Chemical Bonding - Result Question 51

####26. Among KO2,AlO2,BaO2 and NO2+, unpaired electron is present in

(1997 C, 1M)

(a) NO2+and BaO2

(b) KO2 and AlO2

(c) Only KO2

(d) Only BaO2

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Solution:

  1. Molecular orbital electronic configuration are

$KO _2\left(O _2^{-}\right): \sigma 1 s^{2} \stackrel{}{\sigma} 1 s^{2} \sigma 2 s^{2} \stackrel{}{\sigma} 2 s^{2} \sigma 2 p _x^{2}\left|\begin{array}{l|l}\pi 2 p _y^{2} & \stackrel{}{\pi} 2 p _y^{2} \ \pi 2 p _z^{2} & * \ \pi 2 p _z^{1}\end{array}\right| \stackrel{}{\sigma} 2 p _x^{0}$

Has one unpaired electron in π2p orbital.

AlO2has both oxygen in O2 state, therefore, no unpaired electron is present.

Has no unpaired electron.

NO2+has [O=N+=O] bonding, hence no unpaired electron.



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