Question

The correct increasing order of the oxidation state of sulphur in the anions SO₄²⁻, SO₃²⁻, S₂O₄²⁻, S₂O₆²⁻ is
(a) SO₃²⁻ < SO₄²⁻ < S₂O₄²⁻ < S₂O₆²⁻
(b) SO₄²⁻ < S₂O₄²⁻ < S₂O₆²⁻ < SO₃²⁻
(c) S₂O₄²⁻ < SO₃²⁻ < S₂O₆²⁻ < SO₄²⁻
(d) S₂O₆²⁻ < S₂O₄²⁻ < SO₄²⁻ < SO₃²⁻

Answer 1

^_^ OXIDATION NUMBER ^.^

Final Answer : (c)

Steps :
1) Let the Oxidation of number of sulphur be x.
We know,
Oxidation number of Oxygen = -2
Then we have,
Net Charge = Sum of all net Oxidation Number


2) Then,
a) SO4(2-) :
x + 4*(-2) = -2
=> x = 6
ON (SO4(2-)) = 6

b) SO3(2-)
x + 3(-2) = -2
=> x = 4
ON(SO3(2-)) = 4

c) S2O4(2-) :
x*2 + 4*-2 = -2
=> x = 3
ON(S2O4(2-)) = 3

d) S2O6(2-) :
x (2) + 6(-2) = -2
=> x = 5
ON ( S2O6)(2-) = 5

Now, According to above values :
Increasing Order of ON:
S2O4(2-) < SO3(2-)< S2O6(2-) < SO4(2-)

Answer 2

Answer:

C

Explanation:

1) Let the Oxidation of number of sulphur be x.

Since oxidation number of Oxygen = -2

a) SO4(2-)

x + 4×(-2) = -2

x = 6

ON (SO4(2-)) = 6

b) SO3(2-)

x + 3(-2) = -2

x = 4

ON(SO3(2-)) = 4

c) S2O4(2-)

x×2 + 4×-2 = -2

x = 3

ON(S2O4(2-)) = 3

d) S2O6(2-)

x (2) + 6(-2) = -2

=> x = 5

ON ( S2O6)(2-) = 5

So, concluding :

Increasing Order of Oxidation No.:

S2O4(2-) < SO3(2-)< S2O6(2-) < SO4(2-)

Thank you!

Hope it helps!