$(1)$ $C r^{3+} \rightarrow 3 d^1 4 s^2$ 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.
$(a)$ $\left[\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{3} \mathrm{Cl}\right]^{+}$
$(b)$ $\left[\mathrm{Pt}\left(\mathrm{NH}_{3}\right) \mathrm{Cl}_{5}\right]^{-}$
$(c)$ $\left[\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{2} \mathrm{Cl}\left(\mathrm{NO}_{2}\right)\right]$
$(d)$ $\left[\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{ClBr}\right]^{2+}$
$(I)$ $\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right] \mathrm{Br}_{2}$
$(II)$ $ \mathrm{Na}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]$
$(III)$ $\mathrm{Na}_{3}\left[\mathrm{Fe}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{3}\right]\left(\Delta_{0}>\mathrm{P}\right)$
$(IV)$ $\left(\mathrm{Et}_{4} \mathrm{N}\right)_{2}\left[\mathrm{CoCl}_{4}\right]$