Coordination Compounds - Class 12 Chemistry - Chapter 5 - Notes, NCERT Solutions & Extra Questions
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Extra Questions - Coordination Compounds | NCERT | Chemistry | Class 12
The smallest among the following ions is
(A) $\mathrm{Na}^{+}$
(B) $\mathrm{Mg}^{+2}$
(C) $\mathrm{Ba}^{2+}$
(D) $\mathrm{Al}^{3+}$
The correct option is (D) $\mathrm{Al}^{3+}$.
The size of an ion is influenced by its nuclear charge which is the effective charge experienced by electrons. As the nuclear charge increases due to the removal of electrons while maintaining a similar number of protons, the ion becomes smaller due to a stronger attraction between the nucleus and the electrons.
$\mathrm{Al}^{3+}$ has lost three electrons, which increases its effective nuclear charge more significantly compared to other ions listed, drawing its remaining electrons closer and thereby making it the smallest among the given options.
One cationic complex has two isomers A and B. Each has one Co$^{3+}$, five NH$_{3}$, one Br$^{-}$, and one SO$_{4}^{2-}$ stoichiometrically. A gives a white ppt with BaCl$_{2}$ while B gives a yellow ppt with AgNO$_{3}$.
Complexes A and B have similarity in which type of isomerism?
A. Hydrate
B. Ionization
C. Linkage
D. Coordination
The correct answer is B: Ionization
This problem involves understanding both coordination compounds and the colours of salt precipitates typically seen in qualitative inorganic analysis.
Given that we have isomers A and B containing the same entities ($\text{Co}^{3+}$, $\text{NH}_3$, $\text{Br}^-$, $\text{SO}_4^{2-}$) but producing different reactions with test agents, they showcase Ionization isomerism.
Here's the breakdown:
Isomer A: Reacts with $\text{BaCl}_2$ to give a white precipitate, which must be $\text{BaSO}_4$. Therefore, Isomer A could be $\left[\text{Co}(\text{NH}_3)_5\text{Br}\right]\text{SO}_4$.
Isomer B: Forms a yellow precipitate with $\text{AgNO}_3$, identified as $\text{AgBr}$. This indicates that Isomer B could be $\left[\text{Co}(\text{NH}_3)_5\text{SO}_4\right]\text{Br}$.
Thus, Isomer A and B differ by which anionic group ($\text{Br}^-$ vs. $\text{SO}_4^{2-}$) is inside the coordination sphere and which is outside, typical of Ionization isomers.
What is the chemical name of the compound $\mathrm{FeCl}_{3}$?
A. Iron (II) Chloride
B. Ferrous Chloride
C. Iron (III) Chloride
D. Aluminum Chloride
The correct answer is Option C: Iron (III) Chloride.
The chemical compound $\mathrm{FeCl}_3$ consists of iron and chlorine. Here, the subscript '3' next to chlorine indicates that there are three chloride ions, each with a valency of -1. This provides a total of -3 charge contributed by the chloride ions.
The iron in $\mathrm{FeCl}_3$ must balance this charge with a +3, indicating that iron is in the +3 oxidation state. Therefore, the appropriate chemical name for $\mathrm{FeCl}_3$ is Iron (III) Chloride, where the Roman numeral III indicates the +3 oxidation state of the iron. This compound is also known as ferric chloride.
Which one of the following electrolytes has the same value of van't Hoff factor (i) as that of $\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}$ (if all are $100 %$ ionized)?
(A) $\mathrm{Al}\left(\mathrm{NO}_{3}\right)_{3}$
(B) $\mathrm{K}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]$
(C) $\mathrm{K}_{2} \mathrm{SO}_{4}$
(D) $\mathrm{K}_{3}[\mathrm{Fe}(\mathrm{CN})_{6}]$
The correct option is (B) $\mathbf{K}_4[\mathbf{Fe}(\mathbf{CN})_6]$.
To find the correct option with the same value of van't Hoff factor ($i$) as that of $\mathbf{Al}_2(\mathbf{SO}_4)_3$, let's determine the dissociation of each electrolyte first:
$\mathbf{Al}_2(\mathbf{SO}_4)_3$ dissociates as: $$ \mathbf{Al}_2(\mathbf{SO}_4)3 \rightarrow 2 \mathbf{Al}^{3+} + 3 \mathbf{SO}{4}^{2-} $$ Total number of particles ($i$) = $2 + 3 = 5$.
$\mathbf{Al}(\mathbf{NO}_3)_3$ dissociates as: $$ \mathbf{Al}(\mathbf{NO}_3)_3 \rightarrow \mathbf{Al}^{3+} + 3 \mathbf{NO}_3^- $$ Total number of particles ($i$) = $1 + 3 = 4$.
$\mathbf{K}_4[\mathbf{Fe}(\mathbf{CN})_6]$ dissociates as: $$ \mathbf{K}_4[\mathbf{Fe}(\mathbf{CN})_6] \rightarrow 4 \mathbf{K}^+ + [\mathbf{Fe}(\mathbf{CN})_6]^{4-} $$ Total number of particles ($i$) = $4 + 1 = 5$.
$\mathbf{K}_3[\mathbf{Fe}(\mathbf{CN})_6]$ dissociates as: $$ \mathbf{K}_3[\mathbf{Fe}(\mathbf{CN})_6] \rightarrow 3 \mathbf{K}^+ + [\mathbf{Fe}(\mathbf{CN})_6]^{3-} $$ Total number of particles ($i$) = $3 + 1 = 4$.
From above, only $\mathbf{K}_4[\mathbf{Fe}(\mathbf{CN})_6]$ has the same van't Hoff factor ($i$) of 5 as $\mathbf{Al}_2(\mathbf{SO}_4)_3$; hence, option (B) is correct.
Which of the following represent the compound $\mathrm{CuCl}_2$?
A. Copper(I) chloride
B. Copper(II) chloride
C. Cupric chloride
D. Cuprous chloride
The correct options are:
B. Copper(II) chloride
C. Cupric chloride
In the compound $ \mathrm{CuCl}_2 $, copper exhibits a valency of two, while chlorine has a valency of one.
Copper can exhibit two valencies: one or two. The nomenclature for copper's valency utilizes the suffix '-ous' for lower valency and '-ic' for higher valency.
Thus, Cuprous refers to copper with a valency of one (e.g., $\mathrm{CuCl}$), and Cupric refers to copper with a valency of two (e.g., $\mathrm{CuCl}_2$).
Therefore, both Copper(II) chloride and Cupric chloride correctly describe the compound $ \mathrm{CuCl}_2 $.
In which of the following complex ions is the central metal ion $\mathrm{dp}^{2}$ hybridized?
A) $\left[\mathrm{Ni}(\mathrm{CO})_{4}\right]$
B) $\left[\mathrm{NiCl}_{4}\right]^{-2}$
C) $\left[\mathrm{CuCl}_{4}\right]^{-2}$
D) $\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{-2}$
The correct answer is D) $\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{-2}$. This choice can be explained by considering the hybridization of the central metal ion, nickel in its +2 oxidation state ($\mathrm{Ni}^{+2}$). Here’s a detailed breakdown:
Electronic configuration: For $\mathrm{Ni}^{+2}$, which is a $3d^8$ system, the electron configuration in an octahedral field considering strong field ligand effects is pertinent. Since $\mathrm{CN}^-$ is a strong field ligand, it forces the pairing of electrons within the d-orbitals.
Electron pairing:
$\mathrm{Ni}^{+2}$ ($3d^8$ configuration) without any field would be represented as:
$$ \begin{array}{|c|c|c|c|c|} \hline \uparrow \downarrow & \uparrow \downarrow & \uparrow \downarrow & \uparrow & \uparrow \\ \hline \end{array} $$
Under the influence of the strong field created by $\mathrm{CN}^-$ ligands, the electrons rearrange themselves to minimize repulsion:
$$ \begin{array}{|c|c|c|c|c|} \hline \uparrow \downarrow & \uparrow \downarrow & \uparrow \downarrow & \uparrow \downarrow & \\ \hline \end{array} $$
Hybridization: Following $\mathrm{CN}^-$ coordination, the central metal ion $\mathrm{Ni}^{+2}$ utilizes $\mathrm{dsp}^2$ hybridization, which involves the mixing of one d-orbital, one s-orbital, and two p-orbitals of nickel. This leads to a square planar geometry typical for $\mathrm{dsp}^2$ hybridization.
Thus, the $\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{-2}$ complex ion uses $\mathrm{dsp}^2$ hybridization for the central metal atom alignment with the strong field $\mathrm{CN}^-$ ligands causing the pairing of electrons in the d-orbitals.
Which of the following compounds do not follow the Law of Definite Proportion?
A $\mathrm{FeO}$
B $\mathrm{H}{2} \mathrm{SO}{4}$
C $\mathrm{Cr}{1.97} \mathrm{O}{3}$
D $\mathrm{F}_{0.95} \mathrm{O}$
The correct options are:
C $\mathrm{Cr}_{1.97} \mathrm{O}_{3}$
D $\mathrm{F}_{0.95} \mathrm{O}$
The Law of Definite Proportion states that a chemical compound always contains its component elements in fixed ratio and this ratio does not depend on its source and method of preparation. Compounds that always maintain a consistent stoichiometric formula of integers for each element, like $\mathrm{H}_{2}\mathrm{SO}_{4}$, comply with this law. For example, in $\mathrm{FeO}$ (option A), the ratio of Iron (Fe) to Oxygen (O) remains constant at 1:1 under ideal conditions.
However, non-stoichiometric compounds are those where the proportions of elements vary and cannot be expressed in exact whole number ratios due to defects or non-idealities in their structure. For example, $\mathrm{Cr}_{1.97} \mathrm{O}_{3}$ suggests that for nearly every two chromium (Cr) atoms, there are three oxygen (O) atoms, rather than a strict whole number ratio. Similarly, $\mathrm{F}_{0.95} \mathrm{O}$ indicates that the ratio of elements does not conform to whole numbers and the proportion can vary.
Therefore, options C and D are non-stoichiometric and do not follow the Law of Definite Proportion.
Match the catalysts to the correct processes: Catalyst Process
Catalyst | Process |
---|---|
(A) $\mathrm{TiCl}_3$ | (ii) Ziegler-Natta polymerisation |
(B) $\mathrm{PdCl}_2$ | (i) Wacker process |
(C) $\mathrm{CuCl}_2$ | (iv) Deacon's process |
(D) $\mathrm{V}_2\mathrm{O}_5$ | (iii) Contact process |
B (A)-(ii), (B)-(i), (C)-(iv), (D)-(iii)
C (A)-(ii), (B)-(iii), (C)-(iv), (D)-(i)
D (A)-(iii), (B)-(i), (C)-(ii), (D)-(iv)
The correct answer is Option B:
(A)-(ii), (B)-(i), (C)-(iv), (D)-(iii)
$\mathbf{TiCl_3}$ is used in Ziegler-Natta polymerisation for catalyzing the polymerization of alkenes.
$\mathbf{PdCl_2}$ is crucial in the Wacker process, where it converts alkenes into ketones or aldehydes.
$\mathbf{CuCl_2}$ plays a role in the Deacon's process to oxidize $\mathrm{HCl}$ to $\mathrm{Cl_2}$.
$\mathbf{V_2O_5}$ is used as a catalyst in the Contact process for the production of sulfuric acid.
Which of the following compounds are colored due to charge transfer? $$ \mathrm{KMnO}_{4}, \mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{4}, \mathrm{K}_{2} \mathrm{CrO}_{4}, \mathrm{CrO}_{2} \mathrm{Cl}_{2}, \mathrm{HgI}_{2}, \mathrm{BiI}_{3}, \mathrm{PbI}_{2} $$
Coloration in certain compounds often results from charge transfer between different constituents of a molecule. Charge transfer can be broadly classified into two types:
Ligand to Metal Charge Transfer (LMCT)
Metal to Ligand Charge Transfer (MLCT)
For the compounds listed:
$\mathrm{KMnO}_4$ and $\mathrm{K}_2\mathrm{Cr}_2\mathrm{O}_7$ contain $\mathrm{Mn}^{+7}$ and $\mathrm{Cr}^{+6}$ respectively. Both ions have no $\mathrm{d}$ electrons, eliminating the possibility of $\mathrm{d-d}$ transition typically responsible for color in many transition metal complexes. The intense purple and orange colors of these compounds, respectively, result from LMCT.
$\mathrm{K}_2\mathrm{CrO}_4$ also exhibits a yellow color similarly due to LMCT involving the $\mathrm{Cr}^{+6}$ ion.
$\mathrm{CrO}_2\mathrm{Cl}_2$ demonstrates color, attributed to LMCT where the electron density transfers from chloride ions (or possibly oxygen) to the chromium center.
$\mathrm{HgI}_2$, $\mathrm{BiI}_3$, and $\mathrm{PbI}_2$ showcase distinctive colors (red, black, and yellow respectively). These iodides partake in coloration due to LMCT, where iodine ions, having a substantial tendency to donate electrons, facilitate charge transfer to the metal cation.
In conclusion, the compounds $\mathrm{KMnO}_4$, $\mathrm{K}_2\mathrm{Cr}_2\mathrm{O}_7$, $\mathrm{K}_2\mathrm{CrO}_4$, $\mathrm{CrO}_2\mathrm{Cl}_2$, $\mathrm{HgI}_2$, $\mathrm{BiI}_3$, and $\mathrm{PbI}_2$ are colored primarily due to ligand to metal charge transfer mechanisms.
For the given complex $\left[\mathrm{CoCl}_{2}(\mathrm{en})\left(\mathrm{NH}_{3}\right)_{2}\right]^{+}$, the number of geometrical isomers and the number of optical isomers respectively are:
A) 2,2
B) 2,0
C) 2,4
D) 0,2
The correct option is A) 2, 2.
The complex given is $ \left[\mathrm{CoCl}_{2}(\mathrm{en})\left(\mathrm{NH}_{3}\right)_{2}\right]^{+} $, where 'en' is ethylenediamine, a bidentate ligand. The complex can be represented as $\left[M(AB)X_2Y_2\right]$ format, where:
$M = \mathrm{Co}$
$AB$ (bidentate ligand) = en
$X_2$ = $\mathrm{Cl}_2$
$Y_2$ = $(\mathrm{NH}_3)_2$
This complex can exhibit cis and trans geometrical isomers:
In the cis form, both like ligands are adjacent (Cl and NH₃ adjacent).
In the trans form, similar ligands are opposite to each other (Cl opposite Cl, NH₃ opposite NH₃).
Among these, the cis isomer of this complex is optically active due to the non-superimposable nature of its mirror image, and it exists in two enantiomeric forms (d and l). Thus, there are two geometrical isomers and two optical isomers.
"Why do transition metals form interstitial compounds?"
Transition metals form interstitial compounds because in their crystalline lattice, there exist vacant spaces or interstices. These spaces are capable of accommodating small atoms such as hydrogen (H), carbon (C), nitrogen (N), and others. By inserting these smaller atoms into the interstices, transition metals stabilize the structure and form interstitial compounds.
How many glycinate ions are required to make an octahedral complex with a $\mathrm{Ca}^{2+}$ ion?
A. One B. Two C. Six D. Three
The correct option is D Three.
To explain further:
Glycinate ion is a bidentate ligand, which means it forms two coordinate bonds with a central metal ion.
For an octahedral complex, the central ion needs a coordination number of 6, meaning it must form six coordinate bonds in total.
Since each glycinate ion contributes two coordinate bonds, we need three glycinate ions to achieve the coordination number of 6.
Therefore, three glycinate ions are required to form an octahedral complex with a $\mathrm{Ca}^{2+}$ ion.
The oxidation state of molybdenum in $[\text{Mo}_2\text{O}_4(\text{C}_2\text{H}_4)_2\text{H}_2\text{O}_2]^{2-}$ is ______.
To determine the oxidation state of molybdenum in the given complex $[Mo_2O_4(C_2H_4)_2(H_2O_2)]^{2-}$:
Identify the key components of the complex:
$C_2H_4$ (ethylene) and $H_2O_2$ (hydrogen peroxide) are molecules with an oxidation number of 0.
Oxygen in $O^{2-}$ is oxide and has an oxidation number of -2.
Set up the oxidation state equation:
Let $x$ be the oxidation state of molybdenum $(\text{Mo})$.
The complex contains 2 (Mo), 4 (O) (as oxide).
Write out the expression using the known oxidation states and total charge of the complex: $$ 2x + 4(-2) + 2(0) + 0 = -2 $$
Simplify the equation: $$ 2x - 8 = -2 $$
Solve for (x): $$ 2x - 8 = -2 \ 2x = -2 + 8 \ 2x = 6 \ x = \frac{6}{2} \ x = +3 $$
Conclusion: The oxidation state of molybdenum in $[Mo_2O_4(C_2H_4)_2(H_2O_2)]^{2-}$ is +3.
Answer: (+3)
Choose the correct statement:
A $\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right]^{2+}$ is oxidised to diamagnetic $\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right]^{3+}$ by the oxygen in air.
Tetrahedral complexes are more stable than octahedral complexes.
$\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-}$ is stable but $\left[\mathrm{FeF}_{6}\right]^{3-}$ is unstable.
The $\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\right]^{2+}$ ion has a tetrahedral geometry and is diamagnetic.
Option A: Correct
The complex $\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right]^{2+}$ is oxidised to diamagnetic $\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right]^{3+}$ by the oxygen in air.
Option B: Incorrect
Tetrahedral complexes are not more stable than octahedral complexes because of the different ligand field stabilization energies involved.
Option C: Incorrect
While $\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-}$ is stable, the statement $\left[\mathrm{FeF}_{6}\right]^{3-}$ is unstable is not accurate.
Option D: Incorrect
The ion $\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\right]^{2+}$ does not have tetrahedral geometry; instead, it typically forms a square planar structure and is paramagnetic.
Final Answer: A
Among
$$ \begin{array}{l} {\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right]^{3+},\left[\mathrm{CoF}_{6}\right]^{3-},\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}} \\ {\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-},\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+},\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{4-}} \\ {\left[\mathrm{Mn}(\mathrm{CN})_{6}\right]^{4-}} \end{array} $$
the number of inner orbital complexes is
To determine the number of inner orbital complexes among the given complexes, let's analyze each one using the concept of coordination chemistry and Crystal Field Theory (CFT).
List of Complexes:
$$\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]^{3+}$$
$$\left[\mathrm{CoF}_6\right]^{3-}$$
$$\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}$$
$$\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{3-}$$
$$\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+}$$
$$\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{4-}$$
$$\left[\mathrm{Mn}(\mathrm{CN})_6\right]^{4-}$$
Explanation:
$$\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]^{3+}$$:
Cobalt ion (Co^3+) configuration: 3d^6.
Ligand (NH_3): Strong field ligand.
Results in low-spin configuration: Uses inner d-orbitals.
$$\left[\mathrm{CoF}_6\right]^{3-}$$:
Cobalt ion (Co^3+) configuration: 3d^6.
Ligand (F^-): Weak field ligand.
Results in high-spin configuration: Uses outer d-orbitals. Not an inner orbital complex.
$$\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}$$:
Iron ion (Fe^2+) configuration: 3d^6.
Ligand (H_2O): Weak field ligand.
Results in high-spin configuration: Uses outer d-orbitals. Not an inner orbital complex.
$$\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{3-}$$:
Iron ion (Fe^3+) configuration: 3d^5.
Ligand (CN^-): Strong field ligand.
Results in low-spin configuration: Uses inner d-orbitals.
$$\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+}$$:
Iron ion (Fe^3+) configuration: 3d^5.
Ligand (H_2O): Weak field ligand.
Results in high-spin configuration: Uses outer d-orbitals. Not an inner orbital complex.
$$\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{4-}$$:
Iron ion (Fe^2+) configuration: 3d^6.
Ligand (CN^-): Strong field ligand.
Results in low-spin configuration: Uses inner d-orbitals.
$$\left[\mathrm{Mn}(\mathrm{CN})_6\right]^{4-}$$:
Manganese ion (Mn^2+) configuration: 3d^5.
Ligand (CN^-): Strong field ligand.
Results in low-spin configuration: Uses inner d-orbitals.
Conclusion:
There are four inner orbital complexes:
$$\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]^{3+}$$
$$\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{3-}$$
$$\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{4-}$$
$$\left[\mathrm{Mn}(\mathrm{CN})_6\right]^{4-}$$
Thus, the final answer is:
$$ \boxed{4} $$
The coordination number of $ \text{Cr} $ in the complex $ \left[\text{Cr}\left(\text{H}_2 \text{O}\right)_4 \text{Cl}_2\right]^{+} $ is 6.
Given complex: $$ \left[ \text{Cr}\left( \text{H}_2 \text{O} \right)_4 \text{Cl}_2 \right]^{+} $$
To determine the oxidation state of Cr:
Let the oxidation state of Cr be ( a ).
The oxidation state of each $ \text{H}_2 \text{O} $ molecule is ( 0 ).
The oxidation state of each ( \text{Cl} ) is ( -1 ).
The sum of the oxidation states for the complex must equal the overall charge, which is ( +1 ). Thus, we set up the equation:
$$ a + 4(0) + 2(-1) = +1 $$
Simplifying further:
$$ a - 2 = +1 $$
Solving for ( a ):
$$ a = +3 $$
Thus, the oxidation state of Cr in the complex $$ \left[ \text{Cr}\left( \text{H}_2 \text{O} \right)_4 \text{Cl}_2 \right]^{+} $$ is ( +3 ).
Final Answer: B
In a solution of $\mathrm{NH_4CI}$ and $\mathrm{NH_4OH}$, the hydroxide ion concentration is $10^{-6}$ mol/L. Which of the following metal hydroxides will precipitate upon adding $0.1 \mathrm{M}$ of the metal ion?
(I) $\mathrm{Ba(OH)_2}$ ($K_s = 5 \times 10^{-3}$)
(II) $\mathrm{Ni(OH)_2}$ ($K_s = 1.6 \times 10^{-16}$)
(III) $\mathrm{Mn(OH)_2}$ ($K_s = 2 \times 10^{-13}$)
(IV) $\mathrm{Fe(OH)_2}$ ($K_s = 8 \times 10^{-16}$)
Possible answers:
A. I, II, IV
B. IV
C. II and IV
D. II, III, IV
The correct answer is C.
Given that the concentration of hydroxide ions in the solution is ( \left[\mathrm{OH}^{-}\right] = 10^{-6} , \mathrm{M} ) (buffer solution) and the metal ion concentration ( M^{n+} = 0.05 , \mathrm{M} ).
For each metal hydroxide, we calculate the reaction quotient ( Q ):
[ Q = [ \text{Metal ion concentration} ][ \text{Hydroxide ion concentration} ]^2 ]
[ Q = [0.05 , \mathrm{M}][10^{-6} , \mathrm{M}]^2 = 5 \times 10^{-14} ]
Now let's compare ( Q ) with the solubility product ( K_{sp} ) for each metal hydroxide:
Barium Hydroxide, $\mathrm{Ba}(\mathrm{OH})_{2}$:
( K_{sp} = 5 \times 10^{-3} )
( Q = 5 \times 10^{-14} )
( Q < K_{sp} ), no precipitation
Nickel Hydroxide, $\mathrm{Ni}(\mathrm{OH})_{2}$:
( K_{sp} = 1.6 \times 10^{-16} )
( Q = 5 \times 10^{-14} )
( Q > K_{sp} ), precipitation occurs
Manganese Hydroxide, $\mathrm{Mn}(\mathrm{OH})_{2}$:
( K_{sp} = 2 \times 10^{-13} )
( Q = 5 \times 10^{-14} )
( Q < K_{sp} ), no precipitation
Iron(II) Hydroxide, $\mathrm{Fe}(\mathrm{OH})_{2}$:
( K_{sp} = 8 \times 10^{-16} )
( Q = 5 \times 10^{-14} )
( Q > K_{sp} ), precipitation occurs
Hence, $\mathrm{Ni}(\mathrm{OH})_{2}$ and $\mathrm{Fe}(\mathrm{OH})_{2}$ will precipitate in the solution.
Final Answer: C
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Ask Chatterbot AINCERT Solutions - Coordination Compounds | NCERT | Chemistry | Class 12
Explain the bonding in coordination compounds in terms of Werner's postulates.
Werner's theory of coordination compounds postulates several principles regarding the structure and bonding. These postulates can be summarized to explain the bonding in coordination compounds as follows:
Primary and Secondary Valences:
Primary Valences: These are ionizable valences and are usually satisfied by negative ions. They correspond to the oxidation state of the central metal ion. For example, in $\text{CoCl}_3 \cdot 6\text{NH}_3$, cobalt exhibits a primary valence of +3.
Secondary Valences: These are non-ionizable and are satisfied by neutral molecules or negative ions directly bonded to the metal ion. These are also known as coordination numbers and determine the geometry of the complex. For example, cobalt in the same complex has a secondary valence of 6.
Coordination Number and Geometry:
The secondary valence (or coordination number) is fixed for a particular metal and dictates the spatial arrangement of the ligands around the metal ion.
Common geometries for coordination numbers 4 and 6 are tetrahedral or square planar and octahedral, respectively. For instance, complexes like $\left[\text{Co}(\text{NH}_3)_6\right]^{3+}$ are octahedral.
Linkage (Valence) Types:
Primary Valences: These can be ionized and are typically satisfied by negative ions outside the coordination sphere, such as chloride ions in $\left[\text{Co}(\text{NH}_3)_6\right]\text{Cl}_3$.
Secondary Valences: These involve the coordination of ligands within the coordination sphere, forming stable coordination entities that do not dissociate under normal conditions.
Distinct Compounds and Isomerism:
Werner explained the existence of different compounds with the same empirical formula but different properties through isomerism. For example, $\text{CoCl}_3 \cdot 4\text{NH}_3$ can have different structures (isomers) each exhibiting distinct properties.
To illustrate, let’s consider cobalt(III) chloride complexes with ammonia:
$\text{CoCl}_3 \cdot 6\text{NH}_3$ (yellow) forms $\left[\text{Co}(\text{NH}_3)_6\right]\text{Cl}_3$
$\text{CoCl}_3 \cdot 5\text{NH}_3$ (purple) forms $\left[\text{CoCl}(\text{NH}_3)_5\right]\text{Cl}_2$
$\text{CoCl}_3 \cdot 4\text{NH}_3$ (green/violet) forms $\left[\text{CoCl}_2(\text{NH}_3)_4\right]\text{Cl}$
From these observations, Werner concluded that the precise arrangement of ligands bonded directly to the metal ion (secondary valences) and the ionizable counterions (primary valences) are key to understanding the bonding and properties of coordination compounds. This two-tiered valence concept effectively explained the formation, stability, isomerism, and physicochemical properties of coordination compounds.
$\mathrm{FeSO}_{4}$ solution mixed with $\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}$ solution in $1: 1$ molar ratio gives the test of $\mathrm{Fe}^{2+}$ ion but $\mathrm{CuSO}_{4}$ solution mixed with aqueous ammonia in 1:4 molar ratio does not give the test of $\mathrm{Cu}^{2+}$ ion. Explain why?
Scenario 1: $\mathrm{FeSO}_{4}$ and $(\mathrm{NH}_{4})_{2}\mathrm{SO}_{4}$ Solution
When $\mathrm{FeSO}_{4}$ and $(\mathrm{NH}_{4})_{2}\mathrm{SO}_{4}$ solutions are mixed in a $1:1$ molar ratio, they form a double salt known as Mohr's salt, which has the formula $\mathrm{FeSO}_4 \cdot (\mathrm{NH}_4)_2\mathrm{SO}_4 \cdot 6\mathrm{H}_2\mathrm{O}$. A double salt dissociates completely in water to give all its constituent ions. Thus, Mohr's salt dissociates to give $\mathrm{Fe}^{2+}$, $\mathrm{NH}_4^+$, and $\mathrm{SO}_4^{2-}$ ions. This allows the solution to still give a positive test for the $\mathrm{Fe}^{2+}$ ion.
Scenario 2: $\mathrm{CuSO}_{4}$ and Aqueous Ammonia Solution
When $\mathrm{CuSO}_{4}$ solution is mixed with aqueous ammonia in a $1:4$ molar ratio, it forms a complex compound. The reaction can be represented as:
$$\mathrm{CuSO}_{4} + 4\mathrm{NH}_3 \rightarrow \left[\mathrm{Cu}\left(\mathrm{NH}_3\right)_{4}\right]\mathrm{SO}_4$$
The complex ion formed in this case is $\left[\mathrm{Cu}(\mathrm{NH}_3)_4\right]^{2+}$, also known as the tetraamminecopper(II) ion. This complex is quite stable, and as a result, the $\mathrm{Cu}^{2+}$ ion is not freely available in the solution to give its characteristic tests. Hence, $\mathrm{CuSO}_4$ mixed with aqueous ammonia does not give the test for $\mathrm{Cu}^{2+}$ ions.
Conclusion
Double salts (like Mohr’s salt) dissociate completely in solution, allowing the individual ions ($\mathrm{Fe}^{2+}$ in this case) to be detected.
Complex compounds (like $\left[\mathrm{Cu}(\mathrm{NH}_3)_4\right]\mathrm{SO}_4$) do not dissociate in the same way, and the central metal ion ($\mathrm{Cu}^{2+}$) is not freely available for detection.
Thus, Mohr's salt gives the test for $\mathrm{Fe}^{2+}$ ions, while tetraamminecopper(II) sulfate does not give the test for $\mathrm{Cu}^{2+}$ ions due to the formation of a stable complex ion.
Explain with two examples each of the following: coordination entity, ligand, coordination number, coordination polyhedron, homoleptic and heteroleptic.
1. Coordination Entity
A coordination entity constitutes a central metal atom or ion bonded to a fixed number of ions or molecules:
Example 1: $\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{4-}$: In this coordination entity, an iron atom is surrounded by six cyanide ligands.
Example 2: $\left[\mathrm{CoCl}_2(\mathrm{NH}_3)_4\right]^+$: Here, a cobalt ion is coordinated to two chlorides and four ammonia molecules.
2. Ligand
Ligands are ions or molecules bound to the central atom/ion in a coordination entity:
Example 1: $\mathrm{NH}_3$ (Ammonia): A neutral ligand commonly seen in coordination complexes like $\left[\mathrm{Co}(\mathrm{NH}_3)_6\right]^{3+}$.
Example 2: $\mathrm{Cl}^-$ (Chloride): A negatively charged ligand found in complexes like $\left[\mathrm{NiCl}_4\right]^{2-}$.
3. Coordination Number
The coordination number of a metal ion in a complex defines the number of ligand donor atoms to which the metal is directly bonded:
Example 1: 6 in $\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{4-}$: Here, the iron ion is bonded to six cyanide ions.
Example 2: 4 in $\left[\mathrm{Ni}(\mathrm{NH}_3)_4\right]^{2+}$: In this case, the nickel ion is bonded to four ammonia molecules.
4. Coordination Polyhedron
Coordination polyhedron refers to the spatial arrangement of the ligand atoms which are directly attached to the central atom/ion:
Example 1: Octahedral in $\left[\mathrm{Co}(\mathrm{NH}_3)_6\right]^{3+}$: The cobalt ion is surrounded by six ammonia ligands forming an octahedral shape.
Example 2: Tetrahedral in $\left[\mathrm{NiCl}_4\right]^{2-}$: The nickel ion is coordinated to four chloride ions in a tetrahedral geometry.
5. Homoleptic
Homoleptic complexes are those in which a metal is bound to only one kind of donor group:
Example 1: $\left[\mathrm{Co}(\mathrm{NH}_3)_6\right]^{3+}$: Here, cobalt is bound only to ammonia molecules.
Example 2: $\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{4-}$: Iron is coordinated only to cyanide ions.
6. Heteroleptic
Heteroleptic complexes are those in which a metal is bound to more than one kind of donor group:
Example 1: $\left[\mathrm{CoCl}_2(\mathrm{NH}_3)_4\right]^+$: Cobalt is coordinated by both chloride and ammonia ligands.
Example 2: $\left[\mathrm{Pt(NH_3)_2Cl_2\right]$: Platinum is coordinated by both ammonia and chloride ligands, forming a heteroleptic complex.
These examples provide a clear understanding of various concepts related to coordination chemistry.
What is meant by unidentate, didentate and ambidentate ligands? Give two examples for each.
Unidentate Ligands
Unidentate ligands are ligands that have only one donor atom that coordinates to a central metal atom or ion. These ligands can form only one bond with the metal center.
Examples:
Ammonia ($\mathrm{NH}_3$): The nitrogen atom has a lone pair of electrons that can form a single bond with the metal ion.
Chloride ion ($\mathrm{Cl}^-$): The chloride ion uses one of its lone pairs to form a single bond with the metal ion.
Didentate Ligands
Didentate ligands are ligands that have two donor atoms that can simultaneously coordinate to a central metal atom or ion, forming two bonds with the metal center.
Examples:
Ethylenediamine ($\mathrm{H}_2 \mathrm{NCH}_2 \mathrm{CH}_2 \mathrm{NH}_2$): This ligand has two nitrogen atoms, each with a lone pair of electrons, that can bind to the metal ion.
Oxalate ion ($\mathrm{C}_2 \mathrm{O}_4^{2-}$): This ligand has two oxygen atoms that can each donate a lone pair of electrons to the metal ion.
Ambidentate Ligands
Ambidentate ligands are ligands that have two or more donor atoms, but only one of them can bind to the central metal ion at a time.
Examples:
Thiocyanate ion ($\mathrm{SCN}^-$): This ligand can bind to the metal ion either through the sulfur atom or the nitrogen atom.
Nitrite ion ($\mathrm{NO}_2^-$): This ligand can coordinate to the metal ion either through the nitrogen atom (as in $\mathrm{M} - \mathrm{NO}_2$) or through one of the oxygen atoms (as in $\mathrm{M} - \mathrm{ONO}$).
By understanding these types of ligands, you can predict the way they will interact with metal ions in coordination compounds.
Specify the oxidation numbers of the metals in the following coordination entities:
(i) $\left[\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)(\mathrm{CN})(\mathrm{en})_{2}\right]^{2+}$
(iii) $\left[\mathrm{PtCl}_{4}\right]^{2-}$
(v) $\left[\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{3} \mathrm{Cl}_{3}\right]$
(ii) $\left[\mathrm{CoBr}_{2}(\mathrm{en})_{2}\right]^{+}$
(iv) $\mathrm{K}_{3}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]$
Let's determine the oxidation numbers of the metals in each coordination entity manually.
$\left[\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)(\mathrm{CN})(\mathrm{en})_{2}\right]^{2+}$:
$\mathrm{H}_{2} \mathrm{O}$ (neutral ligand) contributes 0.
$\mathrm{CN}^{-}$ contributes -1.
2 $\times$ $\mathrm{en}$ (neutral ligand) contribute 0.
The overall charge on the complex is +2.
Let the oxidation state of $\mathrm{Co}$ be $x$: [ x + 0 - 1 + 0 = +2 \implies x - 1 = +2 \implies x = +3 ] Oxidation number of $\mathrm{Co}$ is +3.
$\left[\mathrm{PtCl}_{4}\right]^{2-}$:
4 $\mathrm{Cl}^{-}$ contribute -4.
The overall charge on the complex is -2.
Let the oxidation state of $\mathrm{Pt}$ be $x$: [ x - 4 = -2 \implies x = +2 ] Oxidation number of $\mathrm{Pt}$ is +2.
$\left[\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{3} \mathrm{Cl}_{3}\right]$:
3 $\mathrm{NH}_{3}$ (neutral ligand) contribute 0.
3 $\mathrm{Cl}^{-}$ (outside the coordination sphere) contribute -3.
The overall charge on the complex is neutral (0).
Let the oxidation state of $\mathrm{Cr}$ be $x$: [ x + 0 - 3 = 0 \implies x - 3 = 0 \implies x = +3 ] Oxidation number of $\mathrm{Cr}$ is +3.
$\left[\mathrm{CoBr}_{2}(\mathrm{en})_{2}\right]^{+}$:
2 $\mathrm{Br}^{-}$ contribute -2.
2 $\mathrm{en}$ (neutral ligand) contribute 0.
The overall charge on the complex is +1.
Let the oxidation state of $\mathrm{Co}$ be $x$: [ x - 2 + 0 = +1 \implies x - 2 = +1 \implies x = +3 ] Oxidation number of $\mathrm{Co}$ is +3.
$\mathrm{K}_{3}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]$:
3 $\mathrm{K}^{+}$ contribute +3 (outside the coordination sphere).
6 $\mathrm{CN}^{-}$ contribute -6.
The overall charge on the coordination sphere $\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]$ is -3.
Let the oxidation state of $\mathrm{Fe}$ be $x$: [ x - 6 = -3 \implies x = +3 ] Oxidation number of $\mathrm{Fe}$ is +3.
Using IUPAC norms write the formulas for the following:
(i) Tetrahydroxidozincate(II)
(ii) Potassium tetrachloridopalladate(II)
(iii) Diamminedichloridoplatinum(II)
(iv) Potassium tetracyanidonickelate(II)
(v) Pentaamminenitrito-O-cobalt(III)
(vi) Hexaamminecobalt(III) sulphate
(vii) Potassium tri(oxalato)chromate(III)
(viii) Hexaammineplatinum(IV)
(ix) Tetrabromidocuprate(II)
(x) Pentaamminenitrito-N-cobalt(III)
Tetrahydroxidozincate(II):
Formula: $$ \mathrm{[Zn(OH)_4]^{2-}} $$
Potassium tetrachloridopalladate(II):
Formula: $$ \mathrm{K_2[PdCl_4]} $$
Diamminedichloridoplatinum(II):
Formula: $$ \mathrm{[Pt(NH_3)_2Cl_2]} $$
Potassium tetracyanidonickelate(II):
Formula: $$ \mathrm{K_2[Ni(CN)_4]} $$
Pentaamminenitrito-O-cobalt(III):
Formula: $$ \mathrm{[Co(NH_3)_5(ONO)]^{2+}} $$
Hexaamminecobalt(III) sulphate:
Formula: $$ \mathrm{[Co(NH_3)_6]_2(SO_4)_3} $$
Potassium tri(oxalato)chromate(III):
Formula: $$ \mathrm{K_3[Cr(C_2O_4)_3]} $$
Hexaammineplatinum(IV):
Formula: $$ \mathrm{[Pt(NH_3)_6]^{4+}} $$
Tetrabromidocuprate(II):
Formula: $$ \mathrm{[CuBr_4]^{2-}} $$
Pentaamminenitrito-N-cobalt(III):
Formula: $$ \mathrm{[Co(NH_3)_5(NO_2)]^{2+}} $$
Each formula has been written in accordance with the IUPAC nomenclature conventions.
Using IUPAC norms write the systematic names of the following:
(i) $\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right] \mathrm{Cl}_{3}$
(ii) $\left[\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{2} \mathrm{Cl}\left(\mathrm{NH}_{2} \mathrm{CH}_{3}\right)\right] \mathrm{Cl}$
(iii) $\left[\mathrm{Ti}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}$
(iv) $\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}\left(\mathrm{NO}_{2}\right)\right] \mathrm{Cl}$
(v) $\left[\mathrm{Mn}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}$
(vi) $\left[\mathrm{NiCl}_{4}\right]^{2-}$
(vii) $\left[\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}\right] \mathrm{Cl}_{2}$
(viii) $\left.[\text { Co(en) })_{3}\right]^{3+}$
(ix) $\left[\mathrm{Ni}(\mathrm{CO})_{4}\right]$
Complex: $\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right] \mathrm{Cl}_{3}$ Name: Hexaamminecobalt(III) chloride
Complex: $\left[\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{2} \mathrm{Cl}\left(\mathrm{NH}_{2} \mathrm{CH}_{3}\right)\right] \mathrm{Cl}$ Name: Diamminechlorido(methylamine)platinum(II) chloride
Complex: $\left[\mathrm{Ti}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}$ Name: Hexaaquatitanium(III) ion
Complex: $\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}\left(\mathrm{NO}_{2}\right)\right] \mathrm{Cl}$ Name: Tetraamminechloridonitrito-N-cobalt(III) chloride
Complex: $\left[\mathrm{Mn}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}$ Name: Hexaaquamanganese(II) ion
Complex: $\left[\mathrm{NiCl}_{4}\right]^{2-}$ Name: Tetrachloridonickelate(II) ion
Complex: $\left[\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}\right] \mathrm{Cl}_{2}$ Name: Hexaamminenickel(II) chloride
Complex: $\left[\mathrm{Co(en)}_{3}\right]^{3+}$ Name: Tris(ethane-1,2-diamine)cobalt(III) ion
Complex: $\left[\mathrm{Ni}(\mathrm{CO})_{4}\right]$ Name: Tetracarbonylnickel(0)
These names are derived according to the IUPAC guidelines which involve identifying the ligands, specifying their quantities, naming the central metal with its oxidation state, and adding any necessary suffixes or counter ions.
List various types of isomerism possible for coordination compounds, giving an example of each.
It seems that the queries provided could not be processed directly. I will use different approaches to illustrate the possible isomers for each type of isomerism.
1. Geometrical Isomerism
Example: $\left[\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{2} \mathrm{Cl}_{2}\right]$
Depictions:
Cis Isomer: Chlorines adjacent
Trans Isomer: Chlorines opposite
2. Optical Isomerism
Example: $\left[\mathrm{Co}\left(\mathrm{en}\right)_{3}\right]^{3+}$
Depictions:
d form
l form
3. Linkage Isomerism
Example: $\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5}\left(\mathrm{NO}_{2}\right)\right]^{2+}$
Depictions:
Nitro ( -N bound)
Nitrito ( -O bound)
4. Coordination Isomerism
Example: $\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right]\left[\mathrm{Cr}\left(\mathrm{CN}\right)_{6}\right]$ and $\left[\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{6}\right]\left[\mathrm{Co}\left(\mathrm{CN}\right)_{6}\right]$
Depictions:
Co primary with $\mathrm{NH}_{3}$, Cr with $\mathrm{CN}$
Cr primary with $\mathrm{NH}_{3}$, Co with $\mathrm{CN}$
5. Ionisation Isomerism
Example: $\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5}\left(\mathrm{SO}_{4}\right)\right]\mathrm{Br}$ and $\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{Br}\right]\mathrm{SO}_{4}$
Depictions:
SO$_4$ as ligand, Br as counter ion
Br as ligand, SO$_4$ as counter ion
6. Solvate Isomerism
Example: $\left[\mathrm{Cr}\left(\mathrm{H}_{2}\mathrm{O}\right)_{6}\right]\mathrm{Cl}_{3}$ and $\left[\mathrm{Cr}\left(\mathrm{H}_{2}\mathrm{O}\right)_{5} \mathrm{Cl}\right]\mathrm{Cl}_{2}\cdot\mathrm{H}_{2}\mathrm{O}$
Depictions:
All water directly bonded
One chloride directly bonded, rest water externally
Illustrations
Since Wolfram|Alpha wasn't able to provide illustrations this time, I recommend referring to standard textbooks or online resources for detailed geometric structures. Here is a typical geometric representation of each type:
Geometrical Isomerism: $\left[\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{2} \mathrm{Cl}_{2}\right]$
Cis:
Trans:
Optical Isomerism: $\left[\mathrm{Co}\left(\mathrm{en}\right)_{3}\right]^{3+}$
d form and l form:
Linkage Isomerism: $\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5}\left(\mathrm{NO}_{2}\right)\right]^{2+}$
Nitro:
Nitrito:
Coordination Isomerism: $\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right]\left[\mathrm{Cr}\left(\mathrm{CN}\right)_{6}\right]$ and $\left[\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{6}\right]\left[\mathrm{Co}\left(\mathrm{CN}\right)_{6}\right]$
Ionisation Isomerism: $\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5}\left(\mathrm{SO}_{4}\right)\right]\mathrm{Br}$ and $\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{Br}\right]\mathrm{SO}_{4}$
Solvate Isomerism: $\left[\mathrm{Cr}\left(\mathrm{H}_{2}\mathrm{O}\right)_{6}\right]\mathrm{Cl}_{3}$ and $\left[\mathrm{Cr}\left(\mathrm{H}_{2}\mathrm{O}\right)_{5} \mathrm{Cl}\right]\mathrm{Cl}_{2}\cdot\mathrm{H}_{2}\mathrm{O}$
These illustrations should give you a good understanding of each type of isomerism in coordination compounds.
How many geometrical isomers are possible in the following coordination entities?
(i) $\left[\mathrm{Cr}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{3}\right]^{3-}$
(ii) $\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{3} \mathrm{Cl}_{3}\right]$
To determine the number of geometrical isomers possible for the given coordination entities, let's analyze each case:
(i) $$\left[\mathrm{Cr}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{3}\right]^{3-}$$
This complex is an octahedral coordination entity where the chromium ion is bonded to three *oxalate* ($$\mathrm{C}_{2} \mathrm{O}_{4}$$) ligands, which are bidentate. Given that all ligands are identical and bidentate, this complex does not present any geometrical isomerism.
Number of geometrical isomers for $$\left[\mathrm{Cr}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{3}\right]^{3-}$$ = 0
(ii) $$\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{3} \mathrm{Cl}_{3}\right]$$
This is a six-coordinate, octahedral complex with three *ammonia* ($$\mathrm{NH}_{3}$$) ligands and three *chloride* ($$\mathrm{Cl}^-$$) ligands. The arrangement of these ligands can lead to two possible geometrical isomers:
Facial (fac) isomer: where three identical ligands occupy adjacent positions at the corners of an octahedral face.
Meridional (mer) isomer: where three identical ligands occupy positions around the meridian of the octahedron.
Number of geometrical isomers for $$\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{3} \mathrm{Cl}_{3}\right]$$ = 2 (fac and mer)
Summary:
$$\left[\mathrm{Cr}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{3}\right]^{3-}$$: 0 geometrical isomers
$$\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{3} \mathrm{Cl}_{3}\right]$$: 2 geometrical isomers (fac and mer)
Draw the structures of optical isomers of:
(i) $\left[\mathrm{Cr}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{3}\right]^{3-}$
(ii) $\left[\mathrm{PtCl}_{2}(\mathrm{en})_{2}\right]^{2+}$
(iii) $\left[\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{2} \mathrm{Cl}_{2}(\mathrm{en})\right]^{+}$
It looks like these queries are still not being understood. Let me guide you in drawing the structures of these optical isomers manually:
1. Optical Isomers of $\left[\mathrm{Cr}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{3}\right]^{3-}$
The complex ion $\left[\mathrm{Cr}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{3}\right]^{3-}$, also known as tris(oxalato)chromate(III), exhibits optical isomerism. Each oxalate ($\mathrm{C}_{2} \mathrm{O}_{4}^{2-}$) ligand is bidentate, meaning it forms two bonds with the central chromium ion. The resulting structure is octahedral, and the different spatial arrangements of the oxalate ligands lead to non-superimposable mirror images.
2. Optical Isomers of $\left[\mathrm{PtCl}_{2}(\mathrm{en})_{2}\right]^{2+}$
The complex ion $\left[\mathrm{PtCl}_{2}(\mathrm{en})_{2}\right]^{2+}$, also known as dichlorobis(ethylenediamine)platinum(II), can show optical isomerism. The ethylenediamine ($\mathrm{en}$) is a bidentate ligand, which leads to the formation of non-superimposable mirror images.
3. Optical Isomers of $\left[\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{2} \mathrm{Cl}_{2}(\mathrm{en})\right]^{+}$
The complex ion $\left[\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{2} \mathrm{Cl}_{2}(\mathrm{en})\right]^{+}$ combines the mono-dentate ammonia and chloride ligands with the bidentate ethylenediamine ligand. This arrangement also produces optical isomers.
For these isomers, typically, the positioning of the ligands creates non-superimposable mirror images. Here's a simplified diagram to represent these:
[ \begin{array}{c} \text{d-Isomer of } \left[\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{2} \mathrm{Cl}_{2}(\mathrm{en})\right]^{+} \ \begin{array}{ccc} \mathrm{NH}_{3} & \quad \mathrm{NH}_{3} \ | & | \ \mathrm{Cr} & \quad \mathrm{Cr} \ | & | \ \mathrm{Cl} & \quad \mathrm{Cl} \end{array} \end{array} ]
These diagrams are overly simplified, but they offer a basic idea of how the optical isomers of these complexes differ in their spatial configurations.
Draw all the isomers (geometrical and optical) of:
(i) $\left[\mathrm{CoCl}_{2}(\mathrm{en})_{2}\right]^{+}$
(ii) $\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right) \mathrm{Cl}(\mathrm{en})_{2}\right]^{2+}$
(iii) $\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{2} \mathrm{Cl}_{2}(\mathrm{en})\right]^{+}$
To illustrate the isomers, let's break down each complex and draw their respective isomers using both geometrical and optical criteria.
(i) $\left[\mathrm{CoCl}_{2}(\mathrm{en})_{2}\right]^{+}$
Geometrical Isomers
Cis Isomer: Chlorides are adjacent to each other.
Trans Isomer: Chlorides are opposite to each other.
Optical Isomers
The cis isomer can exist as a pair of enantiomers (d and l).
Leveraging the power of computational tools, let me generate these isomers:
Cis Isomer (Geometrical and Optical Isomers)
Trans Isomer (Only Geometrical Isomer)
(ii) $\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right) \mathrm{Cl}(\mathrm{en})_{2}\right]^{2+}$
Geometrical Isomers
Cis Isomer: The ammine and chloride are adjacent.
Trans Isomer: The ammine and chloride are opposite each other.
Optical Isomers
The cis isomer can exist as a pair of enantiomers (d and l).
Generating these isomers:
Cis Isomer (Geometrical and Optical Isomers)
Trans Isomer (Only Geometrical Isomer)
(iii) $\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{2} \mathrm{Cl}_{2}(\mathrm{en})\right]^{+}$
Geometrical Isomers
Cis Isomer: The two chlorides are adjacent.
Trans Isomer: The two chlorides are opposite each other.
Generating these isomers:
Cis and Trans Isomers
Note: Optical isomerism is typically not observed for the trans isomer due to the symmetry.
We have successfully visualized all the geometrical and optical isomers for the given coordination compounds.
Write all the geometrical isomers of $\left[\mathrm{Pt}\left(\mathrm{NH}_{3}\right)(\mathrm{Br})(\mathrm{Cl})(\mathrm{py})\right]$ and how many of these will exhibit optical isomers?
To determine the geometrical isomers of $\left[\mathrm{Pt}\left(\mathrm{NH}_{3}\right)(\mathrm{Br})(\mathrm{Cl})(\mathrm{py})\right]$, we need to consider all possible spatial arrangements of these ligands around the platinum(II) ion, which adopts a square planar geometry.
Geometrical Isomers:
In a square planar complex $\left[\mathrm{Pt}\left(\mathrm{NH}_{3}\right)(\mathrm{Br})(\mathrm{Cl})(\mathrm{py})\right]$, we can have the following geometrical isomers:
Isomer 1:
$\mathrm{NH}_{3}$ and $\mathrm{Br}$ are trans.
$\mathrm{Cl}$ and $\mathrm{py}$ are trans.
Isomer 2:
$\mathrm{NH}_{3}$ and $\mathrm{Br}$ are cis.
$\mathrm{Cl}$ and $\mathrm{py}$ are cis.
Isomer 3:
$\mathrm{NH}_{3}$ and $\mathrm{Cl}$ are trans.
$\mathrm{Br}$ and $\mathrm{py}$ are trans.
Isomer 4:
$\mathrm{NH}_{3}$ and $\mathrm{Cl}$ are cis.
$\mathrm{Br}$ and $\mathrm{py}$ are cis.
Isomer 5:
$\mathrm{NH}_{3}$ and $\mathrm{py}$ are trans.
$\mathrm{Br}$ and $\mathrm{Cl}$ are trans.
Isomer 6:
$\mathrm{NH}_{3}$ and $\mathrm{py}$ are cis.
$\mathrm{Br}$ and $\mathrm{Cl}$ are cis.
Optical Isomerism:
In square planar complexes, optical isomerism is rare due to the planar nature of the structure which usually allows for superimposition of the mirror images. Therefore, none of the above geometrical isomers will exhibit optical isomerism.
Summary:
Geometrical Isomers: A total of 6.
Optical Isomers: 0 (none) exhibit optical isomerism.
Here are visual depictions of these isomers for better understanding:
Trans Isomer (1)
Structure: $$\begin{array}{c}\text{NH}_3 \ | \ \text{Pt} \ |\ \ \ \ \ |\ \ \ \ \ |\ \ \ \ | \ \text{Br} \end{array}
\text{trans}\begin{array}{c}\text{Cl} \ | \ \text{py} \end{array}$$
Cis Isomer (2)
Structure: $$\begin{array}{c}\text{NH}_3 \ | \ \ \ \ /\ \ \ \ \ \text{Pt} \ |\ \ \ \ \ |\ \ \ \ \ |\ \ \ \ | \ \text{Br} \end{array}
\text{cis}\begin{array}{c}\text{Cl} \ /\ \ \text{py} \end{array}$$Trans Isomer (3)
Structure: $$\begin{array}{c}\text{NH}_3 \ | \ \text{Pt} \ |\ \ \ \ \ |\ \ \ \ \ |\ \ \ \ | \ \text{Cl} \end{array}
\text{trans}\begin{array}{c}\text{Br} \ | \ \text{py} \end{array}$$
Cis Isomer (4)
Structure: $$\begin{array}{c}\text{NH}_3 \ | \ \ \ \ /\ \ \ \ \ \text{Pt} \ |\ \ \ \ \ |\ \ \ \ \ |\ \ \ \ | \ \text{Cl} \end{array}
\text{cis}\begin{array}{c}\text{Br} \ /\ \ \text{py} \end{array}$$Trans Isomer (5)
Structure: $$\begin{array}{c}\text{NH}_3 \ | \ \text{Pt} \ |\ \ \ \ \ |\ \ \ \ \ |\ \ \ \ | \ \text{py} \end{array}
\text{trans}\begin{array}{c}\text{Br} \ | \ \text{Cl} \end{array}$$
Cis Isomer (6)
Structure: $$\begin{array}{c}\text{NH}_3 \ | \ \ \ \ /\ \ \ \ \ \text{Pt} \ |\ \ \ \ \ |\ \ \ \ \ |\ \ \ \ | \ \text{py} \end{array}
\text{cis}\begin{array}{c}\text{Br} \ /\ \ \text{Cl} \end{array}$$
Aqueous copper sulphate solution (blue in colour) gives:
(i) a green precipitate with aqueous potassium fluoride and
(ii) a bright green solution with aqueous potassium chloride. Explain these experimental results.
The observed experimental results can be explained using the concepts of coordination chemistry and the formation of different coordination compounds.
Reaction with Aqueous Potassium Fluoride
When aqueous copper sulfate reacts with aqueous potassium fluoride, a green precipitate forms. This is due to the formation of a copper(II) fluoride complex:
$$ \mathrm{CuSO}_4 (\text{aq}) + 4 \mathrm{KF} (\text{aq}) \rightarrow [\mathrm{CuF}_4]^{2-} (\text{green}) + 4 \mathrm{K}^{+} (\text{aq}) + \mathrm{SO}_4^{2-} (\text{aq}) $$
Here, the $\mathrm{Cu}^{2+}$ ion coordinates with four $\mathrm{F}^{-}$ ions to form the $\left[\mathrm{CuF}_4\right]^{2-}$ complex, which gives a green precipitate. The formation of this complex causes the precipitate to appear green in color.
Reaction with Aqueous Potassium Chloride
When aqueous copper sulfate reacts with aqueous potassium chloride, a bright green solution forms due to the formation of a copper(II) chloride complex:
$$ \mathrm{CuSO}_4 (\text{aq}) + 4 \mathrm{KCl} (\text{aq}) \rightarrow [\mathrm{CuCl}_4]^{2-} (\text{bright green}) + 4 \mathrm{K}^{+} (\text{aq}) + \mathrm{SO}_4^{2-} (\text{aq}) $$
In this case, the $\mathrm{Cu}^{2+}$ ion coordinates with four $\mathrm{Cl}^{-}$ ions to form the $\left[\mathrm{CuCl}_4\right]^{2-}$ complex, which remains in solution and imparts a bright green color to the solution.
Summary
With Potassium Fluoride: The formation of $\left[\mathrm{CuF}_4\right]^{2-}$ leads to a green precipitate.
With Potassium Chloride: The formation of $\left[\mathrm{CuCl}_4\right]^{2-}$ results in a bright green solution.
These changes occur due to the formation of specific coordination complexes with different ligands, which influence the color of the resulting solutions.
What is the coordination entity formed when excess of aqueous KCN is added to an aqueous solution of copper sulphate? Why is it that no precipitate of copper sulphide is obtained when $\mathrm{H}_{2} \mathrm{~S}(\mathrm{~g})$ is passed through this solution?
When excess aqueous $\mathrm{KCN}$ is added to an aqueous solution of copper sulphate ($\mathrm{CuSO}_{4}$), the coordination entity formed is $\left[\mathrm{Cu}(\mathrm{CN})_{4}\right]^{2-}$ (tetracyanocuprate(II) ion).
Explanation:
Reaction with $\mathrm{KCN}$:$$\mathrm{CuSO}_{4} + 4\mathrm{KCN} \rightarrow \mathrm{K}_{2}\left[\mathrm{Cu}(\mathrm{CN})_{4}\right] + \mathrm{K}_{2}\mathrm{SO}_{4}$$
Coordination Entity:The coordination entity formed is $\left[\mathrm{Cu}(\mathrm{CN})_{4}\right]^{2-}$, where copper is coordinated to four cyanide ions.
Reason for No Precipitate of Copper Sulfide:
When hydrogen sulfide ($\mathrm{H}_{2}\mathrm{S}$) gas is passed through the solution containing $\left[\mathrm{Cu}(\mathrm{CN})_{4}\right]^{2-}$, no precipitate of copper sulfide ($\mathrm{CuS}$) forms. This occurs because the cyanide ions form a very stable complex with copper ions, which prevents the copper ions from reacting with the sulfide ions ($\mathrm{S}^{2-}$) in solution to form copper sulfide ($\mathrm{CuS}$).
In other words, the stability of the $\left[\mathrm{Cu}(\mathrm{CN})_{4}\right]^{2-}$ complex is such that it does not dissociate to release $\mathrm{Cu}^{2+}$ ions, which are required to form the $\mathrm{CuS}$ precipitate.
This phenomenon is a result of the high stability constant of the $\left[\mathrm{Cu}(\mathrm{CN})_{4}\right]^{2-}$ complex, which keeps the copper ions in a tightly bound state, preventing them from interacting with other ions such as sulfide ions.
Discuss the nature of bonding in the following coordination entities on the basis of valence bond theory:
(i) $\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{4-}$
(ii) $\left[\mathrm{FeF}_{6}\right]^{3-}$
(iii) $\left[\mathrm{Co}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{3}\right]^{3-}$
(iv) $\left[\mathrm{CoF}_{6}\right]^{3-}$
Valence Bond Theory Explanation
(i) $\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{4-}$
In this coordination entity, iron is in the +2 oxidation state ($\mathrm{Fe}^{2+}$). The electronic configuration of $\mathrm{Fe}^{2+}$ is $3d^6$.
Hybridization:
The strong field ligand (CN⁻) causes pairing of electrons in the $3d$ orbitals, leading to the use of inner $d$ orbitals for hybridization.
The hybridization involved is $d^2sp^3$, resulting in an octahedral complex.
Structure:
$[ \begin{array}{cccccc} \mathrm{Fe}^{2+} & \rightarrow & 3d & 4s & 4p \ \text{Electronic Configuration} & & \uparrow\downarrow & [\uparrow\downarrow] & [\uparrow\downarrow] \ \text{Hybrid Orbitals} & & d^2 & sp^3 & \ \end{array} ]$
(ii) $\left[\mathrm{FeF}_{6}\right]^{3-}$
In this coordination entity, iron is in the +3 oxidation state ($\mathrm{Fe}^{3+}$). The electronic configuration of $\mathrm{Fe}^{3+}$ is $3d^5$.
Hybridization:
The weak field ligand (F⁻) does not cause pairing of electrons in the $3d$ orbitals, resulting in the use of outer $d$ orbitals for hybridization.
The hybridization involved is $sp^3d^2$, resulting in an octahedral complex.
Structure:
$[ \begin{array}{cccccc} \mathrm{Fe}^{3+} & \rightarrow & 3d & 4s & 4p & 4d \ \text{Electronic Configuration} & & \uparrow \uparrow \uparrow \uparrow \uparrow & & & \ \text{Hybrid Orbitals} & & & sp^3 & d^2 & \ \end{array} ]$
(iii) $\left[\mathrm{Co}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{3}\right]^{3-}$
In this coordination entity, cobalt is in the +3 oxidation state ($\mathrm{Co}^{3+}$). The electronic configuration of $\mathrm{Co}^{3+}$ is $3d^6$.
Hybridization:
The strong field ligand ($\mathrm{C}_{2} \mathrm{O}_{4}^{2-}$) causes pairing of electrons in the $3d$ orbitals, leading to the use of inner $d$ orbitals for hybridization.
The hybridization involved is $d^2sp^3$, resulting in an octahedral complex.
Structure:
$[ \begin{array}{cccccc} \mathrm{Co}^{3+} & \rightarrow & 3d & 4s & 4p \ \text{Electronic Configuration} & & \uparrow\downarrow & [\uparrow\downarrow] & [\uparrow\downarrow] \ \text{Hybrid Orbitals} & & d^2 & sp^3 & \ \end{array} ]$
(iv) $\left[\mathrm{CoF}_{6}\right]^{3-}$
In this coordination entity, cobalt is in the +3 oxidation state ($\mathrm{Co}^{3+}$). The electronic configuration of $\mathrm{Co}^{3+}$ is $3d^6$.
Hybridization:
The weak field ligand (F⁻) does not cause pairing of electrons in the $3d$ orbitals, resulting in the use of outer $d$ orbitals for hybridization.
The hybridization involved is $sp^3d^2$, resulting in an octahedral complex.
Structure:
$[ \begin{array}{cccccc} \mathrm{Co}^{3+} & \rightarrow & 3d & 4s & 4p & 4d \ \text{Electronic Configuration} & & \uparrow \uparrow \uparrow \uparrow \downarrow & & & \ \text{Hybrid Orbitals} & & & sp^3 & d^2 & \ \end{array} ]$
Summary
$\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{4-}$ uses $d^2sp^3$ hybridization, leading to an octahedral structure.
$\left[\mathrm{FeF}_{6}\right]^{3-}$ uses $sp^3d^2$ hybridization, leading to an octahedral structure.
$\left[\mathrm{Co}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{3}\right]^{3-}$ uses $d^2sp^3$ hybridization, leading to an octahedral structure.
$\left[\mathrm{CoF}_{6}\right]^{3-}$ uses $sp^3d^2$ hybridization, leading to an octahedral structure.
Draw figure to show the splitting of $d$ orbitals in an octahedral crystal field.
The splitting of $d$ orbitals in an octahedral crystal field can be visualized as follows:
In the presence of an octahedral field created by six ligands, the five degenerate $d$ orbitals (which are of equal energy in an isolated atom) split into two sets:
$t_{2g}$ orbitals: These are the $d_{xy}$, $d_{yz}$, and $d_{xz}$ orbitals, which experience less repulsion and thus lower energy.
$e_g$ orbitals: These are the $d_{z^2}$ and $d_{x^2 - y^2}$ orbitals, which experience greater repulsion and thus higher energy.
Here's a simplified sketch of the $d$ orbital splitting in an octahedral field:
e_g
_______
/ \
/ \
t_{2g} \__________/
_______
/ \
/ \
The $t_{2g}$ orbitals are lower in energy than the $e_g$ orbitals.
The energy gap between these two sets of orbitals is called the crystal field splitting energy ($\Delta_o$).
For a more precise and accurate depiction, consider using a chemistry textbook or resource that provides visual diagrams of this concept.
What is spectrochemical series? Explain the difference between a weak field ligand and a strong field ligand.
Spectrochemical Series
The spectrochemical series is an experimentally determined sequence of ligands arranged in order of increasing field strength (ability to split the d-orbital energies). This series helps in understanding which ligands produce stronger fields when they bond to a central metal ion, thereby causing greater splitting of the metal's d-orbitals.
Here is the spectrochemical series:
$$ \mathrm{I}^{-}<\mathrm{Br}^{-}<\mathrm{SCN}^{-}<\mathrm{Cl}^{-}<\mathrm{S}^{2-}<\mathrm{F}^{-}<\mathrm{OH}^{-}<\mathrm{C}_{2}\mathrm{O}_{4}^{2-}<\mathrm{H}_{2}\mathrm{O}<\mathrm{NCS}^{-}<\mathrm{edta}^{4-}<\mathrm{NH}_3<\mathrm{en}<\mathrm{CN}^{-}<\mathrm{CO} $$
Weak Field Ligands vs. Strong Field Ligands
Weak Field Ligands:
Characteristics: They produce a small crystal field splitting energy ($\Delta_0$).
Examples: $\mathrm{I}^{-}$, $\mathrm{Br}^{-}$, $\mathrm{SCN}^{-}$, $\mathrm{Cl}^{-}$, $\mathrm{S}^{2-}$.
Effect: The difference in energy levels between ( t_{2g} ) and ( e_g ) orbitals is small; hence, electrons are more likely to occupy the higher energy orbitals rather than pairing up. This leads to high spin complexes.
Orbital Configuration: For example, in a (d^4) system, electrons will occupy all ( t_{2g} ) and at least one ( e_g ), remaining unpaired if $\Delta_0$ is less than the pairing energy ($P$).
Strong Field Ligands:
Characteristics: They produce a large crystal field splitting energy ($\Delta_0$).
Examples: $\mathrm{NH}_3$, $\mathrm{en}$, $\mathrm{CN}^{-}$, $\mathrm{CO}$.
Effect: The larger energy gap prefers electron pairing in the lower ( t_{2g} ) orbitals over occupying the higher ( e_g ) orbitals. This leads to low spin complexes.
Orbital Configuration: For example, in a (d^4) system, electrons will prefer to pair in the ( t_{2g} ) orbitals resulting in fewer unpaired electrons if $\Delta_0$ is greater than the pairing energy ($P$).
Summary: The key difference between weak field and strong field ligands lies in the magnitude of the crystal field splitting energy they produce, affecting how electrons are distributed among the d-orbitals in a coordination complex.
What is crystal field splitting energy? How does the magnitude of $\Delta_{0}$ decide the actual configuration of $d$ orbitals in a coordination entity?
Crystal Field Splitting Energy
Crystal Field Splitting Energy (CFSE) is the energy difference between different sets of $d$ orbitals in a metal ion when it is surrounded by ligands in a coordination compound. This concept arises from Crystal Field Theory (CFT), which treats the interaction between the central metal ion and the surrounding ligands as purely electrostatic.
When ligands approach a metal ion, they create an electrostatic field that interacts with the $d$ orbitals of the metal. Depending on the geometry of the coordination complex (octahedral, tetrahedral, etc.), the degeneracy of the $d$ orbitals is lifted (split) into sets with different energies. This splitting results in energy levels that are no longer equal.
Magnitude of $\Delta_{o}$ and $d$ Orbital Configuration
In octahedral coordination complexes, the $d$ orbitals split into two sets:
Lower energy set ($t_{2g}$): Consists of three orbitals ($d_{xy}, d_{yz}, d_{xz}$).
Higher energy set ($e_g$): Consists of two orbitals ($d_{x^2-y^2}, d_{z^2}$).
The difference in energy between these two sets of orbitals is denoted as $\Delta_{o}$. The magnitude of $\Delta_{o}$ plays a crucial role in determining the actual electron configuration of the $d$ orbitals.
Determining the Electron Configuration
The arrangement of electrons in the $d$ orbitals can follow one of two patterns, depending on the magnitude of $\Delta_{o}$ compared to the pairing energy (P):
High-Spin Configuration (Weak Field Ligands):
If $\Delta_{o}$ is smaller than the pairing energy (P), the complex forms a high-spin configuration.
Electrons prefer to occupy all available orbitals singly before pairing up.
Example configuration for $d^4$ in a high-spin complex: $$ t_{2g}^3 e_g^1 $$
Low-Spin Configuration (Strong Field Ligands):
If $\Delta_{o}$ is larger than the pairing energy (P), the complex forms a low-spin configuration.
Electrons prefer to pair up in the lower energy $t_{2g}$ orbitals before occupying the higher energy $e_g$ orbitals.
Example configuration for $d^4$ in a low-spin complex: $$ t_{2g}^4 e_g^0 $$
Summary
$\Delta_{o}$ is the crystal field splitting energy in octahedral complexes.
Weak field ligands result in a small $\Delta_{o}$, leading to high-spin configurations.
Strong field ligands result in a large $\Delta_{o}$, leading to low-spin configurations.
This understanding of crystal field splitting energies helps explain the electronic structure, magnetic properties, and color of coordination compounds.
$\left[\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{6}\right]^{3+}$ is paramagnetic while $\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{2-}$ is diamagnetic. Explain why?
To understand why $\left[\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{6}\right]^{3+}$ is paramagnetic and $\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{2-}$ is diamagnetic, we need to look at the electronic configurations and the hybridization of these coordination complexes as explained by Valence Bond Theory (VBT) and Crystal Field Theory (CFT).
$\left[\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{6}\right]^{3+}$
Electronic Configuration:
$\mathrm{Cr}^{3+}$ has an electron configuration of $3d^3$ since the ground-state configuration for $\mathrm{Cr}$ is $[Ar]3d^5 4s^1$, and losing three electrons results in $3d^3$.
Ammonia ($\mathrm{NH}_{3}$) is a weak field ligand and thus does not cause significant pairing of the electrons in the $d$ orbitals.
Hybridization and Geometry:
The six $\mathrm{NH}_{3}$ ligands form an octahedral arrangement around the chromium ion.
For an octahedral complex, the $d$-orbitals split into two sets: lower energy $t_{2g}$ and higher energy $e_{g}$ orbitals.
Given $\mathrm{Cr}^{3+}$ has $3d^3$ configuration:
The three electrons will occupy the $t_{2g}$ orbitals, each in a separate orbital before pairing occurs due to Hund's rule.
Paramagnetism:
The complex $\left[\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{6}\right]^{3+}$ is paramagnetic because it has three unpaired electrons in the $t_{2g}$ orbitals.
Paramagnetic materials have unpaired electrons that align with an external magnetic field.
$\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{2-}$
Electronic Configuration:
$\mathrm{Ni}^{2+}$ has an electron configuration of $3d^8$ since the ground-state configuration for $\mathrm{Ni}$ is $[Ar]3d^8 4s^2$, and losing two electrons from $4s$ gives $3d^8$.
Cyanide ($\mathrm{CN}^-$) is a strong field ligand and causes significant pairing of electrons in the $d$ orbitals.
Hybridization and Geometry:
The four $\mathrm{CN}^-$ ligands form a square planar arrangement around the nickel ion.
For a square planar complex, the $d$-orbitals split into different energy levels, with the $d_{x^2-y^2}$ orbital having the highest energy.
Given $\mathrm{Ni}^{2+}$ has $3d^8$ configuration:
The eight electrons will fill the lower energy $d$ orbitals first.
Electrons pair up in the $d$ orbitals due to the strong field ligand $\mathrm{CN}^-$.
Diamagnetism:
The complex $\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{2-}$ is diamagnetic because all the $d$ electrons are paired.
Diamagnetic materials do not have unpaired electrons and thus do not align with an external magnetic field.
Summary:
$\left[\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{6}\right]^{3+}$ is paramagnetic due to the presence of three unpaired electrons in the $t_{2g}$ orbitals.
$\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{2-}$ is diamagnetic because all the $d$ electrons are paired due to the strong field effect of the cyanide ligands.
A solution of $\left[\mathrm{Ni}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}$ is green but a solution of $\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{2-}$ is colourless. Explain.
The color differences between the complexes $\left[\mathrm{Ni}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}$ and $\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{2-}$ can be explained using Crystal Field Theory (CFT).
Crystal Field Theory (CFT) Explanation:
$\left[\mathrm{Ni}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}$:
This is an octahedral complex where water acts as a ligand.
Water is a weak field ligand, creating a small crystal field splitting energy ($Δ_0$).
The d-orbitals of Ni(II) split into two sets: $t_{2g}$ (lower energy) and $e_g$ (higher energy).
The small energy gap ($Δ_0$) allows electrons to absorb energy in the visible light range, leading to transitions between these orbitals. This gives the complex its green color.
$\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{2-}$:
This is a square planar complex where cyanide acts as a ligand.
Cyanide is a strong field ligand, creating a large crystal field splitting energy ($Δ_{sp}$ in the case of square planar).
Due to the strong field strength of CN$^{-}$, the energy gap ($Δ_{sp}$) is so large that it falls outside the visible light range, hence, no absorption occurs within the visible spectrum.
As no visible light is absorbed, the complex appears colorless.
Summary:
$\left[\mathrm{Ni}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}$: Green color due to small crystal field splitting energy allowing visible light absorption.
$\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{2-}$: Colorless due to large crystal field splitting energy preventing visible light absorption.
These differences in color arise due to the different field strengths of the ligands (H$_2$O vs. CN$^-$) and the different geometries of the complexes.
$\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{4-}$ and $\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}$ are of different colours in dilute solutions. Why?
The difference in colors of $\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{4-}$ and $\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}$ in dilute solutions can be explained by Crystal Field Theory (CFT).
Crystal Field Theory Explanation
Ligand Field Strength:
In CFT, ligands are considered to create an electrostatic field that splits the $d$ orbitals of the central metal ion into different energy levels.
Different ligands have different field strengths and thus create different extents of splitting for the $d$ orbitals.
Crystal Field Splitting:
$\mathrm{CN}^{-}$ (cyanide) is a strong field ligand, meaning it causes a large splitting of the $d$ orbitals.
$\mathrm{H}_{2}\mathrm{O}$ (water) is a weak field ligand, causing less splitting of the $d$ orbitals.
Absorption of Light:
The amount of energy required to promote an electron from the lower $t_{2g}$ set of orbitals to the higher $e_{g}$ set of orbitals depends on the extent of this splitting.
In the case of $\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{4-}$, the larger splitting requires higher energy light (shorter wavelength) to be absorbed for electronic transitions.
For $\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}$, the smaller splitting requires lower energy light (longer wavelength) to be absorbed.
Observed Colors:
Because the two complexes absorb different wavelengths of light, the complementary colors that are transmitted and observed will be different.
$\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{4-}$ might absorb light in the ultraviolet/blue region, making it appear pale yellow.
$\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}$ might absorb light in the red/orange region, making it appear light blue or green.
Summary
Strong field ligands (like $\mathrm{CN}^{-}$) cause large split energy differences ($\Delta_0$), resulting in the absorption of higher-energy (shorter wavelength) light, and thus different colors.
Weak field ligands (like $\mathrm{H}_{2}\mathrm{O}$) cause smaller split energy differences, resulting in the absorption of lower-energy (longer wavelength) light.
In essence, the difference in colors arises due to the differing extents of crystal field splitting caused by the different ligands attached to the iron center in each complex.
Visual Representation (optional)
For a visual representation of the crystal field splitting and corresponding light absorption, here are the simplified diagrams:
$$ \text{Splitting in } \left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{4-} \text{ (strong field ligand)}: \quad \text{High } \Delta_0 $$
$$ \text{Splitting in } \left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+} \text{ (weak field ligand)}: \quad \text{Low } \Delta_0 $$
Discuss the nature of bonding in metal carbonyls.
The bonding in metal carbonyls (compounds containing carbonyl ligands only) is distinctive due to the presence of both sigma ($\sigma$) and pi ($\pi$) characteristics, which together contribute to the stability of these complexes through a synergic effect.
Sigma ($\sigma$) Bonding
Formation: The $\sigma$ bond is formed when a lone pair of electrons from the carbon atom in the carbon monoxide ($\mathrm{CO}$) molecule is donated to a vacant orbital of the metal.
Details: This donation occurs through a $\sigma$ overlap between the filled orbital on carbon (which acts as a Lewis base) and an empty metal orbital (which acts as a Lewis acid).
Pi ($\pi$) Bonding
Formation: The $\pi$ bond is formed through a back-donation process, where a filled $d$ orbital of the metal donates electron density to the empty $\pi^*$ antibonding orbital of the CO ligand.
Details: This back-donation helps in reinforcing the bond between the metal and the CO ligand by creating a synergic effect.
Synergic Bonding
This mutual reinforcement between the $\sigma$ donation from the CO ligand and the $\pi$ back-donation from the metal is termed as synergic bonding.
Effect: The net effect of this dual mode of bonding is a stronger and more stable metal-carbonyl bond.
Synergic Interaction Diagram
The figure below shows the synergic bonding interactions in a metal carbonyl complex:
In summary, the nature of bonding in metal carbonyls involves a combination of $\sigma$ donation from the carbon monoxide ligand to the metal and $\pi$ back-donation from the metal to the ligand. This dual interaction leads to a synergic bonding effect, enhancing the overall stability of the metal carbonyl complexes.
Give the oxidation state, $d$ orbital occupation and coordination number of the central metal ion in the following complexes:
(i) $\mathrm{K}_{3}\left[\mathrm{Co}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{3}\right]$
(iii) $\left(\mathrm{NH}_{4}\right)_{2}\left[\mathrm{CoF}_{4}\right]$
(ii) cis- $\left[\mathrm{CrCl}_{2}(\mathrm{en})_{2}\right] \mathrm{Cl}$
(iv) $\left[\mathrm{Mn}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right] \mathrm{SO}_{4}$
It seems that Wolfram Alpha was unable to retrieve data for the requested complexes. Let's analyze and solve each case step by step.
(i) $\mathrm{K}_{3}\left[\mathrm{Co}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{3}\right]$
Oxidation State:
Potassium ions ($\mathrm{K}^+$) contribute a total charge of $+3$.
Let the oxidation state of $\mathrm{Co}$ be $x$.
Each oxalate ion ($\mathrm{C}_{2} \mathrm{O}_4$) carries a charge of $-2$.
The equation is $3 + x - 6 = 0$, solving for $x$ gives $x = +3$.
$d$ Orbital Occupation:
$\mathrm{Co}^{3+}$ with an electron configuration $d^6$
Coordination Number:
Each $\mathrm{C}_{2}\mathrm{O}_{4}^{2-}$ is a bidentate ligand, contributing 2 coordination sites.
Total coordination number is $3 \times 2 = 6$.
(iii) $\left(\mathrm{NH}_{4}\right)_{2}\left[\mathrm{CoF}_{4}\right]$
Oxidation State:
Ammonium ions ($\mathrm{NH}_4^+$) contribute a total charge of $+2$.
Let the oxidation state of $\mathrm{Co}$ be $x$.
Fluoride ions ($\mathrm{F}^-$) contribute a total charge of $-4$.
The equation is $2 + x - 4 = 0$, solving for $x$ gives $x = +2$.
$d$ Orbital Occupation:
$\mathrm{Co}^{2+}$ with an electron configuration $d^7$
Coordination Number:
Each $\mathrm{F}^-$ ion contributes 1 coordination site.
Total coordination number is $4$.
(ii) cis- $\left[\mathrm{CrCl}_{2}(\mathrm{en})_{2}\right] \mathrm{Cl}$
Oxidation State:
The outside chloride ion ($\mathrm{Cl}^-$) contributes a charge of $-1$.
Let the oxidation state of $\mathrm{Cr}$ be $x$.
Each chloride ion ($\mathrm{Cl}^-$) inside the complex contributes a charge of $-1$.
Ethylenediamine (en) is a neutral ligand.
The equation is $x - 1 - 2 = +1$, solving for $x$ gives $x = +3$.
$d$ Orbital Occupation:
$\mathrm{Cr}^{3+}$ with an electron configuration $d^3$
Coordination Number:
Each en ligand is bidentate, contributing 2 coordination sites.
Each $\mathrm{Cl}^-$ ion contributes 1 coordination site.
Total coordination number is $2*2 (en) + 2 (\mathrm{Cl}^-) = 6$.
(iv) $\left[\mathrm{Mn}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right] \mathrm{SO}_{4}$
Oxidation State:
Sulfate ion ($\mathrm{SO}_4^{2-}$) contributes a charge of $-2$.
The overall charge of the complex must be $+2$ since it balances the sulfate ion.
$\mathrm{Mn}$ must be in the $+2$ oxidation state.
$d$ Orbital Occupation:
$\mathrm{Mn}^{2+}$ with an electron configuration $d^5$
Coordination Number:
Each $\mathrm{H}_{2}\mathrm{O}$ molecule contributes 1 coordination site.
Total coordination number is $6$.
In Markdown formatting:
(i) $\mathrm{K}_3[\mathrm{Co}(\mathrm{C}_2\mathrm{O}_4)_3]$
Oxidation State: $+3$
$d$ Orbital Occupation: $d^6$
Coordination Number: $6$
(iii) $(\mathrm{NH}_4)_2[\mathrm{CoF}_4]$
Oxidation State: $+2$
$d$ Orbital Occupation: $d^7$
Coordination Number: $4$
(ii) cis- $[\mathrm{CrCl}_2(\mathrm{en})_2]\mathrm{Cl}$
Oxidation State: $+3$
$d$ Orbital Occupation: $d^3$
Coordination Number: $6$
(iv) $[\mathrm{Mn}(\mathrm{H}_2\mathrm{O})_6]\mathrm{SO}_4$
Oxidation State: $+2$
$d$ Orbital Occupation: $d^5$
Coordination Number: $6$
Write down the IUPAC name for each of the following complexes and indicate the oxidation state, electronic configuration and coordination number. Also give stereochemistry and magnetic moment of the complex:
(i) $\mathrm{K}\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{2}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{2}\right] \cdot 3 \mathrm{H}_{2} \mathrm{O}$
(iii) $\left[\mathrm{CrCl}_{3}(\mathrm{py})_{3}\right]$
(v) $\mathrm{K}_{4}\left[\mathrm{Mn}(\mathrm{CN})_{6}\right]$
(ii) $\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{Cl}.\right] \mathrm{Cl}_{2}$
(iv) $\mathrm{Cs}\left[\mathrm{FeCl}_{4}\right]$
Sure, let's go through each complex step by step.
Part (i): $\mathrm{K}\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{2}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{2}\right] \cdot 3 \mathrm{H}_{2} \mathrm{O}$
IUPAC Name:
Potassium diaquadioxalatochromate(III) trihydrate
Oxidation State:
In this complex, Cr is in the +3 oxidation state.
Electronic Configuration:
For $\mathrm{Cr^{3+}}$, the electronic configuration is $3d^3$.
Coordination Number:
The coordination number is 6 (2 from $\mathrm{H}_2 \mathrm{O}$ and 4 from 2 $\mathrm{C}_2 \mathrm{O}_4^{2-}$ ligands).
Stereochemistry:
The complex is octahedral.
Magnetic Moment:
This complex contains three unpaired electrons, resulting in a magnetic moment. Let's calculate:
$$\mu_{eff} = \sqrt{n(n+2)}$$
Where $n$ is the number of unpaired electrons: $$\mu_{eff} = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87 \text{ BM}$$
Part (ii): $\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{Cl}\right] \mathrm{Cl}_{2}$
IUPAC Name:
Pentaamminechloridocobalt(III) chloride
Oxidation State:
Co is in the +3 oxidation state.
Electronic Configuration:
For $\mathrm{Co^{3+}}$, the electronic configuration is $3d^6$.
Coordination Number:
The coordination number is 6 (5 from $\mathrm{NH}_3$ and 1 from $\mathrm{Cl^-}$).
Stereochemistry:
The complex is octahedral.
Magnetic Moment:
This complex is a low-spin complex due to Co(III) in an octahedral field with strong field $\mathrm{NH}_3$ ligands, resulting in no unpaired electrons. $$\mu_{eff} = 0 \text{ BM}$$
Part (iii): $\left[\mathrm{CrCl}_{3}(\mathrm{py})_{3}\right]$
IUPAC Name:
Trichloridotris(pyridine)chromium(III)
Oxidation State:
Cr is in the +3 oxidation state.
Electronic Configuration:
For $\mathrm{Cr^{3+}}$, the electronic configuration is $3d^3$.
Coordination Number:
The coordination number is 6 (3 from $\mathrm{Cl^-}$ and 3 from $\mathrm{py}$ ligands).
Stereochemistry:
The complex is octahedral.
Magnetic Moment:
This complex contains three unpaired electrons, resulting in a magnetic moment. Let's calculate:
$$\mu_{eff} = \sqrt{n(n+2)}$$
Where $n$ is the number of unpaired electrons: $$\mu_{eff} = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87 \text{ BM}$$
Part (iv): $\mathrm{Cs}\left[\mathrm{FeCl}_{4}\right]$
IUPAC Name:
Caesium tetrachloridoferrate(III)
Oxidation State:
Fe is in the +3 oxidation state.
Electronic Configuration:
For $\mathrm{Fe^{3+}}$, the electronic configuration is $3d^5$.
Coordination Number:
The coordination number is 4 (from 4 $\mathrm{Cl^-}$ ligands).
Stereochemistry:
The complex is tetrahedral.
Magnetic Moment:
For a $d^5$ configuration in a tetrahedral field (high-spin), Fe(III) will have 5 unpaired electrons. Let's calculate:
$$\mu_{eff} = \sqrt{n(n+2)}$$
Where $n$ is the number of unpaired electrons: $$\mu_{eff} = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92 \text{ BM}$$
Part (v): $\mathrm{K}_{4}\left[\mathrm{Mn}(\mathrm{CN})_{6}\right]$
IUPAC Name:
Potassium hexacyanidomanganate(II)
Oxidation State:
Mn is in the +2 oxidation state.
Electronic Configuration:
For $\mathrm{Mn^{2+}}$, the electronic configuration is $3d^5$.
Coordination Number:
The coordination number is 6 (from 6 $\mathrm{CN^-}$ ligands).
Stereochemistry:
The complex is octahedral.
Magnetic Moment:
Since $\mathrm{Mn^{2+}}$ in the presence of strong field $\mathrm{CN^-}$ ligands can form a low-spin complex, it might have fewer unpaired electrons. However, typically it is high-spin with 5 unpaired electrons. Let's calculate:
$$\mu_{eff} = \sqrt{n(n+2)}$$
Where $n$ is the number of unpaired electrons: $$\mu_{eff} = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92 \text{ BM}$$
So, to summarize:
Complex | IUPAC Name | Oxidation State | Electronic Configuration | Coordination Number | Stereochemistry | Magnetic Moment (BM) |
---|---|---|---|---|---|---|
$\mathrm{K}\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{2}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{2}\right] \cdot 3 \mathrm{H}_{2} \mathrm{O}$ | Potassium diaquadioxalatochromate(III) trihydrate | +3 | $3d^3$ | 6 | Octahedral | 3.87 |
$\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{Cl}\right] \mathrm{Cl}_{2}$ | Pentaamminechloridocobalt(III) chloride | +3 | $3d^6$ | 6 | Octahedral | 0 |
$\left[\mathrm{CrCl}_{3}(\mathrm{py})_{3}\right]$ | Trichloridotris(pyridine)chromium(III) | +3 | $3d^3$ | 6 | Octahedral | 3.87 |
$\mathrm{Cs}\left[\mathrm{FeCl}_{4}\right]$ | Caesium tetrachloridoferrate(III) | +3 | $3d^5$ | 4 | Tetrahedral | 5.92 |
$\mathrm{K}_{4}\left[\mathrm{Mn}(\mathrm{CN})_{6}\right]$ | Potassium hexacyanidomanganate(II) | +2 | $3d^5$ | 6 | Octahedral | 5.92 |
Explain the violet colour of the complex $\left[\mathrm{Ti}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}$ on the basis of crystal field theory.
The violet color of the complex $\left[\mathrm{Ti}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}$ can be explained using Crystal Field Theory (CFT). Here’s a step-by-step explanation:
Coordination Environment: $\mathrm{Ti}^{3+}$ ion (with an electronic configuration of d$^1$) is surrounded by six water molecules forming an octahedral complex.
Crystal Field Splitting: In an octahedral field, the five degenerate d-orbitals split into two sets:
Lower energy set ((t_{2g})): $d_{xy}$, $d_{xz}$, and $d_{yz}$.
Higher energy set ((e_g)): $d_{x^2-y^2}$ and $d_{z^2}$.
Electronic Transition: The single electron from $\mathrm{Ti}^{3+}$ occupies one of the lower energy (t_{2g}) orbitals in the ground state. When light is absorbed, this electron gets excited from the (t_{2g}) orbital to one of the higher energy (e_g) orbitals.
The specific energy corresponding to this electronic transition is: $$ \Delta_0 = E(e_g) - E(t_{2g}) $$
Wavelength of Light Absorbed: The energy difference $\Delta_0$ corresponds to the energy of light in the visible spectrum. In this case, the complex absorbs light in the blue-green region of the spectrum.
Using complementary color principles:
The complementary color of blue-green is violet.
Color Observed: Since the complex absorbs blue-green light, the remaining light (which gets transmitted or reflected) is violet, giving the complex its observed violet color.
Thus, the violet color of $\left[\mathrm{Ti}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}$ is due to the d-d transition of electrons in the crystal field created by the octahedral arrangement of water ligands around the titanium ion.
What is meant by the chelate effect? Give an example.
The chelate effect refers to the enhanced stability of coordination compounds where the ligands form rings that include the metal atom. This occurs because the formation of a chelate (a complex involving multidentate ligands) introduces additional entropy and usually involves stronger binding, making the overall complex significantly more stable compared to complexes with similar monodentate ligands.
Example:
One classic example of the chelate effect is the coordination of ethane-1,2-diamine (en) with a metal ion, such as nickel(II), to form a chelate complex:
$$ \text{[Ni(en)_3]^{2+}} $$
In this complex:
Ethane-1,2-diamine (en) is a bidentate ligand, meaning it has two donor atoms that can coordinate to a single metal ion forming a five-membered ring.
For comparison, consider nickel(II) forming a complex with ammonia (which is a monodentate ligand):
$$ \text{[Ni(NH_3)_6]^{2+}} $$
Both complexes have the same metal ion and same overall charge, but the chelated complex with ethane-1,2-diamine is significantly more stable than the complex with ammonia due to the chelate effect. This increased stability is because the entropic gain from binding multidentate ligands and the additional ring formation provides greater thermodynamic stability.
Discuss briefly giving an example in each case the role of coordination compounds in:
(i) biological systems
(iii) analytical chemistry
(ii) medicinal chemistry and
(iv) extraction/metallurgy of metals.
Sure! Let's discuss the roles of coordination compounds in the provided contexts with examples.
(i) Biological Systems
Coordination compounds play vital roles in biological systems. They are present in many enzymes and proteins that facilitate crucial biological processes.
Example: Hemoglobin
Hemoglobin is a coordination compound of iron that is responsible for the transport of oxygen in the blood. The iron in hemoglobin is coordinated to the nitrogen atoms of a porphyrin ring.
\[
\text{Hemoglobin (Fe Complex)}
\]
\]
![](https://www.wolframalpha.com/input?i=hemoglobin+structure)
\]
(ii) Medicinal Chemistry
Coordination compounds are extensively used in medicinal chemistry for their therapeutic properties, including the treatment of metal overload in the body and cancer therapy.
Example: Cisplatin
Cisplatin (([ \text{PtCl}_2(\text{NH}_3)_2 ])) is a widely used chemotherapy drug for treating various types of cancers. It works by forming coordination bonds with DNA, thereby inhibiting the replication of cancer cells.
\[
\text{Cisplatin}
\]
![](https://www.wolframalpha.com/input?i=cisplatin+structure)
\]
(iii) Analytical Chemistry
Coordination compounds are crucial in analytical chemistry for detecting and estimating the presence of various metal ions through complexometric titrations and colorimetric analyses.
Example: EDTA
Ethylene diamine tetraacetic acid (EDTA) forms stable complexes with metal ions like Ca(^2+) and Mg(^2+). It is used in complexometric titrations to determine water hardness by forming coordination complexes with these metal ions.
\[
\text{EDTA Structure}
\]
![](https://www.wolframalpha.com/input?i=EDTA+structure)
\]
(iv) Extraction/Metallurgy of Metals
Coordination compounds are employed in metallurgy for the extraction and purification of metals.
Example: [Ag(CN)(_2)](^-)
In gold extraction, gold is dissolved using a cyanide solution to form the coordination compound (\left[\text{Au(CN}_2\right]{^-}). The gold is then recovered from this solution using zinc or another reducing agent.
\[
\text{Gold Extraction Reaction: Au + CN^- + O_2 + H_2O → [Au(CN)_2^-] + OH^-}
\]
![](https://www.wolframalpha.com/input?i=gold+extraction+reaction)
\]
Summary:
Biological Systems: Hemoglobin
Medicinal Chemistry: Cisplatin
Analytical Chemistry: EDTA
Extraction/Metallurgy: [Au(CN)(_2)](^-)
These examples illustrate the diverse and significant roles that coordination compounds play across various fields.
How many ions are produced from the complex $\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6} \mathrm{Cl}_{2}$ in solution?
(i) 6
(ii) 4
(iii) 3
(iv) 2
To determine the number of ions produced by the complex $\mathrm{Co}(\mathrm{NH}_{3})_{6}\mathrm{Cl}_{3}$ in solution, we need to consider its dissociation in water.
The given coordination complex is $\mathrm{Co}(\mathrm{NH}_{3})_{6}\mathrm{Cl}_{3}$. When dissolved in water, this complex dissociates as follows:
$$ \mathrm{Co}(\mathrm{NH}_{3})_{6}\mathrm{Cl}_{3} \rightarrow \left[\mathrm{Co}(\mathrm{NH}_{3})_{6}\right]^{3+} + 3\mathrm{Cl}^- $$
Here, the complex ion $\left[\mathrm{Co}(\mathrm{NH}_{3})_{6}\right]^{3+}$ remains intact and the three chloride ions $\mathrm{Cl}^-$ dissociate.
Therefore, the total number of ions produced is:
1 $\left[\mathrm{Co}(\mathrm{NH}_{3})_{6}\right]^{3+}$ ion
3 $\mathrm{Cl}^-$ ions
So, the total is 1 + 3 = 4 ions.
Thus, the correct answer is (ii) 4.
Amongst the following ions which one has the highest magnetic moment value?
(i) $\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}$
(ii) $\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}$
(iii) $\left[\mathrm{Zn}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}$
The magnetic moment values for the given ions are:
$\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}$: $$\sqrt{15} \approx 3.87 , \text{BM}$$
$\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}$: $$2\sqrt{6} \approx 4.90 , \text{BM}$$
$\left[\mathrm{Zn}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}$: $$\mu = 0 , \text{BM}$$
Among these values, the ion with the highest magnetic moment is $\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}$ with a magnetic moment of approximately 4.90 BM.
Amongst the following, the most stable complex is
(i) $\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}$
(ii) $\left[\mathrm{Fe}\left(\mathrm{NH}_{3}\right)_{6}\right]^{3+}$
(iii) $\left[\mathrm{Fe}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{3}\right]^{3-}$
(iv) $\left[\mathrm{FeCl}_{6}\right]^{3-}$
To determine the most stable complex among the given options, we can use the concept of *chelate effect*, *ligand field stabilization energy (LFSE)*, and the nature of the ligands involved.
Chelate Effect: Complexes with chelating ligands (ligands that can form multiple bonds to a central metal ion) are generally more stable due to the formation of ring structures, which increases the entropy of the system.
Ligand Field Stabilization Energy (LFSE): Different ligands create varying degrees of crystal field splitting, which contributes to the stability of the complex. Strong field ligands (e.g., $\mathrm{CN}^{-}$, $\mathrm{CO}$) will generally lead to more stable complexes than weak field ligands (e.g., $\mathrm{Cl}^{-}$).
Let's consider the given options:
$\left[\mathrm{Fe}\left(\mathrm{H}_{2}\mathrm{O}\right)_{6}\right]^{3+}$:
$\mathrm{H}_{2}\mathrm{O}$ is a weak field ligand.
No chelate effect.
$\left[\mathrm{Fe}\left(\mathrm{NH}_{3}\right)_{6}\right]^{3+}$:
$\mathrm{NH}_{3}$ is a neutral ligand and a moderate field ligand.
No chelate effect.
$\left[\mathrm{Fe}\left(\mathrm{C}_{2}\mathrm{O}_{4}\right)_{3}\right]^{3-}$:
$\mathrm{C}_{2}\mathrm{O}_{4}^{2-}$ (oxalate) is a chelating ligand (bidentate).
Chelate effect present.
Oxalate is also a relatively strong field ligand.
$\left[\mathrm{FeCl}_{6}\right]^{3-}$:
$\mathrm{Cl}^{-}$ is a weak field ligand.
No chelate effect.
Considering both the chelate effect and ligand field strength, the most stable complex among the given options is:
(iii) $\left[\mathrm{Fe}\left(\mathrm{C}_{2}\mathrm{O}_{4}\right)_{3}\right]^{3-}$.
This complex benefits from the chelate effect provided by the oxalate ligands, making it the most stable among the options.
What will be the correct order for the wavelengths of absorption in the visible region for the following:
$$ \left[\mathrm{Ni}\left(\mathrm{NO}_{2}\right)_{6}\right]^{4-},\left[\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}\right]^{2+},\left[\mathrm{Ni}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+} $$
To determine the order of wavelengths of absorption for the given complexes, we need to refer to the spectrochemical series of ligands, which arranges ligands based on their field strength. This order affects the crystal field splitting energy, and thus the wavelength of light absorbed:
[ \mathrm{I}^{-} < \mathrm{Br}^{-} < \mathrm{SCN}^{-} < \mathrm{Cl}^{-} < \mathrm{S}^{2-} < \mathrm{F}^{-} < \mathrm{OH}^{-} < \mathrm{C}_{2} \mathrm{O}_{4}^{2-} < \mathrm{H}_{2} \mathrm{O} < \mathrm{NCS}^{-} < \mathrm{edta}^{4-} < \mathrm{NH}_{3} < \mathrm{en} < \mathrm{NO}_{2}^{-} < \mathrm{CN}^{-} < \mathrm{CO} ]
Based on the spectrochemical series:
$\mathrm{NO}_{2}^{-}$ (Nitro) is a strong field ligand.
$\mathrm{NH}_{3}$ (Ammonia) is a weaker field ligand compared to $\mathrm{NO}_{2}^{-}$ but stronger than $\mathrm{H}_{2} \mathrm{O}$.
$\mathrm{H}_{2} \mathrm{O}$ (Water) is the weakest field ligand among the three.
[ \text{Crystal field splitting, } \Delta_0 \text{, increases in the following order: } \left[\mathrm{Ni}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+} < \left[\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}\right]^{2+} < \left[\mathrm{Ni}\left(\mathrm{NO}_{2}\right)_{6}\right]^{4-} ]
Thus, the wavelength of light absorbed (inversely related to crystal field splitting) will be in the opposite order:
[ \left[\mathrm{Ni}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+} > \left[\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}\right]^{2+} > \left[\mathrm{Ni}\left(\mathrm{NO}_{2}\right)_{6}\right]^{4-} ]
Therefore, the order for the wavelengths of absorption in the visible region will be: [ \text{(Longest) } \left[\mathrm{Ni}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+} > \left[\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}\right]^{2+} > \left[\mathrm{Ni}\left(\mathrm{NO}_{2}\right)_{6}\right]^{4-} \text{ (Shortest)} ]
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Ask Chatterbot AINotes - Coordination Compounds | Class 12 NCERT | Chemistry
Comprehensive Coordination Compounds Class 12 Notes: Key Concepts, Theories, and Applications
Coordination compounds are a fascinating and integral part of modern inorganic chemistry. These compounds show the unique capabilities of transition metals to form complexes with varied properties and applications. In these notes, we delve deep into the key concepts, theories, and applications surrounding coordination compounds, making them a perfect study guide for Class 12 students.
Introduction to Coordination Compounds
Coordination compounds consist of a central metal atom or ion bonded to surrounding ions or molecules, known as ligands, through coordinate covalent bonds. Unlike simple salts, these complexes often exhibit distinct colours, magnetic properties, and reactivity.
Werner's Theory of Coordination Compounds
Historical Background
Alfred Werner, a Swiss chemist, was the first to propose a comprehensive theory for the structure of coordination compounds in 1898. His work revolutionised our understanding of how metal ions bond with ligands.
Main Postulates
Primary and Secondary Valences:Werner introduced the idea of primary valences (ionisable) and secondary valences (non-ionisable), in which the secondary valences are now known as the coordination number.
Isomers in Coordination Compounds:Werner's theory also explained the existence of isomers—compounds with the same chemical formula but different structures and properties—in coordination chemistry.
Nomenclature of Coordination Compounds
The International Union of Pure and Applied Chemistry (IUPAC) has set specific rules for naming coordination compounds to maintain uniformity.
The cation is named before the anion.
Ligands are named in alphabetical order before the central atom/ion.
Anionic ligands end in "-o" (e.g., chlorido, cyanido), whereas neutral ones retain their names with some exceptions like aqua for $\text{H}_2\text{O}$.
Coordination Number and Geometry
Determining Coordination Number
The coordination number is the count of ligand donor atoms bonded directly to the central atom. For example, in $\left[\text{Co}\left(\text{NH}_3\right)_6\right]^{3+}$, the coordination number is 6.
Common Geometrical Arrangements
Coordination compounds commonly exhibit geometrical shapes such as octahedral, tetrahedral, and square planar, determined by the coordination number and the nature of the ligands.
Ligands: Types and Classifications
Unidentate: Ligands that bond through a single donor atom (e.g., $\text{Cl}^-$, $\text{NH}_3$).
Didentate and Polydentate: Ligands that can attach through two or more donor atoms, often forming more stable chelate complexes, such as ethylenediaminetetraacetate (EDTA).
Valence Bond Theory (VBT) of Coordination Compounds
Principles: VBT explains the formation, magnetic behaviour, and shapes of coordination compounds by considering the hybridisation of metal orbitals.
Limitations: Despite its successes, VBT falls short in predicting magnetic properties and colour, and it makes several assumptions about electron pairing and orbital hybridisation.
Crystal Field Theory (CFT)
Crystal Field Splitting Energy (Δ₀):
When ligands approach a metal ion, the degeneration (equal energy) of the d orbitals splits due to electrostatic interactions, resulting in different energy levels.
In an octahedral field:
The d orbitals split into two sets: lower-energy $t_{2g}$ and higher-energy $e_g$ orbitals.
Field ligands can be strong or weak based on the magnitude of Δ₀, affecting the electronic configuration.
Colour in Coordination Compounds
The colour of coordination compounds is attributed to d-d electronic transitions induced by crystal field splitting. For instance, $\left[\text{Ti}\left(\text{H}_2\text{O}\right)_6\right]^{3+}$ appears violet due to the absorption of light in the blue-green region, promoting an electron from $t_{2g}$ to $e_g$.
Isomerism in Coordination Compounds
Geometric Isomerism
Occurs due to different spatial arrangements of ligands around the central metal. For example, cis and trans isomers in square planar and octahedral complexes.
graph TB
A["Co(NH3)4Cl2"] -->|cis| B{Cis Isomer}
A -->|trans| C{Trans Isomer}
B --> D[4 ammonia and 2 chloride ligands adjacent]
C --> E[4 ammonia and 2 chloride ligands opposite]
Optical Isomerism
Involves non-superimposable mirror images, called enantiomers, often seen in complexes with chiral centres.
Bonding in Metal Carbonyls
Metal carbonyls, such as $\text{Ni}\left(\text{CO}\right)_4$, involve both $\sigma$ and $\pi$ bonding. The $\sigma$ bond forms by electron donation from CO to the metal, and the $\pi$ bond involves back-donation from the metal to CO.
Applications of Coordination Compounds
In Biological Systems
Coordination compounds play crucial roles, such as:
Chlorophyll in photosynthesis (Mg complex).
Haemoglobin for oxygen transport (Fe complex).
Vitamin B₁₂ (cobalamin), essential for human health (Co complex).
Medicinal Uses
Complexes like cisplatin are vital in cancer treatment. EDTA is used to treat heavy metal poisoning by forming stable complexes with toxic metals.
Industrial Applications
Coordination compounds are essential in processes such as:
Metal extraction and purification (e.g., $\text{Ni}\left(\text{CO}\right)_4$).
Catalysis (e.g., rhodium catalysts for hydrogenation).
Electroplating (e.g., [Ag(CN)_2]^- for smoother deposits of silver or gold).
Conclusion
Coordination compounds represent an indispensable area of chemistry, providing profound insights and practical applications in various fields, from biological systems to industrial processes. Understanding the theories and applications of these compounds not only helps ace exams in Class 12 but also serves a solid foundation for advanced studies in chemistry.
Frequently Asked Questions (FAQs)
What are coordination compounds?Coordination compounds consist of a central metal atom or ion surrounded by ligands bonded through coordinate covalent bonds.
Who proposed the first theory of coordination compounds?Alfred Werner proposed the first comprehensive theory of coordination compounds in 1898.
How are primary and secondary valences different in Werner's theory?Primary valences are ionisable and correspond to oxidation states. Secondary valences are non-ionisable and equate to the coordination number.
How are coordination compounds named according to IUPAC?The central cation is named first, followed by ligands in alphabetical order with specific suffixes for anionic ligands, and the oxidation state is indicated in Roman numerals.
What is the coordination number?The coordination number is the count of ligand donor atoms directly bonded to the central atom.
What are the different types of ligands?Ligands can be unidentate, didentate, or polydentate based on the number of donor atoms they have.
Use these comprehensive notes and visual aids to master the topic of coordination compounds, ensuring you grasp all essential concepts and applications effectively.
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