$[{F e (H_{2} O)}_{5} N o]^{2 +}$ is formed as brown ring in ${N O}_{3}^{+}$ test. $F e$ in this complex has _________ unpaired electrons.

one

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

two

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

three

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

four

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

Open in App

Solution

The correct option is C three
The magnetic moment of nitroso ferrous sulphate is 3.87 BM. This indicates presence of 3 unpaired electrons. This is due to $3 d^{7}$ electronic configuration of $F e^{+}$ ion.

Suggest Corrections

Similar questions

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

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

N O_{3}^{-}

N O_{3}^{-}

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

{N O}_{3}^{-}

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

{F e S O}_{4}

H_{2} {S O}_{4}

Join BYJU'S Learning Program