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Diamagnetic species are those which contain no unpaired electrons. Which among the following are diamagnetic ? |
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Answer» `N_(2)` Electronic configuration of `NO_(2) = sigma 1S^(2), sigma^(**) 1s^(2), sigma 2S^(2), sigma^(**) 2s^(2), pi 2p_(x)^(2) = pi 2p_(y)^(2) sigma 2p_(Z)^(2)` It has no unpaired electron indicates DIAMAGNETIC species. Electronic configuration of `N_(2)^(2-) " ion " = sigma 1s^(2), sigma^(**) 1s^(2), sigma 2s^(2), sigma^(**) 2s^(2), pi 2p_(x)^(2) = pi 2p_(y)^(2) sigma 2p_(z)^(2) pi^(**)2p_(x)^(1) = pi^(**) 2p_(x)^(1)` It has two unpaired electrons, so it is PARAMAGNETIC in nature. (C) Electronic configuration of `O_(2) = sigma 1s^(2), sigma^(**) 1s^(2) , sigma 2s^(2), sigma^(**) 2s^(2), sigma^(**) 2p_(z)^(2) , pi 2p_(x)^(2) = pi 2p_(y)^(2) , pi^(**) 2p_(x)^(1) = pi^(**) 2p_(y)^(1)` The presence of two unpaired electrons shows it is paramagnetic nature. (D) Electronic configuration of `O_(2)^(2-) "ion" = sigma 1s^(2), sigma^(**) 1s^(2) , sigma 2s^(2), sigma^(**) 2s^(2), sigma 2p_(z)^(2) , pi 2p_(x)^(2) = pi 2p_(y)^(2) , pi^(**) 2p_(x)^(1) = pi^(**) 2p_(y)^(1)` It contains no unpaired electron, therefore, it is diamagnetic in nature. |
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