Question

\( [NiCl_4]^{2-} \) is paramagnetic, while \( [Ni(CO)_4] \) is diamagnetic although both are tetrahedral. Why?

Hint:

Find Ni oxidation state in each, then use the spectrochemical series: weak \( Cl^- \) leaves unpaired electrons, strong CO pairs them all up.

Answer 1

Step 1: Oxidation state and d-configuration of Ni in each complex.
In \( [NiCl_4]^{2-} \): each \( Cl^- \) is \(-1\), overall charge \(-2\), so Ni is \(+2\). \( Ni^{2+} = 3d^{8} \).
In \( [Ni(CO)_4] \): CO is neutral, so Ni is in the \(0\) oxidation state. \( Ni(0) = 3d^{8}4s^{2} \).

Step 2: Nature of the ligand (spectrochemical series).
\( Cl^- \) is a weak field (high spin) ligand; it cannot pair up the d-electrons.
CO is a strong field ligand; it forces electrons to pair up.

Step 3: \( [NiCl_4]^{2-} \).
\( Ni^{2+} (3d^8) \) with weak \( Cl^- \) keeps 2 unpaired electrons. Hybridisation is \( sp^3 \) (tetrahedral). Since there are unpaired electrons, the complex is paramagnetic \( (\mu = \sqrt{2(2+2)} = \sqrt{8} \approx 2.83 \text{ BM}) \).

Step 4: \( [Ni(CO)_4] \).
Strong CO forces the two \(4s\) electrons into the \(3d\) sub-shell, giving \( 3d^{10}4s^0 \) with no unpaired electrons. The empty \(4s\) and \(4p\) orbitals form \( sp^3 \) hybrids (tetrahedral). With no unpaired electrons the complex is diamagnetic.

Conclusion: Both are \( sp^3 \) tetrahedral, but the ligand field decides the magnetism: weak \( Cl^- \) leaves 2 unpaired electrons (paramagnetic) whereas strong CO pairs all electrons (diamagnetic).
\[\boxed{Cl^-:\ 2\ \text{unpaired (para)};\quad CO:\ 0\ \text{unpaired (dia)}}\]