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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.

  1. A \(0\) or \(1\)
  2. B \(2\)
  3. C \(3\)
  4. D \(4\)
Verified Solution

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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