Q50 / 180 · AIIMS 2006 · Chemistry
Among the following, the species having square planar geometry for central atom are (i) $\mathrm{XeF}_4$, (ii) $\mathrm{SF}_4$, (iii) $\left[\mathrm{NiCl}_4\right]^{2-}$, (iv) $\left[\mathrm{PdCl}_4\right]^{2-}$
- Correct answerA(i) and (iv)
- B(i) and (ii)
- C(ii) and (iii)
- D(iii) and (iv).
Solution
$\mathrm{XeF}_4: s p^3 d^2$ hybridisation, shape is square planar instead of octahedral due to presence of two lone pair of electrons on $\mathrm{Xe}$ atom. $\mathrm{SF}_4: \mathrm{SF}_4$ molecule shows $s p^3 d$ hybridisation but its expected trigonal bipyramidal geometry gets distorted due to presence of a lone pair of electrons and it becomes distorted tetrahedral or see-saw with the bond angles
equal to $89^{\circ}$ and $177^{\circ}$ instead of the expected angles of $90^{\circ}$ and $180^{\circ}$ respectively.
Experiments have, however, shown that 4coordinated complexes of $\mathrm{Pd}$ (II) are diamagnetic $(n=0)$. Thus it is $d s p^2$ hybridisation which is involved in the formation of $\left[\mathrm{PdCl}_4\right]^{2-}$ ion i.e. 4coordinated complexes of Pd(II) have square planar geometry with $n=0$ (diamagnetic).
Experiments have, however, shown that 4coordinated complexes of $\mathrm{Pd}$ (II) are diamagnetic $(n=0)$. Thus it is $d s p^2$ hybridisation which is involved in the formation of $\left[\mathrm{PdCl}_4\right]^{2-}$ ion i.e. 4coordinated complexes of Pd(II) have square planar geometry with $n=0$ (diamagnetic).
Hence, the correct answer is option: 'A'.