NCERT Solutions 3.16, 3.17, 3.18, 3.19, 3.20 | Classification of Elements and Periodicity in Properties Chemistry Class 11 pdf Download

Anand Classes provides complete NCERT solutions for Class 11 Chemistry, covering exercises 3.16, 3.17, 3.18, 3.19, and 3.20 on the topic Classification of Elements and Periodicity in Properties. This post explains key concepts such as ionization enthalpy, electron gain enthalpy, deviations in periodic trends, and factors affecting atomic and ionic properties with clear, step-by-step explanations. Each solution is designed to help students understand periodic behavior and atomic structure effectively. Click the print button to download PDF study material and notes.

NCERT Problems and Solutions of Classification of Elements and Periodicity in Properties Chemistry Class 11 pdf Download

NCERT 3.16 : Among the second period elements, the actual ionization enthalpies are in the order $Li < B < Be < C < O < N < F < Ne$. Explain why (i) Be has higher first ionization enthalpy ($\Delta_1 H$) than B (ii) O has lower $\Delta_1 H$ than N and F

Answer :

(i)

The ionization enthalpy depends, among other factors, on the type of electron to be removed from the same principal shell.

• In the case of $Be$ ($1s^2 2s^2$), the outermost electron is present in the 2s-orbital.
• In $B$ ($1s^2 2s^2 2p^1$), the outermost electron is present in the 2p-orbital.
• Since 2s-electrons are more strongly attracted by the nucleus than 2p-electrons, less energy is required to remove a 2p-electron than a 2s-electron.

Consequently, the first ionization enthalpy ($\Delta_1 H$) of Be is higher than that of B.

(II)

• The electronic configuration of $N$ ($1s^2 2s^2 2p_x^1 2p_y^1 2p_z^1$), where the 2p-orbitals are exactly half-filled, is more stable than the electronic configuration of $O$ ($1s^2 2s^2 2p_x^2 2p_y^1 2p_z^1$), where the 2p-orbitals are neither half-filled nor completely filled.
• Therefore, it is more difficult to remove an electron from N than from O. As a result, $\Delta_1 H$ of N is higher than that of O.
• Further, the electronic configuration of $F$ is $1s^2 2s^2 2p_x^2 2p_y^2 2p_z^1$. Because of the higher nuclear charge (+9), the first ionization enthalpy of F is higher than that of O.

NCERT 3.17 : Explain why the first ionization enthalpy of sodium is lower than that of magnesium but its second ionization enthalpy is higher than that of magnesium

Answer :

The electronic configurations of sodium and magnesium are:

$$
Na : 1s^2 2s^2 2p^6 3s^1
$$

$$
Mg : 1s^2 2s^2 2p^6 3s^2
$$

First Ionization Enthalpy

• In both Na and Mg, the first electron is removed from the 3s-orbital.
• The nuclear charge of Na (+11) is lower than that of Mg (+12), so the attraction of the nucleus on the 3s electron in Na is weaker.
• Hence, the first ionization enthalpy of sodium is lower than that of magnesium.

Second Ionization Enthalpy

• After losing the first electron:

$$
Na^+ : 1s^2 2s^2 2p^6
$$

$$
Mg^+ : 1s^2 2s^2 2p^6 3s^1
$$

• For $Na^+$, the second electron must be removed from a stable inert gas configuration (Ne-like), which is very difficult.
• For $Mg^+$, the second electron is removed from the 3s-orbital, which is easier than removing an electron from a stable configuration.

Conclusion:

• First ionization: $\Delta_1 H$ of Na < $\Delta_1 H$ of Mg
• Second ionization: $\Delta_2 H$ of Na > $\Delta_2 H$ of Mg

This is because ionization enthalpy depends on nuclear charge, orbital stability, and electron configuration.

NCERT 3.18 : Factors causing ionization enthalpy of main group elements to decrease down the group

Answer :

Within the main group elements, the ionization enthalpy decreases regularly as we move down the group due to the following factors:

(i) Atomic Size

• As we move down a group, a new principal energy shell is added for each succeeding element.
• This increases the distance of valence electrons from the nucleus.
• The force of attraction between the nucleus and the valence electrons decreases with distance.
• Result: Ionization enthalpy decreases down the group.

(ii) Screening (Shielding) Effect

• With the addition of new shells, the number of inner electron shells shielding the valence electrons also increases.
• This screening effect reduces the effective nuclear charge experienced by valence electrons.
• Result: The nucleus attracts the valence electrons less strongly, leading to a further decrease in ionization enthalpy down the group.

Conclusion:

The combined effects of increasing atomic size and increasing shielding cause the ionization enthalpy of main group elements to decrease as we move down a group.

NCERT 3.19 : The first ionization enthalpy values (in kJ/mol) of group 13 elements are

How would you explain this deviation from the general trend ?

Answer :

• General Trend: Moving down group 13 from B to Al, the ionization enthalpy decreases as expected due to increase in atomic size and screening effect, which outweigh the effect of increased nuclear charge.
• Deviation in Ga: The first ionization enthalpy of Ga (579 kJ/mol) is slightly higher than that of Al (577 kJ/mol).
• Reason: Al follows immediately after s-block elements, while Ga follows after d-block elements.
• The extra d-electrons in Ga do not shield the outer electrons effectively.
• Therefore, the valence electrons are held more tightly, and more energy is required to remove them.
• Trend from Ga to In: Moving down to In, the increased shielding effect (due to 4d-electrons) outweighs the increased nuclear charge.
• Result: First ionization enthalpy of In (558 kJ/mol) is lower than that of Ga.
• Trend for Tl: Moving further down, Tl has additional 4f and 5d electrons.
• Here, the increase in nuclear charge (+32 units from Ga to Tl) outweighs the shielding effect.
• Result: First ionization enthalpy of Tl (589 kJ/mol) is much higher than that of Al, Ga, and In.

Conclusion:

Deviations from the general trend in group 13 elements are mainly due to the ineffective shielding of d- and f-electrons and the balance between nuclear charge and shielding effects.

NCERT 3.20 : Which of the following pairs of elements would have a more negative electron gain enthalpy? (i) O or F (ii) F or Cl

Answer :

(i) Oxygen (O) vs Fluorine (F)

• Both O and F are in the 2nd period.
• Moving from O to F:
• Atomic size decreases
• Nuclear charge increases
• Both factors increase the attraction of the nucleus for the incoming electron, making the electron gain enthalpy more negative.
• Configuration after gaining one electron:
• $F^- : 1s^2 2s^2 2p^6$ (stable inert gas configuration)
• $O^- : 1s^2 2s^2 2p^5$ (not fully stable)
• Result: Energy released is greater for $F \to F^-$ than $O \to O^-$.
• Electron gain enthalpy values:
  • $F : -328\ \text{kJ mol}^{-1}$
  • $O : -141\ \text{kJ mol}^{-1}$

Conclusion: Fluorine has a more negative electron gain enthalpy than oxygen.

(ii) Fluorine (F) vs Chlorine (Cl)

• Generally, electron gain enthalpy becomes less negative down a group.
• However, chlorine ($Cl$) has $-349\ \text{kJ mol}^{-1}$, more negative than fluorine ($F$, $-328\ \text{kJ mol}^{-1}$).
• Reason:
• Fluorine is very small, so electron–electron repulsions in its compact 2p-subshell are large.
• Incoming electron experiences more repulsion, making it harder to add compared to chlorine.
• Chlorine is larger, so electron-electron repulsion is smaller and the added electron is more easily accommodated.

Conclusion:

Chlorine has a more negative electron gain enthalpy than fluorine due to reduced repulsion in its larger atomic orbitals.

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