KCET · Chemistry · Chemical Bonding and Molecular Structure
The state of hybrid orbitals of carbon in \( \mathrm{CO}_{2}, \mathrm{CH}_{4} \) and \( \mathrm{CO}_{3}{ }^{2-} \) respectively is
- A \( \mathrm{sp}^{3}, \mathrm{sp}^{2} \) and \( \mathrm{sp} \)
- B \( s p^{3}, s p \) and \( s p^{2} \)
- C \( \mathrm{sp}, \mathrm{sp}^{3} \) and \( \mathrm{sp}^{2} \)
- D \( \mathrm{sp}^{2}, \mathrm{sp}^{3} \) and \( \mathrm{sp} \)
Answer & Solution
Correct Answer
(C) \( \mathrm{sp}, \mathrm{sp}^{3} \) and \( \mathrm{sp}^{2} \)
Step-by-step Solution
Detailed explanation
Using the formula
State of hybridisation \(=\frac{(V+M-C+A)}{2}\)
Where,
\(\mathrm{V}=\) No. of valance electrons.
\(\mathrm{M}=\) No. of monovalent atoms.
\(\mathrm{C}=\) Cationic charge.
\(\mathrm{A}=\) Anionic charge.
\(\mathrm{CO}_{2}=\frac{4+0}{2}=\frac{4}{2}=2 \Rightarrow s p h y b r i d i z e d\)
\(\mathrm{CH}_{4}=\frac{4+4}{2}=\frac{8}{2}=4 \Rightarrow s p^{3}\) hybridized.
\(\mathrm{CO}_{3}{ }^{2-}=\frac{4+0+2}{2}=\frac{6}{2} \Rightarrow 3 \Rightarrow s p^{2}\) hybridized.
State of hybridisation \(=\frac{(V+M-C+A)}{2}\)
Where,
\(\mathrm{V}=\) No. of valance electrons.
\(\mathrm{M}=\) No. of monovalent atoms.
\(\mathrm{C}=\) Cationic charge.
\(\mathrm{A}=\) Anionic charge.
\(\mathrm{CO}_{2}=\frac{4+0}{2}=\frac{4}{2}=2 \Rightarrow s p h y b r i d i z e d\)
\(\mathrm{CH}_{4}=\frac{4+4}{2}=\frac{8}{2}=4 \Rightarrow s p^{3}\) hybridized.
\(\mathrm{CO}_{3}{ }^{2-}=\frac{4+0+2}{2}=\frac{6}{2} \Rightarrow 3 \Rightarrow s p^{2}\) hybridized.
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