JEE Mains · Chemistry · STD 11 - 4. Chemical bonding and molecular structure
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 \(1\)
Answer & Solution
Correct Answer
(D) \(1\)
Step-by-step Solution
Detailed explanation
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 \(1\)
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