JEE Main2016ChemistryCoordination CompoundsActual
Identify the correct trend given below: (Atomic No. = Ti : 22, Cr : 24 and Mo : 42)
Options
- A∆ o f C r H 2 O 6 2 + > M o H 2 O 6 2 + and ∆ o f T i H 2 O 6 3 + > T i H 2 O 6 2 +
- B∆ o f C r H 2 O 6 2 + > M o H 2 O 6 2 + and ∆ o f T i H 2 O 6 3 + < T i H 2 O 6 2 +
- C∆ o f C r H 2 O 6 2 + < M o H 2 O 6 2 + and ∆ o f T i H 2 O 6 3 + > T i H 2 O 6 2 +
- D∆ o f C r H 2 O 6 2 + < M o H 2 O 6 2 + and ∆ o f T i H 2 O 6 3 + < T i H 2 O 6 2 +
Correct answer
C. ∆ o f C r H 2 O 6 2 + < M o H 2 O 6 2 + and ∆ o f T i H 2 O 6 3 + > T i H 2 O 6 2 +
Step-by-step solution
As the oxidation state of metal increases the CFSE value increases hence Δ   of   [ Ti ( H 2 O ) 6 ] 3 + > [ Ti ( H 2 O ) 6 ] 2 + Δ increase from 3d to 4d Series. So, Δ ∝ CFSE (Crystal field stabilisation energy) Δ   of   Cr H 2 O 6 2 + < Δ   of   Mo H 2 O 6 2 + Because here Δ depends on Z eff & Z eff of 4d series is more than 3d series