The numbers of skeletal electrons present in the compounds C₂ B₃ H₅, C₂ B₄ H₆, and B₅ H₉ are, respectively · ABC26IN1440

Subject: Inorganic Chemistry · Chapter: Organometallic Compounds · Topic: Cluster Compounds Wade Rules · Exam: CSIR NET DEC 2015 · Marks: 2 · Difficulty: Medium

The numbers of skeletal electrons present in the compounds $\mathrm{C}_{2} \mathrm{B}_{3} \mathrm{H}_{5}, \mathrm{C}_{2} \mathrm{B}_{4} \mathrm{H}_{6}$, and $\mathrm{B}_{5} \mathrm{H}_{9}$ are, respectively
(a)10, 12 and 12
(b)12, 14 and 14
(c)10, 12 and 14
(d)12, 14 and 12
Answer
Answer: B ✓ checked by 4AB · confidence high
Explanation
Skeletal electrons: C₂B₃H₅ = 2×3 + 3×2 = 12; C₂B₄H₆ = 6 + 8 = 14; B₅H₉ = 5×2 + 4 extra H = 14.

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