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Diamagnetic species are those which contain no unapired electrons. Which among the following are diagmagnetic?A. `N_(2)`B. `N_(2)^(2-)`C. `F_(2)^(+)`D. `O_(2)^(-)` |
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Answer» Correct Answer - A::D (a,d) a) Electronic configuration of `N_(2) = sigma1s^(2), sigma^(star)1s^(2), sigma2s^(2), pi2p_(x)^(2)=pi2p_(y)^(2), sigma2p_(z)^(2)`. It has unpaired electron indicates diamagnetic species. b) Electronic configuration of `N_(2)^(2-)` ion =`sigma1s^(2), sigma^(star)s^(2), sigma2s^(2), sigma^(star)2s^(2), pi2p_(x)^(2)=pi2_(y)^(2), sigma2p_(z)^(2), pi^(star)p_(x)^(1)=pi^(star)2p_(y)^(2)` It has two unpaired electrons, paramagnetic in nature. c) Electronic configuration of `O_(2)` = `sigma1s^(2), sigma^(star)2s^(2), pi2p_(s)^(2), sigma^(star)2s^(2), sigma2p_(z)^(2), pi2p_(x)^(2)=pi2p_(y)^(2), pi^(star)2p_(x)^(1)=pi^(star)2p_(y)^(2)` The presence of two unpaired electrons shows its paramagnetic nature. d) Electronic configuration of `O_(2)^(2-) = sigma1s^(2), sigma^(star)1s^(2), sigma2s^(2), sigma2p_(z)^(2), pi2p_(x)^(2)=pi2p_(y)^(2), pi^(star)2p_(x)^(2)=pi^(star)2p_(y)^(2)` It contains no unpaired electron, therefore it is diamagnetic in nature. |
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