Oxidation numbers of P in PO₄³⁻ , of S in SO₄²⁻ and that
of Cr in Cr₂O₇²⁻ are respectively
(a) + 3, + 6 and + 5 (b) + 5, + 3 and + 6
(c) – 3, + 6 and + 6 (d) + 5, + 6 and + 6
Answers
The respective oxidation states are + 5, + 6 and + 6.
Option (D) is correct.
Explanation:
(A) Oxidation numbers of P in PO₄³-
The ion has a (3-) overall charge
The ion is made up of one phosphorus atom and four oxygen atoms
Oxygen's oxidation state in most compounds is −2.
1 x P + 4 x O = -3
P + 4(-2) = -3
P - 8 = - 3
P = 8 - 3 = +5
(B) Oxidation numbers of S in SO₄²⁻
The ion has a (2-) overall charge
The ion is made up of one sulfur atom and four oxygen atoms
Oxygen's oxidation state in most compounds is −2.
1 x S + 4 x O = -2
1 x S + 4 (-2) = -2
S -8 = -2
S = 8 - 2 = + 6
(C) Oxidation numbers of Cr in Cr₂O₇²⁻
The ion has a (2-) overall charge
The ion is made up of two chromium atoms ad seven oxygen atoms
Oxygen's oxidation state in most compounds is −2.
2 x Cr + 7 x O = -2
2 Cr + 7 x(-2) = -2
2 Cr - 14 = -2
2 Cr = 14 - 2 = 12
Cr = 12 / 2 = +6
Thus the respective oxidation states are + 5, + 6 and + 6
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