KCET · Chemistry · Chemical Bonding and Molecular Structure
The percentage of \(s\)-character in the hybrid orbitals of nitrogen in \(\mathrm{NO}_{2}^{+}, \mathrm{NO}_{3}^{-}\)respectively are
- A \(33.3 \%, 50 \%, 25 \%\)
- B \(33.3 \%, 25 \%, 50 \%\)
- C \(50 \%, 33.3 \%, 25 \%\)
- D \(25 \%, 50 \%, 33.3 \%\)
Answer & Solution
Correct Answer
(C) \(50 \%, 33.3 \%, 25 \%\)
Step-by-step Solution
Detailed explanation
\(\mathrm{NO}_{2}^{+}\)shows \(s p\)-hybridisation and as such the \(s\)-character here shall be \(50 \%\).
\(\mathrm{NO}_{3}^{-}\)shows \(s p^{2}\) hybridisation, so the \(s\)-character shall be \(33.3 \%\).
\(\mathrm{NO}_{3}^{-}\)shows \(s p^{2}\) hybridisation, so the \(s\)-character shall be \(33.3 \%\).
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