JEE Mains · Chemistry · STD 11 - 4. Chemical bonding and molecular structure
The number of \(sp ^{3}\) hybridised carbons in an acyclic neutral compound with molecular formula \(C _{4} H _{5} N\) is.
- A \(0\) or \(1\)
- B \(2\)
- C \(3\)
- D \(4\)
Answer & Solution
Correct Answer
(A) \(0\) or \(1\)
Step-by-step Solution
Detailed explanation
\(DU =4+1-\left(\frac{5-1}{2}\right)=3\) \(CH _{2}= C = CH = CH = NH\) Zero \(sp ^{3}\) carbon
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