Oxidation numebr of Ag in $N a_{3} [A g (S_{2} O_{3})_{2}]$ is:

+ 2

No worries! We‘ve got your back. Try BYJU‘S free classes today!

- 2

No worries! We‘ve got your back. Try BYJU‘S free classes today!

0

No worries! We‘ve got your back. Try BYJU‘S free classes today!

+ 1

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D $+ 1$
O.No. of Na = +1
O.No. of $S_{2} O_{3} = - 2$
Let O.No. of Ag = x
$3 (+ 1) + x + 2 (- 2) = 0$
$3 + x - 4$
$x - 1 = 0$
$x = + 1$

Suggest Corrections

Similar questions

A g

{N a}_{3} [A g {(S_{2} O_{3})}_{2}]

N a_{3} [A g (S_{2} O_{3})_{2}]

N a_{2} S_{2} O_{3}

A g N O_{3}

N a_{2} S_{2} O_{3}

A g N O_{3}

N a_{2} [A g (S_{2} O_{3})_{2}]

N a_{3} [C o (N O_{2})_{6}]

Join BYJU'S Learning Program