Which of the following compound is isoelectronic with
$[H_{3} N : B H_{3}]$ ?

B_{2} H_{6}

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C_{2} H_{6}

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C_{2} H_{4}

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C_{3} H_{6}

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Solution

The correct option is B $C_{2} H_{6}$
Number of electrons in $[H_{3} N : B H_{3}] = 10 + 8 = 18$
(a) $B_{2} H_{6}$ :
Number of electrons : $10 + 6 = 16$

(b) $C_{2} H_{6} (12 + 6 = 18)$ has 18 electrons and $[N H_{3} + B H_{3}] = 10 + 8 = 18$ . So, both are isoelectronic with each other.

(c) $C_{2} H_{4}$
Number of electrons : $12 + 4 = 16$

(d) $C_{3} H_{6}$
Number of electrons : $18 + 6 = 24$ .

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