Respuesta :
Answer:
(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
