Consider the following complex ions, P, Q and R.
P = [FeF 6 ] 3– , Q = [V(H 2 O) 6 ] 2+ and R = [Fe(H 2 O) 6 ] 2+ .
The correct order of the complex ions, according to their spin-only magnetic moment values (in B.M.) is
Text Solution
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(b ) P = [FeF 6 ] 3– ox. no. of Fe = +3 , configuration : - 3d 5 4s 0
As F – is weak ligand, pairing does not take place.
so it has 5 unpaired electron
Q = [V(H 2 O) 6 ] 2+ ox. no. of V = + 2, configuration 3d 3 4s 0
It has 3 unpaired electrons.
R = [Fe(H 2 O) 6 ] 2+ , ox. no. of Fe = +2, configuration 3d 6 , 4s 0
As H 2 O is weak ligand, pairing does not take place, so it has 4 unpaired electron
⇒ order of spin only magnetic moment ⇒ Q < R < P
so, answer is .
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