Class 11th Chemistry Chemical Bonding and Molecular Structure Chemistry

Bonding in which of the following diatonic molecule(s) become(s) stronger, on the basis of MO Theory, by removal of an electron?

(A) NO

(B) N₂

(C) O₂

(D) C₂

(E) B₂

Choose the most appropriate answer from the options given below

Option 1 - <p>(A), (B), (C) only            </p>

Option 2 - <p>(B), (C), (E) only  </p>

Option 3 - <p>(A), (C) only</p>

Option 4 - <p>(D) only</p>

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Answered by

Payal Gupta | Contributor-Level 10

8 months ago

Correct Option - 3

Detailed Solution:

Bond strength $\propto$ Bond order

$N_{2} \to σ_{1 s}^{2} σ_{1 s}^{* 2} σ_{2 s}^{2} σ_{2 s}^{* 2} π_{2 p x}^{2} \equiv π_{2 p y}^{2} σ_{2 p z}^{2}$

$O_{2} \to σ_{1 s}^{2} σ_{1 s}^{* 2} σ_{2 s}^{2} σ_{2 s}^{* 2} σ_{2 p z}^{2} π_{2 p x}^{2} \equiv π_{2 p y}^{2} π_{2 p x}^{* 1} \equiv π_{2 p y}^{* 1}$

$C_{2} \to σ_{1 s}^{2} σ_{1 s}^{* 2} σ_{2 s}^{2} σ_{2 s}^{* 2} π_{2 p x}^{2} \equiv π_{2 p y}^{2}$

$B_{2} \to σ_{1 s}^{2} σ_{1 s}^{* 2} σ_{2 s}^{2} σ_{2 s}^{* 2} π_{2 p x}^{1} \equiv π_{2 p y}^{1}$

NO Number of electron = 7 + 8 = 15

B.O. Similar to $N_{2}^{-}$

$N_{2}^{-} \to σ_{1 s}^{2} σ_{1 s}^{* 2} σ_{2 s}^{2} σ_{2 s}^{* 2} π_{2 p x}^{2} \equiv π_{2 p y}^{2} σ_{2 p z}^{2} π_{2 p x}^{* 1} \equiv π_{2 p y}^{*}$

B.O. of N₂ = 3B.O of C₂ = $\frac{8 - 4}{2} = 2$

Removal of e form antibonding molecular orbital increases bond order.

In NO & O₂ has valance e in orbital.

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