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According to molecular orbital theory ,the magnetic characteristics of $F_2^{2+}$ and $Br_2^{2+}$ respectively are

$\begin{array}{1 1}(a)\;\text{paramagnetic and diamagnetic}\\(b)\; \text{diamagnetic and paramagnetic}\\(c)\;\text{both are paramagnetic}\\(d)\;\text{both are diamagnetic}\end{array}$

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According to M.O theory
$F_2(18 e^-):(\sigma 1s)^2(\sigma^*1s)^2 < (\sigma 2s)^2(\sigma^*2s)^2 < (\sigma 2p_z)^2 < \pi2p_x^2=\pi 2p_y^2) < (\pi^*2p_x^2=\pi^*2p_y^2)$
$F_2^{2+}(16 e^-):(\sigma 1s)^2(\sigma^*1s)^2 < (\sigma 2s)^2(\sigma^*2s)^2 < (\sigma 2p_z)^2 < \pi2p_x^2=\pi 2p_y^2) < (\pi^*2p_x^1=\pi^*2p_y^1)$
$Br_2^{2+}$ is analogous to $F_2^{2+}$,except that additional inner shells of electrons are full.
Hence $F_2^{2+}$ and $Br_2^{2+}$ both are paramagnetic.
Hence (c) is the correct answer.
answered Mar 10, 2014 by sreemathi.v
 

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