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In which case electron from bonding molecular orbital is removed?

$\begin{array}{1 1}(a)\;O_2\; to \;O_2^+&(b)\; N_2\; to\; N_2^+\\(c)\; NO\; to\; NO^+&(d)\; O_2\; to\; O_2^-\end{array}$

1 Answer

$N_2$ to $N_2^+$ Electronic configuration of $N_2 = \sigma1s_2,\sigma^\ast 1s^2,\sigma 2s^2,\sigma^\ast 2s^2,\pi^2p^2y=\pi^2p^2z,\sigma^2p^2x$
Electronic configuration of $N_2^+ = \sigma1s^2,\sigma^\ast 1s^2,\sigma^2s^2,\sigma^\ast2s^2,\pi 2p^2y=\pi2p^2z,\sigma^2p^1x$
Hence (b) is the correct answer.
answered Dec 6, 2013 by sreemathi.v
 

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