Coordination Compounds

$\Delta_0$ value of the octahedral complexes can be calculated using the formula CASE $=\left(-0,4 \Delta_0\right) \times$ number of electrons in $f_{2 g}$ orbitals + $\left(+0.6 \Delta_0\right) \times$ number of electrons in eg orbitals.

The ligand CN$^-$ is a strong field ligand and all the complexes form low-spin complex (pairing occurs).

I) $\left[M_N(\mathrm{CN})_6\right]^{3-}$ Oxidation state of $M_n:+3$ $M n$ configuration $\rightarrow d^5 s^2$ $\mathrm{Mn}^{+3}$ configuration $\rightarrow \mathrm{d}^4$

II) $\left[\mathrm{Co}(\mathrm{CN})_6\right]^{4-}$ Oxidation state of $C_0:+2$ $C_0$ configuration $\rightarrow d^7 s^2$ $\mathrm{Co}^{+2}$ configuration $\rightarrow d^7$

III) $\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{4-}$ Oxidation state of $\mathrm{Fe}:+2$ Fe configuration $\rightarrow d^6 s^2$ $\mathrm{Fe}^{+2}$ configuration $\rightarrow \mathrm{d}^6$

IV) $\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{3-}$ Oxidation state of $\mathrm{Fe}:+3$ $F e$ configuration $\rightarrow d^6 s^2$ $\mathrm{Fe}^{+3}$ configuration $\rightarrow \mathrm{d}^5$

I) $\left[\mathrm{Mn}_n(\mathrm{CN})_6\right]^{3-} \rightarrow-1.6 \Delta_0$ II) $\left[\mathrm{Co}(\mathrm{CN})_6\right]^{4-} \rightarrow-1.8 \Delta_0$ III) $\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{4-} \rightarrow-2.4 \Delta_0$ IV) $\left[\mathrm{Fe}\left((\mathrm{N})_6\right]^{3-} \rightarrow-2.0 \Delta_0\right.$ Stability of complexes based on $\Delta_0$ value, greater the CFSE value, more stable the complex.

The more negative the CFSE value, the move stable the complex.

So, complex with CFSE $-2.4 \Delta$ is most stable and complex with CFSE $-1.6 \Delta_0$, is least stable.

Stability order:

(A) $\left[\mathrm{MnBr}_4\right]^{2-}$ $\mathrm{Mn}^{+2} \Rightarrow[\mathrm{Ar}] 3 \mathrm{~d}^5$ In presence of ligand field $\Rightarrow \mathrm{sp}^3$ hybridization, paramagnetic in nature (B) $\left[\mathrm{FeF}_6\right]^{3-}$

In presence of ligand field $\Rightarrow \mathrm{sp}^3 \mathrm{~d}^2$ hybridization, paramagnetic in nature (C) $\left[\mathrm{Co}\left(\mathrm{C}_2 \mathrm{O}_4\right)_3\right]^{3-}$

In presence of ligand field $\Rightarrow \mathrm{d}^2 \mathrm{sp}^3$ hybridization, diamagnetic in nature

In presence of ligand field $\Rightarrow \mathrm{sp}^3$ hybridization, diamagnetic in nature

The magnetic behavior of complexes can provide insights into their geometry and oxidation states.

Given two complexes, $[ \mathrm{NiCl}_4 ]^{2-}$ and $[ \mathrm{Ni}(\mathrm{CO})_4 ]$, we need to analyze the magnetic properties to deduce these characteristics.

$[\mathrm{NiCl}_4]^{2-}$ Oxidation State: In $[\mathrm{NiCl}_4]^{2-}$, nickel is in the +2 oxidation state ($\mathrm{Ni}^{2+}$).

The electron configuration for $\mathrm{Ni}^{2+}$ is $[\mathrm{Ar}]\, 3d^8\, 4s^0$.

Geometry and Hybridization: The complex shows paramagnetic properties, indicating the presence of unpaired electrons.

The hybridization in this case is $\mathrm{sp}^3$, which corresponds to a tetrahedral geometry.

Thus, there are two unpaired electrons leading to its paramagnetism.

$[\mathrm{Ni}(\mathrm{CO})_4]$ Oxidation State: In this complex, nickel is in the zero oxidation state ($\mathrm{Ni}(0)$).

The electron configuration is $[\mathrm{Ar}]\, 3d^{10}\, 4s^0$ after rearrangement.

Geometry and Hybridization: The compound is diamagnetic, indicating all electrons are paired.

Its hybridization is also $\mathrm{sp}^3$, meaning the geometry is tetrahedral, resulting in no unpaired electrons and confirming its diamagnetic nature.

In summary, $[\mathrm{NiCl}_4]^{2-}$ is a tetrahedral complex with nickel in a +2 oxidation state and is paramagnetic.

In contrast, $[\mathrm{Ni}(\mathrm{CO})_4]$ is a tetrahedral complex with nickel in a 0 oxidation state and is diamagnetic.

The crystal field stabilization energy (CFSE) of a complex depends on two main factors: Charge/Oxidation State of the Central Metal Atom: Higher oxidation states generally lead to greater crystal field splitting energy (Δ).

Field Strength of the Ligand: Stronger field ligands also result in greater splitting.

Additionally, chelating ligands typically increase the field strength.

For octahedral complexes, the CFSE is calculated as: $\text{CFSE} = [-0.4 \, t_{2g} + 0.6 \, e_g ] \, \Delta_\circ$ Considering the complexes: $[\text{Co(en)}_3]^{3+}$: Co exists as Co$^{3+}$ with electronic configuration $t_{2g}^6 e_g^0$.

CFSE = $-2.4 \, (\Delta_0)_1$.

$[\text{Co(NH}_3)_6]^{3+}$: Co also as Co$^{3+}$ with $t_{2g}^6 e_g^0$.

CFSE = $-2.4 \, (\Delta_0)_2$.

$[\text{Co(NH}_3)_6]^{2+}$: Co with Co$^{2+}$ configuration $t_{2g}^5 e_g^2$.

CFSE = $-0.8 \, (\Delta_0)_3$.

$[\text{Co(NH}_3)_4]^{2+}$: Co as Co$^{2+}$ with different configuration $e^4 t_2^3$.

CFSE = $-1.2 \, \Delta_t$, where $\Delta_t = \dfrac{4}{9} (\Delta_0)_3$.

The order of crystal field splitting energies, based on our criteria, is: $\Delta_t Therefore, the CFSEs of these complexes compare as follows:$[\text{Co(en)}_3]^{3+}$has the highest CFSE due to the$\Delta_0$of the strongest field ligand and oxidation state.$[\text{Co(NH}_3)_6]^{3+}$follows with similarly high CFSE due to Co$^{3+}$.$[\text{Co(NH}_3)_6]^{2+}$has a moderate CFSE.$[\text{Co(NH}_3)_4]^{2+}$has the lowest CFSE with$\Delta_t$.

This hierarchy of CFSE is determined by both the oxidation state of the cobalt metal and the field strength of the ligands involved.

(A) $[ \text{Fe(CN)}_5 \text{NO} ]^{2-}$: This is a heteroleptic complex with iron in the $\text{Fe}^{2+}$ oxidation state, resulting in a $3d^6$ electronic configuration.

It forms a low spin complex with electron configuration $\text{t}_{2g}^6 \text{e}_g^0$ due to the strong field ligands (SFL).

(B) $[ \text{CoF}_6 ]^{3-}$: This homoleptic complex contains cobalt in the $\text{Co}^{3+}$ oxidation state, leading to a $3d^6$ electron configuration.

It is a high spin complex with an $\text{sp}^3 \text{d}^2$ hybridization due to weak field ligands (WFL).

(C) $[ \text{Fe(CN)}_6 ]^{4-}$: This homoleptic complex has iron in the $\text{Fe}^{2+}$ state, giving a $3d^6$ configuration.

It forms a low spin complex with hybridization $\text{d}^2 \text{sp}^3$ and electron configuration $\text{t}_{2g}^6 \text{e}_g^0$, due to the strong field nature of the cyanide ligands.

(D) $[ \text{Co(NH}_3 )_6 ]^{3+}$: This homoleptic complex contains cobalt in the $\text{Co}^{3+}$ state, resulting in a $3d^6$ electron configuration.

It forms a low spin complex with hybridization $\text{d}^2 \text{sp}^3$ and configuration $\text{t}_{2g}^6 \text{e}_g^0$ due to the strong field ligands.

(E) $[ \text{Cr(H}_2\text{O})_6 ]^{2+}$: This homoleptic complex with chromium in the $\text{Cr}^{2+}$ state results in a $3d^4$ electronic configuration.

It is a high spin complex with hybridization $\text{d}^2 \text{sp}^3$ and configuration $\text{t}_{2g}^3 \text{e}_g^1$ due to weak field ligands.

In summary, the homoleptic complexes that are low spin are: (C) $[ \text{Fe(CN)}_6 ]^{4-}$ (D) $[ \text{Co(NH}_3 )_6 ]^{3+}$

Given the problem, we have the following information: Depression in freezing point, $\Delta T_f = 0.558^\circ \text{C}$ Freezing point depression constant, $k_f = 1.86 \, \text{K kg mol}^{-1}$ Molal concentration of the solution, $0.1 \, \text{mol kg}^{-1}$ Coordination number of Cr is 6.

Using the formula for freezing point depression: $\Delta T_f = i \times k_f \times m$ where $i$ is the van 't Hoff factor, $k_f$ is the freezing point depression constant, and $m$ is the molality of the solution.

Substituting the known values: $0.558 = i \times 1.86 \times 0.1$ Solving for $i$: $i = \dfrac{0.558}{1.86 \times 0.1} = 3$ The van 't Hoff factor $i = 3$ indicates that the complex ionizes into three particles in solution.

Given the coordination number of Cr is 6, we analyze the possible complexes: $\left[\mathrm{Cr}\left(\mathrm{NH}_3\right)_5 \mathrm{Cl}\right] \mathrm{Cl}_2$ $\left[\mathrm{Cr}\left(\mathrm{NH}_3\right)_4 \mathrm{Cl}_2\right] \mathrm{Cl}$ $\left[\mathrm{Cr}\left(\mathrm{NH}_3\right)_6\right] \mathrm{Cl}_3$ $\left[\mathrm{Cr}\left(\mathrm{NH}_3\right)_3 \mathrm{Cl}_3\right]$ The complex that gives three ions upon dissociation could be $\left[\mathrm{Cr}\left(\mathrm{NH}_3\right)_5 \mathrm{Cl}\right] \mathrm{Cl}_2$.

This is because the complex will dissociate as: $\left[\mathrm{Cr}\left(\mathrm{NH}_3\right)_5 \mathrm{Cl}\right]^{2+} + 2 \, \mathrm{Cl}^-$ This results in three ions total, matching the van 't Hoff factor $i = 3$.

Therefore, the complex is $\left[\mathrm{Cr}\left(\mathrm{NH}_3\right)_5 \mathrm{Cl}\right] \mathrm{Cl}_2$.

$\mathrm{Ma}_3 \mathrm{~b}_3$ type complexes show Facial - Meridional isomerism (i) $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_3 \mathrm{Cl}_3\right] \Rightarrow \mathrm{Ma}_3 \mathrm{~b}_3$ (ii) $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_4 \mathrm{Cl}_2\right]^{+} \Rightarrow \mathrm{Ma}_4 \mathrm{~b}_2$ (iii) $\left[\mathrm{Co}(\mathrm{en})_3\right]^{3+} \quad \Rightarrow \mathrm{M}(\mathrm{AA})_3$ (iv) $\left[\mathrm{Co}(\mathrm{en})_2 \mathrm{Cl}_2\right]^{+} \Rightarrow \mathrm{M}(\mathrm{AA})_2 \mathrm{~b}_2$ $\mathrm{a,b},=\mathrm{NH}_3, \mathrm{Cl}^{-}$ $\mathrm{AA}=\mathrm{en}$

$\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_5 \mathrm{Cl}\right] \mathrm{Cl}_2 \rightarrow \underbrace{\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_5 \mathrm{Cl}\right]^{2+}(\text{ aq. })+2 \mathrm{Cl}^{-} \text{(aq.) }}_{3 \text{ ions in water }}$

For $3 d^4$ If ligand is SFL : $\mathrm{t}_{2 \mathrm{~g}}{ }^4 \mathrm{e}_{\mathrm{g}}{ }^0 \quad$ (Low spin) If ligand is WFL : $\mathrm{t}_{2 \mathrm{~g}}{ }^3 \mathrm{e}_{\mathrm{g}}{ }^1$ (High spin)