Coordination Compounds

$ \begin{aligned} & \mathrm{Mn}^{2+} : 3 \mathrm{~d}^5 \\ & \text{In }[\mathrm{MnBr}_4]^{2-}\text{, Mn is in }+2\text{ oxidation state, so configuration is }3d^5. \\ & \text{Br}^-\text{ is a weak field ligand and the complex is tetrahedral, so it is high spin.} \\ & \text{Hence number of unpaired electrons }(n)=5. \\ \\ & \mathrm{Cu}^{2+} : 3 \mathrm{~d}^9 \\ & \text{In }[\mathrm{Cu}(\mathrm{H}_2\mathrm{O})_6]^{2+}\text{, Cu is in }+2\text{ oxidation state, so configuration is }3d^9. \\ & \text{This is an octahedral complex (six water ligands). In }d^9\text{, there is always one unpaired electron.} \\ & \text{So } \mathrm{t}_{2\mathrm{g}}^{2,2,2}\ \mathrm{e}_{\mathrm{g}}^{2,1}\text{ and }(n)=1. \\ \\ & \mathrm{Ni}^{2+} : 3 \mathrm{~d}^8 \\ & \text{In }[\mathrm{Ni}(\mathrm{CN})_4]^{2-}\text{, Ni is in }+2\text{ oxidation state, so configuration is }3d^8. \\ & \text{CN}^-\text{ is a strong field ligand, so electrons pair up and the complex becomes square planar.} \\ & \text{In square planar }d^8\text{, all electrons are paired, so }(n)=0. \\ \\ & \mathrm{Ni}^{2+} : 3 \mathrm{~d}^8 \text { (tetrahedral) }\\ & \text{In }[\mathrm{Ni}(\mathrm{H}_2\mathrm{O})_6]^{2+}\text{, Ni is in }+2\text{ oxidation state, so configuration is }3d^8. \\ & \text{With } \mathrm{H}_2\mathrm{O}\text{ (weak field), the complex is high spin; the distribution gives two unpaired electrons.} \\ & \text{So } \mathrm{e}_{g}^{2,2}\ \mathrm{t}_{2g}^{2,1,1}\text{ and }(n)=2. \end{aligned} $

Option D is correct.

1) $\mathrm{Ni}(\mathrm{CO})_4$

• Oxidation state of Ni: CO is a neutral ligand, so Ni is in $0$ state.

• Electronic configuration: $\mathrm{Ni}(0) \Rightarrow 3d^{10}4s^0$

• Since it is $d^{10}$, all electrons are paired.

• Hence, $\mathrm{Ni}(\mathrm{CO})_4$ is diamagnetic.

2) $\left[\mathrm{Ni}(\mathrm{CN})_4\right]^{2-}$

• Oxidation state of Ni:

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

• So, $\mathrm{Ni}^{2+}$ is $3d^8$.

• $\mathrm{CN}^-$ is a strong field ligand, causes pairing of electrons.

• For $\mathrm{Ni}^{2+}(d^8)$ with strong field ligands, complex becomes square planar (inner orbital), and all electrons get paired.

• Hence, $\left[\mathrm{Ni}(\mathrm{CN})_4\right]^{2-}$ is diamagnetic.

3) $\left[\mathrm{NiCl}_4\right]^{2-}$

• Oxidation state of Ni:

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

• So, again $\mathrm{Ni}^{2+}$ is $3d^8$.

• $\mathrm{Cl}^-$ is a weak field ligand, so it does not cause pairing.

• Such complexes are generally tetrahedral (outer orbital) and remain high spin, leaving 2 unpaired electrons.

• Hence, $\left[\mathrm{NiCl}_4\right]^{2-}$ is paramagnetic.

Therefore:

• $\mathrm{Ni}(\mathrm{CO})_4$ → diamagnetic

• $\left[\mathrm{Ni}(\mathrm{CN})_4\right]^{2-}$ → diamagnetic

• $\left[\mathrm{NiCl}_4\right]^{2-}$ → paramagnetic

So, the correct statement is Option D.

Wavelength of light absorbed increases as C.F.S.E of complex decreases.

$\left[\mathrm{Co}(\mathrm{CN})_6\right]^{3-}$ has maximum CFSE

$\left[\mathrm{CoF}_6\right]^{3-}$ has least CFSE

Ligand field strength ↑ ; C.F.S.E ↑

Correct wavelength order.

V $>$ II $>$ IV $>$ I $>$ III

To decide hybridisation, geometry and magnetic moment, we mainly look at (i) the type of ligands (strong field or weak field) and (ii) the number of unpaired electrons. The spin-only magnetic moment depends on the number of unpaired electrons $n$.

The given results for each complex are summarised below (hybridisation $\Rightarrow$ geometry $\Rightarrow$ number of unpaired electrons $\Rightarrow$ magnetic moment in B.M.).

$\begin{aligned} \text { In } \mathrm{K}_3\left[\mathrm{Co}\left(\mathrm{CO}_3\right)_3\right] & \Rightarrow \mathrm{sp}^3 \mathrm{~d}^2 \text { hybridized, octahedral } \\ & \Rightarrow 4 \text { unpaired electron } \\ & \Rightarrow 4.9 \text { B.M. } \\ {\left[\mathrm{Ni}(\mathrm{CN})_4\right]^{2-} } & \Rightarrow \mathrm{dsp}^2 \text { hybridized, square planar } \\ & \Rightarrow 0 \text { unpaired electron } \\ & \Rightarrow 0 \text { B.M. } \\ {\left[\mathrm{MnBr}_4\right]^{2-} } & \Rightarrow \mathrm{sp}^3 \text { hybridized, tetrahedral } \\ & \Rightarrow 5 \text { unpaired electron } \\ & \Rightarrow 5.9 \text { B.M. } \\ {\left[\mathrm{CoF}_6\right]^{2-} } & \Rightarrow \text { sp } \mathrm{d}^2 \text { hybridized, octahedral } \\ & \Rightarrow 4 \text { unpaired electron } \\ & \Rightarrow 4.9 \text { B.M. }\end{aligned}$

So, the hybridisation/geometry and magnetic moments mentioned in the statements match these standard results (based on unpaired electrons in each case).

Paramagnetic species are those which have one or more unpaired electrons.

Let us check each species one by one.

1. In $\left[\mathrm{CoF}_6\right]^{3-}$, oxidation state of Co is $+3$ (because $6 \times (-1) = -6$ and overall charge is $-3$). So Co is $\mathrm{Co}^{3+}$ i.e. $d^6$. Since $\mathrm{F}^-$ is a weak field ligand, it forms a high spin octahedral complex, so unpaired electrons are present. Hence, $\left[\mathrm{CoF}_6\right]^{3-}$ is paramagnetic.

2. In $\left[\mathrm{TiF}_6\right]^{3-}$, oxidation state of Ti is $+3$. So Ti is $\mathrm{Ti}^{3+}$ i.e. $d^1$. One electron will remain unpaired, so it is paramagnetic.

3. In $\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{3-}$, oxidation state of Fe is $+3$. So Fe is $\mathrm{Fe}^{3+}$ i.e. $d^5$. Since $\mathrm{CN}^-$ is a strong field ligand, it gives a low spin octahedral complex, so only one electron remains unpaired. Hence, it is paramagnetic.

4. In $\mathrm{V}_2 \mathrm{O}_5$, V is in $+5$ oxidation state, so it is $\mathrm{V}^{5+}$ i.e. $d^0$. There are no $d$-electrons, so there cannot be any unpaired electron. Hence, $\mathrm{V}_2 \mathrm{O}_5$ is diamagnetic.

So, the paramagnetic species are :

$\left[\mathrm{CoF}_6\right]^{3-},\left[\mathrm{TiF}_6\right]^{3-},\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{3-}$

And the diamagnetic species is : $\mathrm{V}_2 \mathrm{O}_5$

Now, compare the complexes in Statement II by counting unpaired electrons.

In $\mathrm{K}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right] \Rightarrow$ Fe is $+2$ (because complex anion is $4-$), so $\mathrm{Fe}^{2+}$ is $d^6$. With $\mathrm{CN}^-$ (strong field), it is low spin, so all electrons pair up. Therefore, No. of unpaired electron $=0$

$\mathrm{K}_3\left[\mathrm{Fe}(\mathrm{CN})_6\right] \Rightarrow$ Fe is $+3$ (complex anion is $3-$), so $\mathrm{Fe}^{3+}$ is $d^5$. With $\mathrm{CN}^-$ (strong field), it is low spin, so No. of unpaired electron $=1$

$\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right] \mathrm{SO}_4 \cdot \mathrm{H}_2 \mathrm{O} \Rightarrow$ Here the cation is $\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}$, so Fe is $+2$ i.e. $d^6$. Since $\mathrm{H}_2\mathrm{O}$ is a weak field ligand, it is high spin, so No. of unpaired electron $=4$

$\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right] \mathrm{Cl}_3 \Rightarrow$ Here the cation is $\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+}$, so Fe is $+3$ i.e. $d^5$. With $\mathrm{H}_2\mathrm{O}$ (weak field), it is high spin, so No. of unpaired electron $=5$

For coordination compounds, only the ions outside the coordination sphere dissociate in water and give precipitates.

1) Identify free ions in solution

(i) $\left[\mathrm{Co}(\mathrm{NH}_3)_5\mathrm{SO}_4\right]\mathrm{Br}$

Here $\mathrm{Br}^-$ is outside the bracket, so in water:

$\left[\mathrm{Co}(\mathrm{NH}_3)_5\mathrm{SO}_4\right]\mathrm{Br}\;\to\;\left[\mathrm{Co}(\mathrm{NH}_3)_5\mathrm{SO}_4\right]^+ + \mathrm{Br}^-$

So it gives $1$ mol $\mathrm{Br}^-$ per mol of salt.

(ii) $\left[\mathrm{Co}(\mathrm{NH}_3)_5\mathrm{Br}\right]\mathrm{SO}_4$

Here $\mathrm{SO}_4^{2-}$ is outside the bracket, so:

$\left[\mathrm{Co}(\mathrm{NH}_3)_5\mathrm{Br}\right]\mathrm{SO}_4\;\to\;\left[\mathrm{Co}(\mathrm{NH}_3)_5\mathrm{Br}\right]^{2+} + \mathrm{SO}_4^{2-}$

So it gives $1$ mol $\mathrm{SO}_4^{2-}$ per mol of salt.

2) Moles present in 2 L of mixture X

Total solution volume $=4\,\mathrm{L}$, and we take $2\,\mathrm{L}$, i.e. half.

So in $2\,\mathrm{L}$:

• From $0.4$ mol of $\left[\mathrm{Co}(\mathrm{NH}_3)_5\mathrm{SO}_4\right]\mathrm{Br}$, we have $0.2$ mol of this salt $\Rightarrow 0.2$ mol $\mathrm{Br}^-$.

• From $0.4$ mol of $\left[\mathrm{Co}(\mathrm{NH}_3)_5\mathrm{Br}\right]\mathrm{SO}_4$, we have $0.2$ mol of this salt $\Rightarrow 0.2$ mol $\mathrm{SO}_4^{2-}$.

3) Precipitates formed

With excess $\mathrm{AgNO}_3$ (in 2 L):

$\mathrm{Ag}^+$ precipitates $\mathrm{Br}^-$ as $\mathrm{AgBr}$:

$\mathrm{Ag}^+ + \mathrm{Br}^- \to \mathrm{AgBr}(s)$

So moles of precipitate $Y = 0.2$ mol (as $\mathrm{AgBr}$).

With excess $\mathrm{BaCl}_2$ (in remaining 2 L):

$\mathrm{Ba}^{2+}$ precipitates $\mathrm{SO}_4^{2-}$ as $\mathrm{BaSO}_4$:

$\mathrm{Ba}^{2+} + \mathrm{SO}_4^{2-} \to \mathrm{BaSO}_4(s)$

So moles of precipitate $Z = 0.2$ mol (as $\mathrm{BaSO}_4$).

Correct option

$ \boxed{\text{Option C: } 0.2\text{ mol of } Z \text{ is formed.}} $

(Also, $Y=\mathrm{AgBr}$ and $Z=\mathrm{BaSO}_4$.)

Check each complex using NCERT (VBT + field strength):

A. $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]^{3+}$

Oxidation state of Co: $+3$

So, $\mathrm{Co}^{3+} : 3d^6$

$\mathrm{NH}_3$ is a stronger ligand than $\mathrm{F^-}, \mathrm{Cl^-}$ and with $\mathrm{Co^{3+}}$ (high oxidation state) it causes pairing.

So it forms an inner orbital (low spin) octahedral complex:

• Hybridisation: $d^2sp^3$

✅ A is correct

B. $\left[\mathrm{MnCl}_6\right]^{3-}$

Oxidation state of Mn: $+3$

$\mathrm{Mn}^{3+}: 3d^4$

$\mathrm{Cl^-}$ is a weak field ligand, so no pairing (high spin). Hence it uses outer $d$-orbitals.

So it is an outer orbital octahedral complex:

• Hybridisation: $sp^3d^2$

✅ B is correct

C. $\left[\mathrm{CoF}_6\right]^{3-}$

Oxidation state of Co: $+3$

$\mathrm{Co}^{3+}: 3d^6$

$\mathrm{F^-}$ is a weak field ligand, so complex will be high spin and thus outer orbital.

Therefore hybridisation should be:

• $sp^3d^2$ (not $d^2sp^3$)

❌ C is incorrect

D. $\left[\mathrm{FeF}_6\right]^{3-}$

Oxidation state of Fe: $+3$

$\mathrm{Fe}^{3+}: 3d^5$

$\mathrm{F^-}$ is weak field → no pairing → outer orbital octahedral:

• Hybridisation: $sp^3d^2$

✅ D is correct

E. $\left[\mathrm{Ni}(\mathrm{CN})_4\right]^{2-}$

Oxidation state of Ni: $+2$

$\mathrm{Ni}^{2+}: 3d^8$

$\mathrm{CN^-}$ is a strong field ligand, so it forms square planar complex (pairing occurs).

Square planar hybridisation is:

• $dsp^2$ (not $sp^3$)

❌ E is incorrect

Correct set: A, B and D only → Option D

In this complex, Ni is in the +2 oxidation state (that is, it is present as $\mathrm{Ni}^{2+}$).

A) The nickel dimethylglyoxime complex is a rosy red coloured precipitate, so the statement that it is red in colour is correct.

B) This complex is formed (precipitated) in a basic medium (around $\mathrm{pH}=9$), but it is sparingly soluble in water, not highly soluble. So the statement about high solubility in water at $\mathrm{pH}=9$ is incorrect.

C) $\mathrm{Ni}^{+2}: 3 d^8$ means that $\mathrm{Ni}^{2+}$ has 8 electrons in the d-subshell. In this complex, it forms a square planar $\mathrm{dsp}^2$ complex in which all these electrons get paired. Therefore, the statement that Ni has two unpaired d-electrons is incorrect.

Hybridisation : $\mathrm{dsp}^2$, which leads to a square planar arrangement of the ligands around the Ni ion.

Unpaired $\mathrm{e}^{-}=0$, so the complex is diamagnetic (no unpaired electrons are present).

Geometry : Square planar, which is typical for $\mathrm{dsp}^2$ hybridisation in such $\mathrm{Ni}^{2+}$ complexes.

D) $\mathrm{N}-\mathrm{Ni}-\mathrm{N}$ Bond angle is close to $90^{\circ}$ in a square planar complex because adjacent bond angles in square planar geometry are approximately $90^{\circ}$. So statement D is correct.

E) Each dimethylglyoxime ligand forms one five-membered chelate ring with the Ni ion. Since there are only two such ligands, 2 five-membered metal-containing rings are formed, not four. So statement E is incorrect.

Statement I (True):

The strength of a ligand means how strong it can split the d-orbitals of the metal ion.

Among $\mathrm{Cl}^-$ and $\mathrm{Br}^-$, chloride ($\mathrm{Cl}^-$) is a stronger ligand than bromide ($\mathrm{Br}^-$).

This means for $\left[\mathrm{CoCl}_4\right]^{2-}$, the energy gap between the two sets of d-orbitals ($\Delta_t$) will be larger than in $\left[\mathrm{CoBr}_4\right]^{2-}$.

So, $\left[\mathrm{CoCl}_4\right]^{2-}$ absorbs light of higher energy, and $\left[\mathrm{CoBr}_4\right]^{2-}$ absorbs light of lower energy.

$ \begin{aligned} & \Delta_t : \left[\mathrm{CoCl}_4\right]^{2-} > \left[\mathrm{CoBr}_4\right]^{2-} \\ & \text{Energy absorbed} : \left[\mathrm{CoCl}_4\right]^{2-} > \left[\mathrm{CoBr}_4\right]^{2-} \end{aligned} $

Statement II (False):

Here, we compare $\mathrm{I}^-$ and $\mathrm{Cl}^-$ as ligands. Iodide ($\mathrm{I}^-$) is a weaker ligand than chloride ($\mathrm{Cl}^-$).

This means that in $\left[\mathrm{CoI}_4\right]^{2-}$, the energy difference between the two sets of d-orbitals ($\Delta_t$) is smaller than in $\left[\mathrm{CoCl}_4\right]^{2-}$.

$ \Delta_{\mathrm{t}} : \left[\mathrm{CoI}_4\right]^{2-} < \left[\mathrm{CoCl}_4\right]^{2-} $

For $\left[\mathrm{Ni}\left(\mathrm{PPh}_3\right)_2 \mathrm{Cl}_2\right]$:

1) Oxidation state and $d$-electron count

$\mathrm{PPh}_3$ is neutral and each $\mathrm{Cl}$ is $-1$.

So oxidation state of Ni is $+2$.

$\mathrm{Ni}^{2+} \Rightarrow 3d^8$

2) Geometry using given information (paramagnetic)

For $\mathrm{Ni}^{2+}$ ($d^8$):

• Square planar complexes are usually diamagnetic (all electrons paired).

• Since the complex is paramagnetic, it must be tetrahedral (in NCERT level treatment).

So geometry is tetrahedral and it has 2 unpaired electrons.

Check each statement

A. The complex exhibits geometrical isomerism.

In a tetrahedral complex, all four positions are equivalent, so an $\mathrm{MA}_2\mathrm{B}_2$ type complex does not show cis–trans isomerism.

So A is incorrect.

B. The complex is white in colour.

Tetrahedral $\mathrm{Ni}^{2+}$ complexes are generally coloured due to $d$–$d$ transitions; “white” is not expected.

So B is incorrect.

C. Spin-only magnetic moment is $2.84\ \mathrm{BM}$.

Number of unpaired electrons $n=2$.

$\mu_\text{so}=\sqrt{n(n+2)}=\sqrt{2(2+2)}=\sqrt{8}=2.83\ \mathrm{BM}\approx 2.84\ \mathrm{BM}$

So C is correct.

D. CFSE is $-0.8\Delta_o$.

For tetrahedral splitting, we must use $\Delta_t$, not $\Delta_o$.

For $d^8$ tetrahedral: $e^4t_2^4$

$\text{CFSE}=4(-0.6\Delta_t)+4(+0.4\Delta_t)=(-2.4+1.6)\Delta_t=-0.8\Delta_t$

So writing $-0.8\Delta_o$ is wrong.

So D is incorrect.

E. Geometry similar to $\mathrm{Ni}(\mathrm{CO})_4$.

$\mathrm{Ni}(\mathrm{CO})_4$ is tetrahedral, and this complex is also tetrahedral (as concluded from paramagnetism).

So E is correct.

Incorrect statements: A, B and D

✅ Correct option: Option B (A, B and D Only)

For each ion, first find its $3d$ electron count after losing electrons to form the $3+$ charge.

Since all are low spin octahedral complexes, electrons will first pair up in the lower-energy $t_{2g}$ orbitals before entering the higher-energy $e_g$ orbitals.

$\begin{aligned} & \mathrm{Co}^{3+} \rightarrow 3 \mathrm{~d}^6 \Rightarrow \mathrm{t}_{2 \mathrm{~g}}^{2,2,2} \mathrm{e}_{\mathrm{g}}^{0,0}, \text { unpaired electron }=0 \\ & \mathrm{Fe}^{3+} \rightarrow 3 \mathrm{~d}^5 \Rightarrow \mathrm{t}_{2 \mathrm{~g}}^{2,2,1} \mathrm{e}_{\mathrm{g}}^{0,0} \text { unpaired electron }=1 \\ & \mathrm{Cr}^{3+} \rightarrow 3 \mathrm{~d}^3 \Rightarrow \mathrm{t}_{2 \mathrm{~g}}^{1,1,1} \mathrm{e}_{\mathrm{g}}^{0,0} \text { unpaired electron }=3 \\ & \mathrm{Mn}^{3+} \rightarrow 3 \mathrm{~d}^4 \Rightarrow \mathrm{t}_{2 \mathrm{~g}}^{2,1,1} \mathrm{e}_{\mathrm{g}}^{0,0} \text { unpaired electron }=2\end{aligned}$

Now compare the number of unpaired electrons: $\mathrm{Cr}^{3+}$ has $3$, $\mathrm{Mn}^{3+}$ has $2$, $\mathrm{Fe}^{3+}$ has $1$, and $\mathrm{Co}^{3+}$ has $0$.

So, the decreasing order of unpaired electrons is: $\mathrm{Cr}^{3+} > \mathrm{Mn}^{3+} > \mathrm{Fe}^{3+} > \mathrm{Co}^{3+}$

$\mathrm{Cu} \xrightarrow{\text { di. } \mathrm{HCl}}$ no reaction

Step 1: Identify the metal $\mathrm{M}$

A first row transition metal that does not liberate $\mathrm{H}_2$ gas with dilute HCl is $\mathrm{Cu}$ (copper). So, $\mathrm{M} = \mathrm{Cu}$.

Step 2: Reaction of $\mathrm{MSO}_4$ with excess KCN

So $\mathrm{MSO}_4$ is $\mathrm{CuSO}_4$. When $\mathrm{CuSO}_4$ is treated with excess aqueous KCN, copper forms a very stable (perfect) cyanide complex.

$\mathrm{CuSO}_4 \xrightarrow{\mathrm{KCN}} \underset{\text { perfect complex }}{\left[\mathrm{Cu}(\mathrm{CN})_4\right]^{3-}} \xrightarrow{\mathrm{H}_2 \mathrm{~S}}$ no sulphide ppt

Step 3: Passing $\mathrm{H}_2\mathrm{S}$ through the solution

Because copper is present as the stable complex $\left[\mathrm{Cu}(\mathrm{CN})_4\right]^{3-}$, free $\mathrm{Cu}^{2+}$ ions are not available in solution. Therefore, $\mathrm{H}_2\mathrm{S}$ cannot precipitate copper sulphide.

Final Answer

No metal sulphide $\mathrm{MS}$ is formed. Hence, the amount of $\mathrm{MS}$ formed is $0$ mol.

Statement I:

In $[Cr(H_2O)_6]^{2+}$, $Cr^{2+}$ is $d^4$ and with $H_2O$ (weak field) it is high spin: $t_{2g}^3 e_g^1$.

$ \text{CFSE} = 3(-0.4\Delta_o)+1(+0.6\Delta_o)= -0.6\Delta_o $

In $[Mn(H_2O)_6]^{2+}$, $Mn^{2+}$ is $d^5$ high spin: $t_{2g}^3 e_g^2$.

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

So CFSE of the chromium complex is greater (more stabilizing). Statement I is true.

Statement II:

• $K_3[Fe(CN)_6]$: $Fe^{3+}$ is $d^5$; $CN^-$ is strong field → low spin $t_{2g}^5$, 1 unpaired electron → $\mu_{so}=\sqrt{1(1+2)}=\sqrt{3}$ BM.

• $Na_4[Fe(CN)_6]$: $Fe^{2+}$ is $d^6$; strong field → low spin $t_{2g}^6$, 0 unpaired electrons → $\mu_{so}=0$.

Thus ferricyanide has a greater spin-only magnetic moment. Statement II is true.

Correct option: B (Both Statement I and Statement II are true).

$\left[\mathrm{Cu}\left(\mathrm{NH}_3\right)_4\right]^{2+} \Rightarrow \mathrm{d}^9, \mathrm{dsp}^2$ one unpaired electron

$\left[\mathrm{Ni}(\mathrm{en})_3\right]^{2+} \Rightarrow \mathrm{d}^8, \mathrm{sp}^3 \mathrm{~d}^2$ two unpaired electrons

$\left[\mathrm{Ni}\left(\mathrm{NH}_3\right)_6\right]^{2+} \Rightarrow \mathrm{d}^8, \mathrm{sp}^3 \mathrm{~d}^2$ two unpaired electrons

$\left[\mathrm{Mn}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+} \Rightarrow \mathrm{d}^5, \mathrm{sp}^3 \mathrm{~d}^2$ five unpaired electrons

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

$\left[\mathrm{NiCl}_4\right]^{-2}$ (paramagnetic)

$\left[\mathrm{Ni}(\mathrm{CN})_4\right]^{-2}$ (diamagnetic)

(A) $\left[\mathrm{Ni}(\mathrm{CO})_4\right], \mathrm{Ni}^0:[\mathrm{Ar}] 3 \mathrm{~d}^8 4 \mathrm{~s}^2$

Valence orbitals of $\mathrm{Ni}^0$ in pre-hybridisation state :

Tetrahedral, Diamagnetic, $\mu=0$ B.M.

(B) $\left[\mathrm{Ni}(\mathrm{CN})_4\right]^{2-}, \mathrm{Ni}^{+2}:[\mathrm{Ar}] 3 \mathrm{~d}^8 4 \mathrm{~s}^0$

Valence orbitals of $\mathrm{Ni}^{+2}$ in pre-hybridisation state :

Square planar, Diamagnetic, $\mu=0$ B.M.

(C) $\left[\mathrm{NiCl}_4\right]^{2-}, \mathrm{Ni}^{+2}:[\mathrm{Ar}] 3 \mathrm{~d}^8 4 \mathrm{~s}^0$

Valence orbitals of $\mathrm{Ni}^{+2}$ in ground state :

Tetrahedral, paramagnetic, $\mu=\sqrt{8}=2.8$ B.M.

(D) $\left[\mathrm{MnBr}_4\right]^{2-}, \mathrm{Mn}^{+2}:[\mathrm{Ar}] 3 \mathrm{~d}^5$

Valence orbitals of $\mathrm{Mn}^{+2}$ in ground state :

Tetrahedral, paramagnetic, $\mu=\sqrt{35}=5.9$ B.M.

In $\left[\mathrm{Co}\left(\mathrm{NH}_3\right]_6\right]^{3+}, \mathrm{Co}^{+3}:[\mathrm{Ar}] 3 \mathrm{~d}^6, \mathrm{NH}_3$ is S.F.L

Hybridisation state of $\mathrm{Co}^{3+}$ is $\mathrm{d}^2 \mathrm{sp}^3$

In $\mathrm{SF}_6$, Hybridisation state of sulphur is $\mathrm{sp}^3 \mathrm{~d}^2$

In $\left[\mathrm{CrF}_6\right]^{3-}, \mathrm{Cr}^{+3}:[\mathrm{Ar}] 3 \mathrm{~d}^3$

Hybridisation state of $\mathrm{Cr}^{3+}$ is $\mathrm{d}^2 \mathrm{sp}^3$

$\left[\mathrm{CoF}_6\right]^{3-}, \mathrm{Co}^{+3}:[\mathrm{Ar}] 3 \mathrm{~d}^6 \mathrm{~F}^{-}$is W.F.L

Hybridisation state of $\mathrm{Co}^{3+}$ is $\mathrm{sp}^3 \mathrm{~d}^2$

$\left[\mathrm{Mn}(\mathrm{CN})_6\right]^{3-}, \mathrm{Mn}^{+3}:[\mathrm{Ar}] 3 \mathrm{~d}^4 \mathrm{CN}^{-}$is S.F.L

Hybridisation state of $\mathrm{Mn}^{3+}$ is $\mathrm{d}^2 \mathrm{sp}^3$

$\left[\mathrm{MnCl}_6\right]^{3-}, \mathrm{Mn}^{+3}:[\mathrm{Ar}] 3 \mathrm{~d}^4 \mathrm{Cl}^{-}$is W.F.L

Hybridisation state of $\mathrm{Cl}^{-}$is $\mathrm{sp}^3 \mathrm{~d}^2$

Total number of $\mathrm{sp}^3 \mathrm{~d}^2$ hybridized molecules is 3

Homoleptic complex of type $\left[\mathrm{Ma}_6\right]$ (Where $\mathrm{a} \Rightarrow$ monodentate ligand) cannot show geometrical as well as optical isomerism.

Cis-platin and trans-platin has formula $\left[\mathrm{Pt}\left(\mathrm{NH}_3\right)_2 \mathrm{Cl}_2\right]$ which is a heteroleptic complex of platinum.

Primary valency $=$ Oxidation state

Secondary valency $=$ Co-ordination number

The number of unpaired electrons responsible for the paramagnetic nature of each complex species is calculated as follows:

Complex: $[Fe(CN)_6]^{3-}$

Ion: $Fe^{3+}$

Electronic Configuration: $3d^5$

Orbital Population:

$t_{2g}^{2,2,1}$

$e_g^{0,0}$

Unpaired Electrons: 1

Complex: $[FeF_6]^{3-}$

Ion: $Fe^{3+}$

Electronic Configuration: $3d^5$

Orbital Population:

$t_{2g}^{1,1,1}$

$e_g^{1,1}$

Unpaired Electrons: 5

Complex: $[CoF_6]^{3-}$

Ion: $Co^{3+}$

Electronic Configuration: $3d^6$

Orbital Population:

$t_{2g}^{2,1,1}$

$e_g^{1,0}$

Unpaired Electrons: 4

Complex: $[Mn(CN)_6]^{3-}$

Ion: $Mn^{3+}$

Electronic Configuration: $3d^4$

Orbital Population:

$t_{2g}^{2,1,1}$

$e_g^{0,0}$

Unpaired Electrons: 2

Each complex's paramagnetic nature is due to the specific number of unpaired electrons as indicated above.

In the given compounds, we evaluate which are acidic oxides:

CrO₃ is acidic.

Mn₂O₇ is acidic.

Therefore, the number of acidic oxides, $ X $, is 2.

Next, for the complex $[Co(H₂NCH₂CH₂NH₂)₃]₂(SO₄)₃$, we consider its dissociation:

$ {[\mathop {Co}\limits^{III} {({H_2}NC{H_2}C{H_2}N{H_2})_3}]_2}{(SO_4)_3} \rightleftharpoons 2{[\mathop {Co}\limits^{III} {({H_2}NC{H_2}C{H_2}N{H_2})_3}]^{3+}} + 3SO_4^{2-} $

The primary valency of cobalt (Y) in this context is 3.

Thus, the value of $ X + Y $ is:

$ X + Y = 2 + 3 = 5 $

In this octahedral complex with the molecular composition $\mathrm{Co} \cdot 5 \mathrm{NH}_3 \cdot \mathrm{Cl}^2 \cdot \mathrm{SO}_4$, two isomers, A and B, are present.

Isomer A is represented as $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_5\left(\mathrm{SO}_4\right)\right] \mathrm{Cl}$. When this isomer is dissolved in solution, it reacts with $\mathrm{AgNO}_3$, forming a white precipitate of silver chloride ($\mathrm{AgCl}$).

Isomer B is written as $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_5 \mathrm{Cl}\right] \mathrm{SO}_4$. In solution, it reacts with $\mathrm{BaCl}_2$, forming a white precipitate of barium sulfate ($\mathrm{BaSO}_4$).

These two isomers demonstrate ionisation isomerism. This type of isomerism occurs when the exchange of ions inside and outside the coordination sphere of a complex leads to the formation of isomers with different chemical behaviors in solution.

Most stable is $\left[\mathrm{Fe}\left(\mathrm{C}_2 \mathrm{O}_4\right)_3\right]^{3-}$ due to Chelation effect.

$\mathrm{V}_2 \mathrm{O}_5$ is amphoteric.

$=[-0.4 \times 6+0.6 \times(0)] \Delta_0=-2.4 \Delta_0$

$\begin{aligned} \text{(2) }\quad & {\left[\mathrm{Mn}(\mathrm{SCN})_6\right]^4} \\ & \mathrm{Mn}^{2+} \Rightarrow 3 \mathrm{~d}^5 4 \mathrm{~s}^0 \end{aligned}$

$\begin{aligned} & \mu=\sqrt{35} \text { B.M. }=5.96 \text { B.M. } \\ & \text { CFSE }=(-0.4 \times 3+0.6 \times 2) \Delta_0 \\ & \text { So } \Delta_0=0 \end{aligned}$

(3) $\quad\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{-4} \quad \mathrm{Fe}^{2+} \Rightarrow 3 \mathrm{~d}^6 4 \mathrm{~s}^0$

$\begin{aligned} &\mathrm{CFSE}=-2.4 \Delta_0\\ &\text { (4) }\left[\mathrm{FeF}_6\right]^4 \quad \mathrm{Fe}^{2+} \Rightarrow 3 \mathrm{~d}^6 4 \mathrm{~s}^0 \end{aligned}$

$\begin{aligned} &\mu=\sqrt{24} \text { B.M. }=4.89 \text { B.M. }\\ &\text { CFSE }=(-0.4 \times 4+0.6 \times 2) \Delta_0=-1.2 \Delta_0 \end{aligned}$

Geometrical isomer $=2(1$ cis +1 trans $)$

Optical isomer $=3$ ( 2 optically active +1 optically inactive)

Stereoisomer $=3$

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 $\left[\mathrm{Cr}\left(\mathrm{NH}_3\right)_5 \mathrm{Cl}\right] \mathrm{Cl}_2$.