InterviewSolution
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InterviewSolution
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In terms of the molecular orbital theory , which of the following species will most likely be the one to gain an electron to form thermodynamically more stable species?A. `CN`B. `NO`C. `O_(2)^(2+)`D. `N_(2)` |
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Answer» Correct Answer - 1 `(1)CN(6+7=13)=sigma1s^(2)sigma^(**)1s^(2)sigma2s^(2)sigma^(**)2s^(2)pi2p_(x)^(2)=pi2p_(y)^(2)sigma 2p_(z)^(1),` Bond order `=(9-4)/(2)=2.5` `CN^(-)(6+7+1=14)=sigma1s^(2)sigma^(**)1s^(2)sigma2s^(2)sigma^(**)2s^(2)pi2p_(x)^(2)=pi2p_(y)^(2)sigma2p_(z)^(2),` Bond order `=(10-4)/(2)=3` `(2)NO(7+8=15)=sigma 1s^(2)sigma^(*)1s^(2)sigma2s^(2)sigma^(*)2s^(2)sigma2p_(z)^(2) pi2p_(x)^(2)=pi2p_(y)^(2) pi^(*)2p_(x)^(1)=pi^(*)2p_(y)^(0),` Bond order `=(10-5)/(2)=2.5` `NO^(-)(7+8+1=16)=sigma1s^(2)sigma^(*)1s^(2)sigma2s^(2)sigma^(*)2s^(2)sigma2p_(z)^(2)pi2p_(x)^(2)=pi2p_(y)^(2)pi2p_(x)^(1)=pi^(*)2p_(y)^(1)` Bond order `=(10-6)/(2)2.0` `(3)O_(2)^(2+)(8+8-2=14)=sigma1s^(2)sigma^(*)1s^(2)sigma2s^(2)sigma^(*)2s^(2)sigma2p_(z)^(2)pi2p_(x)^(2)=pi2p_(y)^(2)pi^(*)2p_(y)^(@)=pi^(*)2p_(y)^(@),` Bond order `=(10-4)/(2)=3.0` `O_(2)^(+)(8+8-1=15)=sigma1s^(2)sigma^(*)1s^(2)sigma2s^(2)sigma^(*)2s^(2)sigma^(**)2p_(z)^(2)pi2p_(x)^(2)=pi2p_(y)^(2)pi^(*)2p_(x)^(1)=pi^(*)2p_(y)^(@),` Bond order `=(10-5)/(2)=2.5` `(4)N_(2)(7+7=14)sigma1s^(2)sigma^(*)1s^(2)sigma2s^(2)sigma^(*)2s^(2)pi2p_(x)^(2)=pi2p_(y)^(2)sigma2p_(z)^(2),` Bond order `=(10-4)/(2)=3` `N_(2)^(-)(7+7+1=15)=sigma1s^(2)sigma^(*)1s^(2)sigma2s^(2)sigma^(*)2s^(2)pi2p_(x)^(2)=pi2p_(y)^(2)sigma2p_(z)^(2)pi^(*)2p_(x)^(1),`Bond order `=(10-5)/(2)=2.5` |
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