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NCERT Exercises and Solutions: Classification of Elements and Periodicity in Properties

🎓 Class 11 Chemistry CBSE Theory Ch 3 – Classification of Elements and Periodicity in Properties ⏱ ~8 min
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NCERT Exercises and Solutions: Classification of Elements and Periodicity

📋 Chapter Summary

Classification of Elements & Periodicity — Key Takeaways

  • Genesis: Döbereiner's Triads (1817) → Newlands' Octaves (1865) → Mendeleev's Periodic Table (1869) → Moseley's Modern Periodic Law (1913).
  • Modern Periodic Law: Properties of elements are periodic functions of their atomic numbers (not atomic masses).
  • Periodic Table structure: 7 periods, 18 groups; divided into s, p, d, f blocks based on the orbital being filled.
  • IUPAC nomenclature for Z > 100: digits replaced by Latin/Greek roots + suffix "-ium".
  • Atomic radius: Decreases across period (higher Z, same shell); increases down group (new shell added).
  • Cation < parent atom < anion. Isoelectronic species: more Z = smaller size.
  • Ionisation enthalpy: Energy to remove e⁻. Increases across period; decreases down group. Anomalies: B<Be (s vs p), O<N (half-filled stability).
  • Electron gain enthalpy: Energy on adding e⁻. Generally negative (exothermic). Cl > F (negative) due to F's compactness.
  • Electronegativity: Tendency to attract bonding electrons. F is most EN (4.0); Cs/Fr least.
  • Anomalous Period 2: Small size, high charge density, no d-orbitals → unique chemistry. Diagonal relationship: Li↔Mg, Be↔Al, B↔Si.
  • Reactivity: Highest at extremes (Group 1 metals, Group 17 non-metals); lowest near centre.
  • Oxide character: Basic (left) → Amphoteric (middle) → Acidic (right) across a period.

🔑 Key Terms & Formulas

Modern Periodic LawProperties = periodic function of Z
PeriodHorizontal row; n = principal quantum number
GroupVertical column; same valence shell config
s-blockGroups 1, 2; ns¹⁻²
p-blockGroups 13–18; ns²np¹⁻⁶
d-blockGroups 3–12; (n−1)d¹⁻¹⁰ns⁰⁻²
f-blockLanthanoids, actinoids; (n−2)f¹⁻¹⁴
Atomic radius↓ across period, ↑ down group
Ionisation enthalpy ΔᵢHX(g) → X⁺(g) + e⁻
Electron gain enthalpy ΔₑgHX(g) + e⁻ → X⁻(g)
Electronegativity (Pauling)F = 4.0 (max); Cs = 0.7
Diagonal relationshipLi↔Mg, Be↔Al, B↔Si

📝 NCERT Exercises (Worked Solutions)

Exercise 3.1

What is the basic theme of organisation in the periodic table?

The basic theme is the classification of elements based on similarity in chemical and physical properties. Elements with similar properties are placed in the same vertical column (group), while properties recur periodically across rows (periods). This systematic arrangement (based on increasing atomic number) reveals patterns and trends, making chemistry more predictable.

Exercise 3.2

Which important property did Mendeleev use to classify the elements in his periodic table and did he stick to that?

Mendeleev primarily used atomic mass as the basis. However, he did NOT stick to it strictly:
  • He placed Te (mass 127.6) BEFORE I (mass 126.9), prioritizing chemical similarity over strict mass order.
  • He left gaps for undiscovered elements (eka-Al, eka-Si, eka-B).
This flexibility was actually his genius — chemical behaviour was given priority. The Modern Periodic Law later justified all his choices using atomic number.

Exercise 3.3

What is the basic difference in approach between the Mendeleev's Periodic Law and the Modern Periodic Law?

Mendeleev (1869): Properties of elements are a periodic function of their atomic masses.
Modern (Moseley, 1913): Properties are a periodic function of their atomic numbers.

The shift came after Moseley's X-ray studies showed that the fundamental property of an element is the number of protons (Z), which determines the number of electrons and hence chemistry. Atomic mass varies with isotopes; atomic number is unique to each element.

Exercise 3.4

On the basis of quantum numbers, justify that the sixth period of the periodic table should have 32 elements.

In Period 6 (n = 6), the orbitals filled are: 6s, 4f, 5d, 6p (in Aufbau order — note 4f and 5d come from earlier shells).
Capacities: 6s = 2, 4f = 14, 5d = 10, 6p = 6.

\[\text{Total} = 2 + 14 + 10 + 6 = \boxed{32 \text{ electrons}}\] This means 32 different elements can occupy Period 6, matching the actual count: Cs (Z=55) to Rn (Z=86) — exactly 32 elements.

Exercise 3.5

In terms of period and group where would you locate the element with Z = 114?

Z = 114. Following the Aufbau order, Z = 114 lies in Period 7 (which fills 7s, 5f, 6d, 7p).
Filling: [Rn] 7s² 5f¹⁴ 6d¹⁰ 7p². The 7p² indicates Group 14 (carbon family).

Z = 114 is in Period 7, Group 14. This is now Flerovium (Fl).

Exercise 3.6

Write the atomic number of the element present in the third period and seventeenth group of the periodic table.

Period 3 = n = 3 (outermost shell). Group 17 = halogens, configuration ns²np⁵.
Configuration: [Ne] 3s² 3p⁵. Counting electrons: 2+2+6+2+5 = 17 electrons (so Z = 17).

Element: Chlorine (Cl, Z = 17).

Exercise 3.7

Which element do you think would have been named by (i) Lawrence Berkeley Laboratory; (ii) Seaborg's group?

(i) Lawrence Berkeley Laboratory: Lawrencium (Lr, Z = 103) — named after Ernest Lawrence (founder of LBL, inventor of the cyclotron).

(ii) Seaborg's group: Seaborgium (Sg, Z = 106) — named after Glenn Seaborg, who discovered/co-discovered 10 transuranium elements.

Both teams contributed many super-heavy element discoveries; the IUPAC officially recognized these names.

Exercise 3.8

Why do elements in the same group have similar physical and chemical properties?

Elements in the same group have the same outer electronic configuration (same number of valence electrons in similar orbital types). Since chemical behaviour is determined primarily by valence electrons:
  • They form similar types of bonds (ionic vs covalent)
  • They achieve similar oxidation states
  • They form analogous compounds (e.g., LiCl, NaCl, KCl all are 1:1 ionic chlorides)
Example: All halogens (Group 17, ns²np⁵) need 1 electron to complete octet → they all form X⁻ ions and HX acids.

Exercise 3.9

What does atomic radius and ionic radius really mean to you?

Atomic radius: An estimate of the size of an atom, typically calculated as half the internuclear distance between two atoms in a covalent bond (covalent radius). Concepts include covalent, metallic, and van der Waals radii. The "true" radius doesn't exist because the electron cloud has no sharp boundary.

Ionic radius: The radius of an ion in an ionic crystal, estimated from the inter-atomic distances. Cations are smaller than parent atoms (electron(s) removed); anions are larger (electrons added).

Both are operational definitions — useful for comparing trends and predicting bonding, even though they aren't physically sharp.

Exercise 3.10

How do atomic radius vary in a period and in a group? How do you explain the variation?

Across a period: Atomic radius DECREASES (e.g., Li 152 → F 64 pm).
Reason: As we move left to right, nuclear charge (Z) increases by 1 per element, but electrons are added to the same shell. The increased pull from the nucleus shrinks the atom. Shielding effect of inner electrons remains roughly constant.

Down a group: Atomic radius INCREASES (e.g., Li 152 → Cs 262 pm).
Reason: A new principal shell is added at each step. This extra shell, with greater shielding from intermediate filled shells, more than compensates for the increased nuclear charge.

Exercise 3.11

What do you understand by isoelectronic species? Name a species that will be isoelectronic with each of the following: (a) F⁻ (b) Ar (c) Mg²⁺ (d) Rb⁺.

Isoelectronic species: Atoms or ions that contain the same number of electrons.

(a) F⁻ has 10 electrons. Isoelectronic: Na⁺ (also 10 e⁻; or O²⁻, Mg²⁺, Ne, Al³⁺).
(b) Ar has 18 electrons. Isoelectronic: K⁺ (also 18 e⁻; or S²⁻, Cl⁻, Ca²⁺).
(c) Mg²⁺ has 10 electrons. Isoelectronic: Ne (also 10 e⁻; or F⁻, O²⁻, Na⁺).
(d) Rb⁺ has 36 electrons. Isoelectronic: Kr (also 36 e⁻; or Sr²⁺, Br⁻).

Exercise 3.12

Consider the following species: N³⁻, O²⁻, F⁻, Na⁺, Mg²⁺ and Al³⁺. (a) What is common in them? (b) Arrange them in order of increasing ionic radii.

(a) All have 10 electrons (isoelectronic with Ne). All achieve the noble-gas Ne configuration: 1s² 2s² 2p⁶.

(b) For isoelectronic species, more protons (Z) means smaller ionic radius (electron cloud pulled tighter):
N³⁻ (Z=7) > O²⁻ (Z=8) > F⁻ (Z=9) > Na⁺ (Z=11) > Mg²⁺ (Z=12) > Al³⁺ (Z=13).

Increasing order: Al³⁺ < Mg²⁺ < Na⁺ < F⁻ < O²⁻ < N³⁻.
Radii (pm): 50 < 65 < 95 < 136 < 140 < 171.

Exercise 3.13

Explain why cations are smaller and anions larger in radii than their parent atoms?

Cation smaller: When an atom loses one or more electrons, it becomes a cation. With fewer electrons over the same nuclear charge:
  • Each remaining electron experiences a higher effective nuclear charge → pulled closer to nucleus.
  • Often the OUTERMOST shell is completely emptied (e.g., Na 3s¹ → Na⁺ has only 1s²2s²2p⁶) — drastic reduction in size.
  • Less inter-electronic repulsion → tighter packing.
Example: Na (186 pm) → Na⁺ (95 pm), nearly 50% reduction.

Anion larger: Adding electrons to an atom forms an anion. With more electrons over the same nuclear charge:
  • Each electron experiences less effective pull from nucleus.
  • Increased inter-electronic repulsion expands the electron cloud.
Example: Cl (99 pm) → Cl⁻ (181 pm), almost doubled.

Exercise 3.14

What is the significance of the terms — 'isolated gaseous atom' and 'ground state' while defining the ionization enthalpy and electron gain enthalpy?

"Isolated gaseous atom": Ensures no inter-atomic interactions (bonds, attractions). In gases at low pressure, atoms are far apart. If liquid or solid, neighboring atoms would influence energy values.

"Ground state": The lowest energy electronic configuration. If the atom is in an excited state (electrons in higher orbitals), the energy required to remove or add an electron differs from the standard value.

Together, these conditions define a standard reference state so that ionisation enthalpy values can be compared meaningfully across elements. Both ΔᵢH and ΔₑgH are defined for X(g), ground state.

Exercise 3.15

Energy of an electron in the ground state of the hydrogen atom is −2.18 × 10⁻¹⁸ J. Calculate the ionisation enthalpy of atomic hydrogen in terms of J mol⁻¹.

Ionisation enthalpy = energy required to free one electron from one atom × Avogadro's number.

Energy per atom: |E| = 2.18 × 10⁻¹⁸ J.
\[\Delta_i H = (2.18 \times 10^{-18} \text{ J}) \times (6.022 \times 10^{23} \text{ atoms/mol})\] \[\Delta_i H = 13.13 \times 10^5 \text{ J/mol} = \boxed{1.313 \times 10^6 \text{ J/mol} = 1312 \text{ kJ/mol}}\] This matches the experimental value for hydrogen.

Exercise 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 ΔᵢH than B, (ii) O has lower ΔᵢH than N and F.

(i) Be (899) > B (801) kJ/mol:
  • Be: [He] 2s² — fully filled 2s subshell, extra stability.
  • B: [He] 2s² 2p¹ — one lone p-electron is loosely bound.
  • The s-orbital is more penetrating than p, so the 2s electron in Be is more tightly held than the 2p electron in B.
  • Removing 2p in B is easier than disrupting Be's full 2s² → IE(B) < IE(Be).
(ii) N (1402) > O (1314) kJ/mol:
  • N: [He] 2s² 2p³ — exactly half-filled p-subshell, extra stability (Hund's rule, exchange energy).
  • O: [He] 2s² 2p⁴ — paired p-electron experiences increased electron-electron repulsion.
  • Removing the paired electron in O is easier than disrupting half-filled stability in N → IE(O) < IE(N).

Exercise 3.17

How would you explain the fact that the first ionization enthalpy of sodium is lower than that of magnesium but its second ionization enthalpy is higher than that of magnesium?

First IE:
  • Na: [Ne] 3s¹ — single 3s electron, easily removed. ΔᵢH₁ = 496 kJ/mol.
  • Mg: [Ne] 3s² — fully filled 2s². Removing one electron requires breaking pair-up, plus higher Z. ΔᵢH₁ = 738 kJ/mol.
  • So IE₁(Na) < IE₁(Mg).
Second IE:
  • Na⁺ has [Ne] config (full octet, very stable). Removing 2nd electron means going INTO inner shell — energetically very expensive. ΔᵢH₂ = 4562 kJ/mol.
  • Mg⁺ has [Ne] 3s¹ — still has a loosely held 3s electron, easy to remove. ΔᵢH₂ = 1450 kJ/mol.
  • So IE₂(Na) >> IE₂(Mg).
Conclusion: Na's resistance to ionizing twice is because Na⁺ already has the stable noble-gas configuration; further ionization disrupts this stability.

Exercise 3.18

What are the various factors due to which the ionization enthalpy of the main group elements tends to decrease down a group?

Three main factors:
  1. Increase in atomic size (radius): Outermost electron is farther from nucleus → less attraction → easier to remove.
  2. Increased shielding effect: Inner shells of electrons screen the outermost electron from the nuclear charge. More inner shells = more shielding = lower effective nuclear charge on valence e⁻.
  3. Increase in the number of inner shells: Down the group, n increases by 1 each time, adding new shell(s) of electrons. Although Z also increases, the effect of bigger atom + more shielding outweighs the higher nuclear charge.
Net result: ΔᵢH decreases down a group, even though Z increases.

Example: Li (520) → Na (496) → K (419) → Rb (403) → Cs (376) kJ/mol.

Exercise 3.19

The first ionization enthalpy values (in kJ mol⁻¹) of group 13 elements are: B 801, Al 577, Ga 579, In 558, Tl 589. How would you explain this fact that the first ionization enthalpy of Ga is greater than Al?

Normally, ΔᵢH should DECREASE down a group. Yet Ga (579) > Al (577) — slight reversal.

Reason — Poor shielding of d-electrons:
  • Al has [Ne] 3s² 3p¹ (only s and p inner shells).
  • Ga has [Ar] 3d¹⁰ 4s² 4p¹. Its 3d¹⁰ inner shell shields the 4p¹ electron less effectively than s/p shells of Al.
  • The d-orbitals are diffuse and don't shield well, so Ga's 4p¹ electron experiences higher effective nuclear charge than expected.
  • Result: Ga's ionization is slightly harder than Al's.
This is a small example of the broader phenomenon in transition and post-transition elements — d/f shells provide poor shielding (also called d-block contraction).

Exercise 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.

(i) F > O (more negative ΔₑgH):
  • F: 2s² 2p⁵ + e⁻ → 2s² 2p⁶ (filled, very stable). Energy released: −328 kJ/mol.
  • O: 2s² 2p⁴ + e⁻ → 2s² 2p⁵ (still incomplete). Energy released: −141 kJ/mol.
  • F's gain enthalpy is more negative because adding e⁻ achieves the noble-gas Ne configuration.
(ii) Cl > F (Cl is MORE negative — anomaly!):
  • F: ΔₑgH = −328 kJ/mol; Cl: ΔₑgH = −349 kJ/mol.
  • F's small 2p subshell is compact; adding new electron causes high inter-electronic repulsion, partially offsetting the energy released.
  • Cl's 3p subshell is more spacious; new electron added with less repulsion → more energy released.
  • Hence ΔₑgH (Cl) is more negative than ΔₑgH (F).

Exercise 3.21

Would you expect the second electron gain enthalpy of O as positive, more negative or less negative than the first? Justify your answer.

Second electron gain enthalpy of O is POSITIVE. Reasoning:
  • 1st: O(g) + e⁻ → O⁻(g), ΔₑgH₁ = −141 kJ/mol (energy released).
  • 2nd: O⁻(g) + e⁻ → O²⁻(g). Adding e⁻ to a NEGATIVE ion requires overcoming repulsion between the existing negative charge and incoming electron.
  • So 2nd gain enthalpy is positive (energy must be supplied), about +744 kJ/mol.
This is why isolated O²⁻ ions don't form spontaneously in gas phase — they only exist in ionic crystals where lattice energy stabilizes them.

Exercise 3.22

What is the basic difference between the terms electron gain enthalpy and electronegativity?

Electron gain enthalpy (ΔₑgH):
  • Quantitatively measurable energy change when an isolated gaseous atom gains an electron.
  • Has SI units: kJ/mol.
  • Defined for the FREE atom in gas phase.
Electronegativity (EN):
  • Relative tendency of an atom IN A MOLECULE to attract bonding electrons toward itself.
  • DIMENSIONLESS (a relative scale, e.g., Pauling scale).
  • Defined in the bonded state, not for an isolated atom.
Key difference: ΔₑgH is for free atom + electron; EN is for atom in a chemical bond. They correlate (both increase across period, decrease down group), but are conceptually distinct.

Exercise 3.23

How would you react to the statement that the electronegativity of N on the Pauling scale is 3.0 in all the nitrogen compounds?

The statement is INCORRECT in strict sense. The Pauling EN of 3.0 for N is an AVERAGE/effective value across many compounds. In reality:
  • EN depends on the oxidation state and hybridization of the atom.
  • N in NH₃ (sp³) has slightly different EN than N in HCN (sp).
  • EN also varies with the partner atoms (more EN in NF₃ context vs. less in N₂).
The Pauling scale is a useful approximation, but a single number doesn't capture all situations. More refined scales (Allred-Rochow, Mulliken) give different values, especially for elements in unusual environments.

Exercise 3.24

Describe the theory associated with the radius of an atom as it (a) gains an electron, (b) loses an electron.

(a) Atom gains electron (forms anion):
  • Number of electrons increases, but Z (number of protons) stays the same.
  • Effective nuclear charge per electron decreases.
  • Increased electron-electron repulsion expands the electron cloud.
  • Result: Anion is LARGER than the parent atom.
  • Example: Cl (99 pm) → Cl⁻ (181 pm).
(b) Atom loses electron (forms cation):
  • Number of electrons decreases; Z stays the same.
  • Effective nuclear charge per remaining electron increases.
  • Often the entire OUTERMOST shell is removed (e.g., Na 3s¹ → Na⁺ has only inner shells).
  • Less electron repulsion → tighter packing.
  • Result: Cation is SMALLER than the parent atom.
  • Example: Na (186 pm) → Na⁺ (95 pm).

Exercise 3.25

Would you expect the first ionization enthalpies for two isotopes of the same element to be the same or different? Justify your answer.

The first ionization enthalpies for two isotopes will be essentially the SAME.
  • Isotopes have the SAME atomic number (Z), hence the SAME number of electrons and the SAME electronic configuration.
  • Isotopes differ only in the number of NEUTRONS in the nucleus.
  • Neutrons don't affect electron-electron interactions or the effective nuclear charge.
  • Hence the energy required to remove the outermost electron is essentially the same.
Tiny exception: The slight mass difference can cause a very small "isotope shift" in spectroscopic measurements (the reduced mass affects orbital energies), but this is negligible for chemistry purposes.

Example: ¹H and ²H (deuterium) have IE ≈ 1312 kJ/mol — same to 4 significant figures.

Exercise 3.26

What are the major differences between metals and non-metals?

PropertyMetalsNon-metals
PositionLeft and centre of tableTop right (above stair-step)
State at RTSolid (except Hg, liquid)Solids, liquids, gases
Melting pointGenerally highGenerally low
ConductivityGood (heat & electricity)Poor (insulators), exception: graphite
MechanicalMalleable, ductileBrittle (in solid form)
LustreShiny (lustrous)Dull
Ionisation enthalpyLowHigh
ElectronegativityLowHigh
Tend to formCations (lose e⁻)Anions (gain e⁻)
Oxide characterBasic (mostly)Acidic (mostly)

Exercise 3.27

Use the periodic table to answer the following questions: (a) Identify an element with five electrons in the outer subshell. (b) Identify an element that would tend to lose two electrons. (c) Identify an element that would tend to gain two electrons. (d) Identify the group having metal, non-metal, liquid as well as gas at the room temperature.

(a) Five outer electrons → Group 17 (halogens), e.g., Cl ([Ne] 3s² 3p⁵) or any halogen.

(b) Tends to lose 2 electrons → alkaline earth metal, Group 2, e.g., Mg, Ca.

(c) Tends to gain 2 electrons → Group 16 (chalcogens), e.g., O, S (form O²⁻, S²⁻).

(d) Group containing metal, non-metal, liquid, AND gas at RT: Group 17 (halogens):
- F: gas (yellow); Cl: gas (yellow-green); Br: liquid (red-brown); I: solid (purple); At: solid (radioactive).
Strictly speaking, halogens are all non-metals, so this might be questioned for "metal" — perhaps the question intended to say "diverse states." Group 1 has all metals; only Group 17 has the most diverse range of states at RT among non-metals.

Exercise 3.28

The increasing order of reactivity among group 1 elements is Li < Na < K < Rb < Cs whereas that among group 17 elements is F > Cl > Br > I. Explain.

Group 1 (alkali metals): Reactivity increases down the group (Li < Na < K < Rb < Cs).
  • Reactivity here = tendency to LOSE electron.
  • Down the group: atomic size increases, ionization enthalpy decreases → easier to remove valence electron.
  • Cs gives up its 6s¹ electron most readily → most reactive.
Group 17 (halogens): Reactivity decreases down the group (F > Cl > Br > I).
  • Reactivity here = tendency to GAIN electron.
  • Down the group: atomic size increases → less attraction for incoming electron → less negative electron gain enthalpy → less reactive.
  • F is smallest, has highest tendency to attract an extra electron → most reactive.
Insight: The two groups represent OPPOSITE reactivity drivers — losing vs gaining electrons — and respond differently to atomic size.

Exercise 3.29

Write the general outer electronic configuration of s-, p-, d- and f-block elements.

BlockGeneral outer configurationNumber of valence e⁻
s-blockns¹⁻²1 or 2
p-blockns² np¹⁻⁶3 to 8
d-block(n−1)d¹⁻¹⁰ ns⁰⁻²3 to 12 (variable)
f-block(n−2)f¹⁻¹⁴ (n−1)d⁰⁻¹ ns²2 to 16 (variable)

where n = principal quantum number of outermost shell, increasing across periods.

Exercise 3.30

Assign the position of the element having outer electronic configuration (i) ns² np⁴ for n=3 (ii) (n−1)d² ns² for n=4, and (iii) (n−2) f⁷ (n−1)d¹ ns² for n=6, in the periodic table.

(i) ns² np⁴ for n=3:
  • 3s² 3p⁴ → Period 3, Group 16 (chalcogen, p-block).
  • Total e⁻ to this point: 2+2+6+2+4 = 16. Element: Sulfur (S, Z=16).
(ii) (n−1)d² ns² for n=4:
  • 3d² 4s² → Period 4, d-block, Group 4 (group = d count + s count = 2+2 = 4).
  • Total e⁻: 2+2+6+2+6+2+2 = 22. Element: Titanium (Ti, Z=22).
(iii) (n−2)f⁷ (n−1)d¹ ns² for n=6:
  • 4f⁷ 5d¹ 6s² → Period 6, f-block (lanthanoid).
  • Group: f-block elements are usually placed below period 6 in a separate row; assigned to "Group 3" formally.
  • Total e⁻: 2+2+6+2+6+2+10+6+2+10+6+2+7+1+2 = 64. Wait, let me recount: filling order to 4f⁷ 5d¹ 6s² gives Z = 64. Element: Gadolinium (Gd, Z=64).
🎓 Chapter 3 Complete! You've now mastered classification, periodicity, and the trends governing all of chemistry. The periodic table is your roadmap — every concept in chemistry can be traced back to electron configurations and periodic trends. Next chapter (Chemical Bonding) builds directly on this foundation.

Frequently Asked Questions — NCERT Exercises and Solutions: Classification of Elements and Periodicity

What are the most asked questions in Chapter 3 NCERT exercises?
The most-asked NCERT Class 11 Chemistry Chapter 3 questions are: (1) write electronic configuration of an element and identify its group/period/block; (2) compare atomic/ionic radii of given elements or isoelectronic species; (3) explain trends in ionisation enthalpy with anomalies; (4) predict the most/least electronegative element from a set; (5) name elements with specific configurations; (6) explain diagonal relationship pairs (Li-Mg, Be-Al, B-Si); (7) anomalous behaviour of Period-2 elements. The MyAiSchool exercise set covers every CBSE-aligned question type with stepwise solutions.
How do you compare atomic radii of given elements?
To compare atomic radii in NCERT Class 11 Chemistry exercises: (1) identify the periodic position of each element, (2) apply rules — atomic radius decreases across a period and increases down a group, (3) for isoelectronic species (same number of electrons), radius decreases with increasing Z because greater nuclear charge contracts the cloud. Example: Na > Mg > Al > Si (across period) and Li < Na < K < Rb (down group). For isoelectronic O²⁻ > F⁻ > Na⁺ > Mg²⁺. Always state the reason for full marks in the board exam.
How do you predict the most electronegative element from a set?
To predict the most electronegative element in NCERT Class 11 Chemistry exercises: (1) locate elements in the periodic table, (2) apply trend — electronegativity increases across a period and decreases down a group, (3) the element nearest to fluorine (top right, excluding noble gases) is most electronegative. Example: among C, N, O, F → F is most electronegative; among Cl, Br, I → Cl is most electronegative. Always cite Pauling's scale values when needed: F = 4.0, O = 3.5, N = 3.0, Cl = 3.0. The MyAiSchool exercise set provides full periodic trend tables.
How do you explain ionisation enthalpy anomalies in exercises?
Ionisation enthalpy anomalies in NCERT Class 11 Chemistry Chapter 3 occur due to electronic configuration stability. Common anomalies: (1) IE of B (5) < Be (4) because B loses a 2p electron (easier) while Be loses a 2s electron from a stable s² configuration; (2) IE of O (8) < N (7) because N has a stable half-filled 2p³ while O removes one electron from paired 2p, reducing repulsion; (3) IE of S < P for the same reason. Always state the electronic configuration when explaining. These appear frequently in 2-mark CBSE questions and the MyAiSchool exercises drill all such patterns.
How do you identify the block of an element from its atomic number?
To identify the block of an element from atomic number in NCERT Class 11 Chemistry: (1) write the complete electronic configuration up to Z, (2) identify which subshell the last electron entered — s, p, d or f, (3) that subshell is the block. Examples: Z = 11 (Na, 1s²2s²2p⁶3s¹) → s-block; Z = 17 (Cl, 1s²2s²2p⁶3s²3p⁵) → p-block; Z = 26 (Fe, [Ar]3d⁶4s²) → d-block; Z = 60 (Nd, [Xe]4f⁴6s²) → f-block. Group number can be deduced from valence electrons. Practise this with the periodic table provided in the MyAiSchool exercise set.
What are typical 5-mark questions in Chapter 3 CBSE board exams?
Typical 5-mark CBSE board questions on Class 11 Chemistry Chapter 3 include: (1) discuss the trends in atomic radius, ionisation enthalpy, electron gain enthalpy and electronegativity across a period and down a group with reasons and exceptions; (2) explain the anomalous behaviour of Period-2 elements with examples; (3) describe the diagonal relationship of Li-Mg, Be-Al, B-Si; (4) compare the properties of any two given groups; (5) account for periodic variation in oxidation states across periods. The MyAiSchool exercise set provides model answers for each pattern.
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Chemistry Class 11 Part I – NCERT (2025-26)
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