Step 1: Fix the metal's electron count.
Chloride carries \(-1\) each, so in \([NiCl_4]^{2-}\) nickel is \(Ni^{2+}\) (\(3d^8\)). Carbonyl is neutral, so in \([Ni(CO)_4]\) nickel is \(Ni^0\) (\(3d^8 4s^2\)).
Step 2: Judge ligand strength.
\(Cl^-\) sits low in the spectrochemical series (weak field), whereas CO sits at the top (strong field). Only a strong field ligand supplies enough energy to force electron pairing.
Step 3: Assign structures.
For \([NiCl_4]^{2-}\) the weak \(Cl^-\) leaves the \(3d^8\) set with two unpaired electrons; \(sp^3\) hybridisation gives a tetrahedron, and the unpaired spins make it paramagnetic.
For \([Ni(CO)_4]\) the strong CO drives the \(4s^2\) pair into \(3d\), producing a filled \(3d^{10}\) with zero unpaired electrons; the vacant \(4s\) and \(4p\) orbitals hybridise to \(sp^3\), giving a diamagnetic tetrahedron.
Conclusion: Same \(sp^3\) tetrahedral shape, but ligand field strength (not geometry) controls the number of unpaired electrons and hence the magnetic behaviour.
\[\boxed{\text{Weak } Cl^- \to \text{para};\ \text{Strong CO} \to \text{dia}}\]