Discussion
According to Wade’s rules, the structures of B10C2H12 and B10H12(SEt2)2 are respectively :
Answer: C
The two structures are closo and arachno.
Wade's rules count skeletal electron pairs: each B-H unit gives 2 electrons, each C-H unit 3, each extra hydrogen 1 and each two-electron ligand 2. For n cage atoms, n+1 pairs means closo, n+2 nido and n+3 arachno.
\[ \begin{aligned} \mathrm{B_{10}C_{2}H_{12}} &: 10(2) + 2(3) = 26\ e \\ &= 13\ \text{pairs} = n+1 \ (n=12) \end{aligned} \]
So the dicarbadodecaborane is a closo icosahedron. For B10H12(SEt2)2, the ten B-H units give 20 electrons, the two extra hydrogens 2 and the two SEt2 donors 4, a total of 26 electrons or 13 pairs for 10 cage atoms, which is n+3: arachno.
Options calling the carborane nido would need 14 pairs, and calling the ligand adduct nido ignores the two electron pairs the SEt2 groups add to the parent nido B10H14.
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