Question:medium

The total number of symmetry operations (order, \(h\)) present in the point group of \([\mathrm{PdCl_6}]^{2-}\) is \(x\) and that in trans-\([\mathrm{PdBr_2Cl_4}]^{2-}\) is \(y\). The value of \(x-y\) is (in integer).

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An unsubstituted \(\mathrm{ML_6}\) octahedron is \(O_h\) (order 48); replacing a trans pair of ligands lowers the symmetry to \(D_{4h}\) (order 16).
Updated On: Jul 20, 2026
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Correct Answer: 32

Solution and Explanation

Both ions start from an octahedral ML6 skeleton, so the fastest route is to identify each point group from its ligand pattern and pull the operation count straight from the character table order.

  1. $[\mathrm{PdCl_6}]^{2-}$: All six ligands are identical chlorides, so nothing breaks the symmetry of a perfect octahedron. This is the parent $O_h$ point group. $O_h$ is the point group of highest order among common coordination geometries, with $x = 48$ symmetry operations (identity, rotations, improper rotations, mirror planes and the centre of inversion, all counted together).
  2. trans-$[\mathrm{PdBr_2Cl_4}]^{2-}$: Replacing a trans pair of Cl by Br singles out one of the three fourfold axes of the parent octahedron as special (the Br-Pd-Br axis), while the four remaining Cl ligands stay equivalent in the perpendicular plane. This lowers the symmetry from $O_h$ down to $D_{4h}$, the standard point group of any trans-$MA_4B_2$ octahedral complex, which has $y = 16$ symmetry operations.

A useful check: $D_{4h}$ is a subgroup of $O_h$, and the index of this subgroup, $48/16=3$, equals the number of distinct fourfold axes in the parent octahedron (one for each pair of trans positions the two Br ligands could occupy). This confirms $y=16$ is consistent with $x=48$.

Let's summarize:

  • $[\mathrm{PdCl_6}]^{2-}$: perfect octahedron, point group $O_h$, order 48.
  • trans-$[\mathrm{PdBr_2Cl_4}]^{2-}$: axially disubstituted octahedron, point group $D_{4h}$, order 16.
  • $x - y = 48 - 16 = 32$.

The value of $x-y$ is 32.

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