The shape of XeF_{5}^{-} will be | vedclass

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(C) To determine the shape of $XeF_{5}^{-}$, we calculate the number of electron pairs around the central $Xe$ atom: Number of electron pairs $= \frac

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planar

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(C) To determine the shape of $XeF_{5}^{-}$, we calculate the number of electron pairs around the central $Xe$ atom: Number of electron pairs $= \frac{1}{2} [V + M - C + A] = \frac{1}{2} [8 + 5 - 0 + 1] = 7$. Here, $V = 8$ (valence electrons of $Xe$), $M = 5$ (monovalent atoms), and $A = 1$ (anionic charge). With $7$ electron pairs, the electron geometry is pentagonal bipyramidal ($sp^{3}d^{3}$ hybridization). Out of these $7$ pairs, $5$ are bond pairs and $2$ are lone pairs. The two lone pairs occupy the axial positions to minimize repulsion, leaving the $5$ $Xe-F$ bonds in a single equatorial plane. Therefore, the molecular shape of $XeF_{5}^{-}$ is pentagonal planar.

List-$I$	List-$II$
$(a)$ $PCl_{5}$	$(i)$ Square pyramidal
$(b)$ $SF_{6}$	$(ii)$ Trigonal planar
$(c)$ $BrF_{5}$	$(iii)$ Octahedral
$(d)$ $BF_{3}$	$(iv)$ Trigonal bipyramidal

The shape of $XeF_{5}^{-}$ will be

Similar Questions

In the $NO_3^-$ ion,what are the number of bond pairs and lone pairs of electrons on the $N$ atom,respectively?

The shape of the $XeF_4$ molecule is:

Match List-$I$ with List-$II$.

List-$I$ List-$II$

$(a)$ $PCl_{5}$ $(i)$ Square pyramidal

$(b)$ $SF_{6}$ $(ii)$ Trigonal planar

$(c)$ $BrF_{5}$ $(iii)$ Octahedral

$(d)$ $BF_{3}$ $(iv)$ Trigonal bipyramidal

Choose the correct answer from the options given below.

The correct order of increasing bond angles in the following species is:

Match the following:
$($Molecule$/$Ion$)$ $($Shape$)$
$A. I_3^- \rightarrow 4.$ Linear
$B. ClF_3 \rightarrow 1.$ $T-$shaped
$C. H_2O \rightarrow 3.$ Angular
$D. SF_4 \rightarrow 2.$ See$-$Saw

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