Q. The number of unpaired electrons in the square planar $[Pt(CN)_{4}]^{2-}$ ion is
Chhattisgarh PMTChhattisgarh PMT 2011
Solution:
The electronic configuration of
$P t=[X e] 4 f^{14}, 5 d^{9}, 6 s^{1}$
$\therefore $ $P t^{2+}=[X e] 4 f^{14}, 5 d^{8}, 6 s^{0}$
$\left[P t(C N)_{4}\right]^{2-}=[X e] 4 f^{14}$
$\therefore $ No unpaired electron is present in $\left[Pt(CN)_{4}\right]^{2-}$ ion.
