Match List I with List II List I List II A. Cu I. +3 B. Sc II. +7 C.…

The correct answer is Option (2) → A-IV, B-I, C-II, D-III

Each element is matched with an oxidation state it shows; the states come from the electrons in its 3d and 4s orbitals.

• (A) Cu — This matches (IV) +1. Cu (Z = 29) is [Ar]3d¹⁰4s¹; losing the 4s electron gives Cu⁺ (3d¹⁰). Cu shows +1 and +2, and +2 is used by Fe here.

• (B) Sc — This matches (I) +3. Sc (Z = 21) is [Ar]3d¹4s²; it shows only +3 (3d⁰).

• (C) Mn — This matches (II) +7. Mn (Z = 25) is [Ar]3d⁵4s²; using all seven electrons gives +7, its maximum, as in \( \ce{KMnO4} \).

• (D) Fe — This matches (III) +2. Fe (Z = 26) is [Ar]3d⁶4s²; losing the two 4s electrons gives Fe²⁺, one of its common states.

Shortcut: Sc shows only +3, so B-I; this appears in Options (1) and (2). Mn is the only element here that can show +7 (C-II), which leaves Option (2).

NCERT (2026-27 edition):
Oxidation States – “One of the notable features of a transition elements is the great variety of oxidation states these may show in their compounds. Table 4.3 lists the common oxidation states of the first row transition elements.”

Oxidation States of the first row Transition Metal (Table 4.3)