In the co-ordination compound $N a_{4}$ $[F e) C N)_{5} N O S]$ oxidation state of $F e$ is:

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 B 1
$N a_{4} [F e (C N)_{5} N O]$
$[F e (C N)_{5} N O]^{- 4}$
NO is a natural ligand thus it's
oxidation state is 'O'
Let oxidation state of Fe is 'x'
$(x \times 1) + 5 \times (- 1) = - 4$
$x = - 4 + 5$
$x = + 1$
Oxidation state of Fe is +1 in above
compound.

Suggest Corrections

Similar questions

N a_{4} [F e {(C N)}_{5} (N O S)]

N a_{4} [F e (C N)_{5} (N O S)]

N a_{2} [F e (N O) {(C N)}_{5}]

N a_{4} [F e {(C N)}_{5} (N O S)]

O . S .

F e

N a_{2} [F e (C N)_{5} N O]

N a_{4} [F e (C N)_{5} N O S]

O s - - - (C O)_{5}

Join BYJU'S Learning Program