Which of the following is an incorrect statement about $PCl_5$?
- A$P-Cl$ bond length: $\text{axial} > \text{equatorial}$.
- B$PCl_5$ is an $sp^3d$ hybridized molecule.
- CAxial $P-Cl$ bonds are weaker than equatorial $P-Cl$ bonds.
- D$PCl_5$ is a non-reactive molecule.
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Similar Questions
Match the molecular geometries in List-$I$ with their corresponding molecules in List-$II$:
$A$. Trigonal planar $I$. $PCl_5$ $B$. Tetrahedral $II$. $SF_6$ $C$. Trigonal bipyramidal $III$. $BF_3$ $D$. Octahedral $IV$. $CCl_4$
The correct match for $A, B, C, D$ is:
| $A$. Trigonal planar | $I$. $PCl_5$ |
| $B$. Tetrahedral | $II$. $SF_6$ |
| $C$. Trigonal bipyramidal | $III$. $BF_3$ |
| $D$. Octahedral | $IV$. $CCl_4$ |
The correct match for $A, B, C, D$ is:
What is the shape of the $ClF_3$ molecule?
Medium
View SolutionIn which of the following conversions does a planar compound change into a non-planar compound?
$(a)$ $ClF_3 \rightarrow ClF_4^{-}$
$(b)$ $NH_2^{-} \rightarrow NH_4^{+}$
$(c)$ $I_3^{+} \rightarrow I_3^{-}$
$(d)$ $SO_2 \rightarrow SO_3$
$(a)$ $ClF_3 \rightarrow ClF_4^{-}$
$(b)$ $NH_2^{-} \rightarrow NH_4^{+}$
$(c)$ $I_3^{+} \rightarrow I_3^{-}$
$(d)$ $SO_2 \rightarrow SO_3$
Medium
View SolutionWhich of the following compounds has the smallest bond angle $(X-A-X)$ in each series respectively?
$A$. $OSF_2, OSCl_2, OSBr_2$ $B$. $SbCl_3, SbBr_3, SbI_3$ $C$. $PI_3, AsI_3, SbI_3$
| $A$. $OSF_2, OSCl_2, OSBr_2$ |
| $B$. $SbCl_3, SbBr_3, SbI_3$ |
| $C$. $PI_3, AsI_3, SbI_3$ |
Difficult
View SolutionIn the structure of $SF_{4}$,the lone pair of electrons on $S$ is in ....
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