D- and F-block Elements NEET 2009: Manganese Oxidation States

What is the correct answer to the d- and f-block configuration question from 2009?

Option C, corresponding to manganese, is correct in the d- and f-block elements NEET 2009 question. Its outer electronic configuration is: $3 d^{5} 4 s^{2}$

The question asks which of these outer electronic configurations belongs to the element capable of displaying the widest variety of oxidation states:

This is a 2009 Chemistry MCQ, tagged hard on this question bank’s own difficulty scale. The expected solving time is 60 seconds, based on the short electron-count-and-recall method, not a measured student average.

Count electrons from both outer subshells, then check manganese’s oxidation-state range. Ease of removing one electron does not answer this question.

How do you count the electrons in all four options?

Count both the outer s electrons and the d electrons. The official solution explains variable oxidation states through the participation of electrons from both subshells in bonding. For these configurations, the general subshell labels correspond to:

n s = 4 s, (n - 1) d = 3 d

Each option below gives the configuration, element and relevant outer-electron total:

A: Vanadium

3 d^{3} 4 s^{2} 3 + 2 = 5

B: Chromium

3 d^{5} 4 s^{1} 5 + 1 = 6

C: Manganese

3 d^{5} 4 s^{2} 5 + 2 = 7

D: Titanium

3 d^{2} 4 s^{2} 2 + 2 = 4

Manganese has the largest relevant valence-electron inventory among these choices. That total is not automatically the number of oxidation states available to the element.

Electron counting is the first step, not the complete proof. Identifying the element lets you connect its configuration to its known oxidation-state range. Use that range to decide which option shows the greatest variety, rather than treating the largest electron total as a universal shortcut.

Why does manganese’s oxidation-state range make C correct?

Manganese exhibits oxidation states from positive two through positive seven, as stated in the official worked solution. Its seven valence electrons across the outer s and d subshells support this broad range:

M n : 3 d^{5} 4 s^{2}

Oxidation states: + 2, + 3, + 4, + 5, + 6, + 7

That establishes C as the element with the largest variety among the four options. It does not mean all these states are equally stable or equally common.

Keep the electron count separate from the oxidation-state count. Manganese has seven relevant valence electrons, but the stated range contains six integer oxidation-state values: $7 - 2 + 1 = 6$

“Largest number of oxidation states” asks about variety. “Highest oxidation state” asks about the largest value. The phrases are not interchangeable, even when the same element satisfies both comparisons in a particular set.

The official method combines electron counting with knowledge of manganese chemistry. Do not replace the second step with “more electrons always means more oxidation states.”

Count 3 d and 4 s electrons \to identify Mn \to recall + 2 to + 7 \to select C

Why is B wrong if its lone outer s electron seems easier to remove?

Ease of an initial electron-removal step does not establish oxidation-state variety. A hypothetical incorrect argument is: “Choose B because its lone outer s electron appears easier to remove.” Even if that removal seems favourable, it does not establish how many oxidation states the element can exhibit.

Both B and C contain the same half-filled d subshell. Chromium has one outer s electron, while manganese has two:

B, C r : & 3 d^{5} 4 s^{1}, 5 + 1 = 6 C, M n : & 3 d^{5} 4 s^{2}, 5 + 2 = 7

The half-filled d subshell therefore does not distinguish these options by itself. Nor does focusing on a single electron-removal event.

The deciding evidence is manganese’s stated range from positive two through positive seven. The stem asks about variety across oxidation states, not the first ionisation step.

How do you solve three related d- and f-block practice questions?

Use charge balance for the first two questions and electron-removal order for the third. These are original related practice questions, not verified past-paper questions. Each worked answer applies the method required by that question.

What is manganese’s oxidation state in manganese dioxide?

Original related practice question 1, worked answer: Manganese has oxidation state positive four in manganese dioxide. Oxygen has oxidation state negative two here, and the compound is neutral: $M n O_{2}$

Let manganese’s oxidation state be the unknown. Set the sum of oxidation states to zero: $x + 2 (- 2) = 0$ $x = + 4$

What are manganese’s oxidation states in potassium manganate and potassium permanganate?

Original related practice question 2, worked answer: Manganese has oxidation state positive six in potassium manganate and positive seven in potassium permanganate. Potassium contributes positive one per atom, and oxygen contributes negative two.

For potassium manganate: $K_{2} M n O_{4}$

2 (+ 1) + x + 4 (- 2) = 0

$x = + 6$ For potassium permanganate: $K M n O_{4}$ $(+ 1) + x + 4 (- 2) = 0$ $x = + 7$

The oxygen contribution is unchanged. One fewer potassium atom means manganese must have an oxidation state one unit higher to maintain neutrality.

What is the electronic configuration of the manganese two-plus ion?

Original related practice question 3, worked answer: Remove the two outer s electrons first, leaving the half-filled d subshell. Starting from neutral manganese:

M n = [A r] 3 d^{5} 4 s^{2}

The resulting ion has the configuration:

M n^{2 +} = [A r] 3 d^{5}

This is cation formation, where electron-removal order matters. Removing the outer s electrons first does not contradict the broader participation of both subshells in transition-element bonding.

Before calculating, classify what the question asks:

Oxidation-state variety: Count both subshells, identify the element and recall its range.
Oxidation state in a compound: Use charge balance.
Cation configuration: Apply electron-removal order.

Next step: the past-paper archive on NEET JEEnius AI and search past NEET papers by year, subject or chapter, each with a worked solution (100 free searches a month).

For a worked example of the same idea, see Electromagnetic Induction NEET 2005: Series LCR Solution.

Frequently asked questions

What is the correct answer to the 2009 d- and f-block configuration question?

Option C, 3d⁵ 4s², is correct and corresponds to manganese. Among the four options, manganese shows the widest variety of oxidation states, with the worked solution citing +2 through +7.

Why is chromium not the answer even though it has one 4s electron?

Ease of removing one electron does not establish how many oxidation states an element can display. Both chromium and manganese have a half-filled 3d subshell, so that feature alone cannot distinguish them. Manganese’s stated oxidation-state range from +2 to +7 is the deciding evidence.

Does manganese have seven oxidation states because it has seven valence electrons?

No: manganese has seven relevant valence electrons across 3d and 4s, but the stated range from +2 to +7 contains six integer oxidation-state values. Electron counting helps identify the element; its known oxidation-state range establishes the variety.

What is the electronic configuration of Mn2+?

Mn²⁺ has the electronic configuration [Ar] 3d⁵. Start with neutral manganese, [Ar] 3d⁵ 4s², and remove the two 4s electrons first.