$$\begin{aligned} \text{As we know,}\quad\mu_m & =\sqrt{n(n+2} \mathrm{BM} \\ \text{where,}\quad\mu_m & =\text { magnetic moment } \\ \mu_n & =\text { number of unpaired electrons } \\ \text{It}\quad\mu_m & =1.74 \text { i.e., } n=1 \\ \text{and}\quad\mu_m & =5.92 \text { i.e., } n=5 \end{aligned}$$

$\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{3-}$ involves $d^2 s p^3$ hybridisation with one unpaired electron (as shown by its magnetic moment 1.74 BM ) and $\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+}$ involves $s p^3 d^2$ hybridisation with five unpaired electrons (because magnetic moment equal to 5.92 BM ). $\mathrm{CN}^{-}$is stronger ligand than $\mathrm{H}_2 \mathrm{O}$ according to spectrochemical series. $\Delta_0>P$ for $\mathrm{CN}^{-}$hence, fourth electron will pair itself. Whereas for water pairing will not happen for $\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{3-}$ the electronic configuration of $\mathrm{Fe}^{3+}$ is

One unpaired electron

For $\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+}$ the electronic configuration of $\mathrm{Fe}^{3+}$ is

Five unpaired electron Hence, $\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{3-}$ and $\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+}$ are inner orbital and outer orbital complex respectively.

CFSE is higher when the complex contains strong field ligand. Thus, crystal field splitting energy increases in the order

$$\left[\mathrm{Cr}(\mathrm{Cl})_6\right]^{3-}<\left[\mathrm{Cr}\left(\mathrm{NH}_3\right)_6\right]^{3+}<\left[\mathrm{Cr}(\mathrm{CN})_6\right]^{3-} .$$

Because according to spectrochemical series the order of field strength is

$$\mathrm{Cl}^{-}<\mathrm{NH}_3<\mathrm{CN}^{-}$$

It is due to the presence of weak and strong field ligands in complexes. If CFSE is high, the complex will show low value of magnetic moment and vice-versa, e.g. $\left[\mathrm{CoF}_6\right]^{3-}$ and $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]^{3+}$, the former is paramagnetic, and the latter is diamagnetic because $\mathrm{F}^{-}$is a weak field ligand and $\mathrm{NH}_3$ is a strong field ligand while both have similar geometry.

In $\mathrm{CuSO}_4 \cdot 5 \mathrm{H}_2 \mathrm{O}$, water acts as ligand and causes crystal field splitting. Hence, $d-d$ transition is possible thus $\mathrm{CuSO}_4 \cdot 5 \mathrm{H}_2 \mathrm{O}$ is coloured. In the anhydrous $\mathrm{CuSO}_4$ due to the absence of water (ligand), crystal field splitting is not possible and hence, it is colourless.

Ligand having more than one different binding position are known as ambidentate ligand. e.g., SCN has two different binding positions S and N . Coordination compound containing ambidentate ligands are considered to show linkage isomerism due to presence of two different binding positions.

e.g., (i) $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_5 \mathrm{SCN}\right]^{3+}$ and (ii) $\left[\mathrm{Fe}\left(\mathrm{NH}_3\right)_5\left(\mathrm{NO}_2\right)\right]^{3+}$