Match List I with List II: List I List II A. E^o_M^2+/M for Zn is neg…

Match List I with List II:

List I	List II
A. \(E^o_{M^{2+}/M}\) for \(Zn\) is negative	I. due to almost identical radii
B. \(Cr^{2+}\) is a reducing agent	II. Ionic character decreases as oxidation number of the metal increases
C. \(V_2O_5\) has a low melting point	III. \(Zn^{2+}\) is more stable than \(Zn\)
D.\(Zr\) and \(Hf\) occur together in nature	IV. It attains stable half-filled \(t_{2g}\) level

Choose the correct answer from the options given below:

Answer & explanation

Correct answer: option 4

The correct answer is Option (4) → A-III, B-IV, C-II, D-I

Each fact about these transition metals follows from the stability of a particular electron configuration or from the size of the atom:

(A) E°(M²⁺/M) for Zn is negative — This matches (III) Zn²⁺ is more stable than Zn. Zn (Z = 30) is [Ar]3d¹⁰4s²; removing the two 4s electrons leaves the stable, completely filled 3d¹⁰ configuration, so zinc readily forms \( \ce{Zn^2+} \) (E° = −0.76 V).
(B) Cr²⁺ is a reducing agent — This matches (IV) It attains stable half-filled t_2g level. Cr (Z = 24) is [Ar]3d⁵4s¹, so \( \ce{Cr^2+} \) is d⁴. By giving up one electron it becomes \( \ce{Cr^3+} \), d³ = t_2g³, a half-filled t_2g level.
\( \displaystyle \ce{Cr^2+ -> Cr^3+ + e^-} \)
(C) V₂O₅ has a low melting point — This matches (II) Ionic character decreases as oxidation number of the metal increases. In \( \ce{V2O5} \), 2x + 5(−2) = 0 gives V = +5. A metal in such a high oxidation state polarises the oxide ion strongly, so the oxide is largely covalent and melts at a low temperature.
(D) Zr and Hf occur together in nature — This matches (I) due to almost identical radii. Because of the lanthanoid contraction, Zr (160 pm) and Hf (159 pm) have nearly the same size and very similar chemistry, so they occur together and are hard to separate.

Match List I with List II:

List I	List II
A. \(E^o_{M^{2+}/M}\) for \(Zn\) is negative	I. due to almost identical radii
B. \(Cr^{2+}\) is a reducing agent	II. Ionic character decreases as oxidation number of the metal increases
C. \(V_2O_5\) has a low melting point	III. \(Zn^{2+}\) is more stable than \(Zn\)
D.\(Zr\) and \(Hf\) occur together in nature	IV. It attains stable half-filled \(t_{2g}\) level

Choose the correct answer from the options given below:

Answer & explanation

Correct answer: option 4

The correct answer is Option (4) → A-III, B-IV, C-II, D-I

Each fact about these transition metals follows from the stability of a particular electron configuration or from the size of the atom:

(A) E°(M²⁺/M) for Zn is negative — This matches (III) Zn²⁺ is more stable than Zn. Zn (Z = 30) is [Ar]3d¹⁰4s²; removing the two 4s electrons leaves the stable, completely filled 3d¹⁰ configuration, so zinc readily forms \( \ce{Zn^2+} \) (E° = −0.76 V).
(B) Cr²⁺ is a reducing agent — This matches (IV) It attains stable half-filled t_2g level. Cr (Z = 24) is [Ar]3d⁵4s¹, so \( \ce{Cr^2+} \) is d⁴. By giving up one electron it becomes \( \ce{Cr^3+} \), d³ = t_2g³, a half-filled t_2g level.
\( \displaystyle \ce{Cr^2+ -> Cr^3+ + e^-} \)
(C) V₂O₅ has a low melting point — This matches (II) Ionic character decreases as oxidation number of the metal increases. In \( \ce{V2O5} \), 2x + 5(−2) = 0 gives V = +5. A metal in such a high oxidation state polarises the oxide ion strongly, so the oxide is largely covalent and melts at a low temperature.
(D) Zr and Hf occur together in nature — This matches (I) due to almost identical radii. Because of the lanthanoid contraction, Zr (160 pm) and Hf (159 pm) have nearly the same size and very similar chemistry, so they occur together and are hard to separate.