Classification of Elements and Periodicity in Prop — Class 11 Chemistry NCERT Solutions (Free)
Free step-by-step NCERT solutions for Class 11 Chemistry chapter "Classification of Elements and Periodicity in Prop" — 7 important questions with detailed answers for CBSE board exam preparation.
TL;DR: Free step-by-step NCERT solutions for Class 11 Chemistry chapter "Classification of Elements and Periodicity in Prop" — 7 important questions with det…
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NCERT Solutions for Class 11 Chemistry — Classification of Elements and Periodicity in Prop. Step-by-step answers to all 7 textbook questions from this chapter, written for CBSE board preparation and free to use.
Classification of Elements and Periodicity in Prop — All 7 Questions Solved
Q1. Arrange the following elements in increasing order of ionization enthalpy: Be, Mg, Ca, Sr. Give reason.
Increasing order of ionization enthalpy:
Be < Mg < Ca < Sr
Or in increasing order:
Sr < Ca < Mg < Be
Wait, let me correct this — the question asks for INCREASING order:
Decreasing order of ionization enthalpy:
Be > Mg > Ca > Sr
Increasing order of ionization enthalpy:
Sr < Ca < Mg < Be
Reason:
All these elements belong to Group 2 (Alkaline Earth Metals), moving down the group from Be to Sr.
1. As we move down the group:
→ Atomic number increases
→ Number of electron shells increases
→ Atomic radius increases
→ Electrons are farther from the nucleus
2. As the atomic radius increases:
→ Outermost electrons experience weaker attraction to the nucleus (due to increased distance and shielding by inner electrons)
→ Less energy is required to remove the outermost electron
→ Ionization enthalpy decreases
3. Therefore, as we go down Group 2:
Be (smallest radius, highest IE) > Mg > Ca > Sr (largest radius, lowest IE)
Answer: Sr < Ca < Mg < Be (increasing order of ionization enthalpy)
Q2. What is the periodic law and modern periodic law? How does the modern periodic law differ from Mendeleev's periodic law?
Periodic Law (Mendeleev's, 1869):
The properties of elements are periodic functions of their atomic weights.
When elements are arranged in order of increasing atomic weights, elements with similar properties recur at regular intervals.
Modern Periodic Law (1913):
The properties of elements are periodic functions of their atomic numbers.
When elements are arranged in order of increasing atomic number, elements with similar properties recur at regular intervals.
Key Differences:
1. Basis of Arrangement:
→ Mendeleev: Used atomic weights
→ Modern: Uses atomic numbers (number of protons)
2. Anomalies Resolved:
→ Mendeleev: Some elements seemed out of order by atomic weight (e.g., Co, Ni; Ar, K)
→ Modern: All anomalies resolved because atomic number doesn't have exceptions
3. Reason for Periodicity:
→ Mendeleev: Did not explain why periodicity occurs
→ Modern: Periodicity is due to repetition of electronic configuration after every 2, 8, 18, 32 electrons (relating to shell filling)
4. Accuracy:
→ Mendeleev: Gaps and inconsistencies
→ Modern: Complete and accurate, explains all periodic properties
5. Electron Configuration Connection:
→ Modern periodic law directly relates to electron configuration:
→ Similar elements have similar valence electron configurations
→ This explains the similarity in chemical properties
Example: Noble gases (He, Ne, Ar, Kr, Xe, Rn) all have filled valence shells (ns² np⁶), so they recur periodically.
Conclusion: The modern periodic law is based on atomic number and electron configuration, making it more fundamental and accurate than Mendeleev's law based on atomic weights.
Q3. Compare the effective nuclear charge experienced by 3s and 3p electrons in a sulfur atom (S, Z = 16).
Given: Sulfur atom, Z = 16
Electron configuration of S: 1s² 2s² 2p⁶ 3s² 3p⁴
Effective Nuclear Charge (Zeff) formula:
Zeff = Z - S
where:
Z = actual nuclear charge (atomic number) = 16
S = shielding constant (number of inner electrons)
Part 1: Effective nuclear charge for 3s electron
Shielding for 3s electron:
→ Inner electrons (1s² 2s² 2p⁶): 10 electrons shield completely
→ Other 3s electron: shields 0.35 of nuclear charge
→ 3p electrons: negligible shielding (same shell)
Total shielding (S) ≈ 10 + 0.35 = 10.35
Zeff(3s) = 16 - 10.35 = 5.65
Part 2: Effective nuclear charge for 3p electron
Shielding for 3p electron:
→ Inner electrons (1s² 2s² 2p⁶): 10 electrons shield completely
→ 3s electrons: 2 electrons, each shields 0.85
→ Other 3p electrons: shield 0.35 each (3 electrons)
Total shielding (S) = 10 + (2 × 0.85) + (3 × 0.35)
S = 10 + 1.7 + 1.05 = 12.75
Zeff(3p) = 16 - 12.75 = 3.25
Comparison:
Zeff(3s) = 5.65 > Zeff(3p) = 3.25
Conclusion:
The 3s electron experiences a higher effective nuclear charge than the 3p electron because:
1. 3s electrons penetrate closer to the nucleus and experience less shielding
2. 3p electrons are shielded more by 3s electrons in the same shell
3. Therefore, 3s electrons are more tightly bound and harder to remove than 3p electrons
Q4. Define electron affinity. Explain why noble gases have very low (or near zero) electron affinity.
Electron Affinity Definition:
Electron affinity (EA) is the amount of energy released when a neutral, gaseous atom gains one mole of electrons to form a negative ion.
X(g) + e⁻ → X⁻(g) + ΔE
Units: kJ/mol or eV
More negative EA = greater tendency to gain electrons
Trend in Periodic Table:
→ Generally increases across a period (left to right)
→ Generally decreases down a group
→ Halogens have highest electron affinity
→ Noble gases have nearly zero electron affinity
Why Noble Gases Have Very Low Electron Affinity:
1. Stable Electronic Configuration:
→ Noble gases have completely filled valence shells
→ Configuration: ns² np⁶ (octet rule satisfied)
→ Example: Ne (1s² 2s² 2p⁶) — very stable
2. No Gain in Stability:
→ Adding an extra electron would disrupt the stable, filled valence shell
→ The extra electron must go into a higher, less stable orbital
→ Example: Ne⁻ would be Ne (1s² 2s² 2p⁶ 3s¹) — unstable
3. No Driving Force:
→ Since the atom is already maximally stable, there is no energy advantage to accepting an electron
→ Energy is actually required to add an electron to a higher orbital
→ Electron affinity is near zero or slightly positive (rare)
4. Effective Nuclear Charge:
→ For noble gases, the effective nuclear charge pulling a new electron into a higher shell is very weak
→ The nucleus doesn't strongly attract an electron placed in the next shell
Comparison:
Halogens (Group 17): High negative EA — readily gain one electron to complete their octet
Noble gases (Group 18): Near-zero EA — do not gain electrons; already have stable configuration
Example:
F: EA = -322 kJ/mol (readily gains electron)
Ne: EA ≈ 0 (no tendency to gain electron)
Conclusion: Noble gases have very low electron affinity because they already have a maximally stable valence shell configuration, so accepting an additional electron provides no thermodynamic advantage.
Q5. What is electronegativity? Explain the Pauling scale and trends in electronegativity across the periodic table.
Electronegativity Definition:
Electronegativity is the relative tendency of an atom to attract the electron pair towards itself in a covalent bond.
It is a measure of the ability of an atom to pull electron density in a chemical bond.
Note: Electronegativity ≠ Electron affinity
→ Electronegativity: applies to bonded atoms
→ Electron affinity: applies to isolated atoms gaining electrons
Pauling Scale:
Linus Pauling developed a relative scale (0 to 4) to express electronegativity.
Scales:
→ Fluorine (F) = 4.0 (highest electronegativity)
→ Other elements are scaled relative to fluorine
→ Noble gases are usually not assigned values (rarely form bonds)
→ Range: 0.7 (Cs) to 4.0 (F)
Common Electronegativity Values:
F (4.0) > O (3.44) > N (3.0) > Cl (3.16) > Br (2.96) > C (2.55) > H (2.20) > Na (0.93)
Trends in the Periodic Table:
1. Across a Period (left to right):
→ Electronegativity INCREASES
→ Reason: Atomic radius decreases, effective nuclear charge increases
→ Atom can more strongly attract bonding electrons
→ Example: Na (0.93) < Mg (1.31) < Al (1.61) < Si (1.90) < Cl (3.16)
2. Down a Group (top to bottom):
→ Electronegativity DECREASES
→ Reason: Atomic radius increases, distance from nucleus increases
→ Nucleus has weaker pull on bonding electrons (shielding increases)
→ Example: F (4.0) > Cl (3.16) > Br (2.96) > I (2.66)
3. Exceptions:
→ Some transition metals show irregular trends
→ Noble gases rarely have assigned values (inert)
Periodic Table Pattern:
Highest electronegativity: top-right (F, O, N, Cl)
Lowest electronegativity: bottom-left (Cs, Ba)
Applications:
→ Predicts polarity of covalent bonds
→ Δ EN > 1.7 → ionic bond
→ 0.4 < Δ EN < 1.7 → polar covalent bond
→ Δ EN < 0.4 → nonpolar covalent bond
Example:
H-Cl: Δ EN = 3.16 - 2.20 = 0.96 → polar covalent
H-F: Δ EN = 4.0 - 2.20 = 1.8 → strongly polar covalent (nearly ionic)
Conclusion: Electronegativity increases across a period and decreases down a group, reflecting how easily atoms attract bonding electrons.
Q6. Predict whether the following compounds are ionic or covalent based on electronegativity difference. (a) NaCl, (b) HCl, (c) Cl2, (d) MgO
Criterion for Bond Type (Pauling Scale):
Δ EN = |EN(atom 1) - EN(atom 2)|
Classification:
→ Δ EN ≥ 1.7 → Ionic bond
→ 0.4 < Δ EN < 1.7 → Polar covalent bond
→ Δ EN ≤ 0.4 → Nonpolar covalent bond
Electronegativity Values:
Na = 0.93, Cl = 3.16, H = 2.20, Mg = 1.31, O = 3.44
(a) NaCl:
Δ EN = |0.93 - 3.16| = 2.23
Since Δ EN = 2.23 > 1.7
Bond Type: IONIC
NaCl is an ionic compound (metal + nonmetal)
(b) HCl:
Δ EN = |2.20 - 3.16| = 0.96
Since 0.4 < Δ EN (0.96) < 1.7
Bond Type: POLAR COVALENT
HCl is a covalent compound with significant polarity (H is partially positive, Cl is partially negative)
(c) Cl₂:
Δ EN = |3.16 - 3.16| = 0
Since Δ EN ≤ 0.4
Bond Type: NONPOLAR COVALENT
Cl₂ is a nonpolar covalent molecule (same atoms, electrons equally shared)
(d) MgO:
Δ EN = |1.31 - 3.44| = 2.13
Since Δ EN = 2.13 > 1.7
Bond Type: IONIC
MgO is an ionic compound (metal + nonmetal, large electronegativity difference)
Summary:
(a) NaCl → Ionic
(b) HCl → Polar covalent
(c) Cl₂ → Nonpolar covalent
(d) MgO → Ionic
Q7. Explain the periodic variation of atomic radius in the periodic table. Why does atomic radius decrease across a period?
Atomic Radius Definition:
Atomic radius is half the distance between the nuclei of two adjacent atoms of the same element in a covalent bond.
Unit: Picometer (pm) or Angstrom (Å)
Periodic Variation of Atomic Radius:
1. ACROSS A PERIOD (left to right):
→ Atomic radius DECREASES
→ Example: Na > Mg > Al > Si > P > S > Cl
2. DOWN A GROUP (top to bottom):
→ Atomic radius INCREASES
→ Example: F < Cl < Br < I
Why Does Atomic Radius Decrease Across a Period?
Factor 1: Increasing Effective Nuclear Charge
→ Across a period, atomic number increases (more protons in nucleus)
→ Each new proton increases the positive charge of the nucleus
→ This increased positive charge pulls the valence electrons more strongly
→ Valence electrons are drawn closer to the nucleus
→ Atomic radius decreases
Factor 2: Electrons in Same Shell
→ Electrons added across a period enter the same valence shell
→ Electrons in the same shell provide minimal shielding of nuclear charge
→ Each electron shields only ~0.35 of the effective nuclear charge
→ Net effect: increasing nuclear charge largely unshielded
→ Valence electrons experience greater attraction and move closer
Factor 3: Relatively Constant Shielding
→ Inner electron shells remain the same across a period
→ Shielding effect remains nearly constant
→ Only the effective nuclear charge increases significantly
Numerical Example (Period 3):
Element Electrons Configuration Atomic Radius (pm)
Na 11 [Ne] 3s¹ 186
Mg 12 [Ne] 3s² 160
Al 13 [Ne] 3s² 3p¹ 143
Si 14 [Ne] 3s² 3p² 117
P 15 [Ne] 3s² 3p³ 110
S 16 [Ne] 3s² 3p⁴ 104
Cl 17 [Ne] 3s² 3p⁵ 99
Notice: As atomic number increases (more protons), atomic radius decreases.
Why Atomic Radius Increases Down a Group?
→ New electron shells are added (more distance from nucleus)
→ Outermost electrons enter shells farther from the nucleus
→ Even though nuclear charge increases, distance effect dominates
→ Atomic radius increases
Example (Group 17 — Halogens):
F: 64 pm (smallest)
Cl: 99 pm
Br: 114 pm
I: 133 pm (largest)
Conclusion:
Across a period: Atomic radius decreases due to increasing effective nuclear charge pulling electrons closer (shielding remains constant).
Down a group: Atomic radius increases due to addition of new electron shells (distance effect dominates over nuclear charge increase).
How to Use These Solutions
Attempt each question yourself first and only then compare with the worked answer. Marks in the CBSE board exam are awarded for the METHOD as much as the final result, so reproduce the steps rather than memorising the last line. Where a solution states a law, a formula or a definition, learn that wording — examiners look for it.
Related practice for this chapter: chapter MCQs, previous-year questions, revision notes and sample papers.