D- and f-block Elements NEET 2011: Divalent Ion Stability

Which option correctly ranks the divalent ions in NEET 2011?

Correct answer: B, with the descending stability order below. The d- and f-block elements NEET 2011 question compares the divalent ions of chromium, manganese, iron and cobalt: $C o > M n > F e > C r$

The supplied atomic numbers are Cr 24, Mn 25, Fe 26 and Co 27. The options are:

The deciding evidence is the aqueous reduction-potential order, not a half-filled-shell shortcut. Manganese’s half-filled divalent ion matters, but it does not place manganese above cobalt in this comparison.

The question bank classifies this question as hard, with an expected solving time of 60 seconds. These are the bank’s classification and estimate, not measured student performance.

What does the reduction potential actually compare?

The official method compares each divalent ion with its corresponding trivalent ion in aqueous solution. A more positive standard reduction potential means a greater tendency for the trivalent ion to accept an electron, favouring formation of the divalent product. Write the reduction direction explicitly:

M^{3 +} (a q) + e^{-} ⟶ M^{2 +} (a q)

The trivalent species on the left accepts the electron. The divalent species on the right is the product whose formation is favoured by a more positive potential.

The relevant potential is the trivalent-to-divalent couple:

E^{\circ} (M^{3 +} / M^{2 +})

Do not replace it with the divalent-to-metal couple:

E^{\circ} (M^{2 +} / M)

That second couple describes reduction of a divalent ion to the metal, not the comparison requested here. Check both charge states before using any potential.

Here, “more stable” means more favoured relative to the corresponding trivalent state in the stated aqueous comparison. It does not mean an ion has a fixed stability independent of solvent, ligands or other chemical conditions. Keep the environment and the pair of charge states consistent throughout the ranking.

Why does cobalt rank above manganese in the worked solution?

Cobalt comes first because its trivalent-to-divalent couple has the most positive standard reduction potential. Manganese follows, then iron, while the chromium couple is negative. Apply the same reduction direction to every element rather than ranking isolated electronic configurations.

Step 2: Write the four reductions. All four refer to aqueous ions:

C o^{3 +} + e^{-} & ⟶ C o^{2 +} M n^{3 +} + e^{-} & ⟶ M n^{2 +} F e^{3 +} + e^{-} & ⟶ F e^{2 +} C r^{3 +} + e^{-} & ⟶ C r^{2 +}

Step 3: Use the supplied qualitative potential order. No numerical electrode potentials are needed:

E^{\circ} (C o^{3 +} / C o^{2 +}) > E^{\circ} (M n^{3 +} / M n^{2 +}) > E^{\circ} (F e^{3 +} / F e^{2 +}) > E^{\circ} (C r^{3 +} / C r^{2 +})

Step 4: Rank the divalent products using that order. A more positive potential favours the reduction as written, giving this relative stability order in the question’s comparison:

C o^{2 +} > M n^{2 +} > F e^{2 +} > C r^{2 +}

Mapping those ions back to the printed element symbols gives option B: $C o > M n > F e > C r$

Manganese’s divalent ion does have a half-filled subshell:

M n^{2 +} : 3 d^{5}

That contributes to its stability, but does not override the stronger aqueous tendency of trivalent cobalt to reduce to divalent cobalt. Electronic configuration helps explain the chemistry; the supplied reduction-potential comparison decides this question.

Exam-time sequence: identify the couple, read the reduction direction, order the standard reduction potentials, then match the option.

How can mixing charge states lead to wrong option A?

Option A can result from crediting divalent iron with the half-filled configuration that actually belongs to trivalent iron. This is a possible faulty chain, not a claim about how often students choose it. The resulting order is: $F e > M n > C o > C r$

The faulty reasoning runs like this:

Put iron first because “iron has a stable half-filled configuration.”
Put divalent manganese above divalent cobalt solely because manganese is half-filled.
Leave chromium last.

The first step attaches the configuration to the wrong ion. Iron loses its two outer s-subshell electrons first; removing one further electron from the d subshell gives the trivalent ion:

F e & : [A r] 3 d^{6} 4 s^{2} F e^{2 +} & : [A r] 3 d^{6} F e^{3 +} & : [A r] 3 d^{5}

For manganese, the half-filled configuration genuinely belongs to the divalent ion:

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

The faulty chain makes two separate errors: it mixes iron’s charge states and substitutes a configuration-only ranking for the aqueous comparison. A correct configuration attached to the wrong charge state cannot justify the answer.

Repair rule: attach every configuration to its exact charge state. Then use the same trivalent-to-divalent reduction comparison for all four elements:

E^{\circ} (M^{3 +} / M^{2 +})

How do you apply this method to related questions?

Separate electron counting from the reduction-potential comparison: first identify the exact ion, then check which species accepts the electron. These are original related practice questions, not additional verified PYQs.

Which manganese and iron ions have a half-filled d subshell?

Answer: divalent manganese and trivalent iron. Divalent iron has six d electrons, so it is not half-filled. Starting from the given neutral configurations, remove the outer s-subshell electrons first:

M n : [A r] 3 d^{5} 4 s^{2} & ⟶ M n^{2 +} : [A r] 3 d^{5} F e : [A r] 3 d^{6} 4 s^{2} & ⟶ F e^{2 +} : [A r] 3 d^{6} F e^{2 +} : [A r] 3 d^{6} & ⟶ F e^{3 +} : [A r] 3 d^{5}

For these ions, check the d-electron count by subtracting the ionic charge and the 18-electron argon core from the atomic number:

n_{d} (M n^{2 +}) & = 25 - 2 - 18 = 5 n_{d} (F e^{2 +}) & = 26 - 2 - 18 = 6 n_{d} (F e^{3 +}) & = 26 - 3 - 18 = 5

If chromium’s reduction potential is negative and iron’s is positive, which divalent ion is favoured?

Answer: divalent iron, relative to its trivalent state in this comparison. The given signs apply to these reductions:

C r^{3 +} (a q) + e^{-} & ⟶ C r^{2 +} (a q) F e^{3 +} (a q) + e^{-} & ⟶ F e^{2 +} (a q)

The iron couple is more positive. Its reduction therefore has the greater tendency as written, favouring the divalent iron product.

Which trivalent ion accepts an electron more readily, cobalt or manganese?

Answer: trivalent cobalt, under the stated standard aqueous comparison. The given order is:

E^{\circ} (C o^{3 +} / C o^{2 +}) > E^{\circ} (M n^{3 +} / M n^{2 +})

The more positive cobalt couple means its trivalent ion has the greater tendency to accept an electron. That same reduction favours formation of divalent cobalt.

In your error log, record the exact ion, its configuration and the reduction direction. Use the NEET Chemistry Revision Strategy: A 30-Day Repair Plan to schedule a fresh attempt without looking at option B.

Frequently asked questions

What is the correct divalent ion stability order in NEET 2011?

The correct answer is option B: Co2+ > Mn2+ > Fe2+ > Cr2+. This ranks each divalent ion relative to its corresponding trivalent state using standard aqueous M3+/M2+ reduction potentials.

Why does Co2+ rank above Mn2+ despite Mn2+ being half-filled?

The Co3+/Co2+ couple has a more positive standard aqueous reduction potential than Mn3+/Mn2+, so reduction to Co2+ is more strongly favoured in this comparison. Mn2+ has a half-filled 3d5 subshell, but that configuration alone does not decide the ranking.

Which reduction potential should I use to compare divalent ion stability?

For this question, use E°(M3+/M2+), corresponding to M3+(aq) + e− → M2+(aq). A more positive potential indicates a greater tendency for the trivalent ion to form the divalent product. Do not use E°(M2+/M), which describes reduction to the metal.

Is Fe2+ or Fe3+ half-filled?

Fe3+ has the half-filled configuration [Ar] 3d5; Fe2+ is [Ar] 3d6. Iron loses its two 4s electrons first to form Fe2+, then one 3d electron to form Fe3+. Assigning the half-filled configuration to Fe2+ mixes charge states and cannot justify ranking it first.