Question

Which of the following are paramagnetic tetrahedral species?

• A. ${\left[FeC{l}_4\right]}^{-}$
• B. ${\left[Ni(CN{)}_4\right]}^{2-}$
• C. ${\left[NiC{l}_4\right]}^{2-}$
• D. ${\left[Zn{\left(N{H}_3\right)}_4\right]}^{2+}$

Answer 1

Answer: (A), (C)

${\left[FeC{l}_4\right]}^{-}$
$F{e}^{3+}3d^{5}4s^0.C{l}^{-}$is a weak ligand. Hence $s{p}^3$ hybridization occurs. This gives tetrahedral geometry. It is paramagnetic having 5 unpaired electrons.
${\left[Ni(CN{)}_4\right]}^{2-}$
$N{i}^{2+}3d^{8}4s^0.C{N}^{-}$is a strong ligand. Electron pairing takes place and $ds{p}^2$ hybridization occurs. It is diamagnetic having square planar geometry and has no unpaired electron.
${\left[NiC{l}_4\right]}^{2-}$
$N{i}^{2+}3d^{8}4s^0.C{l}^{-}$is a weak ligand. $s{p}^3$ hybridization occurs. It has tetrahedral geometry and is paramagnetic as it has 2 unpaired electrons.
${\left[Zn{\left(N{H}_3\right)}_4\right]}^{2+}$
$Z{n}^{2+}3d^{10}4s^0.s{p}^3$ hybridization occurs. The complex is tetrahedral but is diamagnetic as it has no unpaired electron.