Chemistry CUET UG 22-05-2023 Shift 3 Paper

Analysis of complexes and hybridization:

• $[\text{Ni}(\text{H}_2\text{O})_6]^{2+}$: $\text{Ni}^{2+} (3d^8)$, weak field $\text{H}_2\text{O}$, outer orbital octahedral complex with $sp^3d^2$ (correct match).
• $[\text{CoF}_6]^{3-}$: $\text{Co}^{3+} (3d^6)$, weak field fluoride $\text{F}^-$, high spin outer orbital complex with $sp^3d^2$ (correct match).
• $[\text{Cu}(\text{NH}_3)_4]^{2+}$: $\text{Cu}^{2+} (3d^9)$, square planar complex with $dsp^2$ hybridization (correct match).
• $[\text{MnCl}_4]^{2-}$: $\text{Mn}^{2+} (3d^5)$, chloride $\text{Cl}^-$ is a weak field ligand and cannot cause pairing of five half-filled $3d$ electrons. It uses one $4s$ and three $4p$ orbitals giving $sp^3$ hybridization with tetrahedral geometry, NOT $dsp^2$. Therefore, match 4 is incorrect.

Calculate the number of unpaired electrons ($n$) and spin-only magnetic moment $\mu = \sqrt{n(n+2)}\text{ BM}$:

• $[\text{NiCl}_4]^{2-}$: $\text{Ni}^{2+} (3d^8)$, weak field $\text{Cl}^- \rightarrow n = 2$, $\mu = \sqrt{2(4)} = 2.83\text{ BM}$.
• $[\text{Mn}(\text{CN})_6]^{3-}$: $\text{Mn}^{3+} (3d^4)$, strong field $\text{CN}^-$ causes pairing to $t_{2g}^4 e_g^0 \rightarrow n = 2$, $\mu = \sqrt{2(4)} = 2.83\text{ BM}$.
• $[\text{Cr}(\text{NH}_3)_6]^{3+}$: $\text{Cr}^{3+} (3d^3)$, $t_{2g}^3 e_g^0 \rightarrow n = 3$, $\mu = \sqrt{3(5)} = 3.87\text{ BM}$.
• $[\text{CoF}_6]^{3-}$: $\text{Co}^{3+} (3d^6)$, weak field $\text{F}^-$, high spin configuration $t_{2g}^4 e_g^2 \rightarrow n = 4$ unpaired electrons, $\mu = \sqrt{4(6)} = 4.90\text{ BM}$.

$[\text{Pt}(\text{NH}_3)_2\text{Cl}_2]$ is a square planar complex of the type $\text{MA}_2\text{B}_2$. It displays geometrical (cis-trans) isomerism:

• Cis-isomer (cisplatin): The two identical chloride ligands (and the two ammine ligands) occupy adjacent positions (at an angle of $90^\circ$). Cisplatin is an effective anti-cancer drug.
• Trans-isomer: The two chloride ligands occupy opposite positions (at an angle of $180^\circ$).
  Because it does not contain ambidentate ligands (no linkage isomerism), has no counter ions (no ionisation isomerism), and has only one coordination sphere (no coordination isomerism), it specifically exhibits geometrical isomerism.

According to NCERT Class 12 Chemistry, Table 4.8 (Colours of some of the first row hydrated transition metal ions):

• $\text{Mn}^{2+} (3d^5)$: Pale pink $\rightarrow$ (A)-(IV).
• $\text{Fe}^{3+} (3d^5)$: Yellow $\rightarrow$ (B)-(I).
• $\text{Ni}^{2+} (3d^8)$: Green $\rightarrow$ (C)-(II).
• $\text{Cu}^{2+} (3d^9)$: Blue $\rightarrow$ (D)-(III).