What is the magnetic moment of $K_{3} [F e F_{6}]$

5.91 B.M.

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

4.89 B.M.

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

3.87 B.M.

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

6.92 B.M.

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

Open in App

Solution

The correct option is A 5.91 B.M.
$K_{3} [F e F_{6}]$
$F e^{3 +} = [A r] 3 d^{5} 4 s^{0}$
$3 d 4 s 4 p$
$\begin{matrix} 1 & 1 & 1 & 1 & 1 \end{matrix} \begin{matrix} \end{matrix} \begin{matrix} \end{matrix}$
Number of unpaired electrons = 5
Magnetic moment = $\sqrt{n (n + 2)} = \sqrt{5 (5 + 2)}$
$= \sqrt{35} = 5.91 B.M.$

Suggest Corrections

Similar questions

[F e F_{6}]^{3 -}

K_{3} [F e {(oxalate)}_{3}]

K_{3} [R u {(oxalate)}_{3}]

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

[F e F_{6}]^{3 -}

{[N i {(C N)}_{4}]}_{4}^{2 -}, {[M n B r_{4}]}_{4}^{2 -}

{[F e F_{6}]}_{6}^{4 -}

K_{3} [C o (C O_{3})_{3}]

Join BYJU'S Learning Program