-FeH2O5NO-2- has O-N- of Fe as -1 and thus magnetic moment of Fe is -15B-M- If true enter 1 else 0-

The correct answer is #160;1The complex [Fe(H_2O)_5NO]^2+ has 3 unpaired electrons. μ = √(n (n+2 )) #160; #160; #160; = √(3 (3+2 )) # ...

[FeH2O5NO]2+ ...

Question

$[F e (H_{2} O)_{5} N O]^{2 +}$ has $O . N .$ of $F e$ as +1 and thus magnetic moment of $F e$ is $\sqrt{15} B . M .$ . If true enter 1 else 0.

Open in App

Solution

Suggest Corrections

Similar questions

[F e (C N)_{6}]^{3 -}

1.73 B M

[F e (H_{2} O)_{6}]^{3 +}

5.92 B M

[F e (H_{2} O)_{6}]^{2 +}

\sqrt{24} B M

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

S c^{2 +}

C u^{2 +}

F e^{3 +}

F e^{2 +}