Match the following
 List – I (complex)                                        List – II (spin only magnetic moment)
a. ${\left[Co{F}_6\right]}^{-3}$                                                  i. 0 B.M
b.  ${\left[Co{\left(C_2{O}_4\right)}_3\right]}^{-3}$                                     ii.   B.M
c.   ${\left[Fe{F}_6\right]}^{-3}$                                                iii.   B.M
d.    ${\left[\mathrm{Mn}{\left(\mathrm{CN}\right)}_6\right]}^{-3}$                                        iv.   B.M
                                                                         v.   B.M
The correct answer is
 

$$\begin{gathered} \left(A\right) {\left[Co{F}_6\right]}^{-3} oxidation state of Co\Rightarrow +3 \\ {27}^{Co\Rightarrow \left[Ar\right]3d^{7}4s^2} C{o}^{+3}\Rightarrow \left[Ar\right]3d^{6}4s^0\Rightarrow t_{2g},e_g \end{gathered}$$

  $\therefore$ (F) is a weak field ligand, hence electrons will not pair up. So, unpaired electrons,  $n=4$

Magnetic moment  $\mu =\sqrt{n(n+2)}BM=\sqrt{4(4+2)}$   $\mu =\sqrt{4\left(6\right)}=\sqrt{24}BM$

(B) ${\left[Co{\left(C_2{O}_4\right)}_3\right]}^{-3}$  similarly, oxidation state of  $Co=+3$

Oxalate is a strong filed ligand, hence electrons will pair up.  $C{o}^{+3}\Rightarrow \left[Ar\right]3d^6\Rightarrow t_{2g}^6,e_g^0$

$\therefore n=0,\mu =0BM$    

(C) ${\left[Fe{F}_6\right]}^{-3},Fe\Rightarrow \left[Ar\right]3d^{6}4s^2$   Oxidation state of Fe in the above complex = +3

(F) is a weak – field ligand, So electrons will not pair up.

$\therefore n=5$   $\mu =\sqrt{5(5+2)}=\sqrt{35}B.M$   

(D)   ${\left[Mn{\left(CN\right)}_6\right]}^{-3},Mn\Rightarrow \left[Ar\right]3d^{5}4s^2$    Oxidation state of Mn in the above complex = +3

(CN) being a strong field ligand, will pair up electrons.

$$\begin{gathered} \therefore n=2 \\ \mu =\sqrt{2(2+2)}=\sqrt{2\left(4\right)}=\sqrt{8}B.M \end{gathered}$$