Assign the oxidation states of all the atoms in each of the following compounds. a) H2CrO4 b) H2SO4 c) K2Cr2O7 d) HNO3 e) Cr2O7 2

(a) H = +1, Cr = +6, O = -2

(b) H = +1, S = +6, O = -2

(c) K = +1, Cr = +6, O = -2

(d) H = +1, N = +5, O = -2

(e) Cr = +6, O = -2

Explanation:

(a)

In H₂CrO₄

The oxidation state of Hydrogen (H) is +1

The oxidation state of Oxygen (O) is -2

To find the oxidation state of Cr,

Let the oxidation state of Cr be x

Then, (+1×2) + x + (-2×4) = 0

2+x-8 = 0

x = +6

Hence, oxidation state of Cr is +6

(b)

In H₂SO₄

The oxidation state of Hydrogen (H) is +1

The oxidation state of Oxygen (O) is -2

To find the oxidation state of Sulphur (S),

Let the oxidation state of Sulphur be x

∴ Then, (+1×2) + x + (-2×4) = 0

2+x-8 = 0

x = +6

Hence, the oxidation state of S is +6

(c)

In K₂Cr₂O₇

The oxidation state of Potassium (K) is +1

The oxidation state of Oxygen (O) is -2

To find the oxidation state of Chromium,

Let the oxidation state of Chromium be x

Then, (+1 ×2) + (x ×2) + (-2×7) = 0

+2 +2x -14 = 0

2x = +12

x = +6

Hence, the oxidation state of Chromium is +6

(d)

In HNO₃

The oxidation state of Hydrogen (H) is +1

The oxidation state of Oxygen (O) is -2

To find the oxidation state of NItrogen (N),

Let the oxidation state of Nitrogen be x

+1 + x + (-2×3) = 0

1 +x - 6 = 0

x = +5

Hence, the oxidation state of Nitrogen is +5

(d)

In Cr₂O₇²⁻

The oxidation state of Oxygen is -2

To the oxidation state of Chromium (Cr(,

Let the oxidation state of Chromium be x

Then, (2×x) + (-2×7) = -2

2x - 14 = -2

2x = +12

x = +6

Hence, the oxidation state of Chromium is +6

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