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. Sc^ 3+ B. Zn^ 2+ C. V^ 2+ D. Fe^ 2+ E. Co^ 2+ Choose the correct answer from the options given below:
C. C and E Only
The highest manganese fluoride is MnF_4, where the oxidation state of Mn is +4. The highest manganese oxide is Mn_2O_7, 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: Sc^ 3+ : [Ar] 3d^0 0 unpaired electrons Zn^ 2+ : [Ar] 3d^ 10 0 unpaired electrons V^ 2+ : [Ar] 3d^3 3 unpaired electrons Fe^ 2+ : [Ar] 3d^6 4 unpaired electrons Co^ 2+ : [Ar] 3d^7 3 unpaired electrons Thus, V^ 2+ (C) and Co^ 2+ (E) have 3 unpaired electrons. Answer: C and E Only
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