Question:
Observe the following statements:
Statement - I: The correct order of O-O bond length in \( O_{2} \), \( H_{2}O_{2} \) and \( O_{3} \) is \( H_{2}O_{2} > O_{3} > O_{2} \).
Statement - II: Hybridisation of carbon in graphite and pyridine is same.
Observe the following statements:
Statement - I: The correct order of O-O bond length in \( O_{2} \), \( H_{2}O_{2} \) and \( O_{3} \) is \( H_{2}O_{2} > O_{3} > O_{2} \).
Statement - II: Hybridisation of carbon in graphite and pyridine is same.
Statement - I: The correct order of O-O bond length in \( O_{2} \), \( H_{2}O_{2} \) and \( O_{3} \) is \( H_{2}O_{2} > O_{3} > O_{2} \).
Statement - II: Hybridisation of carbon in graphite and pyridine is same.
Show Hint
For bond length questions, always convert to bond order first. For hybridisation, quickly count steric number (sigma bonds + lone pairs) instead of memorising structures.
Updated On: Jun 8, 2026
- Both statements I and II are correct
- Statement I is correct, but statement II is not correct
- Statement I is not correct, but statement II is correct
- Both statements I and II are not correct
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The Correct Option is A
Solution and Explanation
Concept:
Bond length is inversely proportional to bond order, and hybridisation depends on the steric number (number of sigma bonds + lone pairs on atom).
Step 1: Check Statement I (O–O bond lengths).
Bond orders: \[ O_2 = 2,\quad O_3 = 1.5,\quad H_2O_2 = 1 \] Since bond length \( \propto \frac{1}{\text{bond order}} \), \[ \text{Bond length order: } H_2O_2 > O_3 > O_2 \] So Statement I is correct. ---
Step 2: Check Statement II (hybridisation).
• In graphite, each carbon is \(sp^2\) hybridised (three \(\sigma\)-bonds in hexagonal planar structure).
• In pyridine, each ring carbon is also \(sp^2\) hybridised (aromatic system with trigonal planar geometry).
Thus, hybridisation of carbon in both is the same: \[ sp^2 \] So Statement II is correct. ---
Step 3: Conclusion.
Both statements are correct: \[ \boxed{\text{(A) Both statements I and II are correct}} \] ---
Step 1: Check Statement I (O–O bond lengths).
Bond orders: \[ O_2 = 2,\quad O_3 = 1.5,\quad H_2O_2 = 1 \] Since bond length \( \propto \frac{1}{\text{bond order}} \), \[ \text{Bond length order: } H_2O_2 > O_3 > O_2 \] So Statement I is correct. ---
Step 2: Check Statement II (hybridisation).
• In graphite, each carbon is \(sp^2\) hybridised (three \(\sigma\)-bonds in hexagonal planar structure).
• In pyridine, each ring carbon is also \(sp^2\) hybridised (aromatic system with trigonal planar geometry).
Thus, hybridisation of carbon in both is the same: \[ sp^2 \] So Statement II is correct. ---
Step 3: Conclusion.
Both statements are correct: \[ \boxed{\text{(A) Both statements I and II are correct}} \] ---
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