The correct order of bond orders of <math> <mrow> <msubsup> <mrow> <mi>C</mi> </mrow> <mrow> <mn>2</mn> </mrow> <mrow> <mn>2</mn> <mo>−</mo> </mrow> </msubsup> <mo>,</mo> <mtext> </mtext> <mtext> </mtext> <msubsup> <mrow> <mi>N</mi> </mrow> <mrow> <mn>2</mn> </mrow> <mrow> <mn>2</mn> <mo>−</mo> </mrow> </msubsup> <mtext> </mtext> <mtext> </mtext> <mi>a</mi> <mi>n</mi> <mi>d</mi> <mtext> </mtext> <mtext> </mtext> <msubsup> <mrow> <mi>O</mi> </mrow> <mrow> <mn>2</mn> </mrow> <mrow> <mn>2</mn> <mo>−</mo> </mrow> </msubsup> </mrow> </math> is, respectively

Bond order of C 2 2 − = N B − N A 2 = 1 0 − 4 2 = 3 Bond order of N 2 2 − = N B − N A 2 = 1 0 − 6 2 = 2 Bond order of O 2 2 − = N B − N A 2 = 1 0 − 8 2 = 1 NB = No. of electrons in bonding molecular orbitals.NA = No. of electron is Anti bonding molecular orbitals

The correct order of bond orders of <math> <mrow> <msubsup> <mrow> <mi>C</mi> </mrow> <mrow> <mn>2</mn> </mrow> <mrow> <mn>2</mn> <mo>−</mo> </mrow> </msubsup> <mo>,</mo> <mtext> </mtext> <mtext> </mtext> <msubsup> <mrow> <mi>N</mi> </mrow> <mrow> <mn>2</mn> </mrow> <mrow> <mn>2</mn> <mo>−</mo> </mrow> </msubsup> <mtext> </mtext> <mtext> </mtext> <mi>a</mi> <mi>n</mi> <mi>d</mi> <mtext> </mtext> <mtext> </mtext> <msubsup> <mrow> <mi>O</mi> </mrow> <mrow> <mn>2</mn> </mrow> <mrow> <mn>2</mn> <mo>−</mo> </mrow> </msubsup> </mrow> </math> &nbsp;is, respectively

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Chemistry Class 11th Chemistry Chemical Bonding and Molecular Structure

The correct order of bond orders of $C_{2}^{2 -}, N_{2}^{2 -} a n d O_{2}^{2 -}$ is, respectively

Option 1 -

$C_{2}^{2 -} < N_{2}^{2 -} < O_{2}^{2 -}$

Option 2 -

$O_{2}^{2 -} < N_{2}^{2 -} - < C_{2}^{2 -}$

Option 3 -

$C_{2}^{2 -} < O_{2}^{2 -} < N_{2}^{2 -}$

Option 4 -

$N_{2}^{2 -} < C_{2}^{2 -} < O_{2}^{2 -}$

2 Views | Posted 2 months ago

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1 Answer

Answered by

alok kumar singh | Contributor-Level 10

2 months ago

Correct Option - 4

Detailed Solution:

Bond order of $C_{2}^{2 -} = \frac{N_{B} - N_{A}}{2} = \frac{10 - 4}{2} = 3$

Bond order of $N_{2}^{2 -} = \frac{N_{B} - N_{A}}{2} = \frac{10 - 6}{2} = 2$

Bond order of $O_{2}^{2 -} = \frac{N_{B} - N_{A}}{2} = \frac{10 - 8}{2} = 1$

N_B = No. of electrons in bonding molecular orbitals.

N_A= No. of electron is Anti bonding molecular orbitals

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The correct order of bond orders of $C_{2}^{2 -}, N_{2}^{2 -} a n d O_{2}^{2 -}$ is, respectively