JEE Main20264 April 2026Evening ShiftChemistryd and f Block ElementsActual
Consider |x| is the difference in oxidation states of Mn in highest manganese fluoride and highest manganese oxide. The ions with |x| number of unpaired electrons from the following are: A. Sc³⁺ B. Zn²⁺ C. V²⁺ D. Fe²⁺ E. Co²⁺ Choose the correct answer from the options given below:
Options
- AA and B Only
- BC, D and E Only
- CC and E Only
- DB and E Only
Correct answer
C. C and E Only
Step-by-step solution
The highest manganese fluoride is MnF₄ , where the oxidation state of Mn is +4 . The highest manganese oxide is Mn₂O₇ , where the oxidation state of Mn is +7 . The difference in oxidation states is |x| = |7 - 4| = 3 . The number of unpaired electrons in the given ions are: Sc³⁺ : [Ar] 3d^0 0 unpaired electrons Zn²⁺ : [Ar] 3d¹⁰ 0 unpaired electrons V²⁺ : [Ar] 3d^3 3 unpaired electrons Fe²⁺ : [Ar] 3d^6 4 unpaired electrons Co²⁺ : [Ar] 3d^7 3 unpaired electrons Thus, V²⁺ (C) and Co²⁺ (E) have 3 unpaired electrons. Ans