JEE Advanced 2019 Paper 2 Chemistry: Paragraph 1

Write the $d$-configurations and fill the $t_{2g}$ and $e_g$ sets according to ligand strength. (P) $\mathrm{Cr^{2+}}$ is $d^4$; with weak-field $\mathrm{H_2O}$ it is high spin, $t_{2g}^3e_g^1$, giving $4$ unpaired electrons, so P→5. (Q) $\mathrm{Fe^{2+}}$ is $d^6$; with strong-field $\mathrm{CN^-}$ it is low spin, $t_{2g}^6$, giving $0$ unpaired electrons, so Q→1. (R) $\mathrm{Co^{2+}}$ is $d^7$; with $\mathrm{NH_3}$ it is high spin, $t_{2g}^5e_g^2$, giving $3$ unpaired electrons, so R→4. (S) $\mathrm{Cu^{2+}}$ is $d^9$, $t_{2g}^6e_g^3$, with $1$ unpaired electron, so S→2. The code P→5, Q→1, R→4, S→2 is therefore correct. The others come from treating Co as low spin (1 unpaired), Fe as high spin (4), or Cr as low spin (2).

Use the same spin assignments as before. (P) $\mathrm{[Cr(H_2O)_6]^{2+}}$ is high-spin $d^4$, $t_{2g}^3e_g^1$: one electron in $e_g$, so P→2 (List-II value 1). (Q) $\mathrm{[Fe(CN)_6]^{4-}}$ is low-spin $d^6$, $t_{2g}^6e_g^0$: no $e_g$ electrons, so Q→1 (value 0). (R) $\mathrm{[Co(NH_3)_6]^{2+}}$ is high-spin $d^7$, $t_{2g}^5e_g^2$: two $e_g$ electrons, so R→3 (value 2). (S) $\mathrm{[Cu(H_2O)_6]^{2+}}$ is $d^9$, $t_{2g}^6e_g^3$: three $e_g$ electrons, so S→4 (value 3). The correct code is P→2, Q→1, R→3, S→4. The distractors arise from taking Co as low spin ($e_g^1$), Fe as high spin ($e_g^2$) or Cr as low spin ($e_g^0$).