The correct structure of the complex \([Ni(CN)_4]^{2-}\) and its magnetic property are
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A very important exception in coordination chemistry is:
\[
[Ni(CN)_4]^{2-}
\]
which is a square planar, diamagnetic complex due to the strong-field nature of \(CN^{-}\).
In contrast,
\[
[NiCl_4]^{2-}
\]
contains the weak-field ligand \(Cl^{-}\), adopts tetrahedral geometry and is paramagnetic.
Step 1: Find the metal oxidation state. In $[Ni(CN)_4]^{2-}$ each $CN^-$ carries $-1$. Four of them give $-4$. The whole ion is $-2$, so nickel must be $+2$.
Step 2: Write the electron count of Ni$^{2+}$. Nickel is $[Ar]3d^8 4s^2$. Removing two electrons gives Ni$^{2+}$ as $3d^8$. So there are 8 electrons in the d orbitals.
Step 3: Note the ligand type. $CN^-$ is a strong field ligand. It causes large splitting and forces electrons to pair up.
Step 4: Pair up the d electrons. In $d^8$ with a strong ligand, the electrons rearrange so that one d orbital is emptied. This frees a d orbital for bonding and leaves no unpaired electrons.
Step 5: Decide the shape. With one $d$ orbital free, nickel uses $dsp^2$ mixing. This gives a square planar shape, not tetrahedral.
Step 6: Decide the magnetism. Since all electrons are paired, the complex has no unpaired electrons, so it is diamagnetic.
Step 7: State the final answer. The structure and property are: \[ \boxed{\text{Square planar, Diamagnetic}} \]
