JEE Chemistry 2026 Question with Solution

Given: Bonds are $\text{C-H}$ ($A$), $\text{C-O}$ ($B$), $\text{C=O}$ ($C$) and $\text{C\equiv N}$ ($D$). We need the increasing order of covalent bond length.

Find: The correct order of bond lengths and the matching option.

Step 1: Understanding the concept Bond length depends on atomic radii and bond order. Higher bond order gives shorter bond length. Hydrogen is much smaller than carbon, oxygen, and nitrogen, so $\text{C-H}$ is exceptionally short.

Step 2: Compare the approximate bond lengths

• $\text{C-H}$ ($A$) is about $109 \, \text{pm}$.
• $\text{C\equiv N}$ ($D$) is about $116 \, \text{pm}$.
• $\text{C=O}$ ($C$) is about $121 \, \text{pm}$.
• $\text{C-O}$ ($B$) is about $143 \, \text{pm}$.

So the increasing order is

$$A < D < C < B$$

Conclude: Therefore, the correct order is $A < D < C < B$, so the correct option is A.

Given: The four bonds are $\text{C-H}$, $\text{C-O}$, $\text{C=O}$, and $\text{C\equiv N}$.

Find: Their increasing order of covalent bond length.

For bonds involving similar-sized atoms, bond length decreases as bond order increases:

$$\text{single bond} > \text{double bond} > \text{triple bond}$$

Thus among $\text{C-O}$, $\text{C=O}$, and $\text{C\equiv N}$, the expected order from longest to shortest is single, then double, then triple.

However, $\text{C-H}$ does not follow this comparison directly because hydrogen belongs to the first shell and has a very small atomic radius. That makes $\text{C-H}$ shorter than even some higher-order bonds involving larger atoms.

Hence the increasing order becomes

$$\text{C-H} < \text{C\equiv N} < \text{C=O} < \text{C-O}$$

which means

$$A < D < C < B$$

Conclude: The correct option is A.