KCET · Chemistry · Chemical Bonding and Molecular Structure
Match List-I with List-II and select the correct option:
\begin{array}{|l|l|l|l|}\hline & \text{List-I (Molecule / ion} & & \text{List-II (Bond order)} \\\hline \text{1}. & \text{NO} & \text{(i)} & \text {15} \\\hline \text{2.} & \text{CO} & \text{(ii)} & \text{2.0} \\\hline \text{3.} & \mathrm{O}_2^{-} & \text{(iii)} & \text{2.5} \\\hline \text{4.} & \mathrm{O}_2 & \text{(iv)} & \text{3.0} \\\hline\end{array}
- A a-iii, b-iv, c-i, d-ii
- B a-i, b-iv, c-iii, d-ii
- C a-ii, b-iii, c-iv, d-i
- D a-iv, b-iii, c-ii, d-i
Answer & Solution
Correct Answer
(A) a-iii, b-iv, c-i, d-ii
Step-by-step Solution
Detailed explanation
Bond order \(=\frac{1}{2}\) [no. of bonding - no. of bonding electrons]
For NO molecule
\(\begin{aligned}
& \mathrm{BO}=(8-3) / 2=2.5 \\
& \mathrm{CO}-\frac{1}{2}[10-4]=3.0 \\
& \mathrm{O}_2^{-1}-\frac{1}{2}[6-3]=1.5 \\
& \mathrm{O}_2-\frac{1}{2}[6-2]=2.0
\end{aligned}\)
For NO molecule
\(\begin{aligned}
& \mathrm{BO}=(8-3) / 2=2.5 \\
& \mathrm{CO}-\frac{1}{2}[10-4]=3.0 \\
& \mathrm{O}_2^{-1}-\frac{1}{2}[6-3]=1.5 \\
& \mathrm{O}_2-\frac{1}{2}[6-2]=2.0
\end{aligned}\)
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