The magnetic moment 5.92 BM shows that there are five unpaired electrons present in the $d$-orbitals of $\mathrm{Mn}^{2+}$ ion. As a result, the hybridisation involved is $s p^3$ rather than $d s p^2$. Thus tetrahedral structure of $\left[\mathrm{MnCl}_4\right]^{2-}$ complex will show 5.92 BM magnetic moment value.

With weak field ligands; $\Delta_0< P$, (pairing energy) so, the electronic configuration of Co (III) will be $t_{2 g}^4 e_g^2$ i.e., it has 4 unpaired electrons and is paramagnetic.

With strong field ligands, $\Delta_0>P$ (pairing energy), so pairing occurs thus, the electronic configuration will be $t_{2 g}^6 e_g^0$. It has no unpaired electrons and is diamagnetic.

In tetrahedral complex, the $d$-orbital is splitting to small as compared to octahedral. For same metal and same ligand $\Delta_t=\frac{4}{9} \Delta_0$.

Hence, the orbital splitting energies are not enough to force pairing. As a result, low spin configurations are rarely observed in tetrahedral complexes.

According to spectrochemical series, ligands can be arranged in a series in the order of increasing field strength i.e., $\mathrm{F}^{-}<\mathrm{NH}_3<\mathrm{CN}^{-}$.

Hence, $\mathrm{CN}^{-}$and $\mathrm{NH}_3$ being strong field ligand pair up the $t_{2 g}$ electrons before filling $e_g$ set. $\left[\mathrm{CoF}_6\right]^{3-} ; \mathrm{Co}^{3+}=\left(d^6\right) t_{2 g}^4 e_g^2$

$\left[\mathrm{Fe}(\mathrm{CN})_6^{4-}, \mathrm{Fe}^{2+}=\left(\mathrm{d}^6\right) t_{2 g}^6 e_g^0\right.$

$\left[\mathrm{Cu}\left(\mathrm{NH}_3\right)_6\right]^{2+}, \mathrm{Cu}^{2+}=\left(d^9\right) t_{3 g}^6 e_g^3$

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

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