Question

The number of unpaired electrons in the square planar $[Pt(CN{)}_4{]}^{2-}$ ion is

• A. 2
• B. 1
• C. 0
• D. 3

Answer 1

Answer: (C)

The electronic configuration of
$Pt=[Xe]4f^{14},5d^9,6s^1$
$\therefore$ $P{t}^{2+}=[Xe]4f^{14},5d^8,6s^0$
${\left[Pt(CN{)}_4\right]}^{2-}=[Xe]4f^{14}$
$\therefore$ No unpaired electron is present in ${\left[Pt(CN{)}_4\right]}^{2-}$ ion.