Question

Which of the following complexes are diamagnetic?
$\left[Pt\right(N{H}_3{)}_4{]}^{2+}\left[Co\right(SCN{)}_4{]}^{2-}\left[Cu\right(en{)}_2{]}^{2+}[Hg{I}_4{]}^{2-}$
Square planar tetrahedral square planar tetrahedral
(i) (ii) (iii) (iv)

• A. (i) and (ii)
• B. (ii) and (iii)
• C. (i) and (iv)
• D. (iii) and (iv)

Answer 1

The correct option is C (i) and (iv)

The square planar $\left[Pt\right(N{H}_3{)}_4{]}^{2+}$ complex has Pt(II) metal with $5d^8$ configuration. In this all electrons are paired. Hence it is a diamagnetic complex.
The tetrahedral $[Hg{I}_4{]}^2$ complex has Hg(II) metal with $5d^{10}$ configuration.In this all electrons are paired. Hence it is a diamagnetic complex.
The tetrahedral $\left[Co\right(SCN{)}_4{]}^{2-}$ and square planar $\left[Cu\right(en{)}_2{]}^{2+}$ complexes have Co(II) and Cu(II) metal ions with $3d^7$ and $3d^9$
configuration respectively. They contain three and one unpaired electrons respectively. Hence they are paramagnetic.