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Question

The Wade’s rules can be applied to the naked clusters of p-block elements. According to Wade’s rules, the number of skeletal electrons in Pb52– cluster is :

This question was previously asked in
MH SET 2023 Paper 2 Computer Science and Applications Question Paper (26-Mar-2023)
The correct answer is

12

In Wade’s rules each main-group vertex atom in a naked cluster keeps one exo-directed lone pair, so it contributes (v – 2) electrons to the skeleton, where v is its number of valence electrons. The charge of the cluster adds to the skeletal electrons.

Lead is in group 14 with v = 4 (6s26p2), so each Pb vertex supplies 4 – 2 = 2 skeletal electrons.

For Pb52–: skeletal electrons = 5 × 2 + 2 = 12, that is 6 skeletal electron pairs.

With n = 5 vertices and n + 1 = 6 pairs the cluster is closo, a trigonal bipyramid, as found in the Zintl ion Pb52–.

The value 22 is the total valence electron count (5 × 4 + 2), which still includes the 10 electrons of the five exo lone pairs, so it is not the skeletal count; 18 and 8 do not arise from any consistent counting.

Check: total valence electrons 22 minus the 10 lone-pair electrons on the five Pb atoms = 12 skeletal electrons, confirming the result.

Hence, 12.

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