The oxidation states of S atoms in S4 O2-6 from left to right respectively are

Question

The oxidation states of S atoms in S4 O2-6 from left to right respectively are (A) + 6, 0, 0, + 6 (B) + 3, + 1, +.1, + 3 (C) + 5, 0, 0, + 5 (D) 4, + 1, + 1, + 4

Tardigrade Admin · Accepted Answer

The structure of S4O2-6 ion is The two S atoms which are linked to each other have O.N. of zero. The O.N. of other two S atoms can be calculated as follows: -2 + x+ 2(-2) + 0 + 0 + x + 2 (-2)- 2 = -2 or 2x- 12 = -2 or 2x = 10 or x = + 5 Thus, option (c) is correct.

Q. The oxidation states of $S$ atoms in $S_{4} O_{6}^{2 -}$ from left to right respectively are

+ 6, 0, 0, + 6

+ 3, + 1, +.1, + 3

+ 5, 0, 0, + 5

4, + 1, + 1, + 4

The structure of $S_{4} O_{6}^{2 -}$ ion is
The two S atoms which are linked to each other have O.N. of zero. The O.N. of other two $S$ atoms can be calculated as follows :
$- 2 + x + 2 (- 2) + 0 + 0 + x + 2 (- 2) - 2 = - 2$ or $2 x - 12 = - 2$ or $2 x = 10$ or $x = + 5$ Thus, option (c) is correct.

Q. The oxidation states of S atoms in S4​O62−​ from left to right respectively are

+ 6, 0, 0, + 6

+ 3, + 1, +.1, + 3

+ 5, 0, 0, + 5

4, + 1, + 1, + 4

The structure of S4​O62−​ ion is The two S atoms which are linked to each other have O.N. of zero. The O.N. of other two S atoms can be calculated as follows : −2+x+2(−2)+0+0+x+2(−2)−2=−2 or 2x−12=−2 or 2x=10 or x=+5 Thus, option (c) is correct.

Q. The oxidation states of $S$ atoms in $S_{4} O_{6}^{2 -}$ from left to right respectively are