Chemical Bonding and Molecular Structure - Class 11 Chemistry - Chapter 4 - Notes, NCERT Solutions & Extra Questions
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Extra Questions - Chemical Bonding and Molecular Structure | NCERT | Chemistry | Class 11
Consider a $\mathrm{p}_{\mathrm{y}}$ orbital of an atom and identify the correct statement.
A s orbital of another atom produces a $\pi$ bond when $y$ is the bond formation axis.
B $\mathrm{p}_{\mathrm{y}}$ orbital of another atom produces a $\sigma$ bond when $\mathrm{x}$ is the bond formation axis.
C $\mathrm{p}_{\mathrm{z}}$ orbital of another atom produces a $\pi$ bond when $\mathrm{x}$ is the bond formation axis.
D $d_{x y}$ orbital of another atom produces a $\pi$ bond when $x$ is the bond formation axis.
The correct answer is D. The $d_{xy}$ orbital of another atom can produce a $\pi$ bond when the bond formation axis is along $x$.
Understanding the Bond Formation:
Option A: An $s$ orbital is spherically symmetric and cannot form $\pi$ bonds because $\pi$ bonds require lateral overlap which $s$ orbitals cannot provide.
Option B: A $\mathrm{p}_{\mathrm{y}}$ orbital forms a $\sigma$ bond when the bond formation axis is along $y$, not $x$ as in the option.
Option C: A $\mathrm{p}{\mathrm{z}}$ orbital along the $z$-axis, and $\mathrm{p}{\mathrm{y}}$ along the $y$-axis are oriented at right angles to each other, preventing effective orbital overlap for $\pi$ bond formation along the $x$ axis.
Option D: The $d_{xy}$ orbital oriented in the $xy$-plane can have its lobes effectively overlap laterally with the lobes of a $\mathrm{p}_{\mathrm{y}}$ orbital to form a $\pi$ bond along the $x$ axis, making it the correct choice.
What is the difference between molecule and covalent bond? If there is no difference, then give similarities between them.
How is $\mathrm{CO}_2$ gas acidic?
Difference between molecule and covalent bond
A molecule is a stable group of two or more atoms that are chemically bonded together. A covalent bond, on the other hand, is the type of bond between atoms within a molecule where pairs of electrons are shared between atoms. Therefore, a molecule could be composed of multiple covalent bonds.
Similarities between a molecule and a covalent bond include:
Both concepts are fundamental in the study of chemistry.
Both involve the interactions of electrons between atoms.
Acidity of $\mathrm{CO}_2$ gas
$\mathrm{CO}_2$ gas itself is not an acid, but it contributes to the acidity of solutions when dissolved in water. This occurs due to its ability to form carbonic acid upon dissolution. The chemical reaction can be described by the equilibrium process:
$$ \mathrm{CO}_2(g) + \mathrm{H}_2\mathrm{O}(l) \rightleftharpoons \mathrm{H}^+(\mathrm{aq}) + \mathrm{HCO}_3^-(\mathrm{aq}) $$
In this reaction, $\mathrm{CO}_2$ dissolves in water to form carbonic acid, which then dissociates into hydrogen ions ($\mathrm{H}^+$), increasing the solution's acidity, and bicarbonate ions ($\mathrm{HCO}_3^-$).
Define delocalization.
Delocalization refers to the phenomenon where electrons are not associated with a single atom or a single covalent bond within a molecule, ion, or solid metal. This term has versatile meanings across different fields:
In organic chemistry, delocalization is observed as resonance in conjugated systems and aromatic compounds, such as benzene (often represented with a circle inside the hexagonal structure to indicate delocalized electrons).
In solid-state physics, it describes the behavior of free electrons that contribute to electrical conduction.
In quantum chemistry, it relates to electrons in molecular orbitals that are spread across several adjacent atoms.
Which of the following statements is incorrect?
A The hydrogen bond in $\mathrm{KHF}_{2}$ is stronger than that in gaseous $\mathrm{HF}$.
B The dipole moment of $\mathrm{CH}{3}\mathrm{F}$ is less than that of $\mathrm{CH}{3}\mathrm{Cl}$.
C $\mathrm{sp}^{2}$ hybrid orbitals have equal $\mathrm{s}$ and $\mathrm{p}$ character.
D Two types of bonds present in $\mathrm{B}{2}\mathrm{H}{6}$ are covalent and three-centre-two-electron bonds.
The incorrect statement is Option C. Here’s a breakdown of each option based on their conceptual understanding:
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Option A: In $\mathrm{KHF}_2$, the hydrogen bond has a dissociation enthalpy of $-212 , \mathrm{kJ/mol}$, significantly stronger than in gaseous $\mathrm{HF}$ with $-28.6 , \mathrm{kJ/mol}$. This is attributed to resonance stabilization in $\mathrm{KHF}_2$, making the statement correct.
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Option B: The dipole moment of $\mathrm{CH}_3\mathrm{F}$ (1.85 D) is less than that of $\mathrm{CH}_3\mathrm{Cl}$ (1.87 D), primarily due to larger charge separation in $\mathrm{CH}_3\mathrm{Cl}$. The statement here holds true.
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Option C: $\mathrm{sp}^2$ hybrid orbitals do not have equal $s$ and $p$ character; they actually consist of 33.3% $s$ character and 66.6% $p$ character. Therefore, this statement is incorrect.
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Option D: In $\mathrm{B}_2\mathrm{H}_6$ (diborane), the terminal $\mathrm{B}-\mathrm{H}$ bonds are covalent, and the bridging $\mathrm{B}-\mathrm{H}$ bonds are classified as three-centre-two-electron bonds, accurately described by the statement.
Thus, the incorrect statement is Option C, as it wrongly credits $\mathrm{sp}^2$ hybrid orbitals with having equal $s$ and $p$ character.
What is the term used for attraction between molecules of the same substance?
A) Adhesion
B) Cohesion
C) Absorption
D) Adsorption
The correct answer is B) Cohesion.
Cohesion refers to the phenomenon where intermolecular forces attract particles of the same substance to one another. This kind of attraction is observed in substances like water, where cohesion leads to the formation of surface tension due to the bonding of water molecules via hydrogen bonds.
The structure and hybridization of $\mathrm{BrF}_{5}$ are
A. Square pyramidal and $sp^{3}d^{2}$, respectively.
B. Trigonal bipyramidal and $sp^{3}d$, respectively.
C. Trigonal and $sp^{2}$, respectively.
D. Square bipyramidal and $sp^{3}d^{2}$, respectively.
Solution
The correct option is A: Square pyramidal and $sp^{3}d^{2}$, respectively.
To determine this, first, deduce the Lewis structure of $\mathrm{BrF}_{5}$. The central bromine atom has five single bonds with fluorine atoms and one lone pair. Thus, the steric number of the central atom is: $$ 5 \text{ bond pairs} + 1 \text{ lone pair} = 6 $$
This corresponds to a square pyramidal structure based on VSEPR theory. The necessary hybridization to accommodate this geometry and bonding is $sp^{3}d^{2}$.
In fibrous proteins, polypeptide chains are held together by:
A Vander Waals forces
B Disulfide linkage
C Electrostatic forces of attraction
D Hydrogen bonds
Solution
The correct answers are:
- B: Disulfide linkage
- D: Hydrogen bonds
Disulfide linkages and hydrogen bonds are the primary forces holding fibrous proteins together. For example, α-keratin, a fibrous protein found in hair, nails, and skin, is primarily stabilized by disulfide bonds which form between cysteine residues on different polypeptide chains.
Similarly, the stability of the beta-pleated sheet structure in silk fibroin largely results from hydrogen bonding between the amide groups across different protein chains. This type of secondary structure exemplifies the role of hydrogen bonds in fibrous proteins.
The correct order of lattice energy of I), II), and III) is: I) $\mathrm{BeCO}{3}$ II) $\mathrm{MgCO}{3}$ III) $\mathrm{CaCO}_{3}$
A) I $<$ II $<$ III
B) I $>$ II $>$ III
C) I $<$ III $<$ II
D) II $<$ I $<$ III
Solution
The correct option is B) I > II > III.
Lattice energy is influenced by two main factors:
- The product of the charges of the ions forming the compound.
- The distance between the nuclei of these ions, often referred to as the interionic distance.
Given that each compound provided in the question (I) $\mathrm{BeCO}_3$, II) $\mathrm{MgCO}_3$, and III) $\mathrm{CaCO}_3$) have the carbonate ion ($\mathrm{CO}_3^{2-}$) and a divalent metal cation ($\mathrm{M}^{2+}$), the charge product is constant across all three. Therefore, the main factor affecting the lattice energy across these compounds is the interionic distance, which is inversely proportional to the lattice energy.
Among the cations, the ionic radius follows the order: $$ \mathrm{Be}^{2+} < \mathrm{Mg}^{2+} < \mathrm{Ca}^{2+} $$ This implies that beryllium ion $\mathrm{Be}^{2+}$ is the smallest and calcium ion $\mathrm{Ca}^{2+}$ is the largest. Since smaller cations can approach anions more closely, they reduce the interionic distance, thereby increasing lattice energy.
Therefore, considering the relation of cation size to lattice energy, we deduce the order of lattice energy as: $$ \mathrm{BeCO}{3} > \mathrm{MgCO}{3} > \mathrm{CaCO}_{3} $$ Or, in the given terms: I > II > III. Thus, the correct answer is B).
I. $\mathrm{CO}$ II. $\mathrm{PCl}{5}$ III. $\mathrm{BF}{3}$ IV. $\mathrm{BCl}{3}$ V. $\mathrm{SiH}{4}$ VI. $\mathrm{SiF}{4}$ VII. $\mathrm{CCl}{4}$ VIII. $\mathrm{BeCl}{2}$ IX. $\mathrm{N}\left(\mathrm{SiH}{3}\right){3}$ X. $\mathrm{H}{2} \mathrm{O}$
The number of molecules showing back bonding among the given molecules.
To determine which of the given molecules exhibit back bonding, it's essential to consider the electron configurations and the presence of lone pairs and vacant orbitals in the bonding atoms. Back bonding typically occurs when an atom with a lone pair can donate electrons to another atom with vacant orbitals. Here's an analysis of the given molecules:
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$\mathrm{CO}$: Both carbon and oxygen have complete octets and there is no back bonding occurring here.
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$\mathrm{PCl}_5$: Phosphorus and chlorine have complete octets and ${\mathrm{PCl}_5}$ does not show back bonding.
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$\mathrm{BF}_3$: Each fluorine atom has three lone pairs which can back-donate to the electron-deficient boron atom, which has vacant $p$-orbitals. Therefore, $\mathrm{BF}_3$ exhibits back bonding.
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$\mathrm{BCl}_3$: Similar to $\mathrm{BF}_3$, each chlorine atom has three lone pairs which can back-donate to boron. Therefore, $\mathrm{BCl}_3$ also shows back bonding.
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$\mathrm{SiH}_4$: Each hydrogen is electron-precise and silicon does not need back bonding. Therefore, no back bonding occurs.
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$\mathrm{SiF}_4$: Similar to $\mathrm{BF}_3$ and $\mathrm{BCl}_3$, the fluorine's lone pairs can back-donate into the vacant $d$-orbitals of silicon. Therefore, $\mathrm{SiF}_4$ exhibits back bonding.
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$\mathrm{CCl}_4$: There is no necessity or possibility for back bonding as chlorines and carbon have complete octets.
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$\mathrm{BeCl}_2$: Although beryllium is electron-deficient, it does not undergo back bonding due to symmetry and lack of appropriate orbital overlap; instead, it forms polymeric structures.
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$\mathrm{N}(\mathrm{SiH}_3)_3$: The lone pairs on nitrogen can back-donate into the vacant $d$-orbitals of silicon, showing back bonding.
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$\mathrm{H}_2O$: Water has lone pairs on oxygen, but there is no deficient electron center to accept these, hence no back bonding occurs.
Summary:
The molecules showing back bonding are:
- $\mathrm{BF}_3$
- $\mathrm{BCl}_3$
- $\mathrm{SiF}_4$
- $\mathrm{N}(\mathrm{SiH}_3)_3$
Thus, a total of four molecules among the given list exhibit back bonding.
The electronic configuration of a sodium atom is 2, 8, 1 and a chlorine atom is 2, 8, 7. The atoms will form a bond by ____ of electrons.
A) covalent B) ionic C) sharing D) transferring
Solution
The correct options are:
- B) Ionic
- D) Transferring
The electronic configurations are as follows: sodium ($\mathrm{Na}$) has $2, 8, 1$, indicating 1 valence electron, while chlorine ($\mathrm{Cl}$) has $2, 8, 7$, indicating 7 valence electrons.
During the bond formation, the sodium atom loses an electron and becomes a sodium ion, $\mathrm{Na}^+$, and the chlorine atom gains this electron, becoming a chloride ion, $\mathrm{Cl}^-$.
These ions, with opposite charges, are then held together by an ionic bond due to the electrostatic force of attraction.
The total sum of the number of $p \pi$-$d \pi$ bonds in the following molecules: $$ \mathrm{SO}{3}, \mathrm{SO}{2}, \mathrm{XeO}{4}, \mathrm{IO}{3}^{-} $$
For each molecule, let's determine the total number of $p \pi$-$d \pi$ bonds.
1. $\mathrm{SO}_3$:
- Hybridization: $\mathrm{sp}^2$
- $\pi$ bonding: 3 electrons contribute to $\pi$ bonding, with 2 electrons participating in $p \pi$-$d \pi$ bonds.
2. $\mathrm{SO}_2$:
- Hybridization: $\mathrm{sp}^2$
- $\pi$ bonding: 2 electrons are involved, with 1 electron contributing to a $p \pi$-$d \pi$ bond.
3. $\mathrm{XeO}_4$:
In $\mathrm{XeO}_4$, the Xe atom utilizes its $p$ and possibly $d$ orbitals for bonding. But based on the general chemistry of $\mathrm{XeO}_4$ and available quantitative data, determining the exact number of $p \pi$-$d \pi$ bonds directly from the information given here is not straightforward without additional assumptions or references.
4. $\mathrm{IO}_3^-$:
This ion involves iodine which can use its $d$ orbitals. However, without explicit details about the bonding or molecular structure provided (like those in the question for $\mathrm{SO}_3$ and $\mathrm{SO}_2$), it is challenging to specify the number of $p \pi$-$d \pi$ bonds.
Conclusion:
The question provides specific details only for $\mathrm{SO}_3$ and $\mathrm{SO}_2$. From these, 3 $p \pi$-$d \pi$ bonds can be summed up: 2 from $\mathrm{SO}_3$ and 1 from $\mathrm{SO}_2$. Without further details or assumptions, we can't definitively compute the number for $\mathrm{XeO}_4$ and $\mathrm{IO}_3^-$.
Choose the correct properties for the following molecules: (I) $\mathrm{CH}{2}\mathrm{F}{2}$ (II) $\mathrm{CHF}{3}$ (III) $\mathrm{CH}{3}\mathrm{F}$
A $\mathrm{C}-\mathrm{F}$ bond length order: $\mathrm{CH}{3}\mathrm{F} > \mathrm{CH}{2}\mathrm{F}{2} > \mathrm{CHF}{3}$
B $\mathrm{C}-\mathrm{H}$ bond length order: $\mathrm{CH}{3}\mathrm{F} > \mathrm{CH}{2}\mathrm{F}{2} > \mathrm{CHF}{3}$
C Shape is not perfectly tetrahedral for the given compounds.
D Dipole moment is non-zero for the given compounds.
Solution Analysis:
Here are the correct answers and explanations for each of the properties:
A. C-F bond length order: $\mathrm{CH}{3}\mathrm{F} > \mathrm{CH}{2}\mathrm{F}{2} > \mathrm{CHF}{3}$
- The electronegative $F$ atom increases the $p$-character in the $C-F$ bond, thereby lengthening it. As more $F$ atoms are added, the $p$-character per $C-F$ bond decreases, which leads to shorter $C-F$ bonds. Thus, $\mathrm{CH}{3}\mathrm{F}$ has the longest $C-F$ bond, followed by $\mathrm{CH}{2}\mathrm{F}{2}$, and $\mathrm{CHF}{3}$ with the shortest.
B. C-H bond length order: $\mathrm{CH}{3}\mathrm{F} > \mathrm{CH}{2}\mathrm{F}{2} > \mathrm{CHF}{3}$
- Since fluorine increases the $s$-character in the $C-H$ bonds, this results in shorter $C-H$ bonds as more fluorine atoms are added. Therefore, $\mathrm{CH}{3}\mathrm{F}$, having the least number of fluorine atoms, has the longest $C-H$ bond, followed by $\mathrm{CH}{2}\mathrm{F}{2}$, and $\mathrm{CHF}{3}$ with the shortest.
C. Shape is not perfectly tetrahedral for the given compounds.
- Due to the different electron-withdrawing effects of the fluorine atoms and the generation of different $s$ and $p$ characters in respective bonds, the ideal 109.5 degrees tetrahedral angle is distorted. This makes none of these compounds perfectly tetrahedral.
D. Dipole moment is non-zero for the given compounds.
- Each molecule has fluorine atoms, which are highly electronegative, creating significant dipole moments due to the difference in electronegativity between carbon and fluorine. Moreover, as the geometry is not symmetric due to the variations in bond angles and the varying number of hydrogen and fluorine atoms, all the molecules are polar.
Hence, all statements A, B, C, and D are correct based on the described properties and molecular arrangements of $\mathrm{CH}{3}\mathrm{F}$, $\mathrm{CH}{2}\mathrm{F}{2}$, and $\mathrm{CHF}{3}$.
Sugar fills in between the intermolecular space in water, as water has gaps between them. But why does sugar fall down when we put it in gas, as gas has a lot more intermolecular space?
The reasoning behind why sugar dissolves in water but falls in gas lies in the nature of molecular interactions and physical states of matter.
When sugar is added to water, it doesn't just occupy the intermolecular space of water. Instead, the process involves intermolecular forces; water molecules attract and interact with sugar molecules, facilitating the sugar dissolved in water despite seeming gaps in the liquid.
On the other hand, gases like air have much larger intermolecular spaces compared to liquids due to their low density and high compressibility. However, the intermolecular forces in gases are considerably weaker than in liquids. Even though there might be more space between gas molecules, the lack of sufficient attractive force prevents sugar from staying suspended or dissolving in the gas. Instead, gravity pulls the sugar grains down as they are denser than the surrounding air.
In summary, while the availability of space is necessary, the presence and strength of intermolecular forces are crucial for the dissolution process. This explains why sugar dissolves in water but sinks in air, highlighting the difference in physical properties between gas and liquid molecular behaviors.
$\mathrm{H}{2}\mathrm{O}$ has a net dipole moment, but $\mathrm{BeF}{2}$ has zero dipole moment. This is because:
A $\mathrm{H}{2}\mathrm{O}$ molecule is linear, while $\mathrm{BeF}{2}$ is bent.
B $\mathrm{BeF}{2}$ molecule is linear, while $\mathrm{H}{2}\mathrm{O}$ is bent.
C Fluorine is more electronegative than oxygen.
D Be is more electronegative than oxygen.
Solution
The correct option is B:
$\mathrm{BeF}_{2}$ (beryllium fluoride) has a linear shape and is sp-hybridized, resulting in symmetrical distribution of the dipole moments. This symmetry causes the dipole moment vectors to cancel each other out, resulting in a zero net dipole moment.
In contrast, $\mathrm{H}_{2}\mathrm{O}$ (water) exhibits a bent shape due to its sp(^3) hybridization and the presence of two lone pairs of electrons on the oxygen atom. This angular structure leads to an asymmetrical charge distribution, giving water a resultant dipole moment of 1.84 D.
Thus, the difference in molecular geometry between $\mathrm{BeF}{2}$ and $\mathrm{H}{2}\mathrm{O}$ explains their differing dipole moments.
Read the following questions and answer as per the directions given below:
Statement I: The first ionization energy of $\mathrm{Be}$ is greater than that of $\mathrm{B}$. Statement II: $2 \mathrm{p}$ orbital is lower in energy than $2 \mathrm{s}$.
A) Statement I is true; Statement II is true; Statement II is the correct explanation for Statement I.
B) Statement I is true; Statement II is true; Statement II is not the correct explanation for Statement I.
C) Statement I is true; Statement II is false.
D) Statement I is false; Statement II is true.
The correct answer is C) Statement I is true; Statement II is false.
Explanation:
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Statement I: The first ionization energy of $\mathrm{Be}$ is greater than that of $\mathrm{B}$. This statement is true because $\mathrm{Be}$, with a full $2s^2$ subshell, has a stable electronic configuration. Removing an electron from a fully filled subshell requires more energy compared to removing one from a partially filled subshell such as in $\mathrm{B}$ ($2s^2 2p^1$).
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Statement II: $2 \mathrm{p}$ orbital is lower in energy than $2 \mathrm{s}$. This statement is false. In atomic orbitals, the energy of an orbital increases with the sum of the principal quantum number $n$ and the azimuthal quantum number $l$. For the $2s$ orbital, $(n+l) = 2+0 = 2$, and for the $2p$ orbital, $(n+l) = 2+1 = 3$. Thus, the $2s$ orbital is lower in energy than the $2p$ orbital.
The ratio of $\sigma$ and $\pi$ bonds in benzene is
A) 2
B) 4
C) 6
D) 8
The correct answer is B) 4.
In a benzene molecule, there are:
- 6 carbon-carbon (C-C) $\sigma$-bonds
- 6 carbon-hydrogen (C-H) $\sigma$-bonds
This gives a total of 12 $\sigma$-bonds in benzene. Additionally, there are 3 $\pi$-bonds between carbon atoms in the six-membered ring, which result from the alternating double bonds characteristic of benzene's structure.
Therefore, the ratio of $\sigma$ bonds to $\pi$ bonds in benzene can be calculated as follows: $$ \frac{\sigma-\text{bonds}}{\pi-\text{bonds}} = \frac{12}{3} = 4 $$
Thus, the ratio of $\sigma$ and $\pi$ bonds in benzene is 4.
The shape of $\mathrm{XeO}{2} \mathrm{~F}{2}$ is:
A) Trigonal Planar. B) See-saw. C) Square Planar. D) Tetrahedral.
Correct Answer: B) See-saw.
Molecules exhibit various shapes depending on the number of bonding pairs and lone pairs around the central atom. In the case of $\mathrm{XeO}_{2} \mathrm{~F}_{2}$, it is represented by the formula $\mathrm{AB}_4 \mathrm{E}$, where $\mathrm{AB}_4$ represents four bonding pairs and $\mathrm{E}$ represents one lone pair.
Electron Geometry: Trigonal Bipyramidal
Because $\mathrm{XeO}_{2} \mathrm{~F}_{2}$ has a total of five electron pair regions around the central atom (Xenon), its electronic geometry is trignonal bipyramidal. However, with one lone pair occupying one of the equatorial positions, the molecular shape adjusts into the See-Saw configuration. This shape minimizes the electron-pair repulsions according to the VSEPR theory.
S - character in $\mathrm{sp}^{3}$ hybridization is:
A) $\frac{3}{4}$
B) $\frac{1}{4}$ C) $\frac{1}{2}$
D) None of these
Solution
The correct option is B, which corresponds to $$ \frac{1}{4} $$
In $\mathrm{sp}^3$ hybridization, we have:
- 1s orbital
- 3p orbitals
This results in the formation of 4 $\mathrm{sp}^3$ hybrid orbitals. The ratio of the $s$-character in any of these orbital is determined by: $$ s\text{-character} = \frac{\text{Number of $s$ orbitals involved}}{\text{Total number of hybrid orbitals}} $$ Given that we use 1 $s$ orbital and generate 4 hybrid orbitals, the $s$-character is: $$ \frac{1}{4} $$ Hence, the $s$-character in $\mathrm{sp}^3$ hybridization is $\frac{1}{4}$.
In which of the following pairs of molecules, the species have identical shape but differ in hybridisation?
(A) $\mathrm{NH}{3}, \mathrm{PF}{3}$ (B) $\mathrm{I}{3}^{-1}, \mathrm{BeCl}{2}$ (C) $\mathrm{XeF}{2}, \mathrm{I}{3}^{-1}$ (D) $\mathrm{NH}{4}^{+}, \mathrm{SF}{4}$
The correct answer is option B: $$ \mathrm{I}_{3}^{-1}, \mathrm{BeCl}_{2} $$ $\mathrm{I}_{3}^{-1}$ is $\mathrm{sp}^3d$ hybridized, while $\mathrm{BeCl}_{2}$ is $\mathrm{sp}$ hybridized. Although both species have a linear shape, their hybridizations differ.
The number of electron pairs (ep) in $\mathrm{ClF}_{3}$ is
A) 4
B) 5
C) 6
D) 7
Solution
The correct answer is Option B: 5
In $\mathrm{ClF}_3$, the total number of electron pairs around the central atom (chlorine) can be determined with the following formula: $$ \mathrm{ep} = \frac{(V + L + A - C)}{2} $$ Where:
- $V$ represents the number of valence electrons of the central atom
- $L$ is the number of ligands (atoms directly bonded to the central atom)
- $A$ denotes the anionic charge
- $C$ represents the cationic charge
For $\mathrm{ClF}_3$:
- Chlorine (Cl) is the central atom with 7 valence electrons ($V = 7$)
- It is bonded to three fluoride (F) atoms, so $L = 3$
- There are no charges, so $A = 0$ and $C = 0$
Substituting these values into the equation, we calculate $\mathrm{ep}$: $$ \mathrm{ep} = \frac{(7 + 3 + 0 - 0)}{2} = \frac{10}{2} = 5 $$
Thus, the number of electron pairs in $\mathrm{ClF}_{3}$ is 5.
$d \pi - p \pi$ bond present in:
(A) $\mathrm{CO}_{3}^{2-}$
(B) $\mathrm{PO}{4}^{3-}$ (C) $\mathrm{NO}{3}^{-}$ (D) $\mathrm{NO}_{2}^{-}$
Solution
The correct answer is (B) $\mathrm{PO}_{4}^{3-}$. To determine the presence of a $d\pi-p\pi$ bond, we need to consider whether the central atom has vacant $d$ orbitals that can participate in bonding.
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Phosphorus (P) in $\mathrm{PO}{4}^{3-}$ is in the third period of the periodic table, which means it has available vacant $3d$ orbitals. This allows phosphorus to form $d\pi-p\pi$ bonding with the $2p$ orbitals of oxygen in $\mathrm{PO}{4}^{3-}$.
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In contrast, Nitrogen (N) and Carbon (C), in $\mathrm{NO}{3}^{-}$ and $\mathrm{CO}{3}^{2-}$ respectively, are from the second period of the periodic table and thus do not possess any vacant $d$ orbitals in their valence shells. Consequently, they cannot participate in $d\pi-p\pi$ bonding.
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The same holds for $\mathrm{NO}_{2}^{-}$, where the central nitrogen does not have the requisite vacant $d$ orbital.
Therefore, $\mathrm{PO}_{4}^{3-}$ is the only ion among the given choices that contains a $d\pi-p\pi$ bond.
Which of the following is true for both beryllium and aluminium?
A) Forms oxide layer to prevent further oxidation
B) Forms chlorine-bridged structure in vapor phase
C) Have strong tendencies to form complexes
D) All of the above.
The correct answer is D) All of the above.
Explanation:
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Both beryllium and aluminium form a protective oxide layer on their surfaces, which prevents further oxidation. This characteristic allows them to resist corrosion by preventing them from being easily attacked by acids.
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In the vapor phase, both elements, when combined with chlorine, form chlorine-bridged structures. This can be specifically observed in compounds such as beryllium chloride and aluminium chloride, which exist as dimers connected by $\mathrm{Cl}^{-}$ bridges.
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Lastly, both beryllium and aluminium have a strong tendency to form complexes. Examples include $\mathrm{BeF}{4}^{2-}$ and $\mathrm{AlF}{6}^{3-}$, demonstrating their ability to coordinate with various ligands to form stable ionic structures.
Thus, all statements A), B), and C) are true for both beryllium and aluminium.
Is there hydrogen bonding in D2O (heavy water)?
Yes, there is hydrogen bonding in D₂O (heavy water). However, it's important to note that the strength of hydrogen bonding varies between protium (H) and deuterium (D).
In hydrogen bonding, strength is directly proportional to the number of moles per mL at a given temperature. For D₂O, the number of moles per mL is 0.0553, while for H₂O, at the same temperature of 4°C, it is slightly higher at 0.0555.
As a result, the hydrogen bonds in regular water (H₂O) are slightly stronger compared to those in heavy water (D₂O). Additionally, while the O-D bond in D₂O is slightly shorter than the O-H bond in H₂O, you might expect a larger number of molecules per mL due to smaller size. However, this is not the case, and hydrogen bonding in D₂O remains slightly weaker than in normal water.
Which bond angle $\theta$ would result in the maximum dipole moment for the triatomic molecule $XY_{2}$?
A) $90^{\circ}$
B) $120^{\circ}$
C) $150^{\circ}$
D) $180^{\circ}$
The correct answer is Option A) $90^{\circ}$.
Explanation:
The dipole moment of a molecule depends largely on the angle between the bonds if the bonds themselves have dipole moments. For a triatomic molecule $XY_2$, where 'X' is the central atom bonded to two 'Y' atoms, the geometry affects the overall dipole moment. The formula for the resultant dipole moment, $\mu$, when two bond dipoles are at an angle $\theta$ to each other is: $$ \mu = \sqrt{X^2 + Y^2 + 2XY \cos \theta} $$ Here, 'X' and 'Y' would represent the magnitudes of the individual dipole moments of the $XY$ bonds.
When $\theta = 90^{\circ}$:
- The value of $\cos 90^{\circ}$ is 0. The equation simplifies to $\mu = \sqrt{X^2 + Y^2}$, which denotes a pure vector addition perpendicular to each other.
As $\theta$ moves towards $180^{\circ}$:
- The $\cos \theta$ becomes negative, reducing the resultant vector length because the two vectors start to oppose each other more directly (i.e., becoming more collinear but in opposite directions).
As $\theta$ moves towards $0^{\circ}$:
- The vectors are fully additive but this scenario is not given as an option.
Thus, to maximize the dipole moment, the optimal scenario is having the two $XY$ dipoles perpendicular to each other. Hence, $\theta = 90^{\circ}$ yields the maximum dipole moment for the molecule $XY_{2}$.
What is the term given for molecular attraction by which the particles within a body remain united throughout mass?
A) Adhesion
B) Cohesion
C) Transpiratory force
D) Capillary force
Correct Answer: B) Cohesion
Cohesion refers to the molecular attraction where particles of the same substance (like water molecules) stick to each other. This attractive force helps maintain unity throughout the mass of the substance. In plants, cohesion allows for a continuous column of water and nutrients to be pulled up through the stem as water evaporates, a process vital for the transport of nutrients and maintaining structural integrity.
The correct increasing order of $\mathrm{C}-\mathrm{O}$ bond lengths among $\mathrm{CO}$, $\mathrm{CO}{3}^{2-}$ and $\mathrm{CO}{2}$ is
(A) $\mathrm{CO}{3}^{2-} < \mathrm{CO}{2} < \mathrm{CO}$
(B) $\mathrm{CO}{2} < \mathrm{CO}{3}^{2-} < \mathrm{CO}$
(C) $\mathrm{CO} < \mathrm{CO}{3}^{2-} < \mathrm{CO}{2}$
(D) $\mathrm{CO} < \mathrm{CO}{2} < \mathrm{CO}{3}^{2-}$
The correct answer is (D) $\mathrm{CO} < \mathrm{CO}{2} < \mathrm{CO}{3}^{2-}$.
To understand the ordering of $\mathrm{C}-\mathrm{O}$ bond lengths in these molecules, consider the type of bonding and resonance structures:
-
Carbon Monoxide ($\mathrm{CO}$): This molecule features a triple bond between carbon and oxygen, including two pi bonds and one sigma bond. Triple bonds are generally shorter than double and single bonds due to greater electron sharing and stronger attraction between the nuclei.
-
Carbon Dioxide ($\mathrm{CO}_2$): This molecule contains two double bonds between carbon and each oxygen atom. The lack of resonance structures here means each oxygen is doubly bonded to carbon, typically resulting in intermediate bond length compared to triple and single bonds.
-
Carbonate Ion ($\mathrm{CO}_3^{2-}$): This ion has a resonance structure where each oxygen atom is singly and doubly bonded to carbon in different resonance forms. The average of these bond situations is a bond order less than a double but more than a single bond, which makes each $\mathrm{C}-\mathrm{O}$ bond in $\mathrm{CO}_3^{2-}$ longer than those in $\mathrm{CO}_2$ and $\mathrm{CO}$.
Hence, the increasing order of bond length (shortest to longest) is $\mathrm{CO}$ (shortest due to triple bond), followed by $\mathrm{CO}{2}$ (double bonds), and then $\mathrm{CO}{3}^{2-}$ (average of single and double bonds due to resonance).
"Why do covalent bonds usually have lower melting points?"
Solution
Covalent bonds are strong bonds that hold atoms together within a molecule to form simple molecular substances. These molecules, however, are held together in a structure by weak intermolecular forces. Due to the weakness of these forces, substances with covalent bonds generally have low melting and boiling points. Furthermore, they do not conduct electricity because there are no free ions or electrons to carry the charge.
The electronic configurations of three elements X, Y, and Z are as follows: $$ \begin{array}{l} X = 2,4 \ Y = 2,7 \ Z = 2,1 \end{array} $$ (a) Which two elements will combine to form an ionic compound? (b) Which two elements will react to form a covalent compound?
Give reasons for your choices.
Solution
(a) Ionic Compound Formation
Elements Y and Z will combine to form an ionic compound. This occurs because element Z, with an electron configuration of $2,1$, can easily donate its one valence electron to element Y, which has an electron configuration of $2,7$. Element Y needs one more electron to achieve a stable noble gas electron configuration of $2,8$. By transferring the electron, Z achieves a configuration of $2$, also stable, resembling the electron configuration of a noble gas. Thus, there is a transfer of an electron from Z to Y, leading to the formation of ions and an ionic bond between them.
(b) Covalent Compound Formation
Elements X and Y will react to form a covalent compound. Here, X with an electron configuration of $2,4$, needs four more electrons to achieve a stable octet configuration (full outermost shell). On the other hand, Y with a $2,7$ configuration, needs one more electron. However, instead of a complete transfer of electrons, these elements tend to share electrons. X can potentially share its four electrons with four atoms of Y, each contributing one electron to the bonding, thereby allowing all related atoms to come closer to achieving or completing their octet. Thus, covalent bonding is more feasible between X and Y due to electron sharing to achieve stability.
"Coordinate bond is also known as:"
A ionic
B dative
C molecular
D acceptor
The correct answer is B) dative.
In the formation of a coordinate bond, the electron pair is donated by one atom but shared by both involved atoms. This type of bond is also referred to as a dative bond. Other names for the coordinate bond include semi-polar bond, donor-acceptor bond, or co-ionic bond.
Which of the following has the greatest electron affinity?
A - O
B - S
C - Se
D - Te
The correct option is B - S.
Electron affinity typically decreases as you move down a group in the periodic table because the number of electron shells increases. This increase in shells leads to a greater distance between the nucleus and the valence shell, which weakens the attractive force exerted on incoming electrons. However, Oxygen (O) is a special case where, despite its higher position in the group, it has a lower electron affinity compared to Sulfur (S). This is because Oxygen's smaller size results in a high electron density, creating stronger inter-electronic repulsions when an additional electron is added. This makes it more difficult for Oxygen to attract and stabilize an extra electron compared to Sulfur.
Which of the following options is not true about Borax?
A. Borax can be obtained from boric acid by neutralizing it with soda ash.
B. In Borax, boron shows both $\mathrm{sp}^{2}$ and $\mathrm{sp}^{3}$ hybridization.
C. In one molecule of Borax, there are 14 $\angle \mathrm{OBO}$ bond angles.
D. When Borax is heated until red hot, a mixture of sodium metaborate and boric anhydride is formed.
Solution
The correct answer is Option C: In one molecule of Borax, there are 14 $\angle \mathrm{OBO}$ bond angles.
-
Option A is true because Borax can indeed be synthesized from boric acid ($\mathrm{H_3BO_3}$) by neutralizing it with soda ash ($\mathrm{Na_2CO_3}$). The reaction is as follows: $$ 4 \mathrm{H}_3\mathrm{BO}_3 + \mathrm{Na}_2\mathrm{CO}_3 \rightarrow \mathrm{Na}_2\mathrm{B}_4\mathrm{O}_7 + \mathrm{CO}_2 + 6\mathrm{H}_2\mathrm{O} $$
-
Option B is true as Borax demonstrates boron atoms in different hybridization states, including both $\mathrm{sp}^2$ and $\mathrm{sp}^3$.
-
Option C is incorrect because a molecule of Borax actually has 18 $\angle \mathrm{OBO}$ bond angles, not 14.
-
Option D is true as upon heating until red hot, Borax decomposes into sodium metaborate ($\mathrm{NaBO_2}$) and boric anhydride ($\mathrm{B_2O_3}$), which can be represented by: $$ \mathrm{Na}_2\mathrm{B}_4\mathrm{O}_7\cdot10\mathrm{H}_2\mathrm{O} \stackrel{\Delta}{\Rightarrow} 2\mathrm{NaBO}_2 + \mathrm{B}_2\mathrm{O}_3 $$
Thus, the statement provided in Option C regarding the number of $\angle \mathrm{OBO}$ bond angles in Borax is false.
Which of the following is not true for $\mathrm{B}{2} \mathrm{H}{6}$?
A) $\mathrm{B}{2} \mathrm{H}{6}$ is an electron-deficient molecule.
B) Hybridization of $\mathrm{B}$ atom is $\mathrm{sp}^{2}$ and molecule is planar.
C) $\mathrm{B}{2} \mathrm{H}{6}$ contains four $2 \mathrm{C}-2 \mathrm{e}^{-}$ bonds and two $3 \mathrm{C}-2 \mathrm{e}^{-}$ bonds.
D) The electronic distribution of the bridge bond has a banana-like appearance.
Solution
The correct answer is B.
- Statement B: "Hybridization of $\mathrm{B}$ atom is $\mathrm{sp}^{2}$ and molecule is planar"
This statement is not true. The hybridization of the $\mathrm{B}$ atoms in $\mathrm{B}{2} \mathrm{H}{6}$ (diborane) is actually $\mathrm{sp}^{3}$. Furthermore, the molecule is non-planar due to its unique bridge structure, consisting of two hydrogen atoms bonding to two boron atoms. This arrangement defies the planar geometry proposed in statement B.
Lowering of vapor pressure due to a solute in a 1 molal aqueous solution at $100^{\circ} \mathrm{C}$ is (assuming solute does not associate).
To solve this query about the lowering of vapor pressure caused by a solute in a 1 molal aqueous solution at $100^\circ \mathrm{C}$, while assuming the solute does not associate, follow these steps:
-
Find the mole fraction of the solute. Let's denote this mole fraction as $x$.
-
Utilize the relationship between mole fraction and molality: $$ \text{molality} = \frac{x \times 1000}{(1 - x) \times M} $$ Here, $M$ is the molecular weight of solvent. For an aqueous solution, the solvent is water, thus $M = 18 \text{ g/mol}$.
-
Given the molality is 1 mol/kg, set up the equation: $$ 1 = \frac{x \times 1000}{(1 - x) \times 18} $$ Solving this equation for $x$, $$ 18(1 - x) = 1000x $$ $$ 18 = 1018x $$ $$ x = \frac{18}{1018} $$
-
The relative lowering of vapor pressure can be expressed as the mole fraction of the solute: $$ \Delta P / P_0 = x $$ where $P_0$ is the initial vapor pressure. Assuming the standard condition, $P_0 = 760 \text{ torr}$.
-
Calculate the change in pressure: $$ \Delta P = 760 \times \frac{18}{1018} \approx 13.44 \text{ torr} $$
Hence, the lowering of vapor pressure due to the solute in this 1 mol/kg aqueous solution at $100^\circ \mathrm{C}$ is about 13.44 torr.
Select the option which is a property of sulfuric acid:
A. The strong affinity of sulfuric acid for water
B. The non-affinity of sulfuric acid for water
C. The hydrophobic nature of sulfuric acid
D. The geometry of the sulfuric acid molecule
The correct option is A. The strong affinity of sulfuric acid for water.
Explanation:
Sulfuric acid, denoted as $H_2SO_4$, possesses several notable properties:
- It turns blue litmus red: indicating its acidic nature.
- Has a great affinity for water: Sulfuric acid readily attracts and absorbs water molecules.
- Acts as a dehydrating agent: It can remove elements of water from other compounds.
- Good oxidizing agent: Demonstrates the ability to accept electrons from other substances.
Given these properties, the statement "The strong affinity of sulfuric acid for water" encapsulates one of the fundamental characteristics of sulfuric acid, making option A the correct choice. The other options either misrepresent the nature of sulfuric acid or are unrelated to its basic properties.
Which of the following triatomic molecules will behave as a diatomic molecule for consideration of degrees of freedom?
A $\mathrm{CO}_{2}$
B $\mathrm{H}_{2}\mathrm{O}$
C $\mathrm{SO}_{2}$
D None of these
Solution
The correct option is A $\mathrm{CO}_{2}$
$\mathrm{CO}{2}$ is a triatomic linear molecule. For linear molecules, regardless of the number of atoms, there are always only two rotational degrees of freedom. Hence, $\mathrm{CO}{2}$ behaves similarly to a diatomic molecule in terms of rotational degrees of freedom.
In contrast, $\mathrm{H}{2}\mathrm{O}$ and $\mathrm{SO}{2}$ are non-linear molecules, each having three rotational degrees of freedom. Thus, they do not behave like diatomic molecules.
Which of the following represent incorrect order of the related properties:
A $\mathrm{BF}{3} < \mathrm{BCl}{3} < \mathrm{BBr}{3} < \mathrm{BI}{3}$ (Acidic character)
B $\mathrm{BaSO}{4} < \mathrm{SrSO}{4} < \mathrm{CaSO}{4} < \mathrm{MgSO}{4}$ (Solubility in water)
C $\mathrm{CHCl}{3} < \mathrm{CHF}{3}$ (Acidic strength)
D $\left(\mathrm{CF}{3}\right){3} \mathrm{N} < \left(\mathrm{CH}{3}\right){3} \mathrm{N}$ (Basic strength)
Solution
The incorrect statement is option C: $\mathrm{CHCl}_3 < \mathrm{CHF}_3$ (Acidic strength).
-
Acidic Character of Boron Halides: The acidity increases as you go from $\mathrm{BF}_3$ to $\mathrm{BI}_3$ because the orbital overlap decreases down the group, making boron more electron deficient. Thus option A is correct: $\mathrm{BF}_3 < \mathrm{BCl}_3 < \mathrm{BBr}_3 < \mathrm{BI}_3$.
-
Solubility of Sulfates: The correct trend is that solubility decreases down the group, hence $\mathrm{BaSO}_4 < \mathrm{SrSO}_4 < \mathrm{CaSO}_4 < \mathrm{MgSO}_4$ is correct, consistent with the general solubility rule in group 2.
-
Acidic Strength of Chloroforms: Contrary to the given order ($\mathrm{CHCl}_3 < \mathrm{CHF}_3$), $\mathrm{CHCl}_3$ is actually more acidic than $\mathrm{CHF}_3$. The reason is due to $\mathrm{Cl}$ being able to exert a $-M$ (mesomeric) effect thanks to its vacant 3$d$ orbital, which helps stabilize the negative charge better than $\mathrm{F}$, which lacks this ability. Thus, the correct order should be $\mathrm{CHF}_3 < \mathrm{CHCl}_3$.
-
Basic Strength of Amines: The N in $(\mathrm{CH}_3)_3\mathrm{N}$ can more easily donate its lone pair compared to $(\mathrm{CF}_3)_3\mathrm{N}$, where the fluorine atoms, due to their high electronegativity, reduce the electron density at the nitrogen atom. Hence, $(\mathrm{CH}_3)_3\mathrm{N}$ is more basic than $(\mathrm{CF}_3)_3\mathrm{N}$, making option D correct.
Therefore, option C displays the incorrect order concerning acidic strength, which correctly should assert $\mathrm{CHCl}_3$ as more acidic than $\mathrm{CHF}_3$.
Transition state structure of the substrate formed during an enzymatic reaction is
A) permanent but unstable
B) transient and unstable
C) permanent and stable
D) transient but stable
The correct answer is B) transient and unstable.
During enzymatic reactions, all substrates go through a transition state. This particular state occurs between the substrate and the product, and it is characterized by having the highest energy of the reaction pathway. Due to this high energy, the transition state is not only short-lived (transient) but also lacking stability (unstable).
The nature of π-bond in perchlorate ion is:
A O(d π)-Cl(p π)
B O(p π)-Cl(d π)
C O(d π)-Cl(d π) D O(d π)-Cl(d π)
Solution
The correct answer is B: O(p π)-Cl(d π).
The ground state outer electronic configuration of chlorine (Cl) is:
$$ 3s^2, 3p^5 $$
When chlorine forms the perchlorate ion (ClO₄⁻), it undergoes sp³ hybridization, using one s orbital and three p orbitals. This configuration leaves an empty $3d$ orbital available for bonding.
In perchlorate ion, chlorine forms four σ bonds with oxygen atoms using the sp³ hybrid orbitals and three π bonds. The π bonds in the perchlorate ion are formed through the overlap of p orbitals of Oxygen (O(p π)) and d orbitals of Chlorine (Cl(d π)). Hence, the nature of the π-bond in the perchlorate ion corresponds to option B: O(p π)-Cl(d π).
The increasing order of bond angles in $\mathrm{H}{2}\mathrm{S}, \mathrm{NH}{3}, \mathrm{BF}{3}$, and $\mathrm{SiH}{4}$ is:
A $\mathrm{H}{2}\mathrm{S} < \mathrm{NH}{3} < \mathrm{SiH}{4} < \mathrm{BF}{3}$
B $\mathrm{NH}{3} < \mathrm{H}{2}\mathrm{S} < \mathrm{SiH}{4} < \mathrm{BF}{3}$ (c) $\mathrm{H}{2}\mathrm{S} < \mathrm{SiH}{4} < \mathrm{NH}{3} < \mathrm{BF}{3}$ (D) $\mathrm{H}{2}\mathrm{S} < \mathrm{NH}{3} < \mathrm{BF}{3} < \mathrm{SiH}{4}$
In order to determine the increasing order of bond angles amongst $\mathrm{H}_2\mathrm{S}$, $\mathrm{NH}_3$, $\mathrm{BF}_3$, and $\mathrm{SiH}_4$, we need to consider the electron pair geometry and presence of lone pairs affecting bond-pair repulsion.
-
$\mathrm{H}_2\mathrm{S}$: This molecule has two lone pairs on the sulfur atom and adopts a bent shape similar to $\mathrm{H}_2\mathrm{O}$. Lone pairs repel more strongly than bonding pairs, leading to smaller bond angles.
-
$\mathrm{NH}_3$: The nitrogen in ammonia is $\mathrm{sp}^3$ hybridized and the molecule has a pyramidal shape with one lone pair. The presence of one lone pair causes the bond angles to be less than $109.5^\circ$, typical for an ideal tetrahedral arrangement.
-
$\mathrm{SiH}_4$: This molecule is also $\mathrm{sp}^3$ hybridized but is a perfect tetrahedral shape with no lone pairs, resulting in bond angles exactly $109.5^\circ$.
-
$\mathrm{BF}_3$: Boron trifluoride is $\mathrm{sp}^2$ hybridized and forms a trigonal planar shape with bond angles of $120^\circ$. There are no lone pairs affecting the geometry in $\mathrm{BF}_3$.
Combining these insights, the increasing order of bond angles, affected by both electron geometry and the presence of lone pairs influencing electron repulsion, is:
$$ \mathrm{H}_2\mathrm{S} < \mathrm{NH}_3 < \mathrm{SiH}_4 < \mathrm{BF}_3 $$
Option A $\mathrm{H}{2}\mathrm{S} < \mathrm{NH}{3} < \mathrm{SiH}{4} < \mathrm{BF}{3}$ is correct.
Which type of bond is present between nitrogen atoms in a nitrogen molecule?
A) Single Bond B) Double Bond C) Triple Bond D) Ionic Bond
The correct answer is C) Triple Bond.
A single nitrogen atom possesses five valence electrons and needs three more electrons to complete its octet for stability. In a nitrogen molecule ($ N_2 $), each nitrogen atom shares three electrons with the other, leading to the formation of a triple covalent bond. This sharing allows both atoms to achieve a stable octet configuration on each side, thus forming a strong triple bond between the two nitrogen atoms.
Identify the correct order of increasing bond order:
(A) $\mathrm{O}{2}^{-} < \mathrm{He}{2}^{2+} < \mathrm{C}{2}^{2-}$ (B) $\mathrm{O}{2}^{-} < \mathrm{C}{2}^{2-} < \mathrm{He}{2}^{2+}$
C) $\mathrm{C}{2}^{2-} < \mathrm{He}{2}^{2+} < \mathrm{O}{2}^{-}$ (D) $\mathrm{He}{2}^{2+} < \mathrm{O}{2}^{-} < \mathrm{C}{2}^{2-}$
Solution
To determine the correct order of increasing bond order among the given molecular ions, we need to calculate the bond orders based on their electronic configurations:
-
$\mathrm{He}_{2}^{2+}$
- Total number of electrons: $4 - 2 = 2$
- Electron configuration: $(\sigma(1s))^2$
- Bond order: $\frac{2 - 0}{2} = 1$
-
$\mathrm{O}_2^-$
- Total number of electrons: $16 + 1 = 17$
- Electron configuration: $(\sigma(1s))^2(\sigma^(1s))^2(\sigma(2s))^2(\sigma^(2s))^2(\sigma(2p_z))^2(\pi(2p_x))^2(\pi(2p_y))^2(\pi^(2p_x))^2(\pi^(2p_y))^1$
- Bond order: $\frac{10 - 7}{2} = 1.5$
-
$\mathrm{C}_2^{2-}$
- Total number of electrons: $12 + 2 = 14$
- Electron configuration: $(\sigma(1s))^2(\sigma^(1s))^2(\sigma(2s))^2(\sigma^(2s))^2 (\pi(2p_x))^2(\pi(2p_y))^2(\sigma(2p_z))^2$
- Bond order: $\frac{10 - 4}{2} = 3$
By comparing the bond orders, we find the correct order of increasing bond order is: $$ \mathrm{He}{2}^{2+} < \mathrm{O}{2}^{-} < \mathrm{C}_{2}^{2-} $$
Hence, the correct answer is (D).
How should we find the number of electron pairs in a molecule like $\mathrm{C}_2\mathrm{H}_2$ using the formula $(V+L+A-C) \div 2$. WHERE V = VALENCE ELECTRONS, L = LIGANDS, A = ANIONIC CHARGE, C = CATIONIC CHARGE...?
To determine the number of electron pairs in a molecule like $\mathrm{C}_2\mathrm{H}_2$ using the formula $$ \frac{V + L + A - C}{2} $$ where the terms represent the following:
- $V$ = number of valence electrons in the main atom(s)
- $L$ = number of ligands, specifically monovalent atoms (like H, F, Cl, Br, I)
- $A$ = anionic charge (if the molecule is an anion)
- $C$ = cationic charge (if the molecule is a cation),
you start by calculating the sum of $V + L + A - C$.
Example Calculation for $\mathbf{C_2H_2}$
-
Valence Electrons ($V$): Carbon has 4 valence electrons, and there are two carbons, contributing a total of $4 \times 2 = 8$ electrons from Carbon. Hydrogen has 1 valence electron, and with two hydrogens, this adds $1 \times 2 = 2$ electrons.
- Thus, $V = 8 + 2 = 10$ valence electrons.
-
Ligands ($L$): In this case, each Hydrogen acts as a ligand and there are 2 Hydrogens.
- So, $L = 2$.
-
Anionic Charge ($A$) and Cationic Charge ($C$): For neutral $\mathrm{C}_2\mathrm{H}_2$, both $A$ and $C$ are 0.
Applying the formula: $$ \text{Total electron pairs} = \frac{V + L + A - C}{2} = \frac{10 + 2 + 0 - 0}{2} = \frac{12}{2} = 6. $$
These 6 total electron pairs include all bonding and non-bonding pairs around central atoms in the molecule $\mathrm{C}_2\mathrm{H}_2$. Note that to find the number of lone pairs specifically (pairs of electrons not involved in bonding), you consider: $$ \text{Lone pairs} = \text{total electron pairs} - \text{number of bonds}. $$ For $\mathrm{C}_2\mathrm{H}_2$, there are typically multiple bonds between the carbons (a triple bond in this case) and one bond to each hydrogen, totaling to three bonds, leaving little to no lone pairs among the central carbon atoms in this particular molecule.
How many structural isomers of the molecular formula $\mathrm{C}{3}\mathrm{H}{5}\mathrm{Br}_{3}$ are there?
The molecular formula $\mathrm{C}_3\mathrm{H}_5\mathrm{Br}_3$ can have several structural isomers depending on how the carbon atoms are connected and how the bromine and hydrogen atoms are distributed. Upon analyzing possible structures, the following five structural isomers can be identified:
- Propyl Tribromide: $\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CBr}_3$, in which one carbon is triply substituted with bromine and the remaining chain is a normal propyl.
- 1-Bromo-2,2-dibromopropane: $\mathrm{CH}_3-\mathrm{CHBr}-\mathrm{CHBr}_2$, featuring a central carbon singly substituted by bromine and the terminal carbon substituted with two bromine atoms.
- 1,3-Dibromo-2-propanol: $\mathrm{CH}_2\mathrm{Br}-\mathrm{CH}_2-\mathrm{CHBr}_2$, with bromine atoms at the first and third carbon positions, and the middle carbon with a hydroxyl substituent (assuming a possibility of tautomerization or a common simplified representation).
- 1-Bromo-1,1-dibromopropane: $\mathrm{CH}_3-\mathrm{CBr}_2-\mathrm{CH}_2\mathrm{Br}$, which alternates the substitution between a singly and doubly substituted carbon.
- 1,1,3-Tribromopropane: $\mathrm{CH}_2\mathrm{Br}-\mathrm{CHBr}-\mathrm{CH}_2\mathrm{Br}$, a symmetric distribution with bromine atoms on both terminal carbons and a singly substituted central carbon.
Each of these structures differ significantly in how the bromine atoms are distributed across the three carbon atoms, reflecting the variety of structural isomers possible with this formula.
One water molecule consists of:
A) two hydrogen atoms and one oxygen atom.
B) two hydrogen atoms and five oxygen atoms.
C) three hydrogen atoms and one oxygen atom.
D) two hydrogen atoms and five oxygen atoms.
The correct answer is Option A - two hydrogen atoms and one oxygen atom.
Water, often known as a universal solvent, is composed of two hydrogen atoms and one oxygen atom, which is chemically represented as $ \mathrm{H}_2\mathrm{O} $.
What is the relation between hydration enthalpy and solubility?
Solution
Hydration enthalpy (also known as hydration energy) refers to the amount of energy released when one mole of ions undergoes hydration, a specific case of solvation where the solvent is water. This is regarded as a type of dissolution energy.
Here's an example to illustrate this relationship: When a salt dissolves in water, the ions on the outermost part of the crystal lattice are liberated and then surrounded by nearby water molecules. If the hydration energy is equal to or greater than the lattice energy (the energy holding the ionic structure together), the salt will be soluble in water. Often, if the hydration energy surpasses the lattice energy, the dissolution process will release energy as heat.
For example:
- CaCl₂ (anhydrous calcium chloride) generates heat when it dissolves in water, indicating that the hydration energy is higher than the lattice energy.
- Conversely, CaCl₂·6H₂O (calcium chloride hexahydrate) actually absorbs heat from the surrounding water when it dissolves, which implies that its hydration energy does not fully counterbalance the lattice energy; energy must then be absorbed from the water to make up for the deficiency in energy release.
In conclusion, if the hydration enthalpy is equal to or greater than the lattice energy, the salt exhibits higher solubility. However, if the hydration enthalpy is less, solubility decreases. This relation between hydration enthalpy and solubility is essential for understanding how different substances dissolve in water.
Metals are considered to be good conductors of heat because their molecules are:
A. tightly packed
B. having strong covalent bonds
C. loosely packed
D. freely bonded
The correct answer is A. tightly packed.
Metals are known as good conductors of heat primarily because their molecules are tightly packed. This arrangement allows for effective transmission of heat energy through the material.
All bonds in benzene are equal due to:
A. Tautomerism
B. Inductive effect
C. Resonance
D. Isomerism
Solution
The correct answer is C. Resonance.
In benzene, all carbon-carbon bonds are of equal length due to the phenomenon known as resonance. The bond length of each carbon-carbon bond in benzene measures approximately $$1.39 \text{ Å}$$, which is a value between that of a single carbon-carbon bond ($$1.54 \text{ Å}$$) and a double carbon-carbon bond ($$1.34 \text{ Å}$$). This intermediate bond length is because the double bond character is distributed over all the carbon atoms in the benzene ring due to resonance, leading to equalization of all bond lengths.
Which of the molecules listed below have an $sp^{3}$ hybridized central atom? (P) PCl3 (Q) COCl2 (R) SF4 A P only B P and R only C Q and R only D P, Q and R
The correct option is A P only.
The central atom in $ \text{PCl}_3 $ is phosphorus (P). Since phosphorus has 3 single bonds with chlorine and one lone pair, it undergoes $ sp^3 $ hybridization.
In $ \text{COCl}_2 $, the central atom is carbon (C). Carbon forms 2 single bonds with chlorine and one double bond with oxygen, leading to $ sp^2 $ hybridization.
For $ \text{SF}_4 $, the central atom is sulfur (S). Sulfur forms 4 single bonds with fluorine and has one lone pair, resulting in $ sp^3d $ hybridization.
Therefore, only $ \text{PCl}_3 $ has an $ sp^3 $ hybridized central atom.
Covalent polymeric hydride among the following is:
(A) $\mathrm{CaH}_{2}$
(B) $\mathrm{SrH}_{2}$
(C) $\mathrm{BeH}_{2}$
(D) $\mathrm{BaH}_{2}$
The correct option is C: $$ \mathrm{BeH}_{2} $$
Covalent polymeric hydride is $\mathrm{BeH}_{2}$.
Identify the correct dot structure of $\mathrm{O}_{2}$ molecule.
The correct option is $\mathbf{D}$.
For the $\mathrm{O}_2$ molecule to achieve stability, it should conform to the octet configuration. This means that each oxygen atom should have a total of 8 electrons in its valence shell after bonding. The accurate dot structure that depicts this configuration is the one shown in option $\mathbf{D}$.
Explain application of valence bond theory with examples.
Application of Valence Bond Theory with Examples
Valence Bond Theory plays a crucial role in explaining covalent bond formation. One of its central principles is the condition of maximum overlap, which leads to the formation of the strongest possible bonds.
Example 1: Fluorine Molecule ($ F_2 $)
In the case of the $ F_2 $ molecule, the $ F-F $ bond is formed by the overlap of the $ p_z $ orbitals of the two fluorine atoms, each containing an unpaired electron. Due to the differing nature of the overlapping orbitals in $ H_2 $ and $ F_2 $ molecules, the bond strengths and bond lengths vary between these molecules.
Example 2: Hydrogen Fluoride (HF) Molecule
In an HF molecule, the covalent bond is formed by the overlap of the $ 1s $ orbital of hydrogen and the $ 2p_z$ orbital of fluorine, each containing an unpaired electron. This mutual sharing of electrons between hydrogen and fluorine results in the formation of a covalent bond in HF.
These examples illustrate how Valence Bond Theory explains the formation and characteristics of covalent bonds through orbital overlap.
Boric acid polymerises due to: Option 1: It is monobasic in nature Option 2: It is acidic in nature Option 3: Hydrogen bonding Option 4: Its geometry
The correct option is Option 3: Hydrogen bonding
Boric acid units are joined together through hydrogen bonds, which leads to polymerisation.
![Boric acid polymerisation](https://englishchatterbox.s3.ap-south-1.amazonaws.com/up-images/a0bbbe76-8861-49d4-9031-f32ce7d2362a.png)