The correct order of decreasing bond angles is

Question

The correct order of decreasing bond angles is (A) NH 3> NH 2-> overset+NH 4 (B)  overset+NH 4> NH 3> NH 2- (C) NH 2-> NH 3> NH 4+ (D) NH 4+> NH 2 - > NH 3

Tardigrade Admin · Accepted Answer

As the number of lone pair of electrons of N increases, bond angle decreases. NH 4+, NH 3 and NH 2- contain 0,1, and 2 lone pairs respectively.

Q. The correct order of decreasing bond angles is

$N H_{3} > N H_{2}^{-} > N + H_{4}$

$N + H_{4} > N H_{3} > N H_{2}^{-}$

$N H_{2}^{-} > N H_{3} > N H_{4}^{+}$

$N H_{4}^{+} > N H_{2}^{-} > N H_{3}$

As the number of lone pair of electrons of $N$ increases, bond angle decreases. $N H_{4}^{+}, N H_{3}$ and $N H_{2}^{-}$ contain $0, 1,$ and 2 lone pairs respectively.

Q. The correct order of decreasing bond angles is

NH3​>NH2−​>N+H4​

N+H4​>NH3​>NH2−​

NH2−​>NH3​>NH4+​

NH4+​>NH2−​>NH3​

As the number of lone pair of electrons of N increases, bond angle decreases. NH4+​,NH3​ and NH2−​ contain 0,1, and 2 lone pairs respectively.

$N H_{3} > N H_{2}^{-} > N + H_{4}$

$N + H_{4} > N H_{3} > N H_{2}^{-}$

$N H_{2}^{-} > N H_{3} > N H_{4}^{+}$

$N H_{4}^{+} > N H_{2}^{-} > N H_{3}$

As the number of lone pair of electrons of $N$ increases, bond angle decreases. $N H_{4}^{+}, N H_{3}$ and $N H_{2}^{-}$ contain $0, 1,$ and 2 lone pairs respectively.