D- and f-block elements NEET 2025: Unpaired Electrons

Do Cr²⁺ and Nd³⁺ have equal unpaired electrons in the NEET 2025 question?

Cr²⁺ has four unpaired electrons; Nd³⁺ has three. That difference decides this d- and f-block elements NEET 2025 question.

The question presents two claims: ferromagnetism represents exceptionally strong paramagnetic behaviour, and Cr²⁺ and Nd³⁺ contain equal numbers of unpaired electrons. The answer is C, because the first claim holds and the second does not.

The options map to Statement I/Statement II as follows:

The question bank classifies this question as hard. That is a difficulty label, not evidence of a measured student failure rate. Solve it as two independent checks: first magnetic behaviour, then ionic electron counting.

Why is the ferromagnetism statement true?

Ferromagnetic substances experience very strong attraction to an external magnetic field. The supplied official solution connects this response to the parallel alignment of magnetic moments. On that basis, it accepts the description of ferromagnetism as an extreme form of paramagnetism, so Statement I is true.

This does not mean that every substance containing unpaired electrons is ferromagnetic. Unpaired electrons alone do not establish the collective alignment required for ferromagnetism.

Keep this conceptual result separate from the ionic comparison. Statement II asks whether the ions have equal numbers of unpaired electrons, so it requires electron removal and counting, not another judgement about magnetic attraction.

How do you derive Cr²⁺ and count its unpaired electrons?

Cr²⁺ has four unpaired electrons under the configuration-based treatment used here. Start from neutral chromium, remove the outer s electron first, and then remove one d electron.

Use the correct neutral chromium configuration:

C r : [A r] 3 d^{5} 4 s^{1}

Do not begin with this incorrect configuration:

[A r] 3 d^{4} 4 s^{2}

For these chromium cations, the 4s electron leaves before a 3d electron. The complete sequence is:

C r & : [A r] 3 d^{5} 4 s^{1} & \overset{remove one 4 s electron}{\to} {C r}^{+} : [A r] 3 d^{5} 4 s^{0} & \overset{remove one 3 d electron}{\to} {C r}^{2 +} : [A r] 3 d^{4} 4 s^{0}

Now apply Hund’s rule. The d subshell contains five orbitals, and its four electrons occupy four separate orbitals before any pairing occurs.

Therefore, Cr²⁺ has four unpaired electrons. The empty outer s subshell contributes none.

How do you derive Nd³⁺ and confirm option C?

Nd³⁺ has three unpaired electrons because forming the tripositive ion requires removing three electrons from neutral neodymium. Removing its two outer s electrons is only the first stage; one further electron must leave the f subshell.

Start with neutral neodymium:

N d : [X e] 4 f^{4} 6 s^{2}

Remove the two 6s electrons:

N d \overset{remove two 6 s electrons}{\to} {N d}^{2 +} : [X e] 4 f^{4} 6 s^{0}

Then remove one 4f electron:

{N d}^{2 +} \overset{remove one 4 f electron}{\to} {N d}^{3 +} : [X e] 4 f^{3} 6 s^{0}

The f subshell contains seven orbitals. By Hund’s rule, three electrons occupy three separate orbitals, giving three unpaired electrons. The closed-shell argon and xenon cores, shown as [Ar] and [Xe], add no unpaired electrons.

Compare the configurations outside those cores:

Cr²⁺: 4 unpaired electrons $3 d^{4} 4 s^{0}$
Nd³⁺: 3 unpaired electrons $4 f^{3} 6 s^{0}$

The counts are unequal: $4 \neq 3$

Thus, Statement II is false. Statement I true plus Statement II false gives option C.

How would stopping at Nd²⁺ incorrectly produce option A?

Removing only neodymium’s two outer s electrons would leave four unpaired f electrons. If that configuration were incorrectly labelled Nd³⁺, it would appear to match Cr²⁺ and make Statement II look true.

The incorrect working is:

N d : [X e] 4 f^{4} 6 s^{2} \overset{remove only two electrons}{\to} \underset{incorrectly labelled {N d}^{3 +}}{\underset{⏟}{[X e] 4 f^{4} 6 s^{0}}}

This configuration actually belongs to Nd²⁺, because only two electrons have been removed. This mistake would produce option A: the mistaken count makes Statement II appear true, while Statement I has already been accepted.

Use charge accounting to catch it. A tripositive ion requires three electrons to be removed, not merely an empty outer s subshell: $Electrons removed = 2 + 1 = 3$

The supplied 75-second expected solving time is a question-bank benchmark, not observed student performance. Keep the charge check in your working even when practising against that benchmark.

Would the unpaired-electron counts match if the ion charges changed?

Both pairs below have equal unpaired-electron counts under the same treatment, unlike the original pair. These are original related practice questions, not additional verified NEET PYQs. Repeat the removal steps for each charge rather than carrying over the original counts.

Would Cr³⁺ and Nd³⁺ have equal numbers of unpaired electrons?

Yes, both have three unpaired electrons. Start from the chromium ion already derived and remove one more d electron:

{C r}^{2 +} : [A r] 3 d^{4} \overset{remove one 3 d electron}{\to} {C r}^{3 +} : [A r] 3 d^{3}

Hund’s rule places these three d electrons in separate orbitals. Nd³⁺ likewise has three singly occupied f orbitals, so the counts are equal:

{N d}^{3 +} : [X e] 4 f^{3}

Would Cr²⁺ and Nd²⁺ have equal numbers of unpaired electrons?

Yes, both have four unpaired electrons. This time, neodymium’s charge requires removing only its two outer s electrons:

N d : [X e] 4 f^{4} 6 s^{2} \overset{remove two 6 s electrons}{\to} {N d}^{2 +} : [X e] 4 f^{4}

The four f electrons occupy four separate orbitals out of seven. They match the four unpaired d electrons in Cr²⁺.

This confirms the error diagnosis: retaining four f electrons is correct for Nd²⁺, not Nd³⁺. On your next question, use this sequence: write the neutral configuration, remove the correct number of electrons, apply Hund’s rule, then compare.

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Frequently asked questions

Do Cr²⁺ and Nd³⁺ have the same number of unpaired electrons?

No. Cr²⁺ has configuration [Ar] 3d⁴ and four unpaired electrons, while Nd³⁺ has configuration [Xe] 4f³ and three. Statement II is therefore false; with Statement I true, the correct answer is option C.

How do you find the electronic configuration of Cr²⁺?

Start with neutral chromium: [Ar] 3d⁵ 4s¹. Remove the 4s electron first, then one 3d electron, giving Cr²⁺ the configuration [Ar] 3d⁴. Under the configuration-based treatment used in this question, Hund’s rule gives four unpaired electrons.

Why is the ferromagnetism statement true in this question?

The supplied official solution accepts ferromagnetism as an extreme form of paramagnetism because ferromagnetic substances show very strong attraction to an external magnetic field, linked to parallel alignment of magnetic moments. This does not mean that every substance with unpaired electrons is ferromagnetic.

Why does Nd³⁺ have three unpaired electrons instead of four?

Neutral neodymium is [Xe] 4f⁴ 6s², and forming Nd³⁺ requires removing three electrons: two from 6s and one from 4f. The resulting [Xe] 4f³ configuration has three unpaired electrons by Hund’s rule. Stopping after removing only the two 6s electrons gives Nd²⁺, not Nd³⁺.

D- and f-block elements NEET 2025: Cr²⁺ vs Nd³⁺