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The bond order and magnetic behaviour of $He_2^{2+}$ are respectively

$\begin{array}{1 1}(a)\;\text{Zero and diamagnetic}&(b)\; \text{One and diamagnetic}\\(c)\;\text{Two and diamagnetic}& (d)\;\text{one and paramagnetic}\end{array}$

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Electronic configuration of $He_2:(\sigma 1s)^2(\sigma ^*1s)^2$
Electronic configuration of $He_2^{2+}:(\sigma 1s)^2(\sigma ^*1s)$
Bond order=$\large\frac{[N_b-N_a]}{2}$
Where $N_b$=No of electrons in bonding orbitals.
$N_a$=No of electrons in antibonding orbitals.
Bond order in $He_2^{2+}=\large\frac{2-0}{2}$$=1$
Molecular diagram of $He_2^{2+}$ is
There is no unpaired electron
$\therefore$Hence it is diamagnetic.
Hence (b) is the correct answer.
answered Mar 6, 2014 by sreemathi.v

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