JEE Main · 2006 · 30 Apr 2006 · Offline · Coordination Compounds

Nickel is $[\mathrm{Ar}]3d^{8}4s^2$, so $\mathrm{N}{\mathrm{i}}^{2+}$ is a $d^8$ ion. The complex $[\mathrm{Ni}{\mathrm{X}}_4{]}^{2-}$ is stated to be paramagnetic, and that fixes its geometry at once: a square-planar $d^8$ nickel would pair all its electrons and be diamagnetic, so the paramagnetism tells us the ion is tetrahedral. In a tetrahedral $d^8$ arrangement the electrons spread out to leave two unpaired. The complex therefore has two unpaired electrons and a tetrahedral geometry.