The correct order regarding the electronegativity of hybrid orbitals of carbon is
- A$sp < sp^2 < sp^3$
- B$sp > sp^2 < sp^3$
- C$sp > sp^2 > sp^3$
- D$sp < sp^2 > sp^3$
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Similar Questions
Find the hybridisation of the $C_1$,$C_2$,and $C_3$ labelled carbon atoms in the given molecule: $H-C \equiv C_1-C_2H-C_3H=CH_2$
What is the $C - C$ bond length in molecules with $sp^2$ hybridization (in $\mathring{A}$)?
Easy
View SolutionObserve the following statements:
Statement-$I$: Hybridisation is not the same in both ${SF}_6$ and ${BrF}_5$.
Statement-$II$: ${BrF}_5$ is square pyramidal while ${SF}_6$ is octahedral in shape.
The correct answer is:
Statement-$I$: Hybridisation is not the same in both ${SF}_6$ and ${BrF}_5$.
Statement-$II$: ${BrF}_5$ is square pyramidal while ${SF}_6$ is octahedral in shape.
The correct answer is:
$A$ molecule is square planar and has no lone pairs. What type of hybridization is it associated with?
Medium
View SolutionBond $X$ is formed by the overlap of which type of hybridized orbitals?
Difficult
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