Chemistry Q62 – JEE-MAIN 2026

Home Chemistry Redox ReactionsQuestion #554

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.

S c^{3 +}

Z n^{2 +}

V^{2 +}

F e^{2 +}

C o^{2 +}

Clue 1 of 3: Determine Oxidation States

Identify the highest stable oxidation state of Manganese in its fluoride and oxide compounds.

🧠 Full Solution Path

Step 1: Determine the value of

| x |

✦ Active

The highest oxidation state of Manganese (Mn) in an oxide is +7, found in $M n_{2} O_{7}$ . The highest stable oxidation state of Mn in a fluoride is +4, found in $M n F_{4}$ . The difference in these oxidation states is $| x | = | 7 - 4 | = 3$ .

Step 2: Determine unpaired electrons for each ion○ Expand

We need to find ions with 3 unpaired electrons. Let's determine the electron configuration and number of unpaired electrons for each given ion:

S c^{3 +} : [A r] ⟹ 0 unpaired electrons

Z n^{2 +} : [A r] 3 d^{10} ⟹ 0 unpaired electrons

V^{2 +} : [A r] 3 d^{3} ⟹ 3 unpaired electrons

F e^{2 +} : [A r] 3 d^{6} ⟹ 4 unpaired electrons

C o^{2 +} : [A r] 3 d^{7} ⟹ 3 unpaired electrons

💡 Teacher's Secret Hint

Remember to remove electrons from the outermost s-orbital first when forming cations of transition metals.

Step 3: Identify ions with

| x |

unpaired electrons○ Expand

Based on the calculations, the ions with 3 unpaired electrons (which is $| x |$ ) are $V^{2 +}$ and $C o^{2 +}$ . This corresponds to options C and E.

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Chemistry Question 62 – JEE-MAIN 2026

Clue 1 of 3: Determine Oxidation States

🧠 Full Solution Path