$(1)$ $C r^{3+} \rightarrow 3 d^1 4 s^2$ no. of unpaired $e^{-}=1$
$(2)$ $M n^{3+} \rightarrow 3 d^3 4 s^2$ no. of unpaired $e^{-}=1$
$(3)$ $F e^{3+} \rightarrow 3 d^3 4 s^2 \quad$ no. of unpaired $e^{-}=1$
$(4)$ $Co ^{3+} \rightarrow 3 d^6 4 s^2$ no. of unpaired $e^{-}=0$
So $C o$ in $\left[\operatorname{Co}(C N)_6\right]^{3-}$ is having less number of unpaired electrons
comparatively i.e 0 . So it is a diamagnetic compound means less paramagnetic comparatively.
$\left[\mathrm{V}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+},\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+},\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+},\left[\mathrm{Ni}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+},\left[\mathrm{Cu}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}$
[ આપેલ પરમાણુ ક્રમાંક : $\mathrm{V}=23, \mathrm{Cr}=24, \mathrm{Fe}=26, \mathrm{Ni}=28 \mathrm{Cu}=29$ ]
$(A)$ $\left. Na _{4}\left[ Fe ( CN )_{5} NOS \right)\right]$
$(B)$ $Na _{4}\left[ FeO _{4}\right]$
$(C)$ $\left[ Fe _{2}( CO )_{9}\right]$