Chemistry - CHEMICAL BONDING Question with Solution | TestHub
ChemistryCHEMICAL BONDINGOxy AcidHard2 minQB
ChemistryHardinteger
A neutral chlorine oxyacid HClOₙ has connectivity HO–Cl(=O)ₙ₋₁, no peroxide and zero formal charges. Let L be central lone pairs and P formal π electrons. Find n+L+P for the member satisfying P=4L. Use chlorine's7 valence electrons and integer n.
Answer:
8
Solution:
Bond-order sum at Cl=2n−1, so L=[7−(2n−1)]/2=4−n. P=2(n−1). Then2(n−1)=4(4−n), giving n=3, L=1, P=4. Sum 8. Counting π bonds rather than π electrons would give the wrong equation.
Stream:JEESubject:ChemistryTopic:CHEMICAL BONDINGSubtopic:Oxy Acid
⏱ 2mℹ️ Source: QB
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