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

(i) $d^{2}s{p}^3$ hybridisation 

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

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

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

(b) ${\left[Co{\left(N{H}_3\right)}_6\right]}^{3+}$
$C{o}^{3+}=3d^{6}4s^0$

(NH${}^3$ pair up the unpaired 3d electons)

(i) $d^{2}s{p}^3$ hybridisation 

(ii) lnner orbital complex because of the invovement of *n-1) d-orbital in bonding.

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

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

(c) ${\left[Cr(H_2{O}_6\right]}^{3+}$
$$

(i) $d^{2}s{p}^3$ hybridisation 

(ii) lnner orbital complex (as(n-1) d-orbital take part)

(iii) Paramagnetic (as three unpaired electrons are present 

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

(d) ${\left[Fe{\left(Cl\right)}_6\right]}^{4-}$
$F{e}^{2+}=3d^6$

(i) $s{p}^3{d}^2$ hybridisation

(ii) Outer orbital complex because nd-orbitals are involved in hybridisation.

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

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