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Ionization Enthalpy & Electron Gain Enthalpy — MH-CET Chemistry MCQs with Solutions

Free MH-CET Chemistry Ionization Enthalpy & Electron Gain Enthalpy MCQs with step-by-step solutions (40 questions). Part of Classification of Elements and Periodicity. Practise online on Prepizo — no login needed.

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Questions with solutions

Q1 — Ionization Enthalpy & Electron Gain Enthalpy · easy · theory
Ionization enthalpy is the energy required to remove the most loosely bound electron from an isolated ______ atom:
A. Liquid
B. Aqueous
C. Gaseous  ✓ Correct
D. Solid
Solution: ΔᵢH refers to removal of an electron from a neutral gaseous atom.
Q2 — Ionization Enthalpy & Electron Gain Enthalpy · easy · theory
Across a period, the first ionization enthalpy generally:
A. Increases  ✓ Correct
B. Stays constant
C. Decreases
D. First increases then decreases sharply
Solution: Rising Z_eff and shrinking size make electron removal harder across a period.
Q3 — Ionization Enthalpy & Electron Gain Enthalpy · easy · theory
Down a group, the first ionization enthalpy generally:
A. First decreases then increases
B. Stays constant
C. Increases
D. Decreases  ✓ Correct
Solution: Larger size and greater shielding make the outer electron easier to remove down a group.
Q4 — Ionization Enthalpy & Electron Gain Enthalpy · easy · theory
The order of successive ionization enthalpies of an element is:
A. All equal
B. ΔᵢH₁ < ΔᵢH₂ < ΔᵢH₃  ✓ Correct
C. ΔᵢH₂ < ΔᵢH₁
D. ΔᵢH₁ > ΔᵢH₂ > ΔᵢH₃
Solution: Removing electrons from an increasingly positive ion needs progressively more energy.
Q5 — Ionization Enthalpy & Electron Gain Enthalpy · medium · theory
Beryllium has a higher first ionization enthalpy than boron because Be has a:
A. Stable, fully-filled 2s² configuration  ✓ Correct
B. Lower nuclear charge
C. Larger size
D. Half-filled 2p subshell
Solution: Removing an electron from the extra-stable 2s² of Be is harder than from the 2p¹ of B.
Q6 — Ionization Enthalpy & Electron Gain Enthalpy · medium · theory
Nitrogen has a higher first ionization enthalpy than oxygen because N has a:
A. Stable half-filled 2p³ configuration  ✓ Correct
B. Fully-filled 2p subshell
C. Larger nuclear charge
D. Smaller size
Solution: The half-filled 2p³ of N is extra stable, so it resists losing an electron more than O's 2p⁴.
Q7 — Ionization Enthalpy & Electron Gain Enthalpy · medium · theory
A large jump between two successive ionization enthalpies indicates:
A. Removal of an electron from a stable noble-gas core (revealing the valency)  ✓ Correct
B. The atom has become larger
C. The presence of isotopes
D. A measurement error
Solution: The sudden jump appears after all valence electrons are gone, so its position reveals the group/valency.
Q8 — Ionization Enthalpy & Electron Gain Enthalpy · easy · theory
Noble gases have the highest ionization enthalpies in their periods because of their:
A. Large size
B. Metallic character
C. Low nuclear charge
D. Stable, completely-filled (octet) configuration  ✓ Correct
Solution: The fully-filled ns²np⁶ shell is very stable, so removing an electron requires the most energy.
Q9 — Ionization Enthalpy & Electron Gain Enthalpy · easy · theory
Electron gain enthalpy is the enthalpy change when an electron is added to a neutral gaseous atom; a negative value means the process is:
A. Exothermic (energy released)  ✓ Correct
B. Endothermic
C. Athermal
D. Impossible
Solution: A negative Δₑ𝓰H means energy is released on adding the electron.
Q10 — Ionization Enthalpy & Electron Gain Enthalpy · medium · theory
Chlorine has a more negative electron gain enthalpy than fluorine because:
A. Cl has fewer protons
B. F is a metal
C. F's small, compact 2p subshell causes greater electron–electron repulsion on the added electron  ✓ Correct
D. F has a larger size
Solution: The incoming electron enters F's very compact 2p, facing high repulsion; Cl's larger 3p accommodates it better — so Δₑ𝓰H(Cl) is more negative.
Q11 — Ionization Enthalpy & Electron Gain Enthalpy · medium · theory
Elements with positive (endothermic) electron gain enthalpy include:
A. Alkali metals
B. Noble gases, Be, Mg and N  ✓ Correct
C. Halogens
D. Chalcogens
Solution: Stable filled/half-filled shells (noble gases, ns² of Be/Mg, half-filled 2p³ of N) resist gaining an electron.
Q12 — Ionization Enthalpy & Electron Gain Enthalpy · medium · theory
The factor that does NOT increase ionization enthalpy is:
A. Increasing nuclear charge
B. Increasing Z_eff
C. Increasing atomic size  ✓ Correct
D. Stable half/fully-filled configuration
Solution: Larger size lowers IE; the others all raise it.
Q13 — Ionization Enthalpy & Electron Gain Enthalpy · medium · theory
The penetration power of orbitals, which affects ionization enthalpy, follows the order:
A. f > d > p > s
B. d > p > s > f
C. p > s > d > f
D. s > p > d > f  ✓ Correct
Solution: s electrons penetrate closest to the nucleus, so they are held most tightly.
Q14 — Ionization Enthalpy & Electron Gain Enthalpy · medium · theory
Across period 2, sulphur (period 3) shows a more negative Δₑ𝓰H than oxygen because:
A. O's compact 2p causes more repulsion on the incoming electron than S's larger 3p  ✓ Correct
B. S has fewer electrons
C. S has a smaller nuclear charge
D. O is a metal
Solution: Like Cl>F, the anomaly S>O arises from the compactness of the 2p subshell in oxygen.
Q15 — Ionization Enthalpy & Electron Gain Enthalpy · medium · theory
Down a group, electron gain enthalpy generally becomes:
A. Zero
B. More negative always
C. Less negative (with the F < Cl exception)  ✓ Correct
D. Positive for all
Solution: Larger size means the added electron is less strongly attracted, so Δₑ𝓰H is less negative down a group (fluorine being the notable exception).
Q16 — Ionization Enthalpy & Electron Gain Enthalpy · medium · theory
Electron gain enthalpy differs from electronegativity in that electron gain enthalpy is:
A. The same thing
B. Always positive
C. A measurable enthalpy for an isolated gaseous atom, while electronegativity is a relative tendency in a bond  ✓ Correct
D. Defined only for metals
Solution: Δₑ𝓰H is a thermodynamic quantity for a free atom; electronegativity describes attraction for shared electrons within a molecule.
Q17 — Ionization Enthalpy & Electron Gain Enthalpy · medium · numerical
The correct order of first ionization enthalpy for Li, Be, B, C (period 2) is:
A. Li < Be < B < C
B. C < B < Be < Li
C. Be < Li < B < C
D. Li < B < Be < C  ✓ Correct
Solution: General rise, but Be(2s²) > B(2p¹): order Li < B < Be < C.
Q18 — Ionization Enthalpy & Electron Gain Enthalpy · easy · numerical
Which element has a higher first ionization enthalpy: N or O?
A. N  ✓ Correct
B. O
C. Cannot be said
D. Both equal
Solution: N's half-filled 2p³ is extra stable, so ΔᵢH(N) > ΔᵢH(O) — a classic anomaly.
Q19 — Ionization Enthalpy & Electron Gain Enthalpy · easy · numerical
Which has the higher first ionization enthalpy: Mg or Al?
A. Both equal
B. Al
C. Depends on temperature
D. Mg  ✓ Correct
Solution: Mg's filled 3s² is stable; Al loses its lone 3p¹ more easily ⇒ ΔᵢH(Mg) > ΔᵢH(Al).
Q20 — Ionization Enthalpy & Electron Gain Enthalpy · hard · numerical
The successive ionization enthalpies (kJ/mol) of an element are 738, 1451, 7733, 10540. The element most likely belongs to group:
A. 1
B. 3
C. 2  ✓ Correct
D. 13
Solution: A huge jump after the 2nd ionization means 2 easily-removed valence electrons ⇒ group 2 (this is Mg).
Q21 — Ionization Enthalpy & Electron Gain Enthalpy · hard · numerical
For an element with ΔᵢH values 496, 4562, 6912 kJ/mol, the group is:
A. 17
B. 1  ✓ Correct
C. 13
D. 2
Solution: A sharp jump after the 1st IE means 1 valence electron ⇒ group 1 (this is Na).
Q22 — Ionization Enthalpy & Electron Gain Enthalpy · medium · numerical
The correct decreasing order of first ionization enthalpy of Na, Mg, Al is:
A. Al > Mg > Na
B. Na > Mg > Al
C. Mg > Al > Na  ✓ Correct
D. Mg > Na > Al
Solution: Across period 3 IE rises (Na<Al) but Mg(3s²) > Al ⇒ Mg > Al > Na.
Q23 — Ionization Enthalpy & Electron Gain Enthalpy · easy · numerical
Which has the more negative (higher magnitude) electron gain enthalpy: F or Cl?
A. Cl  ✓ Correct
B. F
C. Both equal
D. Neither gains electrons
Solution: Cl > F because the incoming electron faces less repulsion in Cl's larger 3p subshell.
Q24 — Ionization Enthalpy & Electron Gain Enthalpy · medium · numerical
The total energy to convert one mole of gaseous Mg to Mg²⁺, given ΔᵢH₁ = 738 and ΔᵢH₂ = 1451 kJ/mol, is:
A. 3640 kJ/mol
B. 1451 kJ/mol
C. 2189 kJ/mol  ✓ Correct
D. 713 kJ/mol
Solution: Total = ΔᵢH₁ + ΔᵢH₂ = 738 + 1451 = 2189 kJ/mol.
Q25 — Ionization Enthalpy & Electron Gain Enthalpy · easy · numerical
The correct order of first ionization enthalpy for the noble gases He, Ne, Ar is:
A. He > Ne > Ar  ✓ Correct
B. Ar > Ne > He
C. He > Ar > Ne
D. Ne > He > Ar
Solution: IE decreases down group 18 as size increases ⇒ He > Ne > Ar.
Q26 — Ionization Enthalpy & Electron Gain Enthalpy · easy · numerical
The energy needed to remove one electron each from a mole of gaseous Al atoms (ΔᵢH₁ = 577 kJ/mol) to form Al⁺ is:
A. 577 kJ/mol  ✓ Correct
B. 1154 kJ/mol
C. 288 kJ/mol
D. 5770 kJ/mol
Solution: Removing the first electron requires ΔᵢH₁ = 577 kJ/mol.
Q27 — Ionization Enthalpy & Electron Gain Enthalpy · medium · numerical
Which of the following has the most negative electron gain enthalpy?
A. I
B. Br
C. F
D. Cl  ✓ Correct
Solution: Chlorine has the most negative Δₑ𝓰H of all elements (Cl > F due to F's compactness).
Q28 — Ionization Enthalpy & Electron Gain Enthalpy · medium · numerical
The pair in which the first element has a HIGHER ionization enthalpy (an anomaly) is:
A. P and S (P > S)  ✓ Correct
B. B and C
C. Li and Be
D. C and N
Solution: P (half-filled 3p³) has a higher ΔᵢH than S — the P > S anomaly.
Q29 — Ionization Enthalpy & Electron Gain Enthalpy · hard · numerical
The successive ΔᵢH (kJ/mol) 578, 1817, 2745, 11577 indicate the element is in group:
A. 13  ✓ Correct
B. 14
C. 3
D. 2
Solution: Large jump after the 3rd IE ⇒ 3 valence electrons ⇒ group 13 (this is Al).
Q30 — Ionization Enthalpy & Electron Gain Enthalpy · hard · numerical
Arrange O, S, Se by increasingly negative electron gain enthalpy correctly (most negative first):
A. S > O > Se
B. S > Se > O  ✓ Correct
C. O > S > Se
D. Se > S > O
Solution: S has the most negative Δₑ𝓰H (S > O anomaly); then Se, with O least due to 2p compactness.