Download the app

Questions  

Spin only magnetic moment is 1.732 B.M for

Unlock the full solution & master the concept.

Get a detailed solution and exclusive access to our masterclass to ensure you never miss a concept
By Expert Faculty of Sri Chaitanya
a
Ti2+
b
Zn2+
c
Co2+
d
Cu2+

Ready to Test Your Skills?

Check Your Performance Today with our Free Mock Tests used by Toppers!

detailed solution

Correct option is D

Spin only magnetic moment, μs=nn+2 B.M ; n=number of unpaired electrons ;One B.M (Bohr Magneton) = 9.273×10-24 A.m2 ;Sc3+, Ti4+, Cu+, Zn2+…..n=0, μs=0 ; DiamagneticTi3+, Cu2+…..n=1, μs=3=1.732 B.M=1.732×9.273×10-24 A.m2 ;Ti2+, V3+, Ni2+…..n=2, μs=8=2.82 B.M=2.82×9.273×10-24 A.m2 ;Cr3+, V2+,Co2+…..n=3, μs=15=3.87 B.M=3.87×9.273×10-24 A.m2 ;Co3+, Fe2+, Mn3+, Cr2+…..n=4, μs=24=4.9 B.M=4.9×9.273×10-24 A.m2 ;Fe3+, Mn2+…..n=5, μs=35=5.9 B.M=5.9×9.273×10-24 A.m2 ;μobs>μcal ......Fe2+, Co2+, Ni2+ ; In these cases orbital contributon cannot be ignored.


Similar Questions

The spin magnetic moment of cobalt in the compound HgCoSCN4 is

Get your all doubts cleared from
Sri Chaitanya experts

counselling
india
+91

whats app icon
phone icon