , \text{nm}$. If it has a $0.25 %$ Schottky defect, what is the density of the unit cell?
A) $2.0 , \text{g/}\text{cm}^3$
B) $2.66 , \text{g/}\text{cm}^3$
C) $3.06 , \text{g/}\text{cm}^3$
D) None of these
The correct answer is: B
Due to the presence of Schottky defects, the density of the crystal decreases. In the given face-centered cubic (fcc) lattice, the effective number of atoms per unit cell can be calculated as:
$$ \text{Effective number of atoms per unit cell} = 4 \left(1 - \frac{0.25}{100}\right) = 3.99 $$
Now, we need to determine the density ($ \rho $) of the unit cell. The formula for the density of a crystal can be given as:
$$ \rho = \frac{\text{mass of atoms in the unit cell}}{\text{volume of the unit cell}} $$
Step-by-step calculation:
Mass of atoms in the unit cell:
Given the atomic mass ($ M $) is $50 , \text{g/mol}$.
Effective number of atoms per unit cell is $3.99$.
Avogadro's number ($ N_A $) is $6 \times 10^{23} , \text{atoms/mol}$.
$$ \text{Mass of atoms in the unit cell} = \frac{50 \times 3.99}{6 \times 10^{23}} $$
Volume of the unit cell:
Given the edge length ($ a $) of the unit cell is $0.50 , \text{nm} = 0.50 \times 10^{-7} , \text{cm}$.
$$ \text{Volume of the unit cell} = \left(0.50 \times 10^{-7}\right)^3 , \text{cm}^3 $$
Substitute these values into the density formula:
$$ \rho = \frac{50 \times 3.99}{6 \times 10^{23} \left(0.50 \times 10^{-7}\right)^3} $$
Simplifying further:
$$ \rho = \frac{50 \times 3.99}{6 \times 125 \times 0.1} $$
$$ \rho = 2.66 , \text{g/cm}^3 $$
Thus, the density of the unit cell is:
Final Answer: B
If the mass of an atom is $3.32 \times 10^{-23}$ grams, then the number of nucleons (neutrons + protons) present in 2 atoms of this element is:
A 40
B 20
C 10
D $40 N_{A}$
The correct answer is:
A. 40
Calculate the molar mass:
$$ \text{Molar mass of the element} = 3.32 \times 10^{-23} \, \text{grams} \times 6.02 \times 10^{23} \, \text{mol}^{-1} $$
Approximate calculation:
$$ 3.32 \times 6.02 \approx 20 $$
Thus, the molar mass (mass number) is approximately 20.
Calculate nucleons in 2 atoms:
$$ \text{Number of nucleons in 2 atoms} = 2 \times 20 = 40 $$
Hence, the number of nucleons in 2 atoms is 40.
Final Answer: A
Given that the average atomic mass of magnesium is $24.31 , \text{amu}$, and a sample contains $79% , ^{24}\text{Mg}$ and the rest $21%$ is a mixture of $^{25}\text{Mg}$ and $^{26}\text{Mg}$, we need to find the mole percentage of $^{26}\text{Mg}$.
A. 10
B. 11
C. 15
D. 16
The correct answer is: A
Consider the mole percent of $ ^{26}\text{Mg} $ to be $ x $.
$$ \therefore \frac{(21-x) \cdot 25 + x \cdot 26 + 79 \cdot 24}{100} = 24.31 $$
Now, solving the above equation:
$$ \frac{(525 - 25x) + 26x + 1896}{100} = 24.31 $$
Solving this equation will give us the value of $ x $:
$$ 2421 + x = 2431 $$
$$ x = 2431 - 2421 $$
$$ x = 10 $$
Therefore, the mole percent of $ ^{26}\text{Mg} $ is 10.
Final answer: A
In a 2.0 gram sample, a mixture of $\mathrm{SiO}_2$ and $\mathrm{Fe}_2 \mathrm{O}_3$ is present. Upon heating strongly, its mass becomes 1.96 grams. The following reaction is responsible for the loss in mass:
$$ \mathrm{Fe}_2 \mathrm{O}_3(\mathrm{s}) \rightarrow \mathrm{Fe}_3 \mathrm{O}_4(\mathrm{s}) + \mathrm{O}_2(\mathrm{g}) \quad (\text{unbalanced equation}) $$
What is the percentage of $\mathrm{SiO}_2$ in the sample?
A) $10%$
B) $20%$
C) $40%$
D) $60%$
First, we balance the chemical reaction responsible for the mass decrease:
$$ 3, \mathrm{Fe}_2 \mathrm{O}_3 (\text{s}) \rightarrow 2, \mathrm{Fe}_3 \mathrm{O}_4 (\text{s}) + \frac{1}{2}, \mathrm{O}_2 (\text{g}) $$
In this reaction, 480 grams of $\mathrm{Fe}_2 \mathrm{O}_3$ produce 16 grams of $\mathrm{O}_2$. Given that the sample's mass loss is 0.04 grams, we can calculate the amount of $\mathrm{Fe}_2 \mathrm{O}_3$ that contributed to this loss:
$$ 0.04, \text{grams of}, \mathrm{O}_2 \times \frac{480, \text{grams of}, \mathrm{Fe}_2\mathrm{O}_3}{16, \text{grams of}, \mathrm{O}_2} = 1.2, \text{grams of}, \mathrm{Fe}_2 \mathrm{O}_3 $$
Next, we need to determine the percentage of $\mathrm{SiO}_2$ in the sample. The total sample originally weighed 2.0 grams, and $\mathrm{Fe}_2\mathrm{O}_3$ alone weighs 1.2 grams. The remaining mass of the sample, which corresponds to $\mathrm{SiO}_2$, is
$$ 2.0, \text{grams} - 1.2, \text{grams} = 0.8, \text{grams} $$
Finally, we calculate the percentage of $\mathrm{SiO}_2$ by mass:
$$ \frac{0.8, \text{grams}}{2.0, \text{grams}} \times 100 = 40% $$
Thus, the percentage of $\mathrm{SiO}_2$ in the sample is ${40%}$ and the correct answer is:
$$ C $$
In the Bohr atomic model, when an electron is excited from $n=1$ to $n=3$, how much energy is emitted/absorbed?
A: $2.15 \times 10^{-21}$ erg
B: $0.1911 \times 10^{-10}$ erg
C: $2.389 \times 10^{-12}$ erg
D: $0.239 \times 10^{-10}$ erg
Correct Answer: B
According to Bohr's model, the energy change $\Delta E$ when an electron transitions from $n=1$ to $n=3$ is given by:
$$ \Delta E = E_1 - E_3 $$
For hydrogen (H), the energy at the first level $E_1$ is:
$$ E_1 = 2.179 \times 10^{-11} \text{ erg} $$
The energy at the third level $E_3$ is:
$$ E_3 = \frac{E_1}{n^2} = \frac{2.179 \times 10^{-11}}{9} \text{ erg} $$
Substituting these values into the equation, we get:
$$ \Delta E = 2.179 \times 10^{-11} - \frac{2.179 \times 10^{-11}}{9} $$ $$ = 2.179 \times 10^{-11} - 0.242 \times 10^{-11} $$ $$ = 1.937 \times 10^{-11} \text{ erg} $$ $$ = 0.1937 \times 10^{-10} \text{ erg} $$
Since the electron is excited from $n=1$ to $n=3$, energy is absorbed.
Thus, the correct answer is:
B: $0.1911 \times 10^{-10} \text{ erg}$
When 4 grams of Fe is converted to ferric oxide at constant pressure, 29.28 kilojoules of heat is released. The enthalpy of ferric oxide is (-) 81.98 kilojoules.
The correct answer is: B
To derive this, let's break it down step-by-step:
Identify the molar relationship:
Ferric oxide is represented as $\mathrm{Fe}_{2}\mathrm{O}_{3}$.
This means 1 mole of $\mathrm{Fe}_{2}\mathrm{O}_{3}$ contains 2 moles of Fe.
Calculate mass of Fe:
Given 2 moles of Fe, the mass can be calculated as: $$ 2 \times 56 \text{ grams} = 112 \text{ grams} $$
Convert per mass energy terms to calculate total heat evolved:
We are given that 4 grams of Fe produces 29.28 kJ of heat.
Therefore, the heat produced by 112 grams of Fe can be calculated using a proportion: $$ \frac{29.28 \text{ kJ} \times 112}{4} = 819.84 \text{ kJ} $$
Determine the enthalpy:
Since the enthalpy of ferric oxide is given as $-81.98$ kJ and based on the above calculated value: $$ -819.84 \text{ kJ} $$
Thus, the enthalpy value calculated aligns with the potential enthalpy provided.
Final Answer: B
Study the information carefully to answer the questions. Seven persons, A, B, C, D, E, F and G are sitting in a straight line (not necessarily in the same order) facing north. Only two persons sit between F and G, and G sits second to the left of B. D sits third to the left of C. E sits exactly between G and B, and B sits at the extreme right end of the row. Who amongst the following sits exactly in the middle of the line?
A. A
B. B
C. C
D. D
E. E
F. F
G. G
The correct option is C.
Given the information, we can deduce the following arrangement:
B is at the extreme right end of the row.
G is second to the left of B.
E sits exactly between G and B.
There are exactly two persons between F and G.
D sits third to the left of C.
By piecing this information together, we get the arrangement as shown below:
A, D, F, C, E, G, B
From the arrangement, it is evident that C sits exactly in the middle of the line.
Therefore, the person who sits exactly in the middle of the line is C.
Column -I | Column-II |
---|---|
A. $2 \mathrm{~s}$ | (P) $n=4, l=2, m=0$ |
B. $2 \mathrm{~p}_{z}$ | (Q) $n=4, l=2, m=-2$ or +2 |
C. $4 \mathrm{~d}_{x^{2}-y^{2}}$ | (R) $n=2, l=1, m=0$ |
D. $4 \mathrm{~d}_{z^{2}}$ | (S) $n=2, l=0, m=0$ |
Match the column -I with column -II
A $A-S, B-R, C-Q, D-P$
B $A-Q, B-P, C-S, D-R$
C $A-R, B-S, C-P, D-Q$
D $A-S, B-R, C-P, D-Q$
The correct option is A: A - S, B - R, C - Q, D - P.
We start by identifying the quantum numbers, where:
$\mathbf{n}$ is the Principal Quantum Number, representing the orbital's energy level.
$\mathbf{l}$ is the Azimuthal Quantum Number, which takes values from $0$ to $n-1$.
$\mathbf{m}$ is the Magnetic Quantum Number, ranging from $-l$ to $+l$.
Match each orbital to its corresponding quantum numbers:
For (2s):
$n=2$
$l=0$
$m=0$
For (2p_z):
$n=2$
$l=1$
$m=0$
For (4d_{x^2 - y^2}):
$n=4$
$l=2$
$m=-2$ or $m=+2$
For (4d_{z^2}):
$n=4$
$l=2$
$m=0$
Thus, we match Column-I with Column-II as follows:
$2s \rightarrow (S) \quad (n=2, l=0, m=0)$
$2p_z \rightarrow (R) \quad (n=2, l=1, m=0)$
$4d_{x^2 - y^2} \rightarrow (Q) \quad (n=4, l=2, m=-2 \text{ or } +2)$
$4d_{z^2} \rightarrow (P) \quad (n=4, l=2, m=0)$
Therefore, the correct choice is:
A - S
B - R
C - Q
D - P
So the correct option is A.
If the radius of $Fe^2+$ is 0.76 Å, the radius of $Fe^3+$ is
(A) 0.88 Å
(B) 0.76 Å
(C) 0.64 Å
(D) 1.08 Å
The correct option is (C) 0.64 Å.
To understand this, consider the atomic structure and electron configurations of $\mathrm{Fe}^{2+}$ and $\mathrm{Fe}^{3+}$. The principal factor here is the nuclear attraction. In $\mathrm{Fe}^{2+}$, the iron ion has lost 2 electrons, while in $\mathrm{Fe}^{3+}$, it has lost 3 electrons.
Because $\mathrm{Fe}^{3+}$ has one less electron compared to $\mathrm{Fe}^{2+}$, the effective nuclear charge experienced by the remaining electrons increases. This happens because the number of protons (nuclear charge) remains the same while the number of electrons decreases. Greater nuclear attraction pulls the remaining electrons closer to the nucleus, resulting in a smaller ionic radius.
Therefore, considering the increased nuclear charge and the consequential smaller size of the $\mathrm{Fe}^{3+}$ ion compared to $\mathrm{Fe}^{2+}$, the radius of $\mathrm{Fe}^{3+}$ is 0.64 Å.
A fraction $f_{1}$ of a radioactive sample decays in one mean life and a fraction $f_{2}$ decays in one half-life. Then
(A) $f_{1}>f_{2}$
(B) $f_{1}>f_{2}$
(C) $f_{1}>f_{2}$
(D) may be (a), (b), or (c) depending on the values of the mean life and half-life
Correct Option: $\mathbf{A} ; f_{1} > f_{2}$
In radioactive decay, the fraction of a sample that decays in one mean life is greater than the fraction that decays in one half-life. The reasoning is based on the definitions and the relationships between mean life and half-life:
Mean Life ($\tau$): The average time it takes for a single radioactive atom to decay.
Half-Life ($T_{1/2}$): The time required for half of the radioactive nuclei in a sample to decay.
Mathematically, the relationship between the two is given by ( \tau = \frac{T_{1/2}}{\ln(2)} ).
Since ( \ln(2) ) is a constant (approximately 0.693), one mean life encompasses a longer time period than one half-life. Therefore, more of the sample will decay in one mean life compared to one half-life. Hence, we conclude that $ f_{1} > f_{2} $.
Among the following complexes, the one which shows zero crystal field stabilization energy (CFSE) is:
A. $\left[\mathrm{Mn}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}$
B. $\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}$
C. $\left[\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}$
D. $\left[\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}$
Correct Option: B
$$ \left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+} $$
Explanation:
The iron ion in the complex, $\mathrm{Fe}^{3+}$, has a $\mathrm{d}^{5}$ electronic configuration. Water ($\mathrm{H}_{2}\mathrm{O}$) is considered a weak field ligand, which means it is incapable of pairing electrons effectively in high-spin complexes. Therefore, in the case of iron in a $\mathrm{d}^{5}$ configuration with water as a ligand, the electrons are distributed in such a way that each ( t_{2g} ) and ( e_g ) orbital are singly occupied.
For such a configuration, the Crystal Field Stabilization Energy (CFSE) is calculated as follows:
$$ \text{CFSE} = \left(-0.4 \times n_{t2g} + 0.6 \times n_{eg}\right) \Delta_0 $$
Given that the electrons are evenly distributed with no additional stabilization:
$$ n_{t2g} = 3 \quad \text{and} \quad n_{eg} = 2 $$
Substituting these values into the CFSE formula:
$$ \text{CFSE} = \left(-0.4 \times 3 + 0.6 \times 2\right) \Delta_0 = (-1.2 + 1.2) \Delta_0 = 0 $$
Hence, the CFSE is zero for this particular complex.
The other complexes (A, C, and D) have configurations and ligand strengths that result in non-zero CFSE:
$\mathrm{Mn}^{3+}$ has a $\mathrm{d}^4$ configuration.
$\mathrm{Co}^{2+}$ has a $\mathrm{d}^7$ configuration.
$\mathrm{Co}^{3+}$ has a $\mathrm{d}^6$ configuration.
Thus, among the given options, the only complex that exhibits zero CFSE is $\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}$.
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