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4.Chemical Bonding and Molecular Structure
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The bond order in $O _2^{2-}$ is
A$2$
B$3$
C$1.5$
D$1$
(KVPY-2010)
Solution
(d)
Total number of electrons in
$O _2^{2-}=18$
The electronic configuration of $O _2^{2-}$ will be
$\sigma 1 s^2 \,\sigma{ }^* 1 s^2 \,\sigma 2 s^2\, \sigma^* 2 s^2\, \sigma 2 p_z^2 \,\pi 2 p_x^2 \,\pi 2 p_y^2\, \pi^* 2 p_x^2 \,\pi^* 2 p_y^2$
Bond order of a diatomic molecule is given by the formula, $BO =\frac{N_b-N_c}{2}$
Bond order of $O _2^{2 \cdot}$ will b $c =\frac{10-8}{2}=1$
Total number of electrons in
$O _2^{2-}=18$
The electronic configuration of $O _2^{2-}$ will be
$\sigma 1 s^2 \,\sigma{ }^* 1 s^2 \,\sigma 2 s^2\, \sigma^* 2 s^2\, \sigma 2 p_z^2 \,\pi 2 p_x^2 \,\pi 2 p_y^2\, \pi^* 2 p_x^2 \,\pi^* 2 p_y^2$
Bond order of a diatomic molecule is given by the formula, $BO =\frac{N_b-N_c}{2}$
Bond order of $O _2^{2 \cdot}$ will b $c =\frac{10-8}{2}=1$
Standard 11
Chemistry
