Q. The number of unpaired electrons in the square planar $\left\lbrack Pt(CN)_{4} \right\rbrack^{2 -}$ ion is
A) 2
B) 1
C) 0
D) 3
Explanation
(c)
The electronic configuration Pt=[Xe] $4\ f^{14},\ 5d^{9},\ 6s^{1}$
$\therefore Pt^{2 +} = \lbrack Xe\rbrack 4f^{14},\ 5d^{8},\ 6s^{0}$
$\left\lbrack Pt(CN)_{4} \right\rbrack^{2 -} = \lbrack Xe\rbrack 4f^{14}$
[image:9_image268.emf]
$\therefore$ No unpaired electron is present in $\left\lbrack Pt(CN)_{4} \right\rbrack^{2 -}$ ion.
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.