Question

What is(are) number(s) of unpaired electrons in the square planar, ${\left[Pt(CN{)}_4\right]}^{2-}$ ion ?

• A. 0
• B. 1
• C. 4
• D. 6

Answer 1

Answer: (A)

In ${\left[Pt(CN{)}_4\right]}^{2-},C{N}^{-}$is a strong field ligand and outer configuration of Pt is $5d^9,6s^1$.
Thus, outer configuration of $P{t}^{2+}$ is $5d^8$.
$C{N}^{-}$will cause pairing of electrons. Hence,

Therefore, number of unpaired electrons in ${\left[Pt(CN{)}_4\right]}^{2-}$ is zero.