Question

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).

Hint:

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).

Answer 1

Correct Answer: 32

Step 1: Find the point group of \([\mathrm{PdCl_6}]^{2-}\).
This ion has six identical chloride ligands around palladium in a regular octahedral arrangement, so it belongs to the highest symmetry octahedral point group, \(O_h\).

Step 2: Find the order of \(O_h\).
The order of a point group is the total number of distinct symmetry operations it contains. \(O_h\) has 48 symmetry operations: \(E\), \(8C_3\), \(6C_2\), \(6C_4\), \(3C_2(=C_4^2)\), \(i\), \(6S_4\), \(8S_6\), \(3\sigma_h\), \(6\sigma_d\), which add up to 48. So
\[ x = 48 \]

Step 3: Find the point group of trans-\([\mathrm{PdBr_2Cl_4}]^{2-}\).
Here two bromide ligands sit trans to each other along one axis, and four chloride ligands occupy the equatorial plane. This is a substituted octahedron with a unique 4-fold axis through the two Br ligands and Pd, a horizontal mirror plane containing the four Cl ligands, plus vertical and dihedral mirror planes: this is the point group \(D_{4h}\), the standard point group for a trans-\(MA_4B_2\) octahedral complex.

Step 4: Find the order of \(D_{4h}\).
\(D_{4h}\) has 16 symmetry operations: \(E\), \(2C_4\), \(C_2\), \(2C_2'\), \(2C_2''\), \(i\), \(2S_4\), \(\sigma_h\), \(2\sigma_v\), \(2\sigma_d\), which add up to 16. So
\[ y = 16 \]

Step 5: Compute \(x-y\).
\[ x - y = 48 - 16 = 32 \]

Final Answer:
\[ \boxed{x-y = 32} \]