$XeF_4$ is square planar while $XeF_6$ has a distorted octahedral structure. What is the correct explanation for this observation?
- ABoth molecules have one lone pair of electrons
- BBoth molecules have two lone pairs of electrons
- C$XeF_4$ does not have any lone pair of electrons; $XeF_6$ has one lone pair of electrons on $Xe$
- D$XeF_4$ has two lone pairs of electrons on $Xe$; $XeF_6$ has one lone pair of electrons on $Xe$
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View SolutionMatch List-$I$ with List-$II$:
List-$I$ List-$II$ $A. XeF_4$ $I. See-saw$ $B. SF_4$ $II. Square planar$ $C. NH_4^{+}$ $III. Bent T-shaped$ $D. BrF_3$ $IV. Tetrahedral$
Choose the correct answer from the options given below:
| List-$I$ | List-$II$ |
| $A. XeF_4$ | $I. See-saw$ |
| $B. SF_4$ | $II. Square planar$ |
| $C. NH_4^{+}$ | $III. Bent T-shaped$ |
| $D. BrF_3$ | $IV. Tetrahedral$ |
Choose the correct answer from the options given below:
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