The geometry and the type of hybrid orbital present about the central atom in $BF_3$ is
- ALinear,$sp$
- BTrigonal planar,$sp^2$
- CTetrahedral,$sp^3$
- DPyramidal,$sp^3$
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Similar Questions
Match the shapes of the compounds given in List-$I$ with the types of hybridization given in List-$II$.
List-$I$ List-$II$ $(1)$ Tetrahedral $(A)$ $sp^2$ $(2)$ Trigonal planar $(B)$ $sp$ $(3)$ Linear $(C)$ $sp^3$
| List-$I$ | List-$II$ |
| $(1)$ Tetrahedral | $(A)$ $sp^2$ |
| $(2)$ Trigonal planar | $(B)$ $sp$ |
| $(3)$ Linear | $(C)$ $sp^3$ |
Easy
View SolutionThe valency of carbon is $4$. On what principle can it be explained in a better way?
Easy
View SolutionWhich set of hybridisation is correct for the following compounds:
$NO_2$,$SF_4$,$PF_6^-$
$NO_2$,$SF_4$,$PF_6^-$
Medium
View SolutionThe bond angle in carbon tetrachloride is approximately $.............. \ ^o$
Medium
View SolutionMatch List-$I$ with List-$II$ :
List-$I$ (Species) List-$II$ (Hybrid Orbitals) $(a)$ $SF_{4}$ $(i)$ $sp^{3}d^{2}$ $(b)$ $IF_{5}$ $(ii)$ $d^{2}sp^{3}$ $(c)$ $NO_{2}^{+}$ $(iii)$ $sp^{3}d$ $(d)$ $NH_{4}^{+}$ $(iv)$ $sp^{3}$ $(v)$ $sp$
Choose the correct answer from the options given below :
| List-$I$ (Species) | List-$II$ (Hybrid Orbitals) |
| $(a)$ $SF_{4}$ | $(i)$ $sp^{3}d^{2}$ |
| $(b)$ $IF_{5}$ | $(ii)$ $d^{2}sp^{3}$ |
| $(c)$ $NO_{2}^{+}$ | $(iii)$ $sp^{3}d$ |
| $(d)$ $NH_{4}^{+}$ | $(iv)$ $sp^{3}$ |
| $(v)$ $sp$ |
Choose the correct answer from the options given below :
DifficultJEE MAIN 2021
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