Which of the following molecules does not hav | vedclass

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(D) To determine the number of lone pairs on the central atom,we look at the valence electrons and the number of bonding pairs:$1$. In $NH_3$,Nitrogen

Vedclass · Accepted Answer

$BF_3$

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(D) To determine the number of lone pairs on the central atom,we look at the valence electrons and the number of bonding pairs:$1$. In $NH_3$,Nitrogen has $5$ valence electrons. It forms $3$ bonds with Hydrogen,leaving $5 - 3 = 2$ electrons,which is $1$ lone pair.$2$. In $H_2O$,Oxygen has $6$ valence electrons. It forms $2$ bonds with Hydrogen,leaving $6 - 2 = 4$ electrons,which is $2$ lone pairs.$3$. In $SO_2$,Sulfur has $6$ valence electrons. It forms $2$ double bonds with Oxygen atoms,using $4$ electrons for bonding,leaving $6 - 4 = 2$ electrons,which is $1$ lone pair.$4$. In $BF_3$,Boron has $3$ valence electrons. It forms $3$ bonds with Fluorine atoms,using all $3$ valence electrons. Thus,there are $0$ lone pairs on the central Boron atom.Therefore,$BF_3$ is the correct answer.

$A$. Trigonal planar	$I$. $PCl_5$
$B$. Tetrahedral	$II$. $SF_6$
$C$. Trigonal bipyramidal	$III$. $BF_3$
$D$. Octahedral	$IV$. $CCl_4$

Which of the following molecules does not have a lone pair of electrons in the valence shell of the central atom?

Similar Questions

The number of lone pairs of electrons present on $Xe$ in $XeF_4$ is

The decreasing order of bond angle in the following molecules is

Out of $CHCl_3$,$CH_4$,and $SF_4$,the molecules that do not have regular geometry are

The number of species below that have two lone pairs of electrons on their central atom is ..... . (Round off to the nearest integer)
$SF_{4}, BF_{4}^{-}, ClF_{3}, AsF_{3}, PCl_{5}, BrF_{5}, XeF_{4}, SF_{6}$

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:

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