Question

Which of the following orders is not correct for the property mentioned against them?

• A. $\text{H}_{2}\text{O} > \text{HF} > \text{NH}_{3} > \text{H}_{2}\text{S}$ (boiling point)
• B. $\text{H}_{2}\text{O} > \text{NH}_{3} > \text{NF}_{3} > \text{CF}_{4}$ (dipole moment)
• C. $\text{C}-\text{C} > \text{N}-\text{O} > \text{C}-\text{O} > \text{C}-\text{H}$ (bond length)
• D. $\text{O}_{2}^{+} > \text{O}_{2} > \text{O}_{2}^{-} > \text{O}_{2}^{2-}$ (bond order)

Hint:

Smaller atoms form tighter, shorter covalent links. Because carbon atoms are larger than nitrogen atoms, a $\text{C}-\text{O}$ link must be longer than an $\text{N}-\text{O}$ link.

Answer 1

The Correct Option is C

Step 1: Concept
Bond length decreases with increasing bond order and decreasing atomic radii of the bonded atoms.

Step 2: Meaning
Let us look at atomic sizes: Carbon (C) is larger than Nitrogen (N), which is larger than Oxygen (O), while Hydrogen (H) is the smallest.

Step 3: Analysis
Let us examine the trends in the options: * *Boiling Point:* Stronger hydrogen bonding makes water higher than HF and ammonia, so $\text{H}_{2}\text{O} > \text{HF} > \text{NH}_{3} > \text{H}_{2}\text{S}$ is correct. * *Dipole Moment:* $\text{H}_{2}\text{O}$ has two lone pairs reinforcing its polar character, while in $\text{NF}_{3}$ the lone pair opposes the polar bonds. This matches option B perfectly. * *Bond Order:* Molecular orbital theory dictates orders of 2.5 ($\text{O}_{2}^{+}$), 2.0 ($\text{O}_{2}$), 1.5 ($\text{O}_{2}^{-}$), and 1.0 ($\text{O}_{2}^{2-}$). This makes the sequence accurate. * *Bond Length:* Comparing $\text{N}-\text{O}$ and $\text{C}-\text{O}$, since Carbon has a larger radius than Nitrogen, a $\text{C}-\text{O}$ single bond is longer than an $\text{N}-\text{O}$ single bond. Thus, the arrangement $\text{N}-\text{O} > \text{C}-\text{O}$ is incorrect.

Step 4: Conclusion
The sequence listed for bond length is incorrect.

Final Answer: (C)