Which molecule is $sp^2$ hybridised?
- A$NCl_3$
- B$N(CH_3)_3$
- C$N(SiH_3)_3$
- DBoth $(A)$ and $(C)$
Explore More
Similar Questions
In which of the following species,$d$-orbitals having $xz$ and $yz$ two nodal planes involved in hybridization of central atom?
Difficult
View SolutionWhich of the following has $sp^3$ hybridization?
Difficult
View SolutionThe species in which the $N$ atom is in a state of $sp$ hybridization is:
Match List-$I$ with List-$II$. List-$I$ (molecule) List-$II$ (hybridization,shape) $A. XeO_3$ $I. sp^3$; pyramidal $B. XeF_2$ $II. sp^3d$; linear $C. XeOF_4$ $III. sp^3d^2$; square pyramidal $D. XeF_6$ $IV. sp^3d^3$; distorted octahedral
Choose the correct answer from the options given below.
| List-$I$ (molecule) | List-$II$ (hybridization,shape) |
| $A. XeO_3$ | $I. sp^3$; pyramidal |
| $B. XeF_2$ | $II. sp^3d$; linear |
| $C. XeOF_4$ | $III. sp^3d^2$; square pyramidal |
| $D. XeF_6$ | $IV. sp^3d^3$; distorted octahedral |
Match 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 |
Vedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo