Coordination Compounds

The energy of absorbed light by a complex can be expressed as $E = hv = \frac{hC}{\lambda}$, where $E$ is the energy, $h$ is Planck's constant, $v$ is the frequency, $C$ is the speed of light, and $\lambda$ is the wavelength of the absorbed light. This formula demonstrates that energy ($E$) is inversely proportional to the wavelength ($\lambda$).

In this context, all cobalt complexes are in the +3 oxidation state. The Crystal Field Stabilization Energy (CFSE) is influenced by the strength of the ligands surrounding the metal ion. As the ligand field strength increases, the CFSE also increases.

The order of field strength for the ligands in this case is:

$ \text{CN}^- > \text{NH}_3 > \text{H}_2\text{O} > \text{Cl}^- $

This implies the following order for CFSE in the given complexes:

$ \text{C} > \text{B} > \text{A} > \text{D} $

Because energy is inversely proportional to wavelength, complexes absorbing at higher energies will absorb at shorter wavelengths. Therefore, the order of the complexes in terms of the wavelength of light absorbed, from longest to shortest wavelength, is:

$ \text{D} > \text{A} > \text{B} > \text{C} $

$\begin{aligned} & \mathrm{PF}_5: 5 \sigma+0 \ell \mathrm{p} \rightarrow \mathrm{sp}^3 \mathrm{~d} \\ & \mathrm{SF}_6: 6 \sigma+0 \ell \mathrm{p} \rightarrow \mathrm{sp}^3 \mathrm{~d}^2 \\ & \mathrm{Ni}(\mathrm{CO})_4: \mathrm{Ni} \rightarrow 0 \end{aligned}$

In presence of ligand field :-

$\begin{aligned} &\left[\mathrm{PtCl}_4\right]^{2-}: \mathrm{Pt} \rightarrow+2\\ &\text { In presence of ligand field :- } \end{aligned}$

Determine the oxidation state of manganese in $\left[\mathrm{MnCl}_6\right]^{3-}:$

Chloride (Cl⁻) has a charge of -1.

With six chloride ions, the total charge contributed by the ligands is $6(-1) = -6.$

Let the oxidation state of Mn be $x.$ Then:

$x - 6 = -3 \quad \Longrightarrow \quad x = +3.$

So, manganese is in the +3 oxidation state.

Find the d-electron count for Mn(III):

The neutral manganese atom has the electronic configuration $[\mathrm{Ar}]\, 3d^5 4s^2.$

Removing three electrons (first from the 4s, then from the 3d) gives:

$\mathrm{Mn}^{3+} : [\mathrm{Ar}]\, 3d^4.$

Therefore, Mn(III) is a $d^4$ system.

Predict the spin state in an octahedral complex:

Chloride (Cl⁻) is a weak-field ligand, meaning it produces a relatively small crystal field splitting ($\Delta_o$).

When $\Delta_o$ is small, the pairing energy is higher than $\Delta_o,$ so electrons prefer to remain unpaired.

For a $d^4$ configuration in a high-spin octahedral complex, the electrons will be distributed as:

Three electrons in the three $t_{2g}$ orbitals (one in each)

One electron in one of the $e_g$ orbitals

This results in a total of 4 unpaired electrons, making the complex paramagnetic.

Determine the type of hybridization:

In octahedral complexes, the two common hybridizations are:

$d^2sp^3$ (inner orbital complex), typically seen in low-spin complexes where inner 3d orbitals are available because the electrons are paired.

$sp^3d^2$ (outer orbital complex), seen in high-spin complexes where the inner $3d$ orbitals are occupied by unpaired electrons.

Since $\left[\mathrm{MnCl}_6\right]^{3-}$ is high-spin (because of Cl⁻ being a weak-field ligand), the inner 3d orbitals are not available for hybridization.

Therefore, the complex uses the outer orbitals (the 4s, 4p, and 4d orbitals), leading to an $sp^3d^2$ hybridization.

Conclusion:

The hybridization of $\left[\mathrm{MnCl}_6\right]^{3-}$ is $sp^3d^2.$

It is paramagnetic with four unpaired electrons.

Thus, the correct answer is:

Option A: $sp^3d^2$, paramagnetic with four unpaired electrons.

Identify the oxidation state and d-electron count for each complex:

For $\left[\mathrm{Co}(\mathrm{en})_3\right]^{3+}$:

Ethylenediamine (en) is neutral.

Oxidation state of cobalt is +3.

Cobalt (atomic number 27) in the +3 state gives a $d^6$ configuration.

With strong/moderate field ligands like en, the complex is low spin, leading to a configuration of $t_{2g}^6 e_g^0$.

For $\left[\mathrm{CoF}_6\right]^{3-}$:

Each fluoride ion is $\mathrm{F}^-$ so, with six of them, Co must be in the +3 state again.

However, since fluoride is a weak field ligand, this complex adopts a high spin configuration for $d^6$, resulting in $t_{2g}^4 e_g^2$.

For $\left[\mathrm{Mn}\left(\mathrm{H}_2\mathrm{O}\right)_6\right]^{2+}$:

Water is neutral.

Manganese in the +2 state gives a $d^5$ configuration.

With weak field water, the $d^5$ configuration is high spin: $t_{2g}^3 e_g^2$.

For $\left[\mathrm{Zn}\left(\mathrm{H}_2\mathrm{O}\right)_6\right]^{2+}$:

Zinc in the +2 state has a $d^{10}$ configuration.

In an octahedral field, all 10 electrons fill the orbitals as $t_{2g}^6 e_g^4$.

Calculate the crystal field stabilization energy (CFSE) for each case:

The CFSE in an octahedral field can be estimated as:

$\mathrm{CFSE} = (n_{t_{2g}}\times -0.4\Delta_o) + (n_{e_g}\times 0.6\Delta_o)$

For $t_{2g}^6 e_g^0$ (low-spin $d^6$ in $\left[\mathrm{Co}(\mathrm{en})_3\right]^{3+}$):

$\mathrm{CFSE} = 6(-0.4\Delta_o) + 0(0.6\Delta_o) = -2.4\Delta_o$

For $t_{2g}^4 e_g^2$ (high-spin $d^6$ in $\left[\mathrm{CoF}_6\right]^{3-}$):

$\mathrm{CFSE} = 4(-0.4\Delta_o) + 2(0.6\Delta_o) = -1.6\Delta_o + 1.2\Delta_o = -0.4\Delta_o$

For $t_{2g}^3 e_g^2$ (high-spin $d^5$ in $\left[\mathrm{Mn}\left(\mathrm{H}_2\mathrm{O}\right)_6\right]^{2+}$):

$\mathrm{CFSE} = 3(-0.4\Delta_o) + 2(0.6\Delta_o) = -1.2\Delta_o + 1.2\Delta_o = 0$

For $t_{2g}^6 e_g^4$ (for $d^{10}$ in $\left[\mathrm{Zn}\left(\mathrm{H}_2\mathrm{O}\right)_6\right]^{2+}$):

$\mathrm{CFSE} = 6(-0.4\Delta_o) + 4(0.6\Delta_o) = -2.4\Delta_o + 2.4\Delta_o = 0$

Compare the CFSE values:

$t_{2g}^6 e_g^0$ gives a CFSE of $-2.4\Delta_o$ (most negative, hence highest stabilization).

The others result in less stabilization (or net zero).

Conclusion:

The complex with the highest CFSE is the one with the configuration $t_{2g}^6 e_g^0$, which corresponds to $\left[\mathrm{Co}(\mathrm{en})_3\right]^{3+}$.

Thus, the answer is:

Option A: $t_{2g}^6 e_g^0$.

In octahedral complexes, which have a coordination number (CN) of 6, the formation of high spin or low spin complexes depends on the relative magnitude of the crystal field splitting energy, denoted as Δ₀, and the pairing energy (P.E.):

If Δ₀ is less than the pairing energy (Δ₀ < P.E.), high spin complexes are formed.

If Δ₀ is greater than the pairing energy (Δ₀ > P.E.), low spin complexes are formed.

In tetrahedral complexes, where the coordination number is 4:

The crystal field splitting energy (Δₜ) is typically less than the pairing energy (Δₜ < P.E.), resulting in the formation of mainly high spin complexes. Low spin complexes are rarely observed in tetrahedral coordination due to this energy consideration.

Homoleptic complexes are complexes with identical ligands attached to the central metal atom.

(A) ${[Fe{O_4}]^{2 - }}$ - Homdeptic

Central metal : Fe

IN this complex, Fe is in +6 oxidation state.

{oxidation state : $x + 4( - 2) = - 2$

$x - 8 = - 2$

$x = - 2 + 6 = + 6$

Configuration of $Fe - {d^6}{s^2}$

Configuration of $Fe( + 6) - c{l^2}$

Iron has even number of d electrons. So, it is not correct.

(B) ${[Fe{(N)_6}]^{3 - }}$ - Homoleptic

Central metal : Fe

In this complex, the oxidation state of Fe is +3

Oxidation state :

$x + 6( - 1) - 3$

$x - 6 = - 3$

$x = - 3 + 6 = + 3$

Configuration of $Fe - {d^6}{s^2}$

Configuration of $Fe( + 3) - {d^5}$

Iron has odd number of d electrons.

So, it is correct.

(C) ${[Fe{(CN)_5}NO]^{2 - }}$ - Not homoleptic

This complex has different type of ligands.

${[CoC{l_4}]^{2 - }}$ - Homoleptic

Central metal - Co

In this complex, the oxidation state of Co is +2

Oxidation state:

$x + 4( - 1) = - 2$

$x - 4 = - 2$

$x = - 2 + 4 = + 2$

Configuration of $Co - {d^7}{s^2}$

Configuration of $Co( + 2) - {d^7}$

Cobalt has odd number of d electrons.

So, it is correct.

(E) $[Co{({H_2}O)_3}{F_3}]$ - Not homoleptic,

It has different type of ligands.

Complexes B and D are homoleptic complexes with odd number of d electrons in the central metal. So, correct answer is

(B) and (D) only.

${K_3}\left[ {Fe{{(OH)}_6}} \right]$

${K_4}\left[ {Fe{{(OH)}_6}} \right]$

Spin-only magnetic moment formula is $\sqrt {n(n + 2)} $, where n is the number of unpaired electrons in the central metal ${K_3}\left[ {Fe{{(OH)}_6}} \right]$

Also written as ${\left[ {Fe{{(OH)}_6}} \right]^{3 - }}$

Oxidation state of $Fe \to + 3$

Configuration of $Fe \to {d^6}{s^2}$

Configuration of $F{e^{3 + }} \to {d^5}$

Since OH$^-$ is a weak ligand, no pairing of electrons occurs.

Unpaired electrons, $n = 5$

So, spin-only magnetic moment

$ = \sqrt {n(n + 2)} $

$ = \sqrt {5(5 + 2)} $

$ = \sqrt {5 \times 7} $

$ = \sqrt {35} $

$ = 5.92$ BM

$3( + 1) + x + 6( - 1) = 0$

$3 + x - 6 = 0$

$x - 3 = 0$

$x = + 3$

${K_4}\left[ {Fe{{(OH)}_6}} \right]$

$4( + 1) + x + 6( - 1) = 0$

$4 + x - 6 = 0$

$x = 6 - 4 = + 2$

Also written as ${\left[ {Fe{{(OH)}_6}} \right]^{4 - }}$

Oxidation state of $Fe \to + 2$

Configuration of $Fe \to {d^6}{s^2}$

Configuration of $F{e^{ + 2}} \to {d^6}$

Since OH$^-$ is a weak field ligand, no pairing of electrons occurs.

Unpaired electrons = 4

Spin-only magnetic moment

$ = \sqrt {n(n + 2)} $

$ = \sqrt {4(4 + 2)} $

$ = \sqrt {4 \times 6} $

$ = \sqrt {24} $

$ = 4.90$ BM

Spin-only magnetic moments of ${K_3}\left[ {Fe{{(OH)}_6}} \right]$ and ${K_4}\left[ {Fe{{(OH)}_6}} \right]$ respectively are 5.92 and 4.90 B.M.

Correct answer is option (3) 5.92 and 4.90 B.M.

$\Delta_0$ 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 $\rightarrow \mathrm{d}^4$

$\begin{array}{r} x+6(-1)=-3 \\ x=-3+b=+3 \end{array}$

$\begin{aligned} C F S E & =-0.4 \Delta_0 \times 4+\left(0.6 \Delta_0\right) \times 0 \\ & =-1.6 \Delta_0 \end{aligned}$

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 $\rightarrow d^7$

$\begin{aligned} \text { CFSE } & =-0.4 \Delta_0 \times 6+\left(0.6 \Delta_0\right) \times 1 \\ & =-2.4 \Delta_0+0.6 \Delta_0=-1.8 \Delta_0 \end{aligned}$

$\begin{aligned} x+6(-1) & =-4 \\ x & =-4+6=+2 \end{aligned}$

$\begin{aligned} \text { CFSE } & =-0.4 \Delta_0 \times 6+\left(0.6 \Delta_{10}\right) \times 1 \\ & =-2.4 \Delta_0+0.6 \Delta_0=-1.8 \Delta_0 \end{aligned}$

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 $\rightarrow \mathrm{d}^6$

$\begin{array}{r} x+(-1) \times 6=-4 \\ x=-4+6 \\ =+2 \end{array}$

$\begin{aligned} C F S E & =-0.4 \Delta_0 \times 6+0.6 \Delta_0 \times 0 \\ & =-2.4 \Delta_0 \end{aligned}$

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 $\rightarrow \mathrm{d}^5$

$\begin{gathered} x+6(-1)=-3 \\ x=-3+6 \\ =+3 \end{gathered}$

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:

$ \begin{aligned} &\begin{aligned} & -1.6 \Delta_0<-1.8 \Delta_0<-2.0 \Delta_0<-2.4 \Delta_0 \\ & {\left[M n(\mathrm{CN})_6\right]^{3-}<\left[\mathrm{C}(\mathrm{CN})_6\right]^{4-}<\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{3-}<\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{4-}} \\ & I < II < IV < III \end{aligned}\\ &\text { Answer: Option (1) } \end{aligned}$

(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) $\left[\mathrm{FeF}_6\right]^{3-}$

$\mathrm{Fe}^{+3} \Rightarrow[\mathrm{Ar}] 3 \mathrm{~d}^5$

In presence of ligand field

$\Rightarrow \mathrm{sp}^3 \mathrm{~d}^2$ hybridization, paramagnetic in nature

(C) $\left[\mathrm{Co}\left(\mathrm{C}_2 \mathrm{O}_4\right)_3\right]^{3-}$

$\mathrm{Co}^{+3} \Rightarrow[\mathrm{Ar}] 3 \mathrm{~d}^6$

In presence of ligand field

$\Rightarrow \mathrm{d}^2 \mathrm{sp}^3$ hybridization, diamagnetic in nature

$\text { (D) }\left[\mathrm{Ni}(\mathrm{CO})_4\right]$

$\mathrm{Ni}^0 \Rightarrow[\mathrm{Ar}] 3 \mathrm{~d}^8 4 \mathrm{~s}^2$

In presence of ligand field

$\Rightarrow \mathrm{sp}^3$ hybridization, diamagnetic in nature

$\begin{aligned} & \text { (A) }\left[\mathrm{CoF}_6\right]^{-3} \\ & \mathrm{Co}^{3+} \rightarrow 3 \mathrm{~d}^6 \end{aligned}$

$\begin{aligned} & \text { (B) }\left[\mathrm{NiCl}_4\right]^{2-} \\ & \quad \mathrm{Ni}^{2+} \rightarrow 3 \mathrm{~d}^8 \end{aligned}$

$\begin{gathered} \text { (C) }\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]^{3+} \\ \mathrm{Co}^{3+} \rightarrow 3 \mathrm{~d}^6 \end{gathered}$

$\begin{aligned} & \text { (D) }\left[\mathrm{Ni}(\mathrm{CN})_4\right]^{2-} \\ & \quad \mathrm{Ni}^{2+} \rightarrow 3 \mathrm{~d}^8 \end{aligned}$

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)

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

$\begin{gathered} \left.\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_5 \mathrm{Cl}\right] \mathrm{Cl}_2 \text { (aq.) }\right)+2 \mathrm{AgNO}_3 \text { (aq.) } \rightarrow \\ {\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_5 \mathrm{Cl}\right]\left(\mathrm{NO}_3\right)_2(\text { aq. })+2 \mathrm{AgCl}(\mathrm{~s})} \end{gathered}$

$\begin{aligned} & \mathrm{Fe}^{+6}=[\mathrm{Ar}] 3 \mathrm{~d}^2 \\ & \mathrm{Mn}^{+3}=[\mathrm{Ar}] 3 \mathrm{~d}^4 \\ & \mathrm{Fe}^{+3}=[\mathrm{Ar}] 3 \mathrm{~d}^5 \\ & \mathrm{Cr}^{+2}=[\mathrm{Ar}] 3 \mathrm{~d}^4 \\ & \mathrm{Ni}^{+4}=[\mathrm{Ar}] 3 \mathrm{~d}^6 \end{aligned}$

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 = \frac{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) $\left[\mathrm{Co}(\mathrm{en})_2 \mathrm{Cl}_2\right]^{+} \Rightarrow \mathrm{M}(\mathrm{AA})_2 \mathrm{~b}_2$

$\mathrm{a,b},=\mathrm{NH}_3, \mathrm{Cl}^{-}$

$\mathrm{AA}=\mathrm{en}$

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 $\Delta_t = \frac{4}{9} (\Delta_0)_3$.

The order of crystal field splitting energies, based on our criteria, is:

$ \Delta_t < (\Delta_0)_3 < (\Delta_0)_2 < (\Delta_0)_1 $

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) $[ \text{Co(NH}_3 )_6 ]^{3+}$

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.

For complex $\mathrm{K}_3\left[\mathrm{Fe}(\mathrm{SCN})_6\right]$

Calculation of CFSE

$ \begin{aligned} & =(-0.4 \times 3+0.6 \times 2) \Delta_0 \\\\ & =0 \Delta_0 \end{aligned} $

The complexes are $\left[\mathrm{Co}(\mathrm{CN})_6\right]^{3-},\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]^{3+}$, $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_5 \mathrm{H}_2 \mathrm{O}\right]^{3+}$ and $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_5 \mathrm{Cl}\right]^{2+}$ It is based on the ligand strength. In spectrochemical series, ligands are weak or strong stronger ligands will split $d$-orbital energies in transition metal complexes. So, strong ligands cause larger splitting. and result in shorter wavelengths of absorbed light. For weak ligands, the splitting is lesser and wavelength is maximum. For the given ligands, $\mathrm{CN}^{-}, \mathrm{NH}_3, \mathrm{H}_2 \mathrm{O}, \mathrm{Cl}^{-}$, the older of increasing field (ligand) strength is $Cl^-<\mathrm{H}_2 \mathrm{O}<\mathrm{NH}_3<\mathrm{CN}^{-}$(splitting is inversely. proportional to wavelength): $C N^{-}$is a strong-field ligand. So, it will cause a larger splitting and absorb light at a shorter wavelength (higher energy). So, the complex $\left[C_0(\mathrm{CN})_6\right]^{3^-}$ has shorter wavelength$\mathrm{NH}_3$ ligand strength is lower than $C{N}^{-}$ but higher than $\mathrm{Cl}^{-}$and $\mathrm{H}_2 \mathrm{O}$. The splitting caused by $\mathrm{NH}_3$ is lesser than $\mathrm{CN}^{-}$. So, the wavelength is higher than $\left[\mathrm{CO}(\mathrm{CN})_6\right]^{3-}$ complex.$ \mathrm{Co}\left(\mathrm{NH}_3\right)_6^{3+} \text { has wavelength higher than }\left[\mathrm{CO}(\mathrm{CN})_6\right]^{3-} \text { - } $$\mathrm{H}_2 \mathrm{O}$ is a weaker ligand than $\mathrm{NH}_3$ so, the complex $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_5 \mathrm{H}_2O\right]^{3+}$ has higher wavelength than $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]^{3+}$Cl is the weak field ligand than $^{}$ $\mathrm{H}_2 \mathrm{O}$ and $\mathrm{NH}_3$ So, the wavelength is maximum for the complex containing $C^{-}$.$ \left[\mathrm{Co}\left(\mathrm{NH}_3\right)_5 \mathrm{Cl}\right]^{2+} \text { has highest wavelength. } $ The correct order of wavelength maxima of the absorption band is. $ \left[\mathrm{Co}(\mathrm{CN})_6\right]^{3-}<\left[\mathrm{C}_0\left(\mathrm{NH}_3\right)_6\right]^{3+}<\left[\mathrm{C}_0\left(\mathrm{NH}_3\right)_5 \mathrm{H}_2 \mathrm{O}\right]^{3+} \leq\left[\mathrm{C}_0\left(\mathrm{NH}_3\right)_5 \mathrm{Cl}\right]^{2+} $$ \text { Answer: Option (A) } $

The number of unpaired electron in the given complex ion are, $[\mathrm{Mn}(\mathrm{CN})]^{3-}$ has two unpaired electrons, $\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{3-}$ has a single unpaired electron, $\left(\mathrm{COF}_{\mathrm{6}}\right)^{3-}$ has four electrons while $\left[\mathrm{Co}\left(\mathrm{C}_2 \mathrm{O}_4\right)_3\right]^{3^-}$ in diamagnetic i.e. has zero unpaired electrons. Thus, the correct order is D, $B, A, C$.

The atomic number of $\mathrm{C}_0$ is 27 . Its electronic configuration is $[\mathrm{Ar}] 3 d^2 4 s^2$.

In $\left[\mathrm{Co}(\mathrm{ox})_3\right]^{3-}$, the oxidation state of Co is +3 .

In the presence of $\mathrm{C}_2 \mathrm{O}_4^{2-}$ (strong field ligand in case of $\mathrm{Co}^{3+}$ ) pairing of electrons takes place.

$ \left[\mathrm{Co}(\mathrm{ox})_3\right]^{3-}= $

It has no unpaired electron, so it is diamagnetic.

The total number of geometrical isomers for the given complexes are

• $\left[\mathrm{NiCl}_4\right]^{2-}$ - No geometrical isomer

• $\left[\mathrm{CoCl}_2\left(\mathrm{NH}_3\right)_4\right]^{+}$- octahedral complex of the type $M A_4 B_2$. So can exist as cis and trans isomers. Thus 2 geometrical isomers are possible.

• [ $\mathrm{Co}\left(\mathrm{NH}_3\right)_3\left(\mathrm{NO}_2\right)_3$ ] - Octahedral complex of the type $M A_3 B_3$. So can exist as facial and meridional isomer. So, 2 geometrical isomers are possible.

• $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_5 \mathrm{Cl}\right]^{2+}$ - Octahedral complex of the type $M A_5 B$. All positions are equivalent relative to the single unique ligand. So no geometrical isomer is possible.

Thus, total 4 geometrical isomers are possible.

The magnetic moment of a complex depends on the number of unpaired electrons present in it.

Formula: The spin-only magnetic moment is given by $ \mu = \sqrt{n(n+2)} $ where $ n $ is the number of unpaired electrons.

Counting Unpaired Electrons for Each Complex:

I. $\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{4-}$: Here, $ n=0 $. This means there are no unpaired electrons.

II. $\left[\mathrm{MnCl}_4\right]^{2-}$: Here, $ n=5 $. There are five unpaired electrons.

III. $\left[\mathrm{Mn}(\mathrm{CN})_6\right]^{4-}$: Here, $ n=1 $. There is one unpaired electron.

IV. $\left[\mathrm{Cr}(\mathrm{NH}_3)_6\right]^{3+}$: Here, $ n=3 $. There are three unpaired electrons.

The complexes with more unpaired electrons will have higher magnetic moments.

So, the order from least to greatest magnetic moment is:

$ \text{I} < \text{III} < \text{IV} < \text{II} $

Moles of ions precipitated.

$ \begin{aligned} n_{\text {ions }} & =\frac{\text { Number of ions }}{\text { Avogadro's number }}=\frac{3.6 \times 10^{22}}{6 \times 10^{23}} \\ n_{\text {ions }} & =0.06 \mathrm{~mol} \\ n_{\text {complex }} & =\text { Molarity } \times \text { Volume }=0.2 \times 0.1=0.02 \mathrm{~mol} \end{aligned} $

Number of $\mathrm{Cl}^{-}$ions per complex units,

$ Y=\frac{\text { Moles of precipitated ions }}{\text { Moles of complex }}=3 $

The complex is $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_x \mathrm{Cl}_{3-y}\right] \mathrm{Cl}_y$.

Coordination number of $\mathrm{Co}=6$.

$ x=6 $

An ambidentate ligand is a ligand that can coordinate to a central metal atom through two different donor atoms, but not at the same time. $\mathrm{SO}_4^{2-}$ is a bidentate ligand.

II and III i.e. $\left[\mathrm{Ti}\left(\mathrm{H}_2 \mathrm{O}\right)_5 \mathrm{Cl}\right]\left(\mathrm{NO}_3\right)_2$ and $\left[\mathrm{Pt}(\mathrm{en})\left(\mathrm{NH}_3\right) \mathrm{Cl}\right] \mathrm{NO}_3$ shows ionisation isomerism.

Ionisation isomerism occur when coordination compound with same empirical formula but different ions in solution are produced.

$\mathrm{C}_2 \mathrm{O}_4^{2-}=$ bidentate ligand

$\mathrm{H}_2 \mathrm{O}=$ monodentate ligand

$\mathrm{SO}_4^{2-}=$ bidentate ligand.

Hence, there is no ambidentate ligand.

An ambidentate ligand is a ligand that

can coordinate to central metal atom

through two different donor atom.

The correct IUPAC name of the complex $\mathrm{K}_3\left[\mathrm{Co}(\mathrm{ox})_3\right]$ is, potassium trioxalatocobaltate (III)

To determine the coordination number of chromium (Cr) in the complex $\mathrm{K}\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_2\left(\mathrm{C}_2 \mathrm{O}_4\right)_2\right]$, we need to identify the ligands and their denticity.

1. Identify the central metal atom: The central metal atom is Chromium (Cr).

2. Identify the ligands:

   • $\mathrm{H}_2 \mathrm{O}$ (aqua)

   • $\mathrm{C}_2 \mathrm{O}_4$ (oxalato, derived from oxalic acid, $\mathrm{C}_2 \mathrm{O}_4^{2-}$)

3. Determine the denticity of each ligand:

   • $\mathrm{H}_2 \mathrm{O}$ (water): This is a monodentate ligand, meaning each water molecule forms one coordinate bond with the central metal atom.

   • $\mathrm{C}_2 \mathrm{O}_4^{2-}$ (oxalate): This is a bidentate ligand, meaning each oxalate ion forms two coordinate bonds with the central metal atom, typically through its two oxygen atoms.

4. Calculate the total number of coordinate bonds:

   • There are two $\mathrm{H}_2 \mathrm{O}$ ligands. Since each is monodentate, they contribute $2 \times 1 = 2$ coordinate bonds.

   • There are two $\mathrm{C}_2 \mathrm{O}_4^{2-}$ ligands. Since each is bidentate, they contribute $2 \times 2 = 4$ coordinate bonds.

5. Sum the coordinate bonds to find the coordination number:

   Coordination number = (bonds from $\mathrm{H}_2 \mathrm{O}$) + (bonds from $\mathrm{C}_2 \mathrm{O}_4^{2-}$)

   Coordination number = $2 + 4 = 6$

Therefore, the coordination number of chromium in the given complex is 6.

The final answer is $\boxed{6}$.

The complex ion $\left[\mathrm{Mn}(\mathrm{Cl})_6\right]^{3-}$ has a spin-only magnetic moment of 4.90 B.M.

Oxidation State and Electron Configuration:

For $\mathrm{Mn}$ in this ion, the oxidation state is +3. This means manganese has lost three electrons.

The electron configuration for $\mathrm{Mn}^{3+}$ is $[\mathrm{Ar}]\,3d^4.$ This tells us the ion has 4 electrons in the 3d orbitals.

Finding the Magnetic Moment:

The formula for the spin-only magnetic moment is:

$ \mu = \sqrt{n(n+2)} \text{ B.M.} $

Here, $n$ is the number of unpaired electrons. For $\mathrm{Mn}^{3+}$, $n = 4$ because there are 4 unpaired electrons in the $3d$ orbitals.

Plug in the value of $n$:

$ \mu = \sqrt{4(4+2)} = \sqrt{24} = 4.90\ \text{B.M.} $

$\left[\mathrm{Mn}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}$ has $t_{2 g}^3 e_9^2$ configuration

$\mathrm{H}_2 \mathrm{O}$ is a weak field ligand, means it does not cause pairing of $e^{-}$.

The $3 d$ electron will occupy the $d$-orbital with 5 -single electrons is each orbital. $d$ splits into $t_{2 g}$ has 3 electrons and $e_g$ has 2 electrons.

We are given the compound:

$ \mathrm{Fe}_x\left[\mathrm{Fe}_y(\mathrm{CN})_6\right]_3 $

and asked to find the values of $x$ and $y$.

Step 1. Identify familiar Prussian blue–type compounds

Prussian blue and related compounds have the general formula:

$ \mathrm{Fe}_4[\mathrm{Fe}(\mathrm{CN})_6]_3 $

This often corresponds to a mixed-valence compound containing both Fe²⁺ and Fe³⁺ ions.

Step 2. Match with the given general form

Given formula: $\mathrm{Fe}_x[\mathrm{Fe}_y(\mathrm{CN})_6]_3$

If we compare:

$ \mathrm{Fe}_4[\mathrm{Fe}(\mathrm{CN})_6]_3 $

we identify:

$ x = 4, \quad y = 1 $

✅ Answer: Option B (4,1)

Option (d) is incorrect regarding magnitude of crystal field splitting $\left.\mathrm{Fe}\left(\mathrm{NH}_3\right)_6\right]^{2+}>\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}$

This is because $\mathrm{NH}_3$ is strong ligand than $\mathrm{H}_2 \mathrm{O}$ and strong ligand causes larger splitting of $d$-orbitals.

Complexes given in

I, II and IV exhibit geometrical isomerism

Geometrical isomerism, also called cis-trans isomerism, occurs in coordination complexes when ligands can occupy different position around the central metal ions.

This is found in square planar and octahedral complex.

$\mathrm{CN}^{-}$is a strong field ligand; it causes pairing in $\mathrm{Ni}^{2+}\left(d^8\right)$.

The $d s p^2$ hybridisation forms a square planar geometry with no unpaired electrons.

The correct match is A-II, B-I, C-IV, D-III.

A. $\mathrm{Ni}(\mathrm{CO})_4$, all ligands neutral. So, oxidation state, $\mathrm{Ni}=0$

B. $\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_5 \mathrm{NO}^{2+}, \mathrm{NO}^{+}=+1, \mathrm{H}_2 \mathrm{O}=0\right.$. So oxidation state, $\mathrm{Fe}=+1$

C. $\left[\mathrm{Co}(\mathrm{CO})_5\right]^{2-}, x+5(0)=-2$, So, oxidation state, $\mathrm{Co}=-2$

D. $\left[\mathrm{Cr}_2(\mathrm{CO})_{10}\right]^{2-}$, each $\mathrm{Cr}=-1($ since total $=-2)$