Explanation

(d)
$\left\lbrack Sc(CN)_{6} \right\rbrack^{3 -}$
Sc=21=$1s^{2},\ 2s^{2},\ 2p^{6},\ 3s^{2},\ 3p^{6},\ 4s^{2},\ 3d^{1}$
3$d$ $4s$ $4d$
[table] | | | | | | | | | | [/table]
$Sc^{3 +} =$
No unpaired electron
$Ni(CN)_{4}^{2 -}$
Ni(G. S.) 3$d$ 4$s$ 4$p$
[table] ⥮ | ⥮ | ⥮ | $\upharpoonleft$ | $\upharpoonleft$ | | ⥮ | | | | [/table]
$Ni^{2 +}$ 3$d$ 4$s$ 4$p$
[table] ⥮ | ⥮ | ⥮ | $\upharpoonleft$ | $\upharpoonleft$ | | | | | | [/table]
[image:9_image147.emf]
It has no unpaired electron hence, diamagnetic.
$\left\lbrack Co(CN)_{6} \right\rbrack^{3 -}$
Co (G.S.) 3$d$ 4$s$ 4$p$
[table] ⥮ | ⥮ | $\upharpoonleft$ | $\upharpoonleft$ | $\upharpoonleft$ | | ⥮ | | | | [/table]
$Co^{3 +}$ 3$d$ 4$s$ 4$p$
[table] ⥮ | $\upharpoonleft$ | $\upharpoonleft$ | $\upharpoonleft$ | $\upharpoonleft$ | | | | | | [/table]
$Co^{3 +}$ 3$d$ 4$s$ 4$p$
[table] ⥮ | ⥮ | ⥮ | | | | | | | | [/table]
(Pairing
due to $CN^{-}$)
$\left\lbrack Cr(CN)_{6} \right\rbrack^{3 -}$
Cr (G.S.) 3$d$ 4$s$ 4$p$
[table] $\upharpoonleft$ | $\upharpoonleft$ | $\upharpoonleft$ | $\upharpoonleft$ | $\upharpoonleft$ | | $\upharpoonleft$ | | | | [/table]
$Cr^{3 +}$ 3$d$ 4$s$ 4$p$
[table] $\upharpoonleft$ | $\upharpoonleft$ | $\upharpoonleft$ | | | | $\upharpoonleft$ | | | | [/table]
$Cr^{3 +}$ 3$d$ 4$s$ 4$p$
[table] ⥮ | $\upharpoonleft$ | | | | | | | | | [/table]
(due to $CN^{-}$)
It has one unpaired electron so, paramagnetic.