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In the change of $NO^+$ to NO,the electron is added to

$\begin{array}{1 1}(a)\;\sigma\;\text {orbital}\\(b)\;\pi\;\text {orbital}\\(c)\;\sigma^*\text {orbital}\\(d)\;\pi^*\text {orbital}\end{array}$

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NO(7+8):$(\sigma 1s)^2 < (\sigma^* 1s)^2 < (\sigma 2s)^2 < (\sigma^* 2s)^2 < (\sigma 2p_z)^2 < \pi 2p_y^2=\pi 2p_x^2) < (\pi ^*2p_x=\pi ^*2p_y)$
In the change of $NO^+$ to NO,the electron is added to $\pi^*$ orbital.
Hence (d) is the correct answer.
answered Mar 6, 2014 by sreemathi.v
 

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