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Bond Angle

III CH2 NH3...

Question

I II
$C H_{2}$ $N H_{3}$
The bond angle is smaller in $N H_{3}$ than in $C H_{4}$ because of the repulsion between the lone pair on N and the bonding pairs. If true enter 1, else enter 0.

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Solution

The increasing order of the repulsion is bond pair - bond pair < bond pair - lone pair < lone pair - lone pair. The bond angle is smaller in $N H_{3}$ than in $C H_{4}$ because of the repulsion between the lone pair on N and the bonding pairs. The statement is true. Hence, the answer is 1.

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Q. If the statement is true enter 1, else enter 0.

Repulsion between bond pair-bond pair electron is lesser than in between lone pair-lone pair.

I	II
$N H_{3}$	$P H_{3}$

The bond angle in

N H_{3}

is larger than, in

P H_{3}

P - H

bonds are longer and the lower electronegativity of P permits electron-density to be displaced towards hydrogen to a greater extent than in the case of

N H_{3}

. Both of these effects diminish the repulsion due to bond-pairs.
If true enter 1, else enter 0.

Q. If true enter 1, else enter 0.
The

H - N - H

{N H}_{3}

is greater than

H - A s - H

A s H_{3}

.

Q. Increasing bond angle:

A s H_{3} < P H_{3} < N H_{3}

If true enter 1, else enter 0.

Q. If true enter 1, else enter 0.
Presence of lone pair on central atom of a molecule although ignored in the identification of shape of molecule but they distort the shape of a molecule so as to reduce lone pair-bond pair repulsion.

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