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CHXI04:CHEMICAL BONDING AND MOLECULAR STRUCTURE

313444 The correct order of the bond angles is

1

N H_{3} > H_{2} O > P H_{3} > H_{2} S

2

N H_{3} > P H_{3} > H_{2} O > H_{2} S

3

N H_{3} > H_{2} S > P H_{3} > H_{2} O

4

P H_{3} > H_{2} S > N H_{3} > H_{2} O

Explanation:

$N H_{3} (10 7^{0}), H_{2} O (104 . 5^{0}), P H_{3} (9 3^{0}), H_{2} S (92 . 5^{0})$
The bond angles depend upon electronegativity of the central atom and number of lone pair of electrons.

CHXI04:CHEMICAL BONDING AND MOLECULAR STRUCTURE

313445 The nitrogen atoms in $N H_{3}, N {H_{2}}^{-} a n d N {H_{4}}^{+}$
are all surrounded by eight electrons. When these species are arranged in order of increasing H – N – H bond angle, what is the correct order?

1

N H_{3}, N {H_{2}}^{-}, N {H_{4}}^{+}

2

N {H_{4}}^{+}, N {H_{2}}^{-}, N H_{3}

3

N H_{3}, N {H_{4}}^{+}, N {H_{2}}^{-}

4

N {H_{2}}^{-}, N H_{3}, N {H_{2}}^{-}

Explanation:

2 lone pairs in $N {H_{2}}^{-}$
1 lone pair in $N H_{3}$
0 lone pairs in $N {H_{4}}^{+}$
Bond angle is inversely proportional to the number of lone pairs.

CHXI04:CHEMICAL BONDING AND MOLECULAR STRUCTURE

313446 The decreasing values of bond angles from $N H_{3} (106^{\circ}) t o S b H_{3} (101^{\circ})$ down group-15 of the periodic table is due to

1 decreasing bp-bp repulsion

2 decreasing electronegativity

3 increasoing bp-bp repulsion

4 increasing lp-bp repulsion

Explanation:

The bond angle decreases on moving down the group due to decrease in bond pair-bond pair repulsion.
$\underset{107^{\circ}}{N H_{3}} \underset{94^{\circ}}{P H_{3}} \underset{92^{\circ}}{A s H_{3}} \underset{91^{\circ}}{S b H_{3}} \underset{90^{\circ}}{B i H_{3}}$
This can also be explained by the fact that as the size of central atom increases and its electronegativity decreases. Thus distance between bond pairs of electron increases and bp-bp repulsion decreases. As a result bond angle decreases from $N H_{3} t o B i H_{3} .$

CHXI04:CHEMICAL BONDING AND MOLECULAR STRUCTURE

313447 Which of the following species has bond angle less than $98^{\circ}$ ?

1

H_{2} O

2

{NH}_{3}

3

{CH}_{4}

4

{PH}_{3}

Explanation:

${In}_{3} {PH}_{3}$ , pure 'p' orbitals participate in bonding, so angle is close to $93^{\circ}$ .

CHXI04:CHEMICAL BONDING AND MOLECULAR STRUCTURE

313444 The correct order of the bond angles is

1

N H_{3} > H_{2} O > P H_{3} > H_{2} S

2

N H_{3} > P H_{3} > H_{2} O > H_{2} S

3

N H_{3} > H_{2} S > P H_{3} > H_{2} O

4

P H_{3} > H_{2} S > N H_{3} > H_{2} O

Explanation:

$N H_{3} (10 7^{0}), H_{2} O (104 . 5^{0}), P H_{3} (9 3^{0}), H_{2} S (92 . 5^{0})$
The bond angles depend upon electronegativity of the central atom and number of lone pair of electrons.

CHXI04:CHEMICAL BONDING AND MOLECULAR STRUCTURE

313445 The nitrogen atoms in $N H_{3}, N {H_{2}}^{-} a n d N {H_{4}}^{+}$
are all surrounded by eight electrons. When these species are arranged in order of increasing H – N – H bond angle, what is the correct order?

1

N H_{3}, N {H_{2}}^{-}, N {H_{4}}^{+}

2

N {H_{4}}^{+}, N {H_{2}}^{-}, N H_{3}

3

N H_{3}, N {H_{4}}^{+}, N {H_{2}}^{-}

4

N {H_{2}}^{-}, N H_{3}, N {H_{2}}^{-}

Explanation:

2 lone pairs in $N {H_{2}}^{-}$
1 lone pair in $N H_{3}$
0 lone pairs in $N {H_{4}}^{+}$
Bond angle is inversely proportional to the number of lone pairs.

CHXI04:CHEMICAL BONDING AND MOLECULAR STRUCTURE

313446 The decreasing values of bond angles from $N H_{3} (106^{\circ}) t o S b H_{3} (101^{\circ})$ down group-15 of the periodic table is due to

1 decreasing bp-bp repulsion

2 decreasing electronegativity

3 increasoing bp-bp repulsion

4 increasing lp-bp repulsion

Explanation:

The bond angle decreases on moving down the group due to decrease in bond pair-bond pair repulsion.
$\underset{107^{\circ}}{N H_{3}} \underset{94^{\circ}}{P H_{3}} \underset{92^{\circ}}{A s H_{3}} \underset{91^{\circ}}{S b H_{3}} \underset{90^{\circ}}{B i H_{3}}$
This can also be explained by the fact that as the size of central atom increases and its electronegativity decreases. Thus distance between bond pairs of electron increases and bp-bp repulsion decreases. As a result bond angle decreases from $N H_{3} t o B i H_{3} .$

CHXI04:CHEMICAL BONDING AND MOLECULAR STRUCTURE

313447 Which of the following species has bond angle less than $98^{\circ}$ ?

1

H_{2} O

2

{NH}_{3}

3

{CH}_{4}

4

{PH}_{3}

Explanation:

${In}_{3} {PH}_{3}$ , pure 'p' orbitals participate in bonding, so angle is close to $93^{\circ}$ .

CHXI04:CHEMICAL BONDING AND MOLECULAR STRUCTURE

313444 The correct order of the bond angles is

1

N H_{3} > H_{2} O > P H_{3} > H_{2} S

2

N H_{3} > P H_{3} > H_{2} O > H_{2} S

3

N H_{3} > H_{2} S > P H_{3} > H_{2} O

4

P H_{3} > H_{2} S > N H_{3} > H_{2} O

Explanation:

$N H_{3} (10 7^{0}), H_{2} O (104 . 5^{0}), P H_{3} (9 3^{0}), H_{2} S (92 . 5^{0})$
The bond angles depend upon electronegativity of the central atom and number of lone pair of electrons.

CHXI04:CHEMICAL BONDING AND MOLECULAR STRUCTURE

313445 The nitrogen atoms in $N H_{3}, N {H_{2}}^{-} a n d N {H_{4}}^{+}$
are all surrounded by eight electrons. When these species are arranged in order of increasing H – N – H bond angle, what is the correct order?

1

N H_{3}, N {H_{2}}^{-}, N {H_{4}}^{+}

2

N {H_{4}}^{+}, N {H_{2}}^{-}, N H_{3}

3

N H_{3}, N {H_{4}}^{+}, N {H_{2}}^{-}

4

N {H_{2}}^{-}, N H_{3}, N {H_{2}}^{-}

Explanation:

2 lone pairs in $N {H_{2}}^{-}$
1 lone pair in $N H_{3}$
0 lone pairs in $N {H_{4}}^{+}$
Bond angle is inversely proportional to the number of lone pairs.

CHXI04:CHEMICAL BONDING AND MOLECULAR STRUCTURE

313446 The decreasing values of bond angles from $N H_{3} (106^{\circ}) t o S b H_{3} (101^{\circ})$ down group-15 of the periodic table is due to

1 decreasing bp-bp repulsion

2 decreasing electronegativity

3 increasoing bp-bp repulsion

4 increasing lp-bp repulsion

Explanation:

The bond angle decreases on moving down the group due to decrease in bond pair-bond pair repulsion.
$\underset{107^{\circ}}{N H_{3}} \underset{94^{\circ}}{P H_{3}} \underset{92^{\circ}}{A s H_{3}} \underset{91^{\circ}}{S b H_{3}} \underset{90^{\circ}}{B i H_{3}}$
This can also be explained by the fact that as the size of central atom increases and its electronegativity decreases. Thus distance between bond pairs of electron increases and bp-bp repulsion decreases. As a result bond angle decreases from $N H_{3} t o B i H_{3} .$

CHXI04:CHEMICAL BONDING AND MOLECULAR STRUCTURE

313447 Which of the following species has bond angle less than $98^{\circ}$ ?

1

H_{2} O

2

{NH}_{3}

3

{CH}_{4}

4

{PH}_{3}

Explanation:

${In}_{3} {PH}_{3}$ , pure 'p' orbitals participate in bonding, so angle is close to $93^{\circ}$ .

CHXI04:CHEMICAL BONDING AND MOLECULAR STRUCTURE

313444 The correct order of the bond angles is

1

N H_{3} > H_{2} O > P H_{3} > H_{2} S

2

N H_{3} > P H_{3} > H_{2} O > H_{2} S

3

N H_{3} > H_{2} S > P H_{3} > H_{2} O

4

P H_{3} > H_{2} S > N H_{3} > H_{2} O

Explanation:

$N H_{3} (10 7^{0}), H_{2} O (104 . 5^{0}), P H_{3} (9 3^{0}), H_{2} S (92 . 5^{0})$
The bond angles depend upon electronegativity of the central atom and number of lone pair of electrons.

CHXI04:CHEMICAL BONDING AND MOLECULAR STRUCTURE

313445 The nitrogen atoms in $N H_{3}, N {H_{2}}^{-} a n d N {H_{4}}^{+}$
are all surrounded by eight electrons. When these species are arranged in order of increasing H – N – H bond angle, what is the correct order?

1

N H_{3}, N {H_{2}}^{-}, N {H_{4}}^{+}

2

N {H_{4}}^{+}, N {H_{2}}^{-}, N H_{3}

3

N H_{3}, N {H_{4}}^{+}, N {H_{2}}^{-}

4

N {H_{2}}^{-}, N H_{3}, N {H_{2}}^{-}

Explanation:

2 lone pairs in $N {H_{2}}^{-}$
1 lone pair in $N H_{3}$
0 lone pairs in $N {H_{4}}^{+}$
Bond angle is inversely proportional to the number of lone pairs.

CHXI04:CHEMICAL BONDING AND MOLECULAR STRUCTURE

313446 The decreasing values of bond angles from $N H_{3} (106^{\circ}) t o S b H_{3} (101^{\circ})$ down group-15 of the periodic table is due to

1 decreasing bp-bp repulsion

2 decreasing electronegativity

3 increasoing bp-bp repulsion

4 increasing lp-bp repulsion

Explanation:

The bond angle decreases on moving down the group due to decrease in bond pair-bond pair repulsion.
$\underset{107^{\circ}}{N H_{3}} \underset{94^{\circ}}{P H_{3}} \underset{92^{\circ}}{A s H_{3}} \underset{91^{\circ}}{S b H_{3}} \underset{90^{\circ}}{B i H_{3}}$
This can also be explained by the fact that as the size of central atom increases and its electronegativity decreases. Thus distance between bond pairs of electron increases and bp-bp repulsion decreases. As a result bond angle decreases from $N H_{3} t o B i H_{3} .$

CHXI04:CHEMICAL BONDING AND MOLECULAR STRUCTURE

313447 Which of the following species has bond angle less than $98^{\circ}$ ?

1

H_{2} O

2

{NH}_{3}

3

{CH}_{4}

4

{PH}_{3}

Explanation:

${In}_{3} {PH}_{3}$ , pure 'p' orbitals participate in bonding, so angle is close to $93^{\circ}$ .