Ionization Enthalpy (Periodic Table) Class 11 Important Conceptual NCERT Question Answers

Anand Classes presents detailed study material on Ionization Enthalpy (Periodic Table) Class 11 Important Conceptual NCERT Question Answers for JEE, NEET, and CBSE Board students. Ionization enthalpy is a fundamental periodic property that explains the energy required to remove an electron from an isolated gaseous atom. Understanding this concept not only strengthens your basics of atomic structure but also helps in solving higher-order conceptual and numerical problems. In this article, you will find well-explained answers to NCERT questions, along with extra tips, solved examples, and exam-oriented insights.

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Successive Ionization Enthalpies of Li and Be

Q.1 : Why first ionization enthalpy (IE₁) of Li is smaller than for Be while the second ionization enthalpy (IE₂) of Li is much greater than for Be ? [NCERT]

The variation in the successive ionization enthalpies of lithium (Li) and beryllium (Be) is an important concept in chemistry. It helps us understand how electronic configurations influence the ease or difficulty of removing electrons.

The observed ionization enthalpy (IE) values are:

Element	IE₁ (kJ mol⁻¹)	IE₂ (kJ mol⁻¹)
Li	520	7298
Be	899	1758

🔹 Electronic configuration explanation in reaction form:

$$\text{Li (1s² 2s¹)} \xrightarrow{\text{IE₁}} \text{Li⁺ (1s²)} \xrightarrow{\text{IE₂}} \text{Li²⁺ (1s¹)}$$

$$\text{Be (1s² 2s²)} \xrightarrow{\text{IE₁}} \text{Be⁺ (1s² 2s¹)} \xrightarrow{\text{IE₂}} \text{Be²⁺ (1s²)}$$

Explanation in Terms of Electronic Configurations

• Lithium (Li) → 1s² 2s¹
• Beryllium (Be) → 1s² 2s²

First Ionization Enthalpy (IE₁)

• In lithium (Li), the first electron is removed from the 2s¹ orbital.
• In beryllium (Be), the first electron is removed from the 2s² orbital, which is more stable due to being a completely filled subshell.

👉 Therefore, IE₁ of Be (899 kJ mol⁻¹) is higher than that of Li (520 kJ mol⁻¹).

Second Ionization Enthalpy (IE₂)

• After losing one electron, Li⁺ has the configuration 1s² (helium configuration), which is a very stable noble gas structure. Removing one more electron from this stable configuration requires very high energy, hence IE₂ of Li is extremely large (7298 kJ mol⁻¹).
• In the case of Be⁺ (1s² 2s¹), the atom still has an extra 2s electron left. Removing this electron is easier compared to disturbing a noble gas core. Thus, IE₂ of Be (1758 kJ mol⁻¹) is much smaller than IE₂ of Li.

Third Ionization Enthalpy (IE₃) of Be

• After removal of two electrons, Be²⁺ has the stable configuration 1s².
• Now, removing the third electron means breaking into the noble gas core, which requires very high energy.
• Hence, IE₃ of Be is also extremely high.

Summary of the Concept

• Li: Low IE₁, but very high IE₂ (due to attainment of stable noble gas configuration after the first ionization).
• Be: Higher IE₁, moderate IE₂, but very high IE₃ (since the noble gas core is disturbed only after the second ionization).

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Question : Which of the following pairs of elements would you expect to have lower ionization enthalpy? Explain. (i) Cl or F (ii) Cl or S (iii) K or Ar (iv) Kr or Xe [NCERT]

Answer :

(i) Cl or F

• Ionization enthalpy trend in a group:
  As we move down a group in the periodic table, the ionization enthalpy decreases.
  This happens because:
  • The atomic size increases down the group due to addition of new electron shells.
  • The nuclear attraction on the outermost electron decreases due to increased distance and greater shielding effect of inner electrons.
  • As a result, it becomes easier to remove the outermost electron.
• Application to F and Cl:
  • Fluorine (F) is above chlorine (Cl) in Group 17 (halogens).
  • F has a very small atomic radius and very strong nuclear attraction on its valence electron.
  • Cl, being larger, has weaker attraction on its valence electron.

👉 Therefore, chlorine (Cl) has lower ionization enthalpy than fluorine (F).

(ii) Cl or S

• Ionization enthalpy trend in a period:
  As we move across a period (left to right), the ionization enthalpy increases because:
  • The nuclear charge increases with increasing atomic number.
  • The atomic radius decreases (no new shells are added, only nuclear pull increases).
  • Valence electrons are held more strongly, making their removal harder.
• Application to Cl and S:
  • Both are in Period 3.
  • Sulfur (S) is in Group 16, while chlorine (Cl) is in Group 17, to the right of S.
  • Cl has higher nuclear charge and smaller atomic size compared to S.

👉 Therefore, sulfur (S) has lower ionization enthalpy than chlorine (Cl).

(iii) K or Ar

• Special case: Alkali metals vs Noble gases
  • Potassium (K) is an alkali metal with configuration [Ar] 4s¹. Its single 4s electron is loosely held and can be removed very easily.
  • Argon (Ar) is a noble gas with configuration 1s² 2s² 2p⁶ 3s² 3p⁶ (completely filled shells).
  • Noble gases have very high ionization enthalpies because their outermost shell is stable, and removing an electron would disturb this stability.

👉 Therefore, potassium (K) has much lower ionization enthalpy than argon (Ar).

(iv) Kr or Xe

• Ionization enthalpy trend in Group 18 (noble gases):
  • As we go down the group, the ionization enthalpy decreases.
  • This is due to:
    • Increasing atomic size.
    • Stronger shielding effect of inner shells.
    • Valence electrons being farther from the nucleus and less tightly bound.
• Application to Kr and Xe:
  • Krypton (Kr) is above xenon (Xe) in Group 18.
  • Xe has a larger atomic radius and weaker attraction on its valence electrons compared to Kr.

👉 Therefore, xenon (Xe) has lower ionization enthalpy than krypton (Kr).

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From each set, choose the atom which has the largest ionization enthalpy:
(a) F, O, N
(b) Mg, P, Ar
(c) B, Al, Ga

(a) F, O, N

• Nitrogen (N):
  • Electronic configuration = 1s² 2s² 2p³.
  • Half-filled p-subshell → extra stability → relatively higher IE.
• Oxygen (O):
  • Configuration = 1s² 2s² 2p⁴.
  • Has a paired electron in 2p orbital, so repulsion makes removal easier.
  • Thus IE of O < IE of N (an anomaly).
• Fluorine (F):
  • Configuration = 1s² 2s² 2p⁵.
  • Highest Z_eff among the three, smallest atomic radius.
  • Very strong attraction on outermost electron → highest IE.

👉 Answer: Fluorine (F) has the largest IE.

(b) Mg, P, Ar

• Magnesium (Mg):
  • Group 2, configuration [Ne] 3s².
  • IE relatively low because it is an alkaline earth metal.
• Phosphorus (P):
  • Group 15, configuration [Ne] 3s² 3p³ (half-filled p-subshell gives some stability).
  • IE higher than Mg, but lower than noble gas.
• Argon (Ar):
  • Group 18, configuration [Ne] 3s² 3p⁶ (completely filled shell).
  • Smallest atomic radius in Period 3, maximum Z_eff.
  • Removing an electron requires disturbing a completely filled, very stable configuration → extremely high IE.

👉 Answer: Argon (Ar) has the largest IE.

(c) B, Al, Ga

• Boron (B):
  • Group 13, Period 2, configuration [He] 2s² 2p¹.
  • Smallest atom among the three, highest Z_eff, lowest shielding → high IE.
• Aluminium (Al):
  • Group 13, Period 3, configuration [Ne] 3s² 3p¹.
  • Larger atomic size, more shielding → lower IE than B.
• Gallium (Ga):
  • Group 13, Period 4, configuration [Ar] 3d¹⁰ 4s² 4p¹.
  • Larger size than Al, so expected to have lower IE.
  • BUT: The poor shielding effect of 3d electrons increases effective nuclear charge, making Ga’s IE almost similar to or slightly greater than Al.
  • Still, both Al and Ga < B.

👉 Answer: Boron (B) has the largest IE.

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How much energy in joules must be needed to convert all the atoms of sodium to sodium ions present in 2.3 mg of sodium vapours? Ionization enthalpy of sodium is 495 kJ/mol (Atomic mass of Na = 23). [NCERT]

According to the definition of ionization enthalpy:

Na(g)  +  IE  ⟶  Na⁺(g)  +  e⁻(g)

Here, IE = 495  kJ/mol

This means the amount of energy needed to ionize 1 mole of sodium vapours = 495 kJ/mol.

Now, Mass of sodium sample = 2.3 mg = 2.3 × 10⁻³ g

Moles of sodium present = Given mass/Molar mass = 2.3 × 10⁻³/23 = 1.0 × 10⁻⁴ mol

Therefore, the energy required to ionize 1.0 × 10⁻⁴ mol of sodium vapours is : 495 × 10⁻⁴ kJ = 0.0495 kJ = 0.0495 × 1000 = 49.5 J

✅ Final Answer:
The energy required = 49.5 J

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The first ionization enthalpy (Δ_iH) values of the third period elements, Na, Mg, and Si are respectively 496, 737, and 786 kJ/mol. Predict whether the first Δ_iH value for Al will be more close to 575 or 760 kJ/mol. Justify your answer.

The first ionization enthalpy of Al will be more close to 575 kJ/mol because its value should be lower than that of Mg.

This decrease is due to the effective shielding of 3p electrons from the nucleus by the inner 3s-electrons.

✅ Final Answer:
The ionization enthalpy of aluminium will be closer to 575 kJ/mol.

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Quick Revision Pointers 🚀

• Across a period → IE ↑ (size ↓, Z_eff ↑).
• Down a group → IE ↓ (size ↑, shielding ↑).
• Anomalies:
  • N > O (due to half-filled stability).
  • Ga ≈ Al (due to poor shielding of 3d).
• Noble gases: highest IE in each period.
• Alkali metals: lowest IE in each period.

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