Engineering
Chemistry
Crystal Field Theory and Valence Bond Theory
Question

Consider the following complex ions, P, Q and R.

P = [FeF6]3–, Q = [V(H2O)6]2+ and  R = [Fe(H2O)6]2+,

the correct order of the complex ions, according to their spin only magnetic moment values (in B.M.) is:

Q < P < R

R < Q < P

R < P < Q

Q < R < P

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Solution

(P) [FeF6]–3   Fe+3   d5 configuration with WFL

unpaired electron = 5

(Q) [V(H2O)6]2+  V+2  d3 configuration

 

unpaired electron = 3

(R) [Fe(H2O)6]2+   Fe+2   d6 configuration with WFL

 

unpaired electron = 4

Spin only magnetic moment (µ) = n(n+2) B.M. (n = unpaired electron)

So,       µ of P > R > Q