The number of lone pairs of electrons on the | vedclass

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(C) To determine the number of lone pairs on the central atom,we use the formula: $	ext{Lone pairs} = \frac{1}{2} (V - N)$,where $V$ is the number of

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$2, 1, 1, 2$

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(C) To determine the number of lone pairs on the central atom,we use the formula: $	ext{Lone pairs} = \frac{1}{2} (V - N)$,where $V$ is the number of valence electrons of the central atom and $N$ is the number of monovalent atoms bonded to it.$1$. For $ClF_3$: $V = 7, N = 3$. $	ext{Lone pairs} = \frac{1}{2} (7 - 3) = 2$.$2$. For $NF_3$: $V = 5, N = 3$. $	ext{Lone pairs} = \frac{1}{2} (5 - 3) = 1$.$3$. For $SF_4$: $V = 6, N = 4$. $	ext{Lone pairs} = \frac{1}{2} (6 - 4) = 1$.$4$. For $XeF_4$: $V = 8, N = 4$. $	ext{Lone pairs} = \frac{1}{2} (8 - 4) = 2$.CompoundLone pairs$ClF_3$$2$$NF_3$$1$$SF_4$$1$$XeF_4$$2$Therefore,the correct sequence is $2, 1, 1, 2$.

List-$I$ (Compound)	List-$II$ (Shape/geometry)
$A$. $NH_3$	$I$. Trigonal Pyramidal
$B$. $BrF_5$	$IV$. Square Pyramidal
$C$. $XeF_4$	$II$. Square Planar
$D$. $SF_6$	$III$. Octahedral

The number of lone pairs of electrons on the central atom of $ClF_3, NF_3, SF_4,$ and $XeF_4$ respectively are:

Similar Questions

In $ClF_3$ molecule,if the lone pair of the central atom occupied the axial position,then which of the following statements is correct?

Explain: The shape of $H_2O$ is $V$ (angular).

What is the shape of the $SF_2Cl_2$ molecule?

The number of lone pairs of electrons on the central atom of $BrF_5, XeO_3, SO_2$ respectively are

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

List-$I$ (Compound) List-$II$ (Shape/geometry)

$A$. $NH_3$ $I$. Trigonal Pyramidal

$B$. $BrF_5$ $IV$. Square Pyramidal

$C$. $XeF_4$ $II$. Square Planar

$D$. $SF_6$ $III$. Octahedral

Choose the correct answer from the options given below:

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