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According to Wade’s rules, the structures of B10C2H12 and B10H12(SEt2)2 are respectively :

This question was previously asked in
MH SET 2021 Paper 2 Mathematical Science Question Paper (26-Sep-2021)
The correct answer is

closo and arachno

Wade's rules count skeletal electron pairs (SEP). A cluster with n vertices and n + 1 pairs is closo, n + 2 pairs is nido and n + 3 pairs is arachno. Each B–H unit gives 2 skeletal electrons, each C–H unit gives 3, each extra H gives 1 and each two-electron donor ligand such as SEt2 gives 2.

Step 1 — B10C2H12 (the carborane C2B10H12): 10 B–H units give 20 electrons and 2 C–H units give 6, total 26 electrons = 13 pairs. With 12 vertices, 13 = n + 1, so it is a closo icosahedron.

Step 2 — B10H12(SEt2)2: 10 B–H units give 20 electrons, the 2 extra H atoms give 2 and the two SEt2 ligands give 4, total 26 electrons = 13 pairs. With only 10 vertices, 13 = n + 3, so the cage is arachno (it is formed from nido B10H14 by adding two donor ligands, each adding a skeletal pair).

The pairings nido and closo, closo and nido, and nido and arachno each misassign at least one of the two clusters: the carborane is not nido, and the bis(diethyl sulfide) adduct has two more skeletal pairs than a closo or nido ten-vertex cage.

closo and arachno option is correct.

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