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Firdavs [7]
4 years ago
12

Select the correct value for the indicated bond angle in each of the following compounds: O-S-O angle of SO2 F-B-F angle of BF3

Cl-S-Cl angle of SCI2 O-C-O angle of CO2 F-P-F angle of PF3 H-C-H angle of CH4
Chemistry
1 answer:
Stels [109]4 years ago
7 0

Answer:

(A) O-S-O bond angle of SO₂ molecule = 119°

(B) F-B-F bond angle of BF₃ molecule = 120°

(C) Cl-S-Cl bond angle of SCl₂ molecule = 103°

(D) O-C-O bond angle of CO₂ molecule = 180°

(E) F-P-F bond angle of PF₃ molecule = 96.3°

(F) H-C-H bond angle of CH₄ molecule = 109.5°

Explanation:

Bond angle refers to the angle between two adjacent chemical bonds in a molecule. The bond angle is different for different molecular geometry.

The Valence shell electron pair repulsion theory predicts the molecular geometry and shape of the given molecule by the number of lone pairs on central atom and number of atoms bonded to central atom.

(A) SO₂ molecule

The number of atoms bonded to S = 2

Number of lone pairs on S = 1

<u>Therefore, the shape of SO₂ molecule is bent and the O-S-O bond angle is 119°.</u>

(B) BF₃ molecule

The number of atoms bonded to B = 3

Number of lone pairs on B = 0

<u>Therefore, the shape of BF₃ molecule is trigonal planar and the F-B-F bond angle is 120°.</u>

(C) SCl₂ molecule

The number of atoms bonded to S = 2

Number of lone pairs on S = 2

<u>Therefore, the shape of SCl₂ molecule is bent and the Cl-S-Cl bond angle is 103°.</u>

(D) CO₂ molecule

The number of atoms bonded to C = 2

Number of lone pairs on C = 0

<u>Therefore, the shape of CO₂ molecule is linear and the O-C-O bond angle is 180°.</u>

(E) PF₃ molecule

The number of atoms bonded to P = 3

Number of lone pairs on P = 1

<u>Therefore, the shape of PF₃ molecule is trigonal pyramidal and the F-P-F bond angle is 96.3°.</u>

(F) CH₄ molecule

The number of atoms bonded to C = 4

Number of lone pairs on C = 0

<u>Therefore, the shape of CH₄ molecule is tetrahedral and the H-C-H bond angle is 109.5°.</u>

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Redox reactions can be balanced using the half-reaction method. It has the following steps:

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<em>Acidic solution</em>

SO₄²⁻(aq) + Sn²⁺(aq) + X ⇄ H₂SO₃(aq) + Sn⁴⁺(aq) + Y

1.

Reduction: SO₄²⁻ ⇒ SO₃²⁻

Oxidation: Sn²⁺ ⇒ Sn⁴⁺

2.

2 H⁺ + SO₄²⁻ ⇒ SO₃²⁻ + H₂O

Sn²⁺ ⇒ Sn⁴⁺

3.

2 H⁺ + SO₄²⁻ + 2 e⁻ ⇒ SO₃²⁻ + H₂O

Sn²⁺ ⇒ Sn⁴⁺ + 2 e⁻

4.

1 x [2 H⁺ + SO₄²⁻ + 2 e⁻ ⇒ SO₃²⁻ + H₂O]

1 x [Sn²⁺ ⇒ Sn⁴⁺ + 2 e⁻]

5.

2 H⁺ + SO₄²⁻ + 2 e⁻ + Sn²⁺ ⇄ SO₃²⁻ + H₂O + Sn⁴⁺ + 2 e⁻

2 H⁺ + SO₄²⁻ + Sn²⁺ ⇄ SO₃²⁻ + H₂O + Sn⁴⁺

Taking this to the general equation:

SO₄²⁻(aq) + Sn²⁺(aq) + 2 H⁺(aq) ⇄ H₂SO₃(aq) + Sn⁴⁺(aq) + H₂O(l)

Since H⁺ are spectator ions, they are not balanced automatically through this method and we have to balance them manually. In this case, we need to add 2 more H⁺ to the left.

SO₄²⁻(aq) + Sn²⁺(aq) + 4 H⁺(aq) ⇄ H₂SO₃(aq) + Sn⁴⁺(aq) + H₂O(l)

<em>Basic solution</em>

MnO₄⁻(aq) + F⁻(aq) + X ⇄ MnO₂(s) + F₂(aq) + Y

1.

Reduction: MnO₄⁻ ⇒ MnO₂

Oxidation: F⁻ ⇒ F₂

2.

2 H₂O + MnO₄⁻ ⇒ MnO₂ + 4 OH⁻

2 F⁻ ⇒ F₂

3.

2 H₂O + MnO₄⁻ + 3 e⁻ ⇒ MnO₂ + 4 OH⁻

2 F⁻ ⇒ F₂ + 2 e⁻

4.

2 × (2 H₂O + MnO₄⁻ + 3 e⁻ ⇒ MnO₂ + 4 OH⁻)

3 × (2 F⁻ ⇒ F₂ + 2 e⁻)

5.

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4 H₂O + 2 MnO₄⁻ + 6 F⁻ ⇄ 2 MnO₂ + 8 OH⁻ + 3 F₂

Taking this to the general equation:

2 MnO₄⁻(aq) + 6 F⁻(aq) + 4 H₂O ⇄ 2 MnO₂(s) + 3 F₂(aq) + 8 OH⁻

This equation is balanced.

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