Match List I with List II:

Choose the correct answer from the options given below:

A-III, B-IV, C-II, D-I

The correct answer is option 4. A-III, B-IV, C-II, D-I.

Let us dive  into each match and explain the reasoning behind it:

A. \(E^o_{M^{2+}/M}\) for \(Zn\) is negative → III. \(Zn^{2+}\) is more stable than \(Zn\)

Standard electrode potential (\(E^o\)) measures the tendency of a metal to lose or gain electrons. For zinc, the standard reduction potential (\(E^o_{Zn^{2+}/Zn}\)) is negative (-0.76 V), indicating that zinc metal is not easily reduced to \(Zn^{2+}\), and instead, it prefers to remain in the oxidized \(Zn^{2+}\) form.

Why is \(Zn^{2+}\) more stable: Zinc in the \(+2\) oxidation state has a completely filled \(d^{10}\) electron configuration, which is very stable because there are no partially filled orbitals. This full d-orbital configuration makes \(Zn^{2+}\) more stable than elemental zinc.

Conclusion: The negative electrode potential reflects the stability of \(Zn^{2+}\) compared to \(Zn\), so the match is A-III.

B. \(Cr^{2+}\) is a reducing agent → IV. It attains a stable half-filled \(t_{2g}\) level

Chromium (Cr) has the electron configuration \([Ar] 3d^5 4s^1\) in the ground state. In the \(+2\) oxidation state (\(Cr^{2+}\)), chromium has the configuration \(3d^4\).

Reducing agent property: \(Cr^{2+}\) is a strong reducing agent because it prefers to lose one more electron to form \(Cr^{3+}\), which has a \(3d^3\) configuration. In the octahedral crystal field, this configuration gives chromium a half-filled \(t_{2g}\) subshell, which is particularly stable due to exchange energy.

Conclusion: Since \(Cr^{2+}\) acts as a reducing agent to attain the more stable \(Cr^{3+}\) with a half-filled \(t_{2g}\) level, the match is B-IV.

C. \(V_2O_5\) has a low melting point → II. Ionic character decreases as oxidation number of the metal increases

Vanadium pentoxide (\(V_2O_5\)) is an oxide of vanadium where vanadium is in the \(+5\) oxidation state. As the oxidation state of a metal increases, its ability to form ionic bonds decreases because higher oxidation states lead to increased polarization of the electron cloud around the anion, leading to more covalent character in the bonds.

Low melting point: Since \(V_2O_5\) has more covalent character due to vanadium's high oxidation state, it has a relatively low melting point compared to more ionic metal oxides.

Conclusion: The low melting point is due to the decreased ionic character as the oxidation number increases, so the match is C-II.

D. \(Zr\) and \(Hf\) occur together in nature → I. due to almost identical radii

Zirconium (Zr) and hafnium (Hf) are chemically and physically very similar, which leads them to be found together in minerals. This similarity is mainly due to the lanthanide contraction, a phenomenon where the filling of the 4f orbitals in the lanthanides results in poor shielding of nuclear charge. As a result, elements following the lanthanides (like Hf) have much smaller atomic radii than expected, making the radius of Hf almost identical to Zr.

Conclusion: Due to their nearly identical radii and chemical properties, Zr and Hf are difficult to separate and often occur together in nature, so the match is D-I.

Therefore, the correct option is option 4: A-III, B-IV, C-II, D-I.