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)

$ \text { } \begin{aligned}(A)\,\, & \mathrm{NiCl}_4^{-2} \\ & \mathrm{Ni}^{2+}=4 \mathrm{~s}^0 3 \mathrm{~d}^8 \end{aligned} $

$ \text { Cl: weak field ligand, high spin complex. } $

$ \begin{aligned}(B) & \mathrm{Ni}\left(\mathrm{NH}_3\right)_6^{+2} \\ & \mathrm{Ni}^{+2} = 4 \mathrm{~s}^0 3 d^8 \end{aligned} $

$\mathrm{NH}_3$ : Weak field ligand, high spin complex.

$ \begin{aligned} &\text { }\\ &\begin{aligned}(C)\,\, & \mathrm{Ni}(\mathrm{CO})_4 \\ & \mathrm{Ni}^0=4 s^2 3 d^8 \xrightarrow[\text { d-system }]{\text { converted }} 4 s^0 3 d^{10} \end{aligned} \end{aligned} $

$ \begin{aligned}(D)\,\, & \mathrm{Ni}(\mathrm{CN})_4^{-2} \\ & \mathrm{Ni}^{+2}=4 \mathrm{~s}^0 3 d^8 \end{aligned} $

CN: strong field ligand, low spin complex

Enthalpy of atomisation is the energy needed to break $1$ mole of a metal (in solid state) into its separate atoms (in gaseous state).

In transition metals, metallic bonding becomes stronger when there are more unpaired electrons. This is because unpaired electrons can take part more effectively in metal–metal bonding.

So, if a transition metal atom has a large number of unpaired electrons, its metallic bonds are stronger, and more energy is required to convert the metal into atoms.

Therefore, more unpaired electrons $\propto$ stronger metallic bonding $\propto$ higher enthalpy of atomisation. Hence, Statement I is correct.

$\left[\mathrm{Fe}\left(\mathrm{H}_2\mathrm{O}\right)_6\right]^{3+}$ is an octahedral complex (coordination number $6$). In an octahedral field, the $d$-orbitals split as:

$\mathrm{d}_{x y}=\mathrm{d}_{x z}=\mathrm{d}_{y z}<\mathrm{d}_{x^2-y^2}=\mathrm{d}_{z^2}$

$\left[\mathrm{Ni}(\mathrm{Cl})_4\right]^{2-}$ is a tetrahedral complex (coordination number $4$). In a tetrahedral field, the splitting is reversed compared to octahedral, i.e.:

$\mathrm{d}_{x^2-y^2}=\mathrm{d}_{z^2}<\mathrm{d}_{x y}=\mathrm{d}_{x z}=\mathrm{d}_{y z}$

So, the given splittings match the correct orbital splitting patterns for these two geometries. Hence, Statement II is also correct.

(A) The stability of a complex increases when the ligand is stronger (i.e., stronger field/stronger bonding ligand gives more stable complex).

The ligand strength order given is:

$ \mathrm{SCN}^{-}<\mathrm{OH}^{-}<\mathrm{C}_2 \mathrm{OH}_4^{-2} $

Also, $\mathrm{C}_2 \mathrm{O}_4^{-2}$ (oxalate) is a bidentate ligand. Because of the chelate effect, complexes with bidentate ligands are generally more stable.

So, $\left[\mathrm{Fe}\left(\mathrm{C}_2 \mathrm{O}_4\right)_3\right]^{3-}$ is the most stable among the given complexes. Hence, statement (A) is true.

(B) en (ethylenediamine) is also a chelating (bidentate) ligand, while $\mathrm{NH}_3$ is monodentate (non-chelating).

Due to the chelate effect, $\left[\mathrm{Cu}(\mathrm{en})_2\right]^{2+}$ is more stable than $\left[\mathrm{Cu}\left(\mathrm{NH}_3\right)_4\right]^{2+}$.

So, statement (B) is false.

$ \begin{aligned}(c)\,\,\, & {\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{-4}} \\ & \mathrm{Fe}^{+2}=4 \mathrm{~s}^{\circ} \text { 3d }{ }^6 \end{aligned} $

       $C N^{-}=$ strong field ligand, so it forms a low spin complex.

Because $\mathrm{CN^-}$ is a strong field ligand, electrons pair up in the $3d$ orbitals. This leads to an inner orbital complex with $d^2sp^3$ hybridisation for Fe.

So, statement (C) is true.

(D) $\left[\mathrm{Fe}\left(\mathrm{NO}_2\right)_3 \mathrm{Cl}_3\right]^{-2}$ and $\left[\mathrm{Fe}(0 \mathrm{NO})_3 \mathrm{Cl}_3\right]^{-2}$

Linkage isomerism is shown when an ambidentate ligand can donate electrons through two different atoms.

Here, $\mathrm{NO_2^-}$ can coordinate through N (nitro) or through O (nitrito). Therefore, the complex shows linkage isomerism.

So, statement (D) is true.

(E) $\mathrm{NO}_2^{-}$ and $\mathrm{SCN}^{-}$ are ambidentate ligands (they can attach through different atoms).

$S$-donor $\leftarrow S=C=N^{\ominus} \rightarrow N$-donor

So, the statement “$\mathrm{NO}_2^{-}$ and $\mathrm{SCN}^{-}$ ligands are NOT ambidentate ligands” is incorrect.

Hence, statement (E) is false.

In metal carbonyls, bonding between the metal and CO ligand is explained by synergic bonding.

Statement A

The metal-carbon bond in metal carbonyls has both $\sigma$ and $\pi$ character.

This is correct.

• CO donates a lone pair from carbon to the metal, forming a $\sigma$ bond.

• The metal donates electron density back to CO into its vacant $\pi^*$ orbital, forming a $\pi$ bond.

So, the bond has both $\sigma$ and $\pi$ character.

Statement B

Due to synergic bonding interactions between metal and CO ligand, the metal-carbon bond becomes weak.

This is incorrect.

Synergic bonding actually strengthens the metal-carbon bond because:

• CO donates electrons to metal by $\sigma$ donation.

• Metal donates electrons back to CO by $\pi$ back bonding.

These two processes reinforce each other, so the M–C bond becomes stronger, not weaker.

Statement C

The metal-carbon $\sigma$ bond is formed by donation of lone pair of electrons on the carbonyl carbon into a vacant orbital of metal.

This is correct.

This is exactly the $\sigma$ donation from CO to the metal.

Statement D

The metal-carbon $\pi$ bond is formed by donation of electrons from filled $d$-orbital of metal into vacant $\pi^*$ orbital of CO.

This is correct.

This is called $\pi$ back bonding or back donation.

So the correct statements are:

$ A,\ C,\ \text{and}\ D $

Hence, the correct answer is:

$ \boxed{\text{Option B}} $

$ \text { Statement I : } $

In an octahedral field, the five $d$-orbitals split into two sets: lower-energy $t_{2g}$ and higher-energy $e_g$.

Each electron placed in the higher-energy $e_g$ orbitals increases (destabilizes) the energy by $+0.6\Delta_o$.

Each electron placed in the lower-energy $t_{2g}$ orbitals decreases (stabilizes) the energy by $-0.4\Delta_o$.

So, when electrons go into $e_g$, the crystal field stabilisation becomes less (energy goes up), and when electrons go into $t_{2g}$, the stabilisation becomes more (energy goes down).

Statement II : In free atomic state of transition metal, degeneracy of d-orbrital is maintained.

In the free (isolated) metal ion, all five $d$-orbitals have the same energy, so they are degenerate.

When ligands approach to form a complex, the metal–ligand interactions are not the same for all $d$-orbitals.

Because of this, the five $d$-orbitals split into two sets (like $t_{2g}$ and $e_g$ in octahedral complexes), and the degeneracy is removed.

For all the three complexes, the metal ion is $ \mathrm{Ni}^{2+} $.

The electronic configuration of Ni is

$ \mathrm{Ni} : [Ar]\,3d^8 4s^2 $

So,

$ \mathrm{Ni}^{2+} : [Ar]\,3d^8 $

Now let us examine each complex.

1. Complex $(A) = [\mathrm{Ni}(\mathrm{en})_3]^{2+}$

Here, $ \mathrm{en} $ is ethane-1,2-diamine, a neutral bidentate ligand.

• Coordination number $= 6$

• Geometry $= $ octahedral

For an octahedral $d^8$ configuration, the arrangement is:

$ t_{2g}^6 e_g^2 $

This gives 2 unpaired electrons.

So, for $(A)$:

$ \text{Number of unpaired electrons} = 2 $

2. Complex $(B) = [\mathrm{NiCl}_4]^{2-}$

• Coordination number $= 4$

• Since $ \mathrm{Cl}^- $ is a weak ligand, this complex is tetrahedral.

For tetrahedral $d^8$, the splitting is small, and the electrons remain high spin.

Hence, it has 2 unpaired electrons.

So, for $(B)$:

$ \text{Number of unpaired electrons} = 2 $

3. Complex $(C) = [\mathrm{Ni}(\mathrm{NH}_3)_6]^{2+}$

• Coordination number $= 6$

• Geometry $= $ octahedral

Again, for octahedral $d^8$:

$ t_{2g}^6 e_g^2 $

So it also has 2 unpaired electrons.

Thus, for $(C)$:

$ \text{Number of unpaired electrons} = 2 $

Frequency of absorption

The frequency of absorbed light depends on crystal field splitting:

$ \Delta \propto \nu $

Greater the splitting, greater the frequency of absorption.

Now compare ligand strengths:

$ \mathrm{en} > \mathrm{NH}_3 > \mathrm{Cl}^- $

Also, tetrahedral complexes have smaller splitting than octahedral complexes.

So,

• $[\mathrm{Ni}(\mathrm{en})_3]^{2+}$ has the highest $ \Delta $

• $[\mathrm{Ni}(\mathrm{NH}_3)_6]^{2+}$ has next

• $[\mathrm{NiCl}_4]^{2-}$ has the least

Therefore, order of frequency of absorption is:

$ (A) > (C) > (B) $

Hence the correct option is:

$ \boxed{\text{Option A}} $

For matching, we use the spin-only magnetic moment formula:

$ \mu = \sqrt{n(n+2)} \text{ BM} $

where $n$ = number of unpaired electrons.

Since $\mathrm{H_2O}$ is a weak field ligand, all these octahedral complexes will be high spin.

Now find oxidation state, electronic configuration, number of unpaired electrons, and then magnetic moment.

A. $\left[\mathrm{Cr}(\mathrm{H_2O})_6\right]^{2+}$

Here, chromium is in $+2$ oxidation state.

Atomic number of Cr = $24$

Neutral Cr:

$ \mathrm{Cr} = [\mathrm{Ar}]\,3d^5 4s^1 $

So,

$ \mathrm{Cr}^{2+} = 3d^4 $

For high spin $d^4$, number of unpaired electrons:

$ n=4 $

Thus,

$ \mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.90 \text{ BM} $

So, A $\to$ III

B. $\left[\mathrm{Co}(\mathrm{H_2O})_6\right]^{2+}$

Cobalt is in $+2$ oxidation state.

Atomic number of Co = $27$

$ \mathrm{Co}^{2+} = 3d^7 $

For high spin $d^7$, number of unpaired electrons:

$ n=3 $

Thus,

$ \mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87 \text{ BM} $

So, B $\to$ I

C. $\left[\mathrm{Cu}(\mathrm{H_2O})_6\right]^{2+}$

Copper is in $+2$ oxidation state.

Atomic number of Cu = $29$

$ \mathrm{Cu}^{2+} = 3d^9 $

For $d^9$, number of unpaired electrons:

$ n=1 $

Thus,

$ \mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73 \text{ BM} $

So, C $\to$ IV

D. $\left[\mathrm{Mn}(\mathrm{H_2O})_6\right]^{2+}$

Manganese is in $+2$ oxidation state.

Atomic number of Mn = $25$

$ \mathrm{Mn}^{2+} = 3d^5 $

For high spin $d^5$, number of unpaired electrons:

$ n=5 $

Thus,

$ \mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92 \text{ BM} $

So, D $\to$ II

Final matching

$ \boxed{\text{A-III, B-I, C-IV, D-II}} $

So the correct option is:

$ \boxed{\text{Option D}} $

Statement I

We need to compare the energy required to remove the third valence electron, that is, the third ionization enthalpy.

First write the electronic configurations:

• $Zn : [Ar]\,3d^{10}4s^2$

• $Mn : [Ar]\,3d^5 4s^2$

• $Sc : [Ar]\,3d^1 4s^2$

• $Cu : [Ar]\,3d^{10}4s^1$

Now, after removal of two electrons:

• $Zn^{2+} : [Ar]\,3d^{10}$

• $Mn^{2+} : [Ar]\,3d^5$

• $Sc^{2+} : [Ar]\,3d^1$

• $Cu^{2+} : [Ar]\,3d^9$

The third electron is now removed from these ions.

Which one will have highest third ionization enthalpy?

For $Zn^{2+}$, the configuration is $3d^{10}$, which is a completely filled and very stable configuration. Removing one electron from this is difficult.

So, third ionization enthalpy is highest for Zn.

Which one will have lowest third ionization enthalpy?

For $Sc^{2+}$, configuration is $3d^1$. Removing this one electron gives $Sc^{3+} : [Ar]$, which is a noble gas configuration, hence this removal is comparatively easier.

So, third ionization enthalpy is lowest for Sc.

Therefore, Statement I is true.

Statement II

We need the order of complexes in terms of CFSE.

Given complexes:

1. $[\text{Co(H}_2\text{O})_6]^{2+}$

2. $[\text{Co(H}_2\text{O})_6]^{3+}$

3. $[\text{Co(en)}_3]^{3+}$

Step 1: Find oxidation state and $d$-configuration

• In $[\text{Co(H}_2\text{O})_6]^{2+}$, cobalt is $Co^{2+}$

  $Co: [Ar]\,3d^7 4s^2$

  $Co^{2+}: 3d^7$

• In $[\text{Co(H}_2\text{O})_6]^{3+}$, cobalt is $Co^{3+}$

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

• In $[\text{Co(en)}_3]^{3+}$, cobalt is also $Co^{3+}$

  so it is also $3d^6$

Step 2: Nature of ligand and spin state

• $H_2O$ is a weak field ligand

• $en$ is a stronger field ligand than $H_2O$

So:

• $[\text{Co(H}_2\text{O})_6]^{2+}$ is octahedral, weak field, $d^7$

  generally high spin

• $[\text{Co(H}_2\text{O})_6]^{3+}$ is octahedral, $d^6$

  because of higher charge on $Co^{3+}$, splitting is larger; this complex is effectively low spin

• $[\text{Co(en)}_3]^{3+}$ is octahedral, $d^6$, strong field

  definitely low spin

Step 3: Calculate CFSE

1. $[\text{Co(H}_2\text{O})_6]^{2+}$ : high-spin $d^7$

Configuration: $t_{2g}^5 e_g^2$

$ CFSE = 5(-0.4\Delta_o) + 2(+0.6\Delta_o) = -2.0\Delta_o + 1.2\Delta_o = -0.8\Delta_o $

Magnitude of CFSE $= 0.8\Delta_o$

2. $[\text{Co(H}_2\text{O})_6]^{3+}$ : low-spin $d^6$

Configuration: $t_{2g}^6 e_g^0$

$ CFSE = 6(-0.4\Delta_o) = -2.4\Delta_o $

Magnitude $= 2.4\Delta_o$

3. $[\text{Co(en)}_3]^{3+}$ : low-spin $d^6$

Also $t_{2g}^6 e_g^0$, so

$ CFSE = -2.4\Delta_o $

But here $en$ is stronger field than $H_2O$, so $\Delta_o$ is larger.

Hence CFSE magnitude is greater for $[\text{Co(en)}_3]^{3+}$ than for $[\text{Co(H}_2\text{O})_6]^{3+}$.

So the order is:

$ [\text{Co(H}_2\text{O})_6]^{2+} < [\text{Co(H}_2\text{O})_6]^{3+} < [\text{Co(en)}_3]^{3+} $

Therefore, Statement II is also true.

Final Answer

$ \boxed{\text{Option A: Both Statement I and Statement II are true}} $

Coordination Isomerism arises from the interchange of ligands between cationic and anionic entities of different metal ions present in a complex. When the complete interchange of ligands takes place, it must result in a new, chemically distinct isomer.

Let us evaluate each of the given choices under this concept:

• A. $[\mathrm{Ag(NH}_3)_2][\mathrm{Ag(CN)}_2]$: Both the cationic and anionic entities contain the same metal ion ($ \text{Ag} $). If we fully interchange the ligands, we get $[\mathrm{Ag(CN)}_2]^- [\mathrm{Ag(NH}_3)_2]^+$, which is exactly the same compound written in reverse order. Further, a partial interchange leads to neutral molecules $[\mathrm{Ag(NH}_3)(\mathrm{CN})]$, which destroys the cation-anion pair. Hence, it does not show coordination isomerism.

• B. $[\mathrm{Co(NH}_3)_6][\mathrm{Cr(CN)}_6]$: This complex contains two different metal ions ($ \text{Co} $ and $ \text{Cr} $). By interchanging the ligands between the cationic and anionic coordination spheres, we get a completely new distinct complex: $[\mathrm{Cr(NH}_3)_6][\mathrm{Co(CN)}_6]$. Hence, it shows coordination isomerism. (This is also the classic example mentioned in NCERT).

• C. $[\mathrm{Co(NH}_3)_6][\mathrm{Co(CN)}_6]$: Both entities contain the same metal ion ($ \text{Co} $). A complete interchange of all ligands gives $[\mathrm{Co(CN)}_6]^{3-} [\mathrm{Co(NH}_3)_6]^{3+}$, which is exactly the same coordination compound. According to the strict NCERT definition requiring different metal ions for complete interchange, it does not exhibit coordination isomerism.

• D. $[\mathrm{Fe(NH}_3)_6][\mathrm{Co(CN)}_6]$: This complex has two different metal ions ($ \text{Fe} $ and $ \text{Co} $). Exchanging the primary coordination spheres yields a distinct coordination isomer: $[\mathrm{Co(NH}_3)_6][\mathrm{Fe(CN)}_6]$. Hence, it shows coordination isomerism.

• E. $[\mathrm{Co(NH}_3)_6][\mathrm{Fe(CN)}_6]$: This complex is exactly the coordination isomer of the complex described in option D. Since it consists of different metal ions ($ \text{Co} $ and $ \text{Fe} $), exchanging its ligands will yield the complex in D. Thus, it also shows coordination isomerism.

Conclusion:

Complexes B, D, and E exhibit coordination isomerism because they possess differently named metal ions in their cationic and anionic parts, allowing the exchange of ligands to produce distinctly new isomers.

Correct Answer:

Option B (B, D and E Only)

The crystal field splitting energy (CFSE), represented by Δ₀, depends on how strong the ligands are and also on chelation (when a ligand forms a ring with the metal ion).

The stronger the ligand, the higher the value of Δ₀. Chelating ligands like ethylenediamine (en) also increase stability and splitting energy.

Among the given ligands, their strength follows this order:

$ \mathrm{CN}^{-} > \text{en} > \mathrm{H_2O} > \mathrm{F}^{-} $

This order means complexes with CN⁻ have the largest Δ₀, and those with F⁻ have the smallest Δ₀.

(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 :-

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