The hybridizations of the central atom in the molecules $BF_3, BeF_2, BrF_3$ are respectively
- A$sp^2, sp, sp^3d$
- B$sp, sp^2, sp^3$
- C$sp^3, sp, sp^3d$
- D$sp^2, sp^3, dsp^2$
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Similar Questions
As the $s$-character of the hybrid orbital decreases,the bond angle ...........
Easy
View SolutionMatch the following List-$I$ (Hybridisation) with List-$II$ (Shape):
List-$I$ (Hybridisation) List-$II$ (Shape) $A$. $dsp^2$ $I$. Square planar $B$. $sp^3$ $II$. Tetrahedral $C$. $d^2sp^3$ $III$. Octahedral $D$. $sp^3d$ $IV$. Trigonal bipyramidal
| List-$I$ (Hybridisation) | List-$II$ (Shape) |
|---|---|
| $A$. $dsp^2$ | $I$. Square planar |
| $B$. $sp^3$ | $II$. Tetrahedral |
| $C$. $d^2sp^3$ | $III$. Octahedral |
| $D$. $sp^3d$ | $IV$. Trigonal bipyramidal |
What is meant by hybridisation? The compound $CH_2 = C = CH_2$ contains $sp$ and $sp^2$ hybridised carbon atoms. Will it be a planar molecule?
Medium
View SolutionThe shapes of methane,ethene and ethyne molecules are,respectively
Medium
View SolutionWhich of the following hybridisation has the highest percentage of $s-$character?
Medium
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