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JEE Advanced 2022 Paper 1 · Chemistry · Coordination Compounds · medium

Match the complex ions in List-I with their crystal field stabilisation energy (CFSE, in units of Δo, pairing energy ignored) in List-II. Use H2O and…

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Match the complex ions in List-I with their crystal field stabilisation energy (CFSE, in units of Δo\Delta_o, pairing energy ignored) in List-II. Use H2O\mathrm{H_2O} and F−\mathrm{F^-} as weak-field ligands and CN−\mathrm{CN^-} and NH3\mathrm{NH_3} as strong-field ligands (Cr 24, Mn 25, Fe 26, Co 27).

| List-I | List-II |
|---|---|
| P. [Mn(H2O)6]2+\mathrm{[Mn(H_2O)_6]^{2+}} | 1. 00 |
| Q. [CoF6]3−\mathrm{[CoF_6]^{3-}} | 2. −0.4 Δo-0.4\,\Delta_o |
| R. [Cr(H2O)6]3+\mathrm{[Cr(H_2O)_6]^{3+}} | 3. −1.2 Δo-1.2\,\Delta_o |
| S. [Fe(CN)6]3−\mathrm{[Fe(CN)_6]^{3-}} | 4. −2.0 Δo-2.0\,\Delta_o |
| | 5. −2.4 Δo-2.4\,\Delta_o |
  1. A. P→1, Q→2, R→3, S→4
  2. B. P→1, Q→2, R→3, S→5
  3. C. P→2, Q→1, R→3, S→4
  4. D. P→1, Q→5, R→3, S→2
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Answer: (A) P→1, Q→2, R→3, S→4

CFSE =(−0.4 nt2g+0.6 neg)Δo=(-0.4\,n_{t_{2g}}+0.6\,n_{e_g})\Delta_o. P: Mn2+\mathrm{Mn^{2+}} is d5d^5, weak-field water gives high spin t2g3eg2t_{2g}^3e_g^2: −1.2+1.2=0-1.2+1.2=0: entry 1. Q: Co3+\mathrm{Co^{3+}} is d6d^6 and F−\mathrm{F^-} is weak field, so high spin t2g4eg2t_{2g}^4e_g^2: −1.6+1.2=−0.4-1.6+1.2=-0.4: entry 2. R: Cr3+\mathrm{Cr^{3+}} is d3d^3, t2g3t_{2g}^3: −1.2-1.2: entry 3. S: Fe3+\mathrm{Fe^{3+}} is d5d^5 and CN−\mathrm{CN^-} is strong field, so low spin t2g5t_{2g}^5: −2.0-2.0: entry 4. Entry 5 (−2.4-2.4) is the value for low-spin d6d^6 such as [Co(NH3)6]3+\mathrm{[Co(NH_3)_6]^{3+}}, not listed. Correct code: P→1, Q→2, R→3, S→4.
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