The Most Probable Estimated Value of the Second Ionization Enthalpy of Mg – JEE Main 2026 PYQ

Nishant Kumar Gupta

Former Faculty · Allen · Aakash · Narayana

Chemistry mentor behind PadhoLikhoJEE and Quantum Chemistry Classes (Arrah, Bihar). Teaches the “why” behind every JEE trap — not just the trick.

Quick Summary

Question Type:

Concept + estimation (successive ionization enthalpy)

Chapter:

Classification of Elements – Periodic Table

Difficulty:

⭐⭐ Easy

Time to Solve:

under 1 minute

Key Concept:

Ionization is endothermic (always positive) and IE₂ > IE₁ — a cation grips its electrons harder

Correct Answer:

(C) +1450 kJ/mol

Why:

It is the only option that is both positive and larger than IE₁ = +737 kJ/mol; Mg⁺ (12 protons, 11 electrons) holds every remaining electron more tightly

The Question

JEE Main 2026 (4 April – Evening Shift) – Periodic Table & Periodicity

The first ionization enthalpy of Mg is +737 kJ/mol. The most probable estimated
value of the second ionization enthalpy of Mg is:

1. (A) -737 kJ/mol
2. (B) -1450 kJ/mol
3. (C) +1450 kJ/mol ✓
4. (D) +59 kJ/mol

Quick Answer

Correct Option: (C) +1450 kJ/mol

Reasoning: Two one-line rules kill three options and crown the survivor:

• The sign rule: ionization consumes energy (you are pulling an electron away from a nucleus), so every ionization enthalpy is positive — options (A) and (B) are eliminated on sight
• The succession rule: IE₂ removes an electron from Mg⁺, where 12 protons pull on only 11 electrons — each electron is held more tightly than in neutral Mg, so IE₂ must be greater than IE₁ = 737
• +59 is positive but smaller than 737 → (D) fails the same test
• Only (C) +1450 kJ/mol survives — roughly double the first value, exactly what the Z_eff argument predicts ✓

Video Solution

If you want the full explanation of this “Octet Emotion Trap” in a clear step-by-step teaching
format, watch the video solution below:

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Understanding the Concept

What Ionization Enthalpy Actually Measures

Ionization enthalpy is the minimum energy required to knock an electron out of an isolated
gaseous atom (or ion) against the pull of the nucleus. The electron does not leave because the
atom “wants” anything — it leaves because you paid energy to overcome electrostatic
attraction. That makes the process endothermic by definition: energy goes in,
the enthalpy change is positive, and any option carrying a minus sign is describing a
different process (electron gain, bond formation) wearing an ionization costume.
This single sign convention settles options (A) and (B) before any calculation begins.

Why Successive Ionization Enthalpies Always Increase

After the first electron leaves, the ion that remains is smaller and more intensely charged:

• Same nuclear charge, fewer electrons: neutral Mg has 12 protons pulling on 12 electrons; Mg⁺ has the same 12 protons pulling on only 11
• Less electron–electron repulsion: the remaining electrons sit closer to the nucleus — the cation’s radius contracts
• Higher effective nuclear charge (Z_eff) per electron: every remaining electron feels a stronger net grip
• Charged target, charged escaping particle: pulling a negative electron away from a positive ion costs more than pulling it from a neutral atom — the attraction you fight is stronger on both counts

Put together, the removal sequence is strictly uphill: IE₁ < IE₂ < IE₃ < …
for every element in the periodic table. There are no exceptions and no “stability discounts”.

The Octet Illusion — Why Noble-Gas Stability Does Not Give a Discount

Here is the emotional trap NTA laid. Students reason: “Mg²⁺ is a super-stable
noble-gas configuration, so losing the second electron must be easy — maybe +59 kJ/mol.”
The logic has the direction of causality backwards. The stable configuration is the
reward that arrives after you pay IE₂, not a subsidy that lowers the bill.
In fact Mg⁺ (configuration [Ne]3s¹) is not yet a noble-gas species at all —
the electron you are removing in the second step is precisely the one whose departure
creates the [Ne] core. Atoms have no desires; the bill is set purely by electrostatics,
and electrostatics says the positively charged, contracted Mg⁺ holds its last valence
electron even more dearly than neutral Mg held its first.

The Key Principle

To estimate any successive ionization enthalpy question of this family:

1. Apply the sign rule: ionization is endothermic → discard every negative option
2. Apply the succession rule: IE_n+1 > IE_n always → discard any positive option smaller than the given value
3. Among survivors, pick a plausible magnitude — for a valence electron, the next ionization is typically 1.5–2.5 times the previous one, not 100 times and not a fraction
4. Cross-check with configuration logic: a giant jump (like 1450 → 7732 for Mg) signals the removal has breached a noble-gas core

Detailed Step-by-Step Solution

Final Answer

Essential Facts for This Topic

The Two Rules That Solve the Question

1. The endothermic rule:
   • Every ionization enthalpy is positive — energy must be supplied to overcome nuclear attraction
   • Negative signs belong to electron gain (electron affinity) or bond formation, never to ionization
2. The successive ionization law:
   • IE₁ < IE₂ < IE₃ < … for every element — a cation (same Z, fewer electrons) grips harder
   • Noble-gas stability of the product never reduces the energy bill; the reward comes after the payment

Values Worth Memorising (kJ/mol)

• Magnesium (Z = 12): IE₁ = 737, IE₂ = 1450, IE₃ = 7732 — the 5.3× leap at IE₃ marks the breach of the [Ne] core
• Sodium (Z = 11): IE₁ = 496, IE₂ = 4562 — the leap comes one step earlier because Na⁺ is already [Ne]
• Aluminium (Z = 13): IE₁ = 578, IE₂ = 1817, IE₃ = 2745, IE₄ = 11577 — the leap sits at IE₄
• Pattern to spot in PYQs: the position of the giant jump = the group number (valence electron count) — the fastest configuration detector in the periodic table

Common Mistakes to Avoid

Key Concept Summary

What You Must Remember

1. Sign first: all ionization enthalpies are positive — negativity is an automatic disqualifier
2. Succession always climbs: IE₁ < IE₂ < IE₃ < …, because each removal leaves a smaller, more tightly bound cation
3. The Z_eff compass: same protons pulling fewer electrons = higher effective charge per electron = higher IE
4. Noble-gas stability is a consequence, not a subsidy: the reward arrives after the energy is paid
5. The giant jump locates the core: Mg 737 → 1450 → 7732 means the [Ne] core breaks at the third removal; Na’s breaks at the second

The Golden Rule for Successive Ionization Questions

“Never assign human emotions to atoms — a cation grips its electrons harder, every ionization bill is positive, and successive bills only ever go up.”

Frequently Asked Questions

Q1: Why is the second ionization enthalpy of Mg greater than the first?

A: After the first electron leaves, Mg+ has 12 protons pulling on only 11 electrons, so the effective nuclear charge felt by every remaining electron increases and the ion shrinks in size. Removing one more electron from this tighter, positively charged ion is harder work than removing it from the neutral atom, so IE2 is always greater than IE1.

Q2: Why can an ionization enthalpy never be negative?

A: Ionization enthalpy is the energy you must supply to pull an electron away from the attraction of the nucleus. Energy goes in, so the process is endothermic and the value is positive by definition. Negative enthalpy changes belong to opposite processes such as electron gain, which is why options like -737 or -1450 kJ/mol are impossible for an ionization step.

Q3: What are the first three ionization enthalpies of magnesium?

A: On the standard scale: IE1 = 737 kJ/mol, IE2 = 1450 kJ/mol and IE3 = 7732 kJ/mol (rounded). The step from IE2 to IE3 is enormous because Mg2+ already has the stable neon configuration, so the third electron must be pulled out of a filled noble-gas shell of a 2+ ion.

Q4: Why is the second ionization enthalpy of Na much larger than that of Mg?

A: Na+ already has the complete neon configuration (2, 8), so the second electron of sodium must be ripped out of a stable noble-gas core, and IE2 of Na is about 4562 kJ/mol. For Mg, IE2 = 1450 kJ/mol only removes the second valence 3s electron and creates the noble-gas core, so the jump for magnesium comes one step later — at IE3.

Q5: How does effective nuclear charge explain that IE2 of Mg is greater than IE1?

A: Neutral Mg has 12 protons pulling on 12 electrons; Mg+ has the same 12 protons pulling on only 11 electrons. With less electron-electron repulsion and no change in nuclear charge, the effective nuclear charge per electron rises and the radius contracts. A more tightly bound electron means a larger ionization enthalpy for the second removal.

Prerequisites to Solve This Question

Before attempting this problem, you should be comfortable with:

1. Definition of ionization enthalpy: minimum energy to remove the most loosely bound electron from an isolated gaseous atom/ion
2. The endothermic sign convention: energy in → positive value, and how that contrasts with electron affinity
3. Effective nuclear charge and screening: how Z_eff, repulsion and radius together set the grip on any electron
4. Electronic configurations: Mg (1s²2s²2p⁶3s²), Mg⁺ ([Ne]3s¹), Mg²⁺ ([Ne])
5. The noble-gas core concept: why removals that breach a filled shell cost dramatically more

After Solving This, You Can:

• Estimate any successive ionization enthalpy in under 30 seconds using sign + succession + magnitude
• Instantly eliminate negative decoy options that dress electron-gain values as ionization values
• Predict where the giant jump sits for any element and read off its valence-electron count (group)
• Explain why Na’s second ionization dwarfs Mg’s — the classic JEE comparison that follows this question everywhere

Study Tips for This Topic

For JEE Main:

1. Memorise the Mg triad: 737 → 1450 → 7732. One triplet answers this question, the Na-vs-Mg comparison, and every “locate the jump” variant.
2. Build a two-step reflex: sign check, then succession check. In this question alone those two lines executed three of the four options.
3. Budget 20–30 seconds: no formula is needed — this is a pure concept-plus-elimination question, the cheapest +4 marks in the periodic table chapter.

Common JEE Variants:

• “Why is IE₂ of Na far larger than IE₂ of Mg?” — answer: Na⁺ is already [Ne]; Mg’s jump waits for IE₃
• “The largest jump in successive IEs of element X lies between IE_n and IE_n+1 — find the group/configuration” — jump position = valence count
• Order-of-IE comparisons across a period or between Be/B and N/O (anomaly questions)
• Estimation traps with mixed signs for halogen electron affinities vs alkali ionizations
• IE data used to identify an unknown element among four options — same jump-reading skill

Difficulty Rating & Exam Frequency

Difficulty Level: ⭐⭐ (2/5) – Easy (the physics is one line — endothermic, and uphill; the negative mark lives only in the emotional “octet discount” and the sign decoys)

JEE Main Frequency: High — ionization enthalpy concepts appear in almost every periodic-table session, and estimation/successive-IE formats repeat regularly

JEE Advanced Frequency: Low to Medium — direct estimation is rare, but IE data as a configuration-detection tool appears inside structural and periodicity problems

Topic Importance: Very High — ionization enthalpy is the gateway concept for periodic trends, bond formation, oxidation behaviour and half of inorganic reasoning

Written by Nishant Kumar Gupta
Former Faculty · Allen · Aakash · Narayana — Quantum Chemistry Classes, Arrah (Bihar)
Last Updated: 31 Aug 2026 ·
Question Source: JEE Main 2026 (4 April – Evening Shift) Previous Year Question ·
Topic: Classification of Elements and Periodicity – Successive Ionization Enthalpies of Magnesium