The oxidation state of sulphur in Na2S4O6 is

Question

The oxidation state of sulphur in Na2S4O6 is (A) + 6 (B) (+3/2) (C) (+5/2) (D) -2

Tardigrade Admin · Accepted Answer

Oxidation state of Na =+1 Oxidation state of O =-2 The average oxidation state of Sulphur ⇒ 2( Na +)+4( S )+6( O 2-)=0 ⇒ 2(+1)+4( x )+6(-2)=0 ⇒ x =(+5/2)

Q. The oxidation state of sulphur in $N a_{2} S_{4} O_{6}$ is

+ 6

$\frac{+ 3}{2}$

$\frac{+ 5}{2}$

-2

Oxidation state of $N a = + 1$
Oxidation state of $O = - 2$
The average oxidation state of Sulphur
$\Rightarrow 2 (N a^{+}) + 4 (S) + 6 (O^{2 -}) = 0$
$\Rightarrow 2 (+ 1) + 4 (x) + 6 (- 2) = 0$
$\Rightarrow x = \frac{+ 5}{2}$

Q. The oxidation state of sulphur in Na2​S4​O6​ is

+ 6

2+3​

2+5​

-2

Oxidation state of Na=+1 Oxidation state of O=−2 The average oxidation state of Sulphur ⇒2(Na+)+4(S)+6(O2−)=0 ⇒2(+1)+4(x)+6(−2)=0 ⇒x=2+5​

Q. The oxidation state of sulphur in $N a_{2} S_{4} O_{6}$ is

$\frac{+ 3}{2}$

$\frac{+ 5}{2}$

Oxidation state of $N a = + 1$
Oxidation state of $O = - 2$
The average oxidation state of Sulphur
$\Rightarrow 2 (N a^{+}) + 4 (S) + 6 (O^{2 -}) = 0$
$\Rightarrow 2 (+ 1) + 4 (x) + 6 (- 2) = 0$
$\Rightarrow x = \frac{+ 5}{2}$