The oxidation state of Fe in the 'Brown ring' complex $[F e (H_{2} O)_{5} N O] S O_{4}$ is:

0.0

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

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

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

Open in App

Solution

The correct option is C 1
$[F e ↓ (H_{2} O)_{5} N O] S O_{4}$
$x + 5 (0) + 1 - 2 = 0$
$\Rightarrow x - 1 = 0 \Rightarrow x = + 1$

Suggest Corrections

Similar questions

F e

[F e (H_{2} O)_{5} N O] S O_{4}

[F e (H_{2} O)_{5} N O] S O_{4}

The oxidation state of Fe in browns ring complex $[F e (H_{2} O)_{5} N O^{+}] S O_{4}$ is:

[F e (H_{2} O)_{5} N O^{+}] S O_{4}

[{F e (H_{2} O)}_{5} N O] {S O}_{4}

Join BYJU'S Learning Program