The spin only magnetic moment of - ZCl 4-2- is 3-87 BM where Z isA- MnB- NiC- CoD- Cu

The correct option is C CoThe spin only magnetic moment is given in quantum mechanics as µ = √(n(n+2)) BM where n is the number of unpaired e and BM (Bohr mag ...

You visited us 4 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Chemistry

Crystal Field Split in Octahedral Complexes

The spin only...

Question

The spin only magnetic moment of $[Z C l_{4}]^{2 -}$ is 3.87 BM where Z is

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

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 C Co
The spin only magnetic moment is given in quantum mechanics as µ = $\sqrt{n (n + 2)}$ BM where n is the number of unpaired e- and BM (Bohr magnetons)
So, $\sqrt{n (n + 2)} = 3.87$
⟹ $n (n + 2) = (3.87)^{2} = 15$
⟹ $n^{2} + 2 n - 15 = 0$
⟹ n = -5 or 3.

So, No. of unpaired electrons are 3
Hence $Z^{2 +} = d^{7}$

Suggest Corrections

Similar questions

Q. The spin only magnetic moment of

[Z C l_{4}]^{2 -}

3.87 B M

where

Z

is:

Q. The calculated spin only magnetic moment of

C u^{2 +}

ion is:

Q. Calculate the 'spin only' magnetic moment of

M^{2 +} (a q)

ion (Z=27).

Spin only magnetic moment of the compound $H g [C o (S C N)_{4}]$ is

Q. Spin only magnetic moment of

F e^{2 +}

is:

Join BYJU'S Learning Program

The spin only magnetic moment of [ZCl4]2− is 3.87 BM where Z is

The correct option is C CoThe spin only magnetic moment is given in quantum mechanics as µ = √n(n+2) BM where n is the number of unpaired e- and BM (Bohr magnetons) So, √n(n+2)=3.87 ⟹ n(n+2)=(3.87)2=15 ⟹ n2+2n−15=0 ⟹ n = -5 or 3. So, No. of unpaired electrons are 3 Hence Z2+=d7

The spin only magnetic moment of $[Z C l_{4}]^{2 -}$ is 3.87 BM where Z is