Step 1: Understanding the Concept:
This problem deals with the Crystal Field Theory (CFT) and Valence Bond Theory (VBT) applied to coordination complexes. The magnetic property (paramagnetic or diamagnetic) and the count of unpaired electrons depend on the oxidation state of the central transition metal ion, its $d$-electron configuration, and the field strength of the surrounding ligands (strong field ligands induce electron pairing, while weak field ligands do not).
Step 2: Key Formula or Approach:
- Identify the oxidation state and $d^n$ configuration of the central metal in each complex.
- Determine whether the ligand is a Weak Field Ligand (WFL) or a Strong Field Ligand (SFL) using the spectrochemical series.
- Fill the $d$-orbitals according to the geometry to count the unpaired electrons ($n$).
- Species with $n>0$ are paramagnetic; species with $n = 0$ are diamagnetic.
Step 3: Detailed Explanation:
Analysis of Statement I:
1. $[\text{Cu}(\text{NH}_3)_4]^{2+}$: $\text{Cu}$ is in the $+2$ oxidation state. Configuration of $\text{Cu}^{2+}$ is $[ \text{Ar} ] 3d^9$. A $3d^9$ system always has exactly 1 unpaired electron.
2. $[\text{Ni}(\text{NH}_3)_6]^{2+}$: $\text{Ni}$ is in the $+2$ oxidation state. Configuration of $\text{Ni}^{2+}$ is $[ \text{Ar} ] 3d^8$. In an octahedral field, $3d^8$ fills as $t_{2g}^6 e_g^2$. It has 2 unpaired electrons.
3. $[\text{Mn}(\text{H}_2\text{O})_6]^{2+}$: $\text{Mn}$ is in the $+2$ oxidation state. Configuration of $\text{Mn}^{2+}$ is $[ \text{Ar} ] 3d^5$. Since $\text{H}_2\text{O}$ is a weak field ligand, it forms a high-spin octahedral complex ($t_{2g}^3 e_g^2$). It contains 5 unpaired electrons.
Comparing the values, $[\text{Mn}(\text{H}_2\text{O})_6]^{2+}$ clearly has the maximum number of unpaired electrons (5). Therefore, Statement I is true.
Analysis of Statement II:
Let's find the magnetic character of each individual component:
1. $[\text{NiCl}_4]^{2-}$: $\text{Ni}^{2+}$ ($3d^8$). $\text{Cl}^-$ is a weak field ligand causing tetrahedral geometry without pairing. It contains 2 unpaired electrons, making it paramagnetic.
2. $[\text{Ni}(\text{CO})_4]$: $\text{Ni}$ is in the $0$ oxidation state ($3d^8 4s^2$). $\text{CO}$ is an exceptionally strong field ligand, which forces the two $4s$ electrons to move into the $3d$ subshell, generating a completely filled $3d^{10}$ configuration. It contains 0 unpaired electrons, making it diamagnetic.
3. $[\text{Ni}(\text{CN})_4]^{2-}$: $\text{Ni}^{2+}$ ($3d^8$). $\text{CN}^-$ is a strong field ligand that forces pairing in a square planar configuration ($d_{xy}^2 d_{yz}^2 d_{xz}^2 d_{z^2}^2 d_{x^2-y^2}^0$). It contains 0 unpaired electrons, making it diamagnetic.
Now evaluate the given pairs for "only diamagnetic species":
- Pair 1: $\{[\text{NiCl}_4]^{2-} \text{ (Para)}, [\text{Ni}(\text{CO})_4] \text{ (Dia)}\}$ $\rightarrow$ Contains a paramagnetic species.
- Pair 2: $\{[\text{NiCl}_4]^{2-} \text{ (Para)}, [\text{Ni}(\text{CN})_4]^{2-} \text{ (Dia)}\}$ $\rightarrow$ Contains a paramagnetic species.
- Pair 3: $\{[\text{Ni}(\text{CO})_4] \text{ (Dia)}, [\text{Ni}(\text{CN})_4]^{2-} \text{ (Dia)}\}$ $\rightarrow$ Both are diamagnetic.
There is only one pair containing exclusively diamagnetic species. Statement II claims there are two such pairs, which makes Statement II is false.
Step 4: Final Answer:
Statement I is true but Statement II is false, matching option (4).