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-$I$	List-$II$
$A$. $BF_3$	$II$. Trigonal planar
$B$. $ClF_3$	$III$. $T$-shape
$C$. $NH_3$	$IV$. Trigonal pyramidal
$D$. $NH_4^+$	$I$. Tetrahedral

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

Match the following:
List-$I$ List-$II$
$A$. $BF_3$ $II$. Trigonal planar
$B$. $ClF_3$ $III$. $T$-shape
$C$. $NH_3$ $IV$. Trigonal pyramidal
$D$. $NH_4^+$ $I$. Tetrahedral

The correct match is:

Which of the following compounds have the number of lone pairs on the central $Xe$ atom equal to zero?
$XeO_{3}, XeO_{2}F_{2}, XeO_{4}, XeO_{3}F_{2}, Ba_{2}XeF_{4}$

Identify the set of molecules in which the central atom has only one lone pair of electrons in their valence shells.

In both water $(H_{2}O)$ and dimethyl ether $((CH_{3})_{2}O)$,the oxygen atom is the central atom and has the same hybridization,yet they have different bond angles. Which one has a greater bond angle? Give a reason.

Molecular bromine on reaction with three moles of molecular fluorine produces an interhalogen compound. The total number of lone pairs at the central halogen atom is

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