Question

The oxidation number of Sb in$K_{2}H_{2}Sb_{2}O_{7}$ is

• A.
  • 2
• B.
  • 5
• C. – 2
• D. – 5

Answer 1

Answer: (B)

• 5

Answer 2

As $K_{2}H_{2}Sb_{2}O_{7}$ is a neutral species, therefore the sum of the oxidation number of all the atoms present in $K_{2}H_{2}Sb_{2}O_{7}$ = 0

Let oxidation number per Sb atom = x

$\therefore$ $2 + 2 + 2x + ( - 2)7 = 0$

$\lbrack\because$Oxidation number of K, H and O are +1, +1 and –2 respectively]

or $4 + 2x - 14 = 0$or $2x - 10 = 0$or $x = + 5$