Explanation

(c)
Diamagnetic substances have all paired electron.
$\left\lbrack Fe(CN)_{6} \right\rbrack^{3 -}$ Oxidation state of Fe=+3
$Fe^{3 +}$ 3$d$ 4$s$ 4$p$
[table] $\upharpoonleft$ | $\upharpoonleft$ | $\upharpoonleft$ | $\upharpoonleft$ | $\upharpoonleft$ | | | | | | [/table]
[image:9_image184.emf]
It has one unpaired electron and is paramagnetic.
$\left\lbrack NiCl_{4} \right\rbrack^{2 -}$ Oxidation state of Ni=+2
[image:9_image185.emf]
It has two unpaired electrons and is paramagnetic
$Ni(CO)_{4}$ Oxidation state of Ni=0
Ni (G.S.) 3$d$ 4$s$ 4$p$
[table] ⥮ | ⥮ | ⥮ | $\upharpoonleft$ | $\upharpoonleft$ | | | | | | [/table]
[image:9_image186.emf]
It has no unpaired electron and is diamagnetic
$\left\lbrack MnCl_{4} \right\rbrack^{2 -}\ Oxidation\ state\ of\ Mn = + 2$
[image:9_image187.emf]
$\therefore$ It is paramagnetic as it has five unpaired electrons.