JEE Main & Advanced Sample Paper JEE Main Sample Paper-47

  • question_answer
    Considering \[NO,\,N{{O}^{+}}\] and \[N{{O}^{-}}\] we say that

    A)  the three, the one isoelectronic with \[{{N}_{2}}\] alone is diamagnetic

    B)  among the three, the one isoelectronic with \[{{O}_{2}}\] alone is paramagnetic

    C)  all the three species are paramagnetic

    D)  the three overlaps are stronger than \[{{N}_{3}}\] or \[{{O}_{2}}\]

    Correct Answer: A

    Solution :

     \[N{{O}^{+}}(14{{e}^{-}})\] is diamagnetic and isoelectronic with \[{{N}_{2}}\] \[NO\,(7+8=15)\sigma 1{{s}^{2}},\,\,\overset{*}{\mathop{\sigma }}\,1{{s}^{2}},\,\,\sigma 2{{s}^{2}},\,\overset{*}{\mathop{\sigma }}\,2{{s}^{2}},\,\sigma 2p_{z}^{2}\] \[\left\{ \begin{align}   & \pi 2p_{x}^{2}\,\,\overset{*}{\mathop{\pi }}\,2p_{x}^{1} \\  & \pi 2p_{y}^{2}\,\,\overset{*}{\mathop{\pi }}\,2p_{y}^{0} \\ \end{align} \right.\] NO = one unpaired electron\[N{{O}^{-}}\] two unpaired electrons - is electronic with \[{{O}_{2}}\]


You need to login to perform this action.
You will be redirected in 3 sec spinner