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Chemistry NCERT Exemplar Solutions Class 11th Chapter Four Molecular Orbital Theory Chemistry Chemical Bonding and Molecular Structure Chemistry Class 11th

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 -

(A), (B), (C) only

Option 2 -

(B), (C), (E) only

Option 3 -

(A), (C) only

Option 4 -

(D) only

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1 Answer

Answered by

Raj Pandey | Contributor-Level 9

a month 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 = 3 B.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 p orbital.

Bond strength <math> <mrow> <mo>∝</mo> </mrow> </math>  Bond order <math> <mrow> <msub> <mrow> <mi>N</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msub> <mo>→</mo> <msubsup> <mrow> <mi>σ</mi> </mrow> <mrow> <mn>1</mn> <mi>s</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msubsup> <msubsup> <mrow> <mi>σ</mi> </mrow> <mrow> <mn>1</mn> <mi>s</mi> </mrow> <mrow> <mo>*</mo> <mn>2</mn> </mrow> </msubsup> <msubsup> <mrow> <mi>σ</mi> </mrow> <mrow> <mn>2</mn> <mi>s</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msubsup> <msubsup> <mrow> <mi>σ</mi> </mrow> <mrow> <mn>2</mn> <mi>s</mi> </mrow> <mrow> <mo>*</mo> <mn>2</mn> </mrow> </msubsup> <msubsup> <mrow> <mi>π</mi> </mrow> <mrow> <mn>2</mn> <mi>p</mi> <mi>x</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msubsup> <mo>≡</mo> <msubsup> <mrow> <mi>π</mi> </mrow> <mrow> <mn>2</mn> <mi>p</mi> <mi>y</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msubsup> <msubsup> <mrow> <mi>σ</mi> </mrow> <mrow> <mn>2</mn> <mi>p</mi> <mi>z</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msubsup> </mrow> </math>   <math> <mrow> <msub> <mrow> <mi>O</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msub> <mo>→</mo> <msubsup> <mrow> <mi>σ</mi> </mrow> <mrow> <mn>1</mn> <mi>s</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msubsup> <msubsup> <mrow> <mi>σ</mi> </mrow> <mrow> <mn>1</mn> <mi>s</mi> </mrow> <mrow> <mo>*</mo> <mn>2</mn> </mrow> </msubsup> <msubsup> <mrow> <mi>σ</mi> </mrow> <mrow> <mn>2</mn> <mi>s</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msubsup> <msubsup> <mrow> <mi>σ</mi> </mrow> <mrow> <mn>2</mn> <mi>s</mi> </mrow> <mrow> <mo>*</mo> <mn>2</mn> </mrow> </msubsup> <msubsup> <mrow> <mi>σ</mi> </mrow> <mrow> <mn>2</mn> <mi>p</mi> <mi>z</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msubsup> <msubsup> <mrow> <mi>π</mi> </mrow> <mrow> <mn>2</mn> <mi>p</mi> <mi>x</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msubsup> <mo>≡</mo> <msubsup> <mrow> <mi>π</mi> </mrow> <mrow> <mn>2</mn> <mi>p</mi> <mi>y</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msubsup> <msubsup> <mrow> <mi>π</mi> </mrow> <mrow> <mn>2</mn> <mi>p</mi> <mi>x</mi> </mrow> <mrow> <mo>*</mo> <mn>1</mn> </mrow> </msubsup> <mo>≡</mo> <msubsup> <mrow> <mi>π</mi> </mrow> <mrow> <mn>2</mn> <mi>p</mi> <mi>y</mi> </mrow> <mrow> <mo>*</mo> <mn>1</mn> </mrow> </msubsup> </mrow> </math>  <math> <mrow> <msub> <mrow> <mi>C</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msub> <mo>→</mo> <msubsup> <mrow> <mi>σ</mi> </mrow> <mrow> <mn>1</mn> <mi>s</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msubsup> <msubsup> <mrow> <mi>σ</mi> </mrow> <mrow> <mn>1</mn> <mi>s</mi> </mrow> <mrow> <mo>*</mo> <mn>2</mn> </mrow> </msubsup> <msubsup> <mrow> <mi>σ</mi> </mrow> <mrow> <mn>2</mn> <mi>s</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msubsup> <msubsup> <mrow> <mi>σ</mi> </mrow> <mrow> <mn>2</mn> <mi>s</mi> </mrow> <mrow> <mo>*</mo> <mn>2</mn> </mrow> </msubsup> <msubsup> <mrow> <mi>π</mi> </mrow> <mrow> <mn>2</mn> <mi>p</mi> <mi>x</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msubsup> <mo>≡</mo> <msubsup> <mrow> <mi>π</mi> </mrow> <mrow> <mn>2</mn> <mi>p</mi> <mi>y</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msubsup> </mrow> </math>   <math> <mrow> <msub> <mrow> <mi>B</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msub> <mo>→</mo> <msubsup> <mrow> <mi>σ</mi> </mrow> <mrow> <mn>1</mn> <mi>s</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msubsup> <msubsup> <mrow> <mi>σ</mi> </mrow> <mrow> <mn>1</mn> <mi>s</mi> </mrow> <mrow> <mo>*</mo> <mn>2</mn> </mrow> </msubsup> <msubsup> <mrow> <mi>σ</mi> </mrow> <mrow> <mn>2</mn> <mi>s</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msubsup> <msubsup> <mrow> <mi>σ</mi> </mrow> <mrow> <mn>2</mn> <mi>s</mi> </mrow> <mrow> <mo>*</mo> <mn>2</mn> </mrow> </msubsup> <msubsup> <mrow> <mi>π</mi> </mrow> <mrow> <mn>2</mn> <mi>p</mi> <mi>x</mi> </mrow> <mrow> <mn>1</mn> </mrow> </msubsup> <mo>≡</mo> <msubsup> <mrow> <mi>π</mi> </mrow> <mrow> <mn>2</mn> <mi>p</mi> <mi>y</mi> </mrow> <mrow> <mn>1</mn> </mrow> </msubsup> </mrow> </math>  NO ® Number of electron = 7 + 8 = 15B.O. Similar to  <math> <mrow> <msubsup> <mrow> <mi>N</mi> </mrow> <mrow> <mn>2</mn> </mrow> <mrow> <mo>−</mo> </mrow> </msubsup> </mrow> </math>   <math> <mrow> <msubsup> <mrow> <mi>N</mi> </mrow> <mrow> <mn>2</mn> </mrow> <mrow> <mo>−</mo> </mrow> </msubsup> <mo>→</mo> <msubsup> <mrow> <mi>σ</mi> </mrow> <mrow> <mn>1</mn> <mi>s</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msubsup> <msubsup> <mrow> <mi>σ</mi> </mrow> <mrow> <mn>1</mn> <mi>s</mi> </mrow> <mrow> <mo>*</mo> <mn>2</mn> </mrow> </msubsup> <msubsup> <mrow> <mi>σ</mi> </mrow> <mrow> <mn>2</mn> <mi>s</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msubsup> <msubsup> <mrow> <mi>σ</mi> </mrow> <mrow> <mn>2</mn> <mi>s</mi> </mrow> <mrow> <mo>*</mo> <mn>2</mn> </mrow> </msubsup> <msubsup> <mrow> <mi>π</mi> </mrow> <mrow> <mn>2</mn> <mi>p</mi> <mi>x</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msubsup> <mo>≡</mo> <msubsup> <mrow> <mi>π</mi> </mrow> <mrow> <mn>2</mn> <mi>p</mi> <mi>y</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msubsup> <msubsup> <mrow> <mi>σ</mi> </mrow> <mrow> <mn>2</mn> <mi>p</mi> <mi>z</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msubsup> <msubsup> <mrow> <mi>π</mi> </mrow> <mrow> <mn>2</mn> <mi>p</mi> <mi>x</mi> </mrow> <mrow> <mo>*</mo> <mn>1</mn> </mrow> </msubsup> <mo>≡</mo> <msubsup> <mrow> <mi>π</mi> </mrow> <mrow> <mn>2</mn> <mi>p</mi> <mi>y</mi> </mrow> <mrow> <mo>*</mo> </mrow> </msubsup> </mrow> </math>  B.O. of N2 = 3     B.O of C2 = <math> <mrow> <mfrac> <mrow> <mn>8</mn> <mo>−</mo> <mn>4</mn> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> <mo>=</mo> <mn>2</mn> </mrow> </math>  Removal of e- form antibonding molecular orbital increases bond order.In NO & O2 has valance e- in p orbital.

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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:

1 Answer

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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:

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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: