Anand Classes provides NCERT Exemplar Solutions for questions 41 to 48, including Matching Columns and Assertion-Reason Type Questions, from the chapter Classification of Elements and Periodicity in Properties for Class 11 Chemistry. These solutions are designed to help students understand periodic trends, element classification, and reasoning-based problems with detailed explanations, examples, and practice questions for thorough preparation and improved problem-solving skills. Click the print button to download study material and notes.
Q. 41 The radius of Na⁺ cation is less than that of Na atom. Give reason.
Answer:
When an atom loses an electron to form a cation, its radius always decreases.
- In Na atom: 11 protons attract 11 electrons.
- In Na⁺ ion: 11 protons attract only 10 electrons.
This means the effective nuclear charge per electron increases in Na⁺.
As a result, electrons are pulled closer to the nucleus and the ionic size decreases.
Reaction:
$$ \text{Na} \; \longrightarrow \; \text{Na}^+ + e^- $$
Comparison:
- Electrons: 11 → 10
- Nuclear charge: 11 → 11 (same)
- Atomic radius of Na: $186 \, \text{pm}$
- Ionic radius of Na⁺: $95 \, \text{pm}$
Final Answer:
$$ \boxed{\text{Radius of Na⁺ is smaller than Na due to increased effective nuclear charge.}} $$
Q. 42 Among alkali metals which element do you expect to be least electronegative and why?
Answer:
Electronegativity decreases down the group because atomic size increases and the attraction between nucleus and valence electrons becomes weaker.
Francium (Fr) lies at the bottom of Group 1 and has the largest atomic size.
Thus, its nucleus has the weakest pull on bonding electrons.
Final Answer:
$$ \boxed{\text{Francium (Fr) is the least electronegative alkali metal.}} $$
Q. 43 Match the correct atomic radius with the element.
| Element | Atomic Radius (pm) |
|---|---|
| O | 66 |
| N | 74 |
| C | 77 |
| B | 88 |
| Be | 111 |
Answer:
All elements belong to the 2nd period. Across a period, atomic radius decreases due to increasing effective nuclear charge.
Order:
$$ O < N < C < B < Be $$
Final Answer:
$$ \boxed{O=66,\; N=74,\; C=77,\; B=88,\; Be=111} $$
Q. 44 Match the correct ionisation enthalpies and electron gain enthalpies of the following elements.
| Element Type | ΔiH₁ (kJ/mol) | ΔiH₂ (kJ/mol) | ΔegH (kJ/mol) |
|---|---|---|---|
| (i) Most reactive non-metal | 1681 | 3374 | –328 |
| (ii) Most reactive metal | 419 | 3051 | –48 |
| (iii) Least reactive element | 2372 | 5251 | +48 |
| (iv) Metal forming halide | 738 | 1451 | –40 |
Answer :
Explanation:
- Most reactive non-metals (halogens) → have high ionisation enthalpy and very negative electron gain enthalpy (e.g., Fluorine).
- Most reactive metals (alkali metals) → have low first ionisation enthalpy but very high second ionisation enthalpy (because the second electron must be removed from noble gas configuration).
- Noble gases (least reactive elements) → have very high ionisation enthalpies and positive electron gain enthalpy (because adding electrons is highly unfavorable).
- Binary halide forming metals (alkaline earth metals) → intermediate values.
Final Answer:
$$ \boxed{(i) \, B, \; (ii) \, A, \; (iii) \, D, \; (iv) \, C} $$
Q. 45 Electronic configuration of some elements is given in Column I and their electron gain enthalpies are given in Column II. Match the electronic configuration with electron gain enthalpy
| Configuration (Column I) | Electron Gain Enthalpy (Column II) |
|---|---|
| A. $1s^2 2s^2 2p^6$ | +48 kJ/mol |
| B. $1s^2 2s^2 2p^6 3s^1$ | –53 kJ/mol |
| C. $1s^2 2s^2 2p^5$ | –328 kJ/mol |
| D. $1s^2 2s^2 2p^4$ | –141 kJ/mol |
Answer :
A. This electronic configuration corresponds to the noble gas i.e., neon. Since, noble gases have positive $Δ_{eg}H$ values, therefore, electronic configuration (A) corresponds to the $Δ_{eg}H$ = +48 kJ/mol.
B. This electronic configuration corresponds to the alkali metal i.e., potassium. Alkali metals have small negative $Δ_{eg}H$ values, hence, electronic configuration (B) corresponds to $Δ_{eg}H$ = –53 kJ/mol.
C. This electronic configuration corresponds to the halogen i.e., fluorine. Since, halogens have high negative $Δ_{eg}H$ values, therefore, electronic configuration (C) corresponds to $Δ_{eg}H$ = 328 kJ/mol.
D. This electronic configuration corresponds to the chalcogen i.e., oxygen. Since, chalcogens have $Δ_{eg}H$ values less negative than those of halogens, therefore, electronic configuration (D) corresponds to $Δ_{eg}H$ = – 141kJ/mol.
Explanation:
A: Neon (noble gas) → positive $Δ_{eg}H$
B: Sodium (alkali metal) → small negative $Δ_{eg}H$
C: Fluorine (halogen) → very high negative $Δ_{eg}H$
D: Oxygen (chalcogen) → less negative than halogens
Final Answer:
$$ \boxed{A \to 4, \; B \to 1, \; C \to 2, \; D \to 3} $$
Q. 46 Assertion and Reason:
Assertion (A): Generally, ionisation enthalpy increases from left to right in a period.
Reason (R): When successive electrons are added in the same principal quantum level, shielding effect does not increase much compared to the increase in nuclear attraction of the electron to the nucleus.
Answer:
Both assertion and reason are true, and the reason correctly explains the assertion. lonisation enthalpy increases along a period because effective nuclear charge increases and atomic size decreases
Final Answer:
$$ \boxed{(b) \; \text{Both correct, reason explains assertion.}} $$
Q. 47 Assertion and Reason:
Assertion (A): Boron has a smaller first ionisation enthalpy than beryllium.
Reason (R): The 2s electron penetrates closer to the nucleus than the 2p electron, hence 2p electron is more shielded by the inner core of electrons than the 2s electrons.
Answer:
Both statements are true, and the reason correctly explains the assertion.
Boron has a smaller first ionisation enthalpy than beryllium because the penetration of a 2s electron to the nucleus is more than the 2p electron. Hence, 2p electron is more shielded by the inner core of electron than the 2s electron
Explanation:
- Be: $1s^2 2s^2$ (stable fully filled 2s orbital) → higher I.E.
- B: $1s^2 2s^2 2p^1$ → easier to remove 2p electron (less penetrating, more shielded).
Final Answer:
$$ \boxed{(c) \; \text{Both correct, reason explains assertion.}} $$
Q. 48 Assertion and Reason:
Assertion (A): Electron gain enthalpy becomes less negative as we go down a group.
Reason (R): Atomic size increases, so the added electron is farther from the nucleus.
Answer:
Both assertion and reason are true, and the reason correctly explains the assertion.
Electron gain enthalpy becomes less negative as the size of an atom increases down the group. This is because within a group screening effect increases on going downward and the added electron would be farther away from the nucleus.
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