Matching Chemical Properties and Explanations

This question requires us to match specific chemical properties or observations related to elements like Zinc (Zn), Chromium (Cr), Vanadium (V), Zirconium (Zr), and Hafnium (Hf) with the correct underlying explanations.

Let's analyze each statement from List-I and find the most appropriate explanation from List-II.

Analysis of Each Matching Pair

(A) E°_M²⁺/M for Zn is negative

• The standard electrode potential ($\text{E}^\circ$) for $\text{M}^{2+}/\text{M}$ couple indicates the tendency of the metal ion to be reduced to the metal. A negative value means the metal has a higher tendency to get oxidized to the metal ion than hydrogen.
• For Zinc, Zn → $\text{Zn}^{2+}$ + 2e^-. The electronic configuration of Zn is [Ar] $3\text{d}^{10} 4\text{s}^2$. The electronic configuration of $\text{Zn}^{2+}$ is [Ar] $3\text{d}^{10}$. The $\text{Zn}^{2+}$ ion has a completely filled d-orbital, which makes it very stable.
• This stability of $\text{Zn}^{2+}$ is the reason why Zn readily loses electrons to form $\text{Zn}^{2+}$, resulting in a negative $\text{E}^\circ_{\text{Zn}^{2+}/\text{Zn}}$ value.
• Therefore, statement (A) matches with explanation (III) Zn²⁺ is more stable than Zn.

(B) Cr²⁺ is a reducing agent

• A reducing agent is a substance that loses electrons (gets oxidized) and causes another substance to gain electrons (get reduced).
• $\text{Cr}^{2+}$ has the electronic configuration [Ar] $3\text{d}^4$. When $\text{Cr}^{2+}$ acts as a reducing agent, it is oxidized, typically to $\text{Cr}^{3+}$.
• $\text{Cr}^{3+}$ has the electronic configuration [Ar] $3\text{d}^3$. In an octahedral crystal field (common in aqueous solutions and complexes), the d-orbitals split into lower energy $\text{t}_{2\text{g}}$ and higher energy $\text{e}_{\text{g}}$ sets. The $3\text{d}^3$ configuration in $\text{Cr}^{3+}$ has three electrons in the $\text{t}_{2\text{g}}$ orbitals ($\text{t}_{2\text{g}}^3 \text{e}_{\text{g}}^0$), which corresponds to a stable half-filled $\text{t}_{2\text{g}}$ level (or a particularly stable arrangement due to crystal field splitting energy).
• The tendency to achieve this stable $\text{d}^3$ configuration in $\text{Cr}^{3+}$ makes $\text{Cr}^{2+}$ (d⁴) readily lose an electron, hence acting as a reducing agent.
• Therefore, statement (B) matches with explanation (IV) It attains stable half-filled t_2g level.

(C) V₂O₅ has a low melting point

• $\text{V}_2\text{O}_5$ is Vanadium(V) oxide, where Vanadium is in the +5 oxidation state.
• In general, as the oxidation state of a metal increases, the bonding in its oxides tends to become more covalent. This is because the highly charged metal ion polarizes the oxide anion more effectively.
• Ionic compounds typically have high melting points due to strong electrostatic forces between ions. Covalent compounds, especially polymeric or molecular ones, tend to have lower melting points.
• $\text{V}_2\text{O}_5$ is known to have significant covalent character compared to oxides with lower oxidation states or oxides of more electropositive metals. This increased covalent nature contributes to its relatively low melting point compared to ionic oxides.
• Therefore, statement (C) matches with explanation (II) Ionic character decreases as oxidation number of metal increases.

(D) Zr and Hf occur together in nature

• Zirconium (Zr) is a 4d transition metal in Group 4, and Hafnium (Hf) is a 5d transition metal also in Group 4.
• Normally, elements in the same group but different periods show a significant increase in atomic and ionic radii down the group. However, between the 4d and 5d series, there is a phenomenon called the Lanthanide Contraction.
• The Lanthanide elements (atomic numbers 58-71) are placed before Hf (atomic number 72) in the periodic table. The poor shielding effect of the intervening 4f electrons causes a greater effective nuclear charge experienced by the outer electrons of the elements following the Lanthanides, resulting in smaller atomic and ionic radii than might be expected.
• Consequently, the atomic and ionic radii of Hf are very close to those of Zr. This similarity in size, along with similar electronic configurations ($4\text{d}^2 5\text{s}^2$ for Zr and $5\text{d}^2 6\text{s}^2$ for Hf), leads to very similar chemical properties.
• Because their chemical properties are so alike, Zr and Hf are very difficult to separate and thus are found together in the same mineral deposits.
• Therefore, statement (D) matches with explanation (I) due to almost identical radii.

Summary of Matching Pairs

The correct matching is (A)-(III), (B)-(IV), (C)-(II), (D)-(I).

Revision Table: Key Chemical Concepts

Additional Information on Chemical Properties

Understanding trends in chemical properties of transition metals is crucial. The properties are influenced by factors like electronic configuration, ionization enthalpies, hydration enthalpies, and lattice energies. The standard electrode potential (E°) is a combined result of sublimation enthalpy, ionization enthalpy, and hydration enthalpy for the $\text{M}^{2+}/\text{M}$ couple. For Zn, while ionization energy is relatively high, the high stability of $\text{Zn}^{2+}$ contributes significantly to the negative E° value.

The reducing nature of species like $\text{Cr}^{2+}$ or $\text{V}^{2+}$ in aqueous solutions is often explained by the relative stability of the resulting ions in the crystal field. The $\text{d}^3$ configuration in octahedral complexes is particularly stable, making species that can attain this configuration by losing electrons strong reducing agents.

The nature of bonding in oxides changes across periods and down groups, and also with the oxidation state of the metal. Higher oxidation states favor covalent bonding, while lower oxidation states of electropositive metals favor ionic bonding. This directly impacts physical properties like melting point.

The Lanthanide contraction is a key concept explaining the similarities between 4d and 5d transition metals, particularly in Group 4 (Zr and Hf), Group 5 (Nb and Ta), and Group 6 (Mo and W). These similarities make their separation challenging in metallurgy.