(a) $\left[\mathrm{CoF}_6\right]^{3-}$.

$\mathrm{F}^{-}$is a weak field ligand.

Configuration of $\mathrm{Co}^{3+}=3 d^6$ (or $t_{2 g}^4 e_g^2$)

Number of unpaired electrons $(n)=4$

Magnetic moment $(\mu)=\sqrt{n(n+2)}=\sqrt{4(4+2)}=\sqrt{24}=4.9 \mathrm{BM}$

$\left[\mathrm{Co}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}$, $\mathrm{H}_2 \mathrm{O}$ is a weak field ligand.

Configuration of $\mathrm{Co}^{2+}=3 d^7\left(\right.$ or $\left.t_{2 g}^5 e_g^2\right)$

Number of unpaired electrons $(n)=3$

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

$\left[\mathrm{Co}(\mathrm{CN})_6\right]^{3-}$ i.e., $\mathrm{Co}^{3+}$

$\because \quad \mathrm{CN}$ is strong field ligand.

$$\mathrm{Co}^{3+}=3 \mathrm{~d}^6\left(\mathrm{or} t_{2 g}^6 \mathrm{e}_g^0\right)$$

There is no unpaired electron, so it is diamagnetic.

$$\mu=0$$

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

$\mathrm{Fe}^{3+}=3 d^5\left(\right.$ or $\left.t_{2 g}^3 e_g^2\right)$

Number of unpaired electrons, $n=5$

$$\begin{aligned} \mu & =\sqrt{5(5+2)} \\ & =\sqrt{35}=5.92 \mathrm{BM} \\ {\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+} } & \end{aligned}$$

$$\mathrm{Fe}^{2+}=3 d^6\left(\operatorname{or} t_{2 g}^4 e_g^2\right)$$

Number of unpaired electrons, $n=4$

$$\begin{aligned} \mu & =\sqrt{4(4+2)} \\ Z & =\sqrt{24} \\ & =4.98 \mathrm{BM} \\ {\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{4-} } & \end{aligned}$$

Since, $\mathrm{CN}^{-}$is a strong field ligand, all the electrons get paired.

$$\mathrm{Fe}^{2+}=3 d^6\left(\operatorname{or} t_{2 g}^6 e_g^0\right)$$

Because there is no unpaired electron, so it is diamagnetic in nature.

(a) $\left[\mathrm{Mn}(\mathrm{CN})_6\right]^{3-}$

(i) $d^2 s p^3$ hybridisation

(ii) Inner orbital complex because $(n-1) d$-orbitals are used.

(iii) Paramagnetic, as two unpaired electrons are present.

(iv) Spin only magnetic moment $(\mu)=\sqrt{2(2+2)}=\sqrt{8}=2.82 \mathrm{BM}$

$$\begin{aligned} & \text{(b) }\quad{\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]^{3+}} \\ & \mathrm{Co}^{3+}=3 d^6 4 s^0 \end{aligned}$$

( $\mathrm{NH}_3$ pair up the unpaired $3 d$ electrons.)

(i) $d^2 s p^3$ hybridisation

(ii) Inner orbital complex because of the involvement of $(n-1) d$-orbital in bonding.

(iii) Diamagnetic, as no unpaired electron is present.

(iv) $\mu=\sqrt{n(n+2)}=\sqrt{0(0+2)}=0$ (Zero)

(c) $\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+}$

(i) $d^2 s p^3$ hybridisation

(ii) Inner orbital complex (as $(n-1) d$-orbital take part.)

(iii) Paramagnetic (as three unpaired electrons are present.)

(iv) $\mu=\sqrt{n(n+2)}=\sqrt{3(3+2)}=\sqrt{15}=3.87 \mathrm{BM}$

$\begin{aligned} & \text { (d) }\left[\mathrm{Fe}(\mathrm{Cl})_6\right]^{4-} \\ & \mathrm{Fe}^{2+}=3 d^6\end{aligned}$

(i) $s p^3 d^2$ hybridisation

(ii) Outer orbital complex because $n d$-orbitals are involved in hybridisation.

(iii) Paramagnetic (because of the presence of four unpaired electrons).

(iv) $\mu=\sqrt{n(n+2)}=\sqrt{4(4+2)}=\sqrt{24}=4.9 \mathrm{BM}$

' $A$ ' gives precipitate with $\mathrm{AgNO}_3$, so in it Cl is present outside the coordination sphere. ' $B$ ' gives precipitate with $\mathrm{BaCl}_2$, so in it $\mathrm{SO}_4^{2-}$ is present outside the coordination sphere.

(a) $\mathrm{So}, \mathrm{A}-\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_5 \mathrm{SO}_4\right] \mathrm{Cl}$

$B-\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_5 \mathrm{Cl}\right] \mathrm{SO}_4$

(b) Ionisation isomerism (as give different ions when subjected to ionisation.)

(c) $[A]$, Pentaamminesulphatocobalt (III) chloride.

[B], Pentaamminechloridocobalt (III) sulphate.

When white light falls on the complex, some part of it is absorbed. Higher the crystal field splitting energy, lower will be the wavelength absorbed by the complex. The observed colour of complex is the colour generated from the wavelength left over. e.g., if green light is absorbed, the complex appears red.

In terms of crystal field theory, suppose there is an octahedral complex with empty $e_g$ level and unpaired electrons in the $t_{2 g}$ level in ground level. If the unpaired electron absorbs light corresponding is blue-green region, it will excite to $e_g$ level and the complex will appear violet in colour.

In absence of ligand, crystal field splitting does not occur and the substance is colourless. e.g., anhydrous $\mathrm{CuSO}_4$ is while, but $\mathrm{CuSO}_4 \cdot 5 \mathrm{H}_2 \mathrm{O}$ is blue in colour.

Extent of splitting of $d$-orbitals is different in octahedral and tetrahedral field. CFSE in octahedral and tetra federal field are closely related as.

$$\Delta_t=\left(\frac{4}{9}\right) \Delta_o$$

where, $\quad \Delta_t=$ crystal field splitting energy in tetrahedral field

$\Delta_0=$ crystal field splitting energy in octahedral field

Wavelength of light and CFSE are related to each other by formula

$$\begin{aligned} \Delta_0=E & =\frac{h c}{\lambda} \\ E & \propto \frac{1}{\lambda} \end{aligned}$$

So, higher wavelength of light is absorbed in octahedral complexes than tetrahedral complexes for same metal and ligands. Thus, different colours are observed.