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4.Chemical Bonding and Molecular Structure
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In the molecular orbital diagram for the molecular ion, $N_2^ + $ , the number of electrons in the ${\sigma _{2p}}$ molecular orbital is
A
$0$
B
$2$
C
$3$
D
$1$
(JEE MAIN-2018)
Solution
Total electrons in $N_2^ + = (7 \times 2) – 1 = 13$
$N_2^ + \to {\sigma _{1{s^2}}},\sigma _{1{s^2}}^*,{\sigma _{2{s^2}}},\sigma _{2{s^2}}^*,[\pi _{2{p_x}}^2 = \pi _{2{p_y}}^2]\sigma _{2{p_z}}^1$
Number of electron in ${\sigma _2}_{{p_z}}$ is $1$
Standard 11
Chemistry
Similar Questions
Match each of the diatomic molecules in Column $I$ with its property / properties in Column $II$.
Column $I$ | Column $II$ |
$(A)$ $\mathrm{B}_2$ | $(p)$ Paramagnetic |
$(B)$ $\mathrm{N}_2$ | $(q)$ Undergoes oxidation |
$(C)$ $\mathrm{O}_2^{-}$ | $(r)$ Undergoes reduction |
$(D)$ $\mathrm{O}_2$ | $(s)$ Bond order $\geq 2$ |
$(t)$ Mixing of ' $\mathrm{s}$ ' and ' $\mathrm{p}$ ' orbitals |