Question

The magnetic moment of complex anion [Cr(NO)(NH3)(CN)4]2- is

Answer 1

Find the d electron count on the metal center (i.e. find it's oxidation state), then use the formula:

\(\sf \mu=\sqrt{n(n+2)}\)

where \(\mu\) is the magnetic moment in Bohr Magnetons, and \(n\) the number of unpaired electrons (i.e. \(d\) electrons).

(this is assuming you're only using the spin-spin coupling term, if you're using the spin-orbit coupling you have to extent that formula a bit)

Answer 2

But here we have ligands too (ligands are strong) and Cr is in +2 oxidation state that means 3d4 so pairing will occur in d orbital and hence we will be having the 4 electrons paired up in just two orbitals and no unpaired electron is left . so dipole moment should be 0

Answer 3

@priyar

Answer 4

i got 1.73 as answer ?

Answer 5

How?

Answer 6

is it right ?

Answer 7

Shayad

Answer 8

NO is positively charged

Answer 9

NO=Nitrosonium ion

Answer 10

Shi h ans tumhara

Answer 11

Bt nitrosyl is a neutral ligand NO
@mayankdevnani

Answer 12

Even negative too

Answer 13

there exists both properties :-
NO can be neutral and positively charged

Answer 14

nope not negative !

Answer 15

but i have a question !

Answer 16

Nitrosyl is NO
Nitrosonium is NO+
Nitroso is NO-

Answer 17

how can we came to know that NO is positively charged in this case ?

Answer 18

Well !..
We simply can't make it out whether what charge it might have out of the three...

Answer 19

yep! that's the problem

Answer 20

so we can tick answer 0 too

Answer 21

@chmvijay @Zale101 @Photon336 @aaronq

Answer 22

Yep...ryt...