The spin-only magnetic moments of $[M n (C N)_{6}]^{4 -}$ and $[M n B r_{4}]^{2 -}$ in Bohr Magnetons, respectively, are:

5.92

and

5.92

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

4.89

and

1.73

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

1.73

and

5.92

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

1.73

and

1.73

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

Open in App

Solution

The correct option is C $5.92$ and $5.92$
$M n^{2 +} = [A r] (3 d)^{5}$
There is one electron in each of the five orbitals.
$C N^{-}$ is a strong field ligand it will pair up the electrons so magnetic moment will be 1.73 BM
$B r^{-}$ is a weak field ligand it will not pair up the electrons so magnetic moment will be 5.92 BM

Suggest Corrections

Similar questions

C r (C O)_{6}

_{26} F e^{+ 3}

[P d (F) (C l) (B r) (I)]^{2 -}

[C F S E]

[F e (C N_{6}]^{n - 6}

Match the following sets of number of unpaired electrons and magnetic moments.

Set -I Set -II

A) Sc $^{3 +}$ 1) 5, 5.92 B.M

B) V $^{2 +}$ 2) 1, 1.73 B.M.

C) Fe $^{3 +}$ 3) 0, 0 B.M.

D) Cu $^{2 +}$ 4) 3, 3.87 B.M

The correct matching for A B C D is:

[M n B r_{4}]^{2 -}

Join BYJU'S Learning Program