Q1. Match List-I with List-II:
List-I
(А) Em²+/м for Zn is negative
(B) Cr2+ is a reducing agent
(C)

Q: What is the correct answer for CUET UG Chemistry Q1?

The correct answer is d. Let's match the statements:(A) Em²+/M for Zn is negative: This indicates that Zn prefers to remain in its oxidized state (Zn²⁺) rather than being reduced to Zn. This is because Zn²⁺ has a stable d¹⁰ electronic configuration. So, (A) matches (III) Zn²⁺ is more stable than Zn.(B) Cr²⁺ is a reducing agent: Cr²⁺ has a d⁴ configuration. It can readily lose an electron to form Cr³⁺ (d³), which has a sta

Q: Q1. Match List-I with List-II: List-I (А) Em²+/м for Zn is negative (B) Cr2+ is a reducing agent (C) V2O5 has a low melting point (D) Zr and Hf occur together in nature List-II (I) due to almost…

The correct answer is d. Let's match the statements:(A) Em²+/M for Zn is negative: This indicates that Zn prefers to remain in its oxidized state (Zn²⁺) rather than being reduced to Zn. This is because Zn²⁺ has a stable d¹⁰ electronic configuration. So, (A) matches (III) Zn²⁺ is more stable than Zn.(B) Cr²⁺ is a reducing agent: Cr²⁺ has a d⁴ configuration. It can readily lose an electron to form Cr³⁺ (d³), which has a sta

Q: What are the options for this CUET UG Chemistry question?

A. (A)-(III), (B)-(I), (C)-(IV), (D)-(II) · B. (A)-(III), (B)-(II), (C)-(IV), (D)-(I) · C. (A)-(IV), (B)-(III), (C)-(I), (D)-(II) · D. (A)-(III), (B)-(IV), (C)-(II), (D)-(I)

Question

Q1. Match List-I with List-II:
List-I
(А) Em²+/м for Zn is negative
(B) Cr2+ is a reducing agent
(C) V2O5 has a low melting point
(D) Zr and Hf occur together in nature
List-II
(I) due to almost identical radii
(II) Ionic character decreases as oxidation number of metal increases.
(III) Zn2+ is more stable than Zn.
(IV) It attains stable half-filled t2g level.
Choose the correct answer from the options given below:

FirstInTest · Accepted Answer

Let's match the statements:(A) Em²+/M for Zn is negative: This indicates that Zn prefers to remain in its oxidized state (Zn²⁺) rather than being reduced to Zn. This is because Zn²⁺ has a stable d¹⁰ electronic configuration. So, (A) matches (III) Zn²⁺ is more stable than Zn.(B) Cr²⁺ is a reducing agent: Cr²⁺ has a d⁴ configuration. It can readily lose an electron to form Cr³⁺ (d³), which has a stable half-filled t₂g level (in octahedral complexes). Thus, Cr²⁺ acts as a reducing agent. So, (B) matches (IV) It attains stable half-filled t₂g level.(C) V₂O₅ has a low melting point: V₂O₅, where Vanadium is in a +5 oxidation state, has significant covalent character. As the oxidation number of a metal increases, the ionic character of its oxide decreases (and covalent character increases), leading to weaker intermolecular forces and thus a lower melting point compared to ionic oxides. So, (C) matches (II) Ionic character decreases as oxidation number of metal increases.(D) Zr and Hf occur together in nature: Zirconium (Zr) and Hafnium (Hf) are elements of the same group but different periods. Due to the lanthanoid contraction, they have almost identical atomic and ionic radii, leading to very similar chemical properties and their occurrence together in nature. So, (D) matches (I) due to almost identical radii.Combining these matches: (A)-(III), (B)-(IV), (C)-(II), (D)-(I). This corresponds to option (d).

Answer

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

Answer

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

Answer

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

Answer

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

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