The H-C-H,H-N-H,and H-O-H bond angles (in deg | vedclass

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(B) The bond angle in a molecule is affected by the number of lone pairs and the electronegativity of the central atom.In $CH_4$ (methane),the central

Vedclass · Accepted Answer

$109.5, 107.1, 104.5$

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(B) The bond angle in a molecule is affected by the number of lone pairs and the electronegativity of the central atom.In $CH_4$ (methane),the central carbon atom has $4$ bonding pairs and $0$ lone pairs,resulting in a tetrahedral geometry with a bond angle of $109.5^{\circ}$.In $NH_3$ (ammonia),the central nitrogen atom has $3$ bonding pairs and $1$ lone pair. The lone pair-bond pair repulsion reduces the bond angle to $107.1^{\circ}$.In $H_2O$ (water),the central oxygen atom has $2$ bonding pairs and $2$ lone pairs. The greater lone pair-lone pair repulsion further reduces the bond angle to $104.5^{\circ}$.Therefore,the bond angles are $109.5^{\circ}, 107.1^{\circ}, 104.5^{\circ}$ respectively.

Molecule	$P-X$ (axial) bond length / $P-X$ (equatorial) bond length
$PF_5$	$a / b$
$PF_4CH_3$	$c / d$
$PF_3(CH_3)_2$	$e / f$
$PCl_5$	$g / h$

The $H-C-H$,$H-N-H$,and $H-O-H$ bond angles (in $degrees$) in methane,ammonia,and water are respectively,closest to

Similar Questions

Assertion : Lone pair-lone pair repulsive interactions are greater than lone pair-bond pair and bond pair-bond pair interactions.
Reason : The space occupied by lone pair electrons is more as compared to bond pair electrons.

Which among the following species has unequal bond lengths?

Which of the following molecules is symmetrical?

Consider the following information $(X = F$ or $Cl)$:

Molecule $P-X$ (axial) bond length / $P-X$ (equatorial) bond length

$PF_5$ $a / b$

$PF_4CH_3$ $c / d$

$PF_3(CH_3)_2$ $e / f$

$PCl_5$ $g / h$

According to the given information,choose the incorrect order of bond lengths.

Identify the molecular geometry of $XeF_4$.

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