The correct order of bond orders of is, respectively
The correct order of bond orders of is, respectively
Option 1 - <p><span class="mathml" contenteditable="false"> <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> <mo><</mo> <msubsup> <mrow> <mi>O</mi> </mrow> <mrow> <mn>2</mn> </mrow> <mrow> <mn>2</mn> <mo>−</mo> </mrow> </msubsup> </mrow> </math> </span></p>
Option 2 - <p><span class="mathml" contenteditable="false"> <math> <mrow> <msubsup> <mrow> <mi>O</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> <mo>−</mo> <mo><</mo> <mtext> </mtext> <mtext> </mtext> <msubsup> <mrow> <mi>C</mi> </mrow> <mrow> <mn>2</mn> </mrow> <mrow> <mn>2</mn> <mo>−</mo> </mrow> </msubsup> </mrow> </math> </span></p>
Option 3 - <p><span class="mathml" contenteditable="false"> <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>O</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> </mrow> </math> </span></p>
Option 4 - <p><span class="mathml" contenteditable="false"> <math> <mrow> <msubsup> <mrow> <mi>N</mi> </mrow> <mrow> <mn>2</mn> </mrow> <mrow> <mn>2</mn> <mo>−</mo> </mrow> </msubsup> <mo><</mo> <mtext> </mtext> <mtext> </mtext> <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>O</mi> </mrow> <mrow> <mn>2</mn> </mrow> <mrow> <mn>2</mn> <mo>−</mo> </mrow> </msubsup> </mrow> </math> </span></p>
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4 months ago
Correct Option - 4
Detailed Solution:
Bond order of
Bond order of
Bond order of
NB = No. of electrons in bonding molecular orbitals.
NA = No. of electron is Anti bonding molecular orbitals
Similar Questions for you
CH3COOH + NaOH → CH3COONa + H2O
ΔH = –50.6 kJ/mol
NaOH + SA [HCl] → NaCl + H2O
ΔH = –55.9 kJ/mol
the value of ΔH for ionisation of CH3COOH
⇒ ΔH = +55.9 – 50.6
5.3 kJ/mol
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