Correct Answer - Option 1 : C
2
Concept:
Bond order is a measurement of the number of electrons involved in bonds between two atoms in a molecule. It is used as an indicator of the stability of a chemical bond. Usually, the higher the bond order, the stronger the chemical bond.
\(Bond\;order = \frac{{No.\;of\;bonding\;{e^ - } - No.\;of\;antibonding\;{e^ - }}}{2}\)
Calculation:
Configuration of C2
= σ1s2 σ* 1s2 σ2s2 σ* 2s2 π2px2 = π2py2
Configuration of C2-
= (σ1s) 2 (σ* 1s)2 (σ2s) 2 (σ* 2s) 2 (π2px2 = π2py2)
Bond order \( = \frac{1}{2}\left( {9 - 4} \right) = 2.5\)
Configuration of \(O_2^ - \)
=(σ1s)2 (σ* 1s) 2 (σ2s) 2 (σ* 2s) 2 (σ2pz) 2 (π2px2 = π2py2)(π* 2px2 = π* 2py1)
Bond order \( = \frac{1}{2}\left( {10 - 7} \right) = 1.5\)
Configuration of F2-
= (σ1s) 2 (σ* 1s) 2 (σ2s) 2 (σ* 2s) 2 (σ2pz) 2 (π2px2 = π2py2)(π* 2px2 = π* 2p y2) σ* 2Pz1
Bond Order \( = \frac{1}{2}\left( {10 - 9} \right) = 0.5\)
Configuration of \(N{O^ - }\)
= (σ1s) 2 (σ* 1s) 2 (σ2s) 2 (σ* 2s) 2 (σ2pz) 2 (π2px2 = π2py2)(π* 2px1 = π* 2py1)
Bond Order \( = \frac{1}{2}\left( {10 - 6} \right) = 2.0\)
Bond Order ∝ Stability
In case of only C2, incoming electron will enter in the bonding molecular orbital which increases the bond order and stability too. Whereas rest of all takes electron in their antibonding molecular orbital which decreases bond order and stability.
C2 has s-p mixing and the HOMO is π2px = π2py and LUMO is σ2pz. So, the extra electron will occupy bonding molecular orbital and this will lead to an increase in bond order.
C2- has more bond order than C2.
The increase in bond order in C2- increases its stability.
Thus, the
molecule expected to be stabilized by anion formation is C2
