For the [Cr(H 2 O) 6 ] 2+ ion, the mean pairing energy P is found to be 23500 cm –1 . The magnitude of Δ 0 is 13900 cm –1 . Calculate the C.F.S.E (cm –1 ) for this complex ion corresponding to high spin state (x) and low spin state (y). Write your answer as 
Text Solution
Verified by ExpertsCHECK THE SOLUTION.
= 96
Sol. For a d 4 ion in a high spin state.
CFSE = – 0.6 Δ o = – 0.6 × (13,900 cm –1 ) = – 8340 cm –1
For a d 4 ion in a low spin state, the net CFSE is,
= – 1.6 Δ o + P = – 1.6 × (13,900 cm –1 ) + 23500 cm –1 = + 1,260 cm –1
Since Δ o (= 13,900 cm –1 ) < P (= 23,500 cm – 1 ), the high spin configuration would be more stable.
Prepare Smarter with CGP Edu
Get practice questions, solutions, and test series in one place.
Write a Review
Share your experience with this question and solution.
Commentary
Send your comment, doubt, correction, or feedback to admin.
Similar Questions
Explore conceptually related problems