4. In which pair of substances does the first underlined atom have a lower oxidation number than the second? A. NH3OH+ NH4– B. H2O H2O2 C. SO3 SO42– D. HCHO C

Option B is correct.

Only this option has the first underlined element with a lower oxidation number than the second amongst the options.

Explanation:

Complete Question

In which pair of substances does the first underlined atom have a lower oxidation number than the second?

A. NH₃OH⁺, NH₄⁻ (N is underlined)

B. H₂O, H₂O₂ (O is underlined)

C. SO₃, SO₄²⁻ (S is underlined)

D. HCHO, C (C is underlined)

Solution

In determination of the oxidation number of an atom in a compound, we first name the unknown oxidation number x.

Then, the total oxidation number of the atoms in the compound is equal to the charge of on the compound (or radical).

So, elements in their neutral state have no charge and no oxidation number.

A. NH₃OH⁺, NH₄⁻ (N is underlined)

N in NH₃OH⁺

Oxidation number of N = x

Oxidation number of H = +1

Oxidation number of O = -2

x + (3×+1) + (-2) + (+1) = +1

x - 3 - 2 + 1 = 1

x = +5

N in NH₄⁻

Oxidation number of N = x

Oxidation number of H = +1

x + (4×1) = -1

x + 4 = -1

x = -1 - 4 = -5

First underlined element has a greater oxidation number than the second. So, this doesn't qualify.

B. H₂O, H₂O₂ (O is underlined)

O in H₂O

Oxidation number of H = +1

Oxidation number of O = x

(2×1) + x = 0

2 + x = 0

x = -2

H₂O₂

Oxidation number of H = +1

Oxidation number of O = x

(2×1) + (2×x) = 0

2 + 2x = 0

2x = -2

x = (-2/2) = -1.

First underlined element has a lower oxidation number than the second. So, this qualifies.

C. SO₃, SO₄²⁻ (S is underlined)

S in SO₃

Oxidation number of S = x

Oxidation number of O = -2

x + (3×-2) = 0

x - 6 = 0

x = +6

SO₄²⁻

Oxidation number of S = x

Oxidation number of O = -2

x + (4×-2) = -2

x - 8 = -2

x = 8 - 2 = +6

First underlined element has the same oxidation number as the second. So, this doesn't qualify.

D. HCHO, C (C is underlined)

C in HCHO

Oxidation number of H = +1

Oxidation number of C = x

Oxidation number of O = -2

+1 + x + 1 - 2 = 0

x = 0

C in C

Oxidation number of C = x

x = 0

First underlined element has the same oxidation number as the second. So, this doesn't qualify.

Hope this Helps!!!

answer: your mm

explanation: