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Question

A complex with maximum spin angular momentum

Options :

  1. [FeF6]3–

  2. [Fe(CN)6]3–

  3. [Fe(H2O)6]2+

  4. [V(H2O)6]2+

Show Answer

Answer :

[FeF6]3–

Solution :

F⊝ with Fe+3 behaves as WFL, Hence pairing does not take place, so it forms high spin complex.

[FeF6]3– ⇒ sp3d2 hybridisation

Number of unpaired electron = 5 [Fe(CN)6]3- ⇒ d2sp3 hybridisation

Fe+3 = 3d5 Number of unpaired electron = 1

[Fe(H2O)6]2+ ⇒ sp3d2

Fe+2 = 3d6

Number of unpaired electron = 4

[V(H2O)6]+2 ⇒ d2sp3 hybridisation

V+2 = 3d3

Number of unpaired electron = 3

Spin angular momentum = S ( S + 1 ) h 2 π

S = total spin quantum no.

More the number of unpaired electron more will be spin angular momentum. [FeF6]3– has 5 unpaired electron hence maximum spin angular momentum value.

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