Classification of Elements and Periodicity in Properties
Class 11 Chemistry Chapter 3 notes, in-text problems (3.13.10), NCERT exercises (3.1 to 3.40), revision formula sheets, and 3-level chapter tests.
Introduction: Why the Periodic Table?
"The Periodic Table is arguably the most important concept in chemistry" — Glenn T. Seaborg. It organises all known elements, shows trends and families, suggests research avenues, and is the everyday support for students.
What You Will Be Able to Do
Trace how grouping elements by properties led to the Periodic Table.
State the Periodic Law and its modern (atomic number) form.
Name elements with Z > 100 by IUPAC nomenclature.
Classify elements into s, p, d, f blocks and know their characteristics.
Recognise periodic trends in atomic/ionic radii, ionization enthalpy, electron gain enthalpy, electronegativity, valence.
Relate ionization enthalpy to metallic character and correlate reactivity with occurrence in nature.
3.1 Why Do We Need to Classify Elements?
1800: 31 elements known. 1865: 63 elements. Present: 114 elements (some man-made). Studying each element and its innumerable compounds individually is impossible → scientists needed a systematic classification that rationalises known facts and predicts new ones.
Mnemonic: Element Count
"31 → 63 → 114" (1800 → 1865 → now). ⚡ In one line: classification makes chemistry predictable, not just memorable.
3.2 Genesis of Periodic Classification
Systematic classification evolved through several attempts — Triads → Octaves → Mendeleev's Table → Modern Periodic Law.
Evolution of the Periodic Table — Historical Roadmap
3.2.1 Dobereiner's Triads (18171829)
German chemist Johann Dobereiner grouped elements into triads (sets of three): the atomic weight of the middle element ≈ average of the other two, and its properties were in between.
Triad 1
At. wt.
Triad 2
At. wt.
Triad 3
At. wt.
Li
7
Ca
40
Cl
35.5
Na
23
Sr
88
Br
80
K
39
Ba
137
I
127
Limitation
The Law of Triads worked only for a few elements — dismissed as coincidence.
3.2.2 Newlands' Law of Octaves (1865)
English chemist John Alexander Newlands arranged elements in increasing atomic weight and found that every 8th element resembled the first — just like every 8th note resembles the first in music.
1
2
3
4
5
6
7
H
Li
Be
B
C
N
O
F
Na
Mg
Al
Si
P
S
Cl
K
Ca
Limitation
Held true only up to calcium (Ca). Not accepted then, but Newlands was awarded the Davy Medal (1887).
3.2.3 de Chancourtois' Cylindrical Table (1862)
French geologist A.E.B. de Chancourtois arranged elements by increasing atomic weight on a cylindrical table to show periodic recurrence of properties — but it did not attract much attention.
3.2.4 Mendeleev's Periodic Table (1869)
Dmitri Mendeleev (Russia) & Lothar Meyer (Germany), working independently (1869), proposed the Periodic Law:
The properties of the elements are a periodic function of their atomic weights.
Features:
Elements arranged in increasing atomic weight in horizontal rows & vertical columns; similar-property elements in the same group.
Ignored atomic-weight order when needed — e.g., I (lower at. wt. than Te) placed in Group VII (halogens) due to similar properties.
Left gaps for undiscovered elements: Eka-aluminium (→ Ga) and Eka-silicon (→ Ge) — whose properties he predicted correctly.
Used empirical formulas of compounds (oxides, chlorides) for classification.
Property
Eka-aluminium (predicted)
Gallium (found)
Eka-silicon (predicted)
Germanium (found)
Atomic weight
68
70
72
72.6
Density (g/cm³)
5.9
5.94
5.5
5.36
Melting point
Low
302.93 K
High
1231 K
Formula of oxide
E₂O₃
Ga₂O₃
EO₂
GeO₂
Formula of chloride
ECl₃
GaCl₃
ECl₄
GeCl₄
Mendeleev's Genius
His bold quantitative predictions (eka-aluminium & eka-silicon) were verified when gallium & germanium were discovered — making him and his Table famous.
3.3 Modern Periodic Law & the Long Form
Henry Moseley (1913) studied characteristic X-ray spectra: plotting √ν of X-rays against atomic number (Z) gave a straight line (not against atomic mass). Thus atomic number is more fundamental:
Moseley's Law: √ν vs Atomic Number (Z) vs Atomic Mass
The physical & chemical properties of elements are periodic functions of their atomic numbers. (Modern Periodic Law)
Long form of the Periodic Table: 7 horizontal periods + 18 vertical groups (IUPAC numbering 118, replacing IAVIIA, VIII, IBVIIB, 0). The period number = highest principal quantum number (n) of elements in it. Period sizes: 2, 8, 8, 18, 18, 32, 32 (7th incomplete). The 14 lanthanoids & 14 actinoids are placed in separate panels below.
Period
1
2
3
4
5
6
7
No. of elements
2
8
8
18
18
32
32 (incomplete)
Mnemonic: Period Sizes
"2, 8, 8, 18, 18, 32, 32" → "Two Great Eights, Eighteen Again, Thirty-Two". ⚡ In one line: Modern law = properties depend on atomic number (Z), not atomic mass.
3.4 Nomenclature of Elements with Z > 100
Problem: super-heavy elements are extremely unstable (minute quantities, competitive claims — e.g., American "Rutherfordium" vs Soviet "Kurchatovium" for element 104).
IUPAC solution: temporary systematic nomenclature derived directly from the atomic number, made from numerical roots + suffix "ium":
Digit
Root
Abbr.
Digit
Root
Abbr.
0
nil
n
5
pent
p
1
un
u
6
hex
h
2
bi
b
7
sept
s
3
tri
t
8
oct
o
4
quad
q
9
enn
e
How to build: join roots of each digit in order (e.g., 101 = un + nil + un = unnilunium, symbol Unu). Later, IUPAC gives the permanent name & symbol (e.g., element 106 = Seaborgium, Sg).
Mnemonic: IUPAC Roots
"N-Un-Bi-Tri-Quad-Pent-Hex-Sept-Oct-Enn" for 09 → ⚡ In one line: Z = 120 → Un + Bi + Nil = Unbinilium (Ubn).
3.1
What would be the IUPAC name and symbol for the element with atomic number 120?
Roots for 1, 2 and 0 are un, bi, nil. Hence the name = unbinilium, symbol = Ubn.
Name: Unbinilium (Ubn)
3.5 Electronic Configurations & the Periodic Table
(a) Electronic Configurations in Periods
Period number = n of the valence shell. Number of elements in a period = 2 × number of orbitals being filled (2n²).
1st period (n = 1): 1s fills → H (1s¹), He (1s²) → 2 elements.
2nd (n = 2): 2s + 2p fill → Li to Ne → 8 elements.
3rd (n = 3): 3s + 3p fill → Na to Ar → 8 elements.
4th (n = 4): 4s, 3d, 4p fill → K to Kr → 18 elements (includes the 3d transition series, Sc to Zn).
5th (n = 5): 5s, 4d, 5p → Rb to Xe → 18 elements (4d transition series, Y to Cd).
Elements are classified into four blocks based on the orbital being filled last:
The Periodic Table Blocks — 3D View
3.6.1 The s-Block Elements
Groups 1 (alkali metals, ns¹) and 2 (alkaline earth metals, ns²). Reactive metals with low ionization enthalpies; form M⁺ (Gr 1) or M²⁺ (Gr 2) ions. Metallic character & reactivity increase down the group; never found free in nature. Compounds are predominantly ionic (except Li & Be).
3.6.2 The p-Block Elements
Groups 1318; with s-block they form the Representative (Main Group) Elements. Valence configuration ns²np¹ → ns²np⁶. Ends in noble gases (closed ns²np⁶ — very low reactivity). Halogens (Group 17) & chalcogens (Group 16) have highly negative electron gain enthalpies. Non-metallic character increases left → right; metallic character increases down a group.
3.6.3 The d-Block Elements (Transition Elements)
Groups 312; inner d orbitals are being filled. General configuration (n−1)d¹⁻¹⁰ns⁰⁻² (exception: Pd = 4d¹⁰5s⁰). All metals; form coloured ions, show variable oxidation states, paramagnetism, and act as catalysts. Zn, Cd, Hg (d¹⁰ns²) are exceptions. They bridge s-block and p-block elements ("transition").
3.6.4 The f-Block Elements (Inner-Transition Elements)
Lanthanoids (Ce, Z=58 → Lu, Z=71) & Actinoids (Th, Z=90 → Lr, Z=103); outer configuration (n−2)f¹⁻¹⁴(n−1)d⁰⁻¹ns². All metals with very similar properties within each series. Actinoids are radioactive; elements after uranium = transuranium elements.
3.3
The elements Z = 117 and 120 have not yet been discovered. In which family/group would you place these elements and also give the electronic configuration in each case?
Z = 117: Group 17 (halogen family); configuration = [Rn] 5f¹⁴ 6d¹⁰ 7s² 7p⁵. Z = 120: Group 2 (alkaline earth metals); configuration = [Uuo] 8s².
117 → Group 17 (7p⁵); 120 → Group 2 (8s²)
3.6.5 Metals, Non-metals and Metalloids
Metals (78% of elements): left side; solids at room temperature (Hg exception; Ga, Cs melt ~30 °C); high m.p./b.p.; good conductors; malleable & ductile.
Non-metals: top right; usually gases or low-melting solids (B, C exceptions); poor conductors; brittle.
Metalloids / semi-metals: along the zig-zag line (Si, Ge, As, Sb, Te) — properties of both.
Metallic character decreases left → right across a period; increases down a group.
Mnemonic: Blocks & Their Homes
"s = Shell out (12), p = Party (1318), d = Dance floor (312), f = Downstairs". ⚡ In one line: block = last filled subshell (l value): s(0), p(1), d(2), f(3).
3.4
Considering the atomic number and position in the periodic table, arrange the following elements in the increasing order of metallic character: Si, Be, Mg, Na, P.
Metallic character increases down a group and decreases across a period left → right.
Order of increasing metallic character: P < Si < Be < Mg < Na
3.7 Periodic Trends in Properties of Elements
Properties vary periodically with atomic number. Key trends: atomic & ionic radii, ionization enthalpy, electron gain enthalpy, electronegativity, valence, chemical reactivity.
3.7.1 (a) Atomic Radius
Definition (exam answer): atomic radius = half the distance between the nuclei of two atoms in a molecule / crystal — covalent radius (Cl₂ bond 198 pm → r = 99 pm) for non-metals, metallic radius (Cu internuclear 256 pm → r = 128 pm) for metals.
Measurement of Atomic Radii: Covalent vs Metallic Radius
Why Size Decreases Across a Period
Same valence shell, but effective nuclear charge (Zeff) increases → electrons pulled in → size decreases.
Why Size Increases Down a Group
n increases → new shells, valence electrons farther from nucleus + shielding by inner electrons → size increases.
Period II (pm)
Li 152
Be 111
B 88
C 77
N 74
O 66
F 64
Period III (pm)
Na 186
Mg 160
Al 143
Si 117
P 110
S 104
Cl 99
Group I
Li
Na
K
Rb
Cs
Radius (pm)
152
186
231
244
262
Noble gases not included — their non-bonded (van der Waals) radii are very large.
3.7.1 (b) Ionic Radius
Cation < parent atom (fewer electrons, same nuclear charge): Na 186 pm → Na⁺ 95 pm.
Anion > parent atom (extra electrons → repulsion): F 64 pm → F⁻ 136 pm.
Isoelectronic species (same electron count, e.g., O²⁻, F⁻, Na⁺, Mg²⁺ all have 10 e⁻): size decreases with increasing nuclear charge → order from largest to smallest: O²⁻ > F⁻ > Na⁺ > Mg²⁺.
Isoelectronic Series Size Progression (All 10 Electrons)
Mnemonic: Ion Sizes
"Cat's Small, Anion's Huge" → Cation small (loses shell), Anion huge (gains electrons). For isoelectronic: more Z⁺ → smaller. ⚡ In one line: ions with same e⁻ count: Mg²⁺ < Na⁺ < F⁻ < O²⁻.
3.5
Which of the following species will have the largest and the smallest size? Mg, Mg²⁺, Al, Al³⁺.
Atomic radii decrease across a period; cations are smaller than their parent atoms; among isoelectronic species the one with larger positive charge is smaller.
Largest: Mg; Smallest: Al³⁺
3.7.1 (c) Ionization Enthalpy
Definition (exam answer): the energy required to remove the most loosely bound electron from an isolated gaseous atom in its ground state:
$$\text{X(g)} \rightarrow \text{X}^+\text{(g)} + e^- \qquad \Delta_i H_1$$
Units: kJ mol⁻¹. Always positive. Second ionization enthalpy > first (removing an electron from a cation is harder).
Across a period: generally increases (Zeff ↑, shielding ~constant). Maxima at noble gases; minima at alkali metals.
Down a group:decreases (n ↑, shielding ↑ outweighs nuclear charge).
Exceptions: Be > B (2s more penetrating than 2p, so B's 2p-electron is easier to remove); N > O (O has electronelectron repulsion in a doubly-occupied 2p orbital).
Period 2 First Ionization Enthalpy Profile (Showing Be > B and N > O Anomalies)
Mnemonic: Ionization Trends
"Across = up, Down = down" (like climbing stairs): ΔiH rises left→right, falls top→bottom. ⚡ In one line: small atoms (high Zeff) hold electrons tight → high ionization enthalpy.
3.6
The first ionization enthalpy (ΔᵢH) values of the third period elements, Na, Mg and Si are respectively 496, 737 and 786 kJ mol⁻¹. Predict whether the first ΔᵢH value for Al will be more close to 575 or 760 kJ mol⁻¹? Justify your answer.
Al follows Mg in the same period. Its 3p-electron is shielded by 3s-electrons, so it is easier to remove than Mg's 3s-electron.
ΔᵢH(Al) ≈ 575 kJ mol⁻¹ — lower than Mg (737) because of effective shielding of the 3p electron by 3s electrons.
3.7.1 (d) Electron Gain Enthalpy
Definition (exam answer): the enthalpy change when an electron is added to a neutral isolated gaseous atom to form a negative ion:
$$\text{X(g)} + e^- \rightarrow \text{X}^-\text{(g)} \qquad \Delta_{eg}H$$
Usually negative (energy released). Halogens have very negative ΔegH (attain noble-gas configuration); noble gases have positive ΔegH (electron enters next higher level).
Across a period: becomes more negative (Zeff ↑).
Down a group: becomes less negative (atom bigger; added electron farther away).
Exceptions:O & F are less negative than S & Cl — the added electron goes into the small n = 2 level and suffers strong electronelectron repulsion. Hence Cl has the most negative electron gain enthalpy.
Gr 17
F
Cl
Br
I
At
ΔegH (kJ mol⁻¹)
−328
−349
−325
−295
−270
Mnemonic: Cl > F in Electron Gain
"Cl beats F" → despite F being more electronegative, Cl (−349) has the most negative electron gain enthalpy in the whole table. ⚡ In one line: F and O miss the crown because of 2p repulsion — Cl and S win.
3.7
Which of the following will have the most negative electron gain enthalpy and which the least negative? P, S, Cl, F. Explain your answer.
ΔegH becomes more negative across a period (P → S → Cl) and less negative down a group (F → Cl). But adding an electron to F's small 2p-level causes strong repulsion.
Most negative: Cl • Least negative: P
3.7.1 (e) Electronegativity
Definition (exam answer): the qualitative measure of the ability of an atom in a chemical compound to attract shared electrons towards itself (Pauling scale: F = 4.0 highest).
Across a period:increases (smaller atom → stronger attraction).
Down a group:decreases (larger atom → weaker attraction). Trend is like ionization enthalpy.
Electronegativity ↑ = non-metallic character ↑ and metallic character ↓.
Period II
Li 1.0
Be 1.5
B 2.0
C 2.5
N 3.0
O 3.5
F 4.0
Period III
Na 0.9
Mg 1.2
Al 1.5
Si 1.8
P 2.1
S 2.5
Cl 3.0
Definition Catch
Electronegativity is not measurable; it is a qualitative scale used to predict bond nature. It varies with the element bonded to the atom.
Mnemonic: All Five Trends in One Line
Across a period: r ↓, ΔiH ↑, ΔegH more −ve, electronegativity ↑. Down a group: r ↑, ΔiH ↓, ΔegH less −ve, electronegativity ↓. ⚡ In one line: size and electronegativity are opposites everywhere (small = greedy).
Periodic Trend Arrows — 3D Quick Map
3.7.2 (a) Periodicity of Valence / Oxidation States
Valence of representative elements = number of valence electrons, or = 8 − valence electrons:
Group
1
2
13
14
15
16
17
18
Valence electrons
1
2
3
4
5
6
7
8
Valence
1
2
3
4
3, 5
2, 6
1, 7
0, 8
Oxidation state = charge an atom acquires considering electronegativity (e.g., in OF₂: F = −1 each, O = +2; in Na₂O: Na = +1 each, O = −2). Transition elements & actinoids show variable valency.
3.7.2 (b) Anomalous Properties of Second Period Elements
Li, Be, and BF differ from the rest of their groups:
Small size, high charge/radius ratio, high electronegativity → covalent rather than ionic compounds (Li & Be resemble Mg & Al — diagonal relationship).
Diagonal Relationship in 2nd and 3rd Period Elements
Only 4 valence orbitals (2s, 2p) → maximum covalency = 4 (e.g., BF₄⁻ exists, AlF₆³⁻ also forms using 3d orbitals).
Ability to form pπpπ multiple bonds (C=C, C≡C, N≡N, C=O) — stronger than for heavier members.
Gr 1
Li
Na
Gr 2
Be
Mg
Metallic radius (pm)
152
186
111
160
Ionic radius (pm)
76
102
31
72
3.8
Using the Periodic Table, predict the formulas of compounds which might be formed by the following pairs of elements: (a) silicon and bromine (b) aluminium and sulphur.
(a) Si (Gr 14, valence 4) + Br (Gr 17, valence 1) → SiBr₄.
(b) Al (Gr 13, valence 3) + S (Gr 16, valence 2) → Al₂S₃.
(a) SiBr₄ • (b) Al₂S₃
3.7.3 Periodic Trends & Chemical Reactivity
Across a period: reactivity is highest at the two extremes and lowest in the centre. Left extreme (alkali metals) easily lose electrons (low ΔᵢH) → form cations; right extreme (halogens) easily gain electrons → form anions.
Metallic character: decreases left → right; non-metallic character increases. Metals on the left form basic oxides (Na₂O); non-metals on the right form acidic oxides (Cl₂O₇); middle elements form amphoteric (Al₂O₃) or neutral (CO, NO) oxides.
Down a group: metallic character increases (size ↑, ΔᵢH ↓). Transition metals: radii change little across 3d series; ΔᵢH intermediate.
High reactivity ⇒ elements never found free in nature (occur as compounds).
Mnemonic: Oxide Nature & Position
"Left = Base, Right = Acid, Middle = Both (amphoteric), Centre-adjacent = Neutral". ⚡ In one line: Na₂O basic + H₂O → NaOH; Cl₂O₇ acidic + H₂O → 2HClO₄.
3.9
Are the oxidation state and covalency of Al in [AlCl(H₂O)₅]²⁺ same?
Oxidation state of Al = +3. The complex has 6 ligands around Al (1 Cl + 5 H₂O).
No. Oxidation state = +3; covalency = 6.
3.10
Show by a chemical reaction with water that Na₂O is a basic oxide and Cl₂O₇ is an acidic oxide.
Na₂O + H₂O → 2NaOH (strong base — turns red litmus blue). Cl₂O₇ + H₂O → 2HClO₄ (strong acid — turns blue litmus red).
Their acidic/basic nature can be tested with litmus paper.
Na₂O → 2NaOH (basic); Cl₂O₇ → 2HClO₄ (acidic)
Chapter Summary (Unit 3)
Mendeleev's table was based on atomic mass; the Modern table is based on atomic number (Moseley).
7 periods (2, 8, 8, 18, 18, 32, 32) × 18 groups (IUPAC 118); period number = n of valence shell.
Four blocks: s, p, d, f by last-filled subshell; metals (78%), non-metals (<20), metalloids on the zig-zag line.
Trends: atomic/ionic radii ↓ across, ↑ down; ΔᵢH ↑ across, ↓ down; ΔegH more −ve across, less −ve down (Cl most −ve); electronegativity ↑ across, ↓ down (F = 4.0).
Reactivity highest at period extremes; left → basic oxides, right → acidic oxides.
Practice Quiz (25 Questions)
Q01 Which scientist proposed that properties of elements are a periodic function of their atomic weights?
Incorrect: Dobereiner proposed the Law of Triads.
Incorrect: Newlands proposed the Law of Octaves.
Correct Answer: Mendeleev published the first Periodic Law: properties are a periodic function of atomic weights.
Incorrect: Moseley modernised the law using atomic number.
Q02 According to Newlands' Law of Octaves, every _____ element had properties similar to the first.
Incorrect: That describes Dobereiner's triads.
Correct Answer: Like musical octaves, the 8th element resembled the 1st (true only up to calcium).
Incorrect: Not related to Newlands' pattern.
Incorrect: The resemblance was at every 8th element.
Q03 Mendeleev left gaps in his table for undiscovered elements he called:
Incorrect: Transuranium elements come after uranium (Z > 92).
Incorrect: Noble gases were not predicted by Mendeleev's gaps in that way.
Incorrect: These are Si, Ge, As etc. along the diagonal.
Correct Answer: Ga (eka-Al) and Ge (eka-Si) were predicted with correct properties before discovery.
Q04 The IUPAC systematic name of the element with atomic number 120 is:
Incorrect: Unnill... Unb would correspond to 102 (1-0-2 → un-nil-bi).
Correct Answer: Roots of 1, 2, 0 = un, bi, nil → unbinilium, symbol Ubn.
Incorrect: Uub is element 112 (un-un-bi).
Incorrect: Unq corresponds to element 104 (1-0-4).
Q05 Which block of elements has the general outer configuration (n−1)d¹⁻¹⁰ ns⁰⁻²?
Incorrect: s-block is ns¹ or ns².
Incorrect: p-block is ns²np¹⁻⁶.
Correct Answer: d-block (transition elements, Groups 312) fill inner d orbitals.
Incorrect: f-block is (n−2)f¹⁻¹⁴(n−1)d⁰⁻¹ns².
Q06 The 3d transition series begins at scandium (Z=21). At which element does it end?
Incorrect: Cu has 3d¹⁰ but the series formally ends at Zn.
Incorrect: Ga is the first p-block element after the 3d series.
Incorrect: 3d fills completely at Zn, not Ni.
Correct Answer: The 3d orbitals are completely filled at Zn (3d¹⁰4s²), ending the 3d transition series.
Q07 Which element is an exception in the d-block because its configuration is 4d¹⁰5s⁰?
Correct Answer: Pd has a completely filled 4d¹⁰ and an empty 5s — breaking the (n−1)d¹⁻¹⁰ns⁰⁻² general pattern.
Incorrect: Cu is 3d¹⁰4s¹.
Incorrect: Zn is 3d¹⁰4s².
Incorrect: Cr is 3d⁵4s¹.
Q08 Across a period (left to right), the atomic radius generally:
Incorrect: Atoms get smaller across a period.
Correct Answer: Same shell, but effective nuclear charge increases → electrons pulled in → radius decreases.
Incorrect: Radius changes regularly across a period.
Incorrect: The trend is a steady decrease left → right.
Q09 Which species is LARGEST among O²⁻, F⁻, Na⁺ and Mg²⁺ (all isoelectronic, 10 e⁻)?
Incorrect: Highest nuclear charge → smallest.
Incorrect: More protons than F⁻ and O²⁻ → smaller.
Incorrect: F⁻ has 9 protons; O²⁻ has only 8 protons for the same 10 electrons.
Correct Answer: Among isoelectronic species, size decreases with increasing nuclear charge: O²⁻ (Z=8) is the largest.
Q10 The first ionization enthalpy of the second period has the anomaly Be > B, N > O. Why?
Incorrect: There are no d-orbitals in the second period.
Incorrect: B and O are not radioactive.
Correct Answer: B's 2p-electron is shielded by 2s; O has a doubly-occupied 2p orbital with extra repulsion — both lower ΔᵢH.
Q22 The maximum covalency of the first member of a group (e.g., boron) is 4 because it has only four valence orbitals (2s + 2p). Which ion shows this?
Incorrect: Al can expand to 6 using 3d orbitals.
Incorrect: Si uses 3d orbitals for covalency 6.
Correct Answer: B has only 2s, 2p orbitals → maximum covalency 4 (BF₄⁻).
Incorrect: Al can also form AlF₆³⁻ (covalency 6).
Q23 Diagonal relationship — Li resembles which element?
Correct Answer: Li (Gr 1, period 2) resembles Mg (Gr 2, period 3) — same charge/radius ratio. Similarly BeAl, BSi.
Incorrect: Na is in the same group, not diagonally related.
Incorrect: Ca is in Group 2, period 4.
Incorrect: Be is diagonal to Al, not Li.
Q24 Which of the following statements about the modern periodic table is INCORRECT?
Incorrect: This is correct — p can hold 6 electrons → 6 columns.
Correct Answer: A d-subshell holds 10 electrons → the d-block has 10 columns (Groups 312), not 8.
Incorrect: This statement is correct (s:2, p:6, d:10, f:14).
Incorrect: This statement is correct.
Q25 Which factor does NOT affect the chemistry of the valence shell?
Incorrect: n definitely affects the valence shell.
Incorrect: Z is the most important factor.
Incorrect: Core electrons produce shielding, affecting valency.
Correct Answer: Nuclear mass does not affect valence electrons (isotopes of an element have identical chemistry).
NCERT Exercises (3.1 3.40)
3.1
What is the basic theme of organisation in the periodic table?
Basic theme: elements with similar chemical and physical properties are grouped together in the same vertical column (group), and properties vary periodically with atomic number across periods.
3.2
Which important property did Mendeleev use to classify the elements in his periodic table and did he stick to that?
He used atomic weight as the basis. No — he did not always stick to it: he ignored the atomic-weight order when needed to place similar-property elements together (e.g., iodine placed after tellurium), and left gaps for undiscovered elements.
3.3
What is the basic difference in approach between the Mendeleev's Periodic Law and the Modern Periodic Law?
Mendeleev's law used atomic weight as the periodic function; the Modern Periodic Law (Moseley) uses the more fundamental atomic number (Z), since √ν of X-rays vs Z gave a straight line.
3.4
On the basis of quantum numbers, justify that the sixth period of the periodic table should have 32 elements.
For n = 6, orbitals filled are 6s, 4f, 5d, 6p → 1 + 7 + 5 + 3 = 16 orbitals. Max electrons = 16 × 2 = 32. Hence the 6th period has 32 elements (incl. the 14 lanthanoid elements in the 4f series).
3.5
In terms of period and group where would you locate the element with Z = 114?
Configuration: [Rn] 5f¹⁴ 6d¹⁰ 7s² 7p² → valence principal quantum number n = 7 → 7th period; ends in 7p² → Group 14, p-block (carbon family).
3.6
Write the atomic number of the element present in the third period and seventeenth group of the periodic table.
3rd period, Group 17 = chlorine, Z = 17 (3s² 3p⁵).
3.7
Which element do you think would have been named by (i) Lawrence Berkeley Laboratory (ii) Seaborg's group?
(i) Berkelium (Bk, Z=97) — named after Berkeley. (ii) Seaborgium (Sg, Z=106) — named in honour of Glenn T. Seaborg (only element named after a living person at the time).
3.8
Why do elements in the same group have similar physical and chemical properties?
Elements in a group have the same valence-shell electronic configuration (same number and distribution of outermost electrons), which governs their properties.
3.9
What does atomic radius and ionic radius really mean to you?
Atomic radius: half the distance between the nuclei of two like atoms bonded together (covalent radius) or adjacent atoms in a metal (metallic radius). Ionic radius: effective radius of an ion, estimated from cationanion distances in ionic crystals.
3.10
How do atomic radius vary in a period and in a group? How do you explain the variation?
Across a period: decreases — Zeff increases in the same shell, pulling electrons closer. Down a group: increases — n increases, new shells are added and inner electrons shield the outer ones.
3.11
What do you understand by isoelectronic species? Name a species that will be isoelectronic with each of the following atoms or ions: (i) F⁻ (ii) Ar (iii) Mg²⁺ (iv) Rb⁺.
Isoelectronic species have the same number of electrons. (i) F⁻ (10 e⁻) → Ne, Na⁺, Mg²⁺; (ii) Ar (18 e⁻) → Cl⁻, K⁺, Ca²⁺; (iii) Mg²⁺ (10 e⁻) → Na⁺, F⁻; (iv) Rb⁺ (36 e⁻) → Kr, Sr²⁺, Br⁻.
3.12
Consider the following species: N³⁻, O²⁻, F⁻, Na⁺, Mg²⁺ and Al³⁺. (a) What is common in them? (b) Arrange them in the order of increasing ionic radii.
(a) All are isoelectronic (10 electrons each). (b) Higher Z → smaller radius:
Al³⁺ < Mg²⁺ < Na⁺ < F⁻ < O²⁻ < N³⁻
3.13
Explain why cations are smaller and anions larger in radii than their parent atoms?
Cation: loses electrons → fewer electrons, same nuclear charge → stronger pull → smaller. Anion: gains electrons → more electronelectron repulsion and less Zeff → larger.
3.14
What is the significance of the terms — 'isolated gaseous atom' and 'ground state' while defining the ionization enthalpy and electron gain enthalpy?
Both definitions are for comparison purposes: isolated gas-phase atom removes any intermolecular effects, and ground state is the minimum energy state — so measured enthalpies are intrinsic properties of the atom alone, comparable across elements.
3.15
Energy of an electron in the ground state of the hydrogen atom is −2.18×10⁻¹⁸ J. Calculate the ionization enthalpy of atomic hydrogen in terms of J mol⁻¹.
Ionization enthalpy per atom = +2.18×10⁻¹⁸ J. Per mole = 2.18×10⁻¹⁸ × 6.022×10²³ = 1.313 × 10⁶ J mol⁻¹
= 1313 kJ mol⁻¹ ≈ 1.31 × 10³ kJ mol⁻¹
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 (2s²): removing a 2s electron (highly penetrating). B (2p¹): removing a 2p electron, shielded by 2s → easier → Be > B. (ii) N (2p³): half-filled, stable (Hund). O (2p⁴): the 4th 2p electron must pair up → repulsion makes it easy to remove → N > O; F (2p⁵) has higher Zeff than O → N < F too.
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?
ΔᵢH₁: Na (3s¹) < Mg (3s²) because Mg has higher Z and a filled orbital. ΔᵢH₂: removing the 2nd electron from Na⁺ (1s²2s²2p⁶) breaks a stable noble-gas core → very high. For Mg⁺ (3s¹), the 2nd electron is a normal valence electron → its second ΔᵢH is much lower. Hence Na's ΔᵢH₂ > Mg's ΔᵢH₂.
3.18
What are the various factors due to which the ionization enthalpy of the main group elements tends to decrease down a group?
(i) Increase in atomic size (outer electron farther from nucleus). (ii) Increase in shielding/screening by inner electrons, which outweighs the increasing nuclear charge.
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 deviation from the general trend?
General: ΔᵢH decreases down a group. Ga ≈ Al because of d-block contraction (poor shielding by 3d¹⁰). In < Al normally; Tl rises (589) because of f-block (lanthanoid) contraction — 4f¹⁴ electrons screen poorly, increasing Zeff on the 6p electron.
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 (−328) more negative than O (−141) — higher Zeff, same row. (ii) Cl (−349) more negative than F (−328) — F's added electron suffers 2p repulsion; Cl's 3p accepts it easily.
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.
Strongly positive. O⁻ (already negative) → O²⁻ requires adding an electron to a negatively charged ion against the electronelectron repulsion; energy must be supplied. (O → O⁻ = −141; O⁻ → O²⁻ = +744 kJ mol⁻¹.)
3.22
What is the basic difference between the terms electron gain enthalpy and electronegativity?
Electron gain enthalpy is a measurable energy change for adding an electron to an isolated gaseous atom (kJ mol⁻¹). Electronegativity is a qualitative scale describing the tendency of an atom in a compound to attract shared electrons (no units).
3.23
How would you react to the statement that the electronegativity of N on Pauling scale is 3.0 in all the nitrogen compounds?
Disagree. Electronegativity of an element varies with the element it is bonded to (bond polarity). So N's effective electronegativity differs in different nitrogen compounds; the value 3.0 is only an average reference.
3.24
Describe the theory associated with the radius of an atom as it (a) gains an electron (b) loses an electron.
Would you expect the first ionization enthalpies for two isotopes of the same element to be the same or different? Justify your answer.
Same. ΔᵢH depends on the electron cloud and nuclear charge (Z), not on the number of neutrons (mass). Isotopes have identical Z and electronic configuration → identical first ΔᵢH.
3.26
What are the major differences between metals and non-metals?
Metals: solids (Hg except.), high m.p./b.p., good conductors, malleable & ductile, form basic oxides, lose electrons (low ΔᵢH). Non-metals: solids/gases, low m.p., poor conductors, brittle, form acidic oxides, gain electrons (negative ΔegH).
3.27
Use the periodic table to answer: (a) an element with five electrons in the outer subshell (b) an element that would tend to lose two electrons (c) an element that would tend to gain two electrons (d) the group having metal, non-metal, liquid as well as gas at room temperature.
(a) P (phosphorus), 3s²3p³ (5 outer electrons). (b) Mg / Ca (Group 2 → M²⁺) or O. (c) O / S (Group 16 → gain 2 e⁻ → O²⁻/S²⁻). (d) Group 17: F, Cl (gases); Br (liquid); I (solid) — includes non-metals; metals also present in group... actually classify as Group 17 has gas/liquid/solid non-metals; Group 1 has metal, and (Cs, Fr). The NCERT answer: Group 17 contains non-metals incl. gas (F, Cl), liquid (Br), solid (I).
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.
Metals react by losing electrons; down Group 1, ΔᵢH decreases (size ↑) → reactivity increases. Non-metals (halogens) react by gaining electrons; down Group 17, ΔegH becomes less negative and size increases → reactivity decreases.
3.29
Write the general outer electronic configuration of s-, p-, d- and f- block elements.
Assign the position of the element having outer electronic configuration (i) ns²np⁴ for n=3 (ii) (n−1)d²ns² for n=4 (iii) (n−2)f⁷(n−1)d¹ns² for n=6.
(i) n=3, 3s²3p⁴ → Period 3, Group 16 (S). (ii) n=4 → 3d²4s² → Period 4, Group 4 (Ti), d-block. (iii) n=6 → 4f⁷5d¹6s² → Period 6, Group 3, f-block (Gd/Eu region — specifically Gd), inner-transition element.
3.31
The first (ΔᵢH₁) and second (ΔᵢH₂) ionization enthalpies (kJ mol⁻¹) and electron gain enthalpy (ΔegH) of a few elements are given: I: 520, 7300, −60; II: 419, 3051, −48; III: 1681, 3374, −328; IV: 1008, 1846, −295; V: 2372, 5251, +48; VI: 738, 1451, −40. Which is (a) the least reactive element (b) the most reactive metal (c) the most reactive non-metal (d) the least reactive non-metal (e) the metal which can form a stable binary halide of the formula MX₂ (f) the metal which can form a predominantly stable covalent halide of the formula MX?
(a) V — very high ΔᵢH₁ (2372) + positive ΔegH (+48) → noble gas (He/Ne-like). (b) II — lowest ΔᵢH₁ (419) → most reactive metal (K-like). (c) III — very high ΔᵢH₁ (1681) + most negative ΔegH (−328) → most reactive non-metal (F-like). (d) IV — high ΔᵢH₁ (1008), ΔegH −295 → reactive non-metal but less than III (Cl-like). (e) VI — ΔᵢH₁ 738, ΔᵢH₂ 1451 (both moderate) → forms MX₂ (Mg-like). (f) I — ΔᵢH₁ 520, huge ΔᵢH₂ 7300 (noble-gas core after 1st loss) → forms MX (Li-like, covalent).
3.32
Predict the formulas of the stable binary compounds formed by: (a) Lithium and oxygen (b) Magnesium and nitrogen (c) Aluminium and iodine (d) Silicon and oxygen (e) Phosphorus and fluorine (f) Element 71 and fluorine.
In the modern periodic table, the period indicates the value of: (a) atomic number (b) atomic mass (c) principal quantum number (d) azimuthal quantum number.
(c) principal quantum number (n)
— the period number equals the n of the valence shell.
3.34
Which of the following statements related to the modern periodic table is incorrect? (a) The p-block has 6 columns (b) The d-block has 8 columns because a maximum of 8 electrons can occupy all the orbitals in a d-subshell (c) Each block contains a number of columns equal to the number of electrons that can occupy that subshell (d) The block indicates value of azimuthal quantum number (l) for the last subshell that received electrons.
(b) is incorrect
— a d-subshell holds 10 electrons, so the d-block has 10 columns (Groups 312), not 8.
3.35
Anything that influences the valence electrons will affect the chemistry of the element. Which one of the following factors does not affect the valence shell? (a) Valence principal quantum number (n) (b) Nuclear charge (Z) (c) Nuclear mass (d) Number of core electrons.
(c) Nuclear mass
— neutrons don't affect the electron cloud; isotopes have identical chemistry.
3.36
The size of isoelectronic species — F⁻, Ne and Na⁺ is affected by: (a) nuclear charge (Z) (b) valence principal quantum number (n) (c) electron-electron interaction in the outer orbitals (d) none of the factors because their size is the same.
(a) nuclear charge (Z)
— same n and same electron count, but Z increases F⁻(9) < Ne(10) < Na⁺(11), so size decreases.
3.37
Which one of the following statements is incorrect in relation to ionization enthalpy? (a) Ionization enthalpy increases for each successive electron (b) The greatest increase in ionization enthalpy is experienced on removal of electron from core noble gas configuration (c) End of valence electrons is marked by a big jump in ionization enthalpy (d) Removal of electron from orbitals bearing lower n value is easier than from orbital having higher n value.
(d) is incorrect
— electrons with lower n are closer to the nucleus and harder to remove; higher n electrons are easier to remove.
3.38
Considering the elements B, Al, Mg, and K, the correct order of their metallic character is: (a) B > Al > Mg > K (b) Al > Mg > B > K (c) Mg > Al > K > B (d) K > Mg > Al > B.
(d) K > Mg > Al > B
— metallic character increases down a group (K > ...) and decreases across a period (Mg > Al > B).
3.39
Considering the elements B, C, N, F, and Si, the correct order of their non-metallic character is: (a) B > C > Si > N > F (b) Si > C > B > N > F (c) F > N > C > B > Si (d) F > N > C > Si > B.
(c) F > N > C > B > Si
— non-metallic character increases across a period (B→F) and decreases down a group (C > Si).
3.40
Considering the elements F, Cl, O and N, the correct order of their chemical reactivity in terms of oxidizing property is: (a) F > Cl > O > N (b) F > O > Cl > N (c) Cl > F > O > N (d) O > F > N > Cl.
(b) F > O > Cl > N
— oxidizing power follows electron gain tendency: F (most negative ΔegH among these, small size) > O > Cl > N.
Periodic Trends at a Glance
Property
Across a Period (→)
Down a Group (↓)
Reason
Atomic radius
Decreases
Increases
Zeff ↑ across; n & shielding ↑ down
Ionic radius
Decreases
Increases
Same as atomic radius
Ionization enthalpy
Increases
Decreases
Zeff ↑ across; size ↑ down
Electron gain enthalpy
More negative
Less negative
Zeff ↑ across; size ↑ down
Electronegativity
Increases
Decreases
Size ↓ across; size ↑ down
Metallic character
Decreases
Increases
ΔᵢH ↑ across; ΔᵢH ↓ down
Non-metallic character
Increases
Decreases
Opposite of metallic
Oxide nature
Basic → Acidic
Acidic → Basic
Metallic character change
Key Facts & Values
Fact
Value / Detail
Period sizes
2, 8, 8, 18, 18, 32, 32
Groups (IUPAC)
118 (replaces IAVIIA, VIII, IBVIIB, 0)
Most negative ΔegH
Chlorine (−349 kJ mol⁻¹)
Highest electronegativity
Fluorine (4.0, Pauling)
Lowest ΔᵢH (metals)
Cs (alkali metals, bottom of Group 1)
d-block exception
Pd = 4d¹⁰5s⁰
Metals %
> 78% of known elements
Metalloids
Si, Ge, As, Sb, Te (zig-zag line)
Lanthanoids
Ce (Z=58) Lu (Z=71), 4f series
Actinoids
Th (Z=90) Lr (Z=103), 5f series
Transuranium elements
Elements after uranium (Z > 92)
🧠 Mnemonics Cheat-Sheet (Quick Revision)
Concept
Mnemonic
Key Point
Element count
31 → 63 → 114
1800 → 1865 → now
Period sizes
Two Great Eights, Eighteen Again, Thirty-Two
2, 8, 8, 18, 18, 32, 32
IUPAC roots
N-Un-Bi-Tri-Quad-Pent-Hex-Sept-Oct-Enn
09; Z=120 → Unbinilium (Ubn)
Blocks
s = Shell out, p = Party, d = Dance floor, f = Downstairs
s(12), p(1318), d(312), f(bottom)
Ion sizes
Cat's Small, Anion's Huge
Cation smaller, anion larger than atom
Isoelectronic size
More Z⁺ → Smaller
Al³⁺ < Mg²⁺ < Na⁺ < F⁻ < O²⁻ < N³⁻
Ionization trend
Across = up, Down = down
ΔᵢH ↑ across, ↓ down
Electron gain
Cl beats F
Cl (−349) most negative ΔegH
Oxides
Left = Base, Right = Acid
Na₂O basic; Cl₂O₇ acidic; Al₂O₃ amphoteric
Diagonal relationship
LiMg, BeAl, BSi
Similar charge/radius ratio
Chapter Tests (3 Levels)
Test your mastery of periodic trends, electronic configurations, and chemical reactivity with three graded difficulty tiers.
L1.1
According to the Modern Periodic Law (Moseley), the physical and chemical properties of elements are periodic functions of their:
Incorrect: Atomic weight was the basis of Mendeleev's periodic table.
Incorrect: Mass number (A) includes neutrons and does not govern periodic recurrence.
Correct Answer: Moseley's X-ray spectra showed √ν is directly proportional to atomic number Z.
Incorrect: Density is a physical property, not the periodic fundamental variable.
L1.2
What is the IUPAC systematic symbol for the superheavy element with atomic number Z = 119?
Incorrect: Uun is Ununnilium (Z = 110).
Correct Answer: Roots for 1, 1, 9 are un + un + enn → Ununennium, symbol Uue.
Incorrect: Uuo is Ununoctium (Z = 118, Oganesson).
Incorrect: Une is Unnilennium (Z = 109, Meitnerium).
L1.3
Which block in the periodic table contains the alkali metals and alkaline earth metals?
Correct Answer: Group 1 (ns¹) and Group 2 (ns²) form the s-block.
Incorrect: Groups 13 to 18 constitute the p-block.
Incorrect: Groups 3 to 12 form the d-block (transition metals).
Incorrect: Lanthanoids and actinoids form the f-block.
L1.4
Which of the following elements has the highest electronegativity on the Pauling scale?
Incorrect: Oxygen has electronegativity 3.5.
Incorrect: Chlorine has electronegativity 3.0.
Incorrect: Nitrogen has electronegativity 3.0.
Correct Answer: Fluorine is the most electronegative element with a value of 4.0 on the Pauling scale.
L1.5
How many elements are accommodated in the 4th period of the long form of the periodic table?
Incorrect: The 2nd and 3rd periods contain 8 elements each.
Correct Answer: The 4th period fills 4s (2), 3d (10), and 4p (6) orbitals → 2 + 10 + 6 = 18 elements.
Incorrect: The 6th period contains 32 elements.
Incorrect: 10 is the number of d-block transition elements in the 3d series.
Result
0 / 5
L2.1
Why does beryllium (Be, 1s²2s²) have a higher first ionization enthalpy (ΔᵢH₁) than boron (B, 1s²2s²2p¹)?
Incorrect: Higher nuclear charge would tend to increase ΔᵢH, not lower it.
Incorrect: Beryllium is larger than boron, which normally lowers ΔᵢH.
Correct Answer: Beryllium has a fully filled stable 2s² subshell with greater penetration; removing Boron's 2p¹ electron requires less energy because it is well shielded by 2s² electrons.
Incorrect: Boron is 2p¹, not half-filled (nitrogen is 2p³).
L2.2
Which of the following elements has the most negative electron gain enthalpy (ΔegH)?
Incorrect: Fluorine has ΔegH = −328 kJ mol⁻¹ because the added electron suffers strong interelectronic repulsion in the compact 2p subshell.
Correct Answer: Chlorine has ΔegH = −349 kJ mol⁻¹, the most negative in the entire periodic table, as the incoming electron enters the larger 3p subshell with less repulsion.
Incorrect: Bromine is −325 kJ mol⁻¹ due to increased atomic size down the group.
Incorrect: Oxygen is only −141 kJ mol⁻¹.
L2.3
What is the correct order of increasing ionic radii for the isoelectronic series: N³⁻, O²⁻, F⁻, Na⁺, Mg²⁺, Al³⁺?
Incorrect: Anions are larger than cations in an isoelectronic series.
Incorrect: This is the decreasing order with inverted comparison symbols.
Incorrect: Al³⁺ is smaller than Na⁺.
Correct Answer: For isoelectronic ions (10 e⁻), radius decreases as nuclear charge (Z) increases: Al (Z=13) attracts 10 e⁻ most tightly, while N (Z=7) has the lowest Z and largest radius.
L2.4
In the compound oxygen difluoride (OF₂), the oxidation state of oxygen is:
Correct Answer: Fluorine is more electronegative than oxygen and is assigned −1; hence oxygen must be +2 for neutrality [x + 2(−1) = 0 → x = +2].
Incorrect: Oxygen is −2 when bonded to less electronegative elements (like H in H₂O or Na in Na₂O).
Incorrect: −1 is the oxidation state in peroxides (e.g., H₂O₂).
Incorrect: Oxygen is +1 in dioxygen difluoride (O₂F₂).
L2.5
Which pair of elements demonstrates a diagonal relationship due to similar charge-to-radius ratio and electronegativity?
Incorrect: Li and Na belong to the same group (congener pair).
Incorrect: B and Al are in Group 13.
Correct Answer: Beryllium (Group 2, Period 2) and Aluminium (Group 13, Period 3) show strong diagonal resemblance (amphoteric oxides, covalent chlorides, polymeric hydrides).
Incorrect: The diagonal relationship pairs are LiMg, BeAl, and BSi.
Result
0 / 5
L3.1
The first ionization enthalpy (ΔᵢH₁) values for Group 13 elements are: B (801), Al (577), Ga (579), In (558), and Tl (589 kJ mol⁻¹). Why does Tl have a higher ΔᵢH₁ than In?
Incorrect: 3d orbitals explain Ga's slight deviation, not Tl.
Correct Answer: Tl is preceded by 14 4f electrons which provide very poor shielding; the effective nuclear charge Z_eff increases markedly, pulling the outer 6p¹ electron more tightly.
Incorrect: Tl has a [Xe] 4f¹⁴ 5d¹⁰ 6s² 6p¹ configuration.
Incorrect: In is 5s² 5p¹, which is not half-filled.
L3.2
An element has successive ionization enthalpies (in kJ mol⁻¹): ΔᵢH₁ = 738, ΔᵢH₂ = 1451, ΔᵢH₃ = 7733, ΔᵢH₄ = 10540. To which group of the periodic table does this element belong?
Correct Answer: The huge jump between ΔᵢH₂ and ΔᵢH₃ (from 1451 to 7733 kJ mol⁻¹) indicates removal of the 3rd electron from a stable noble-gas core; hence the atom has 2 valence electrons (Group 2, like Mg).
Incorrect: Group 1 elements show a gigantic jump after ΔᵢH₁ (e.g., Na: 496 to 4562 kJ mol⁻¹).
Incorrect: Group 13 elements show the large jump after ΔᵢH₃.
Incorrect: Group 14 elements show the large jump after ΔᵢH₄.
L3.3
The second electron gain enthalpy (ΔegH₂) of oxygen is +780 kJ mol⁻¹ (positive / endothermic), even though O²⁻ achieves a stable neon configuration. What is the fundamental reason?
Incorrect: Nuclear charge (Z = 8) remains invariant.
Incorrect: The electron enters the 2p orbital, not d orbitals.
Incorrect: Z_eff decreases but does not become zero.
Correct Answer: Adding an electron to an already negatively charged ion (O⁻) encounters massive electrostatic repulsion; energy must be supplied to force the electron into the anion.
L3.4
What is the maximum covalency of the second-period elements (such as B, C, N), and why are third-period elements (such as Al, P, S) able to expand their covalency beyond this limit?
Incorrect: 4f orbitals are not available in the 3rd period.
Incorrect: Second period elements can form up to 4 bonds (e.g., NH₄⁺, BF₄⁻).
Correct Answer: For n = 2, only one 2s and three 2p orbitals exist (total 4 orbitals → max covalency 4, e.g., BF₄⁻). Third period elements have vacant 3d orbitals allowing expanded octets (e.g., AlF₆³⁻, PF₅, SF₆).
Incorrect: 2nd period elements cannot achieve covalency 8.
L3.5
Arrange the following oxides in order of increasing acidic nature: Al₂O₃, Cl₂O₇, Na₂O, SO₃, MgO.
Incorrect: This is the decreasing acidic order (or increasing basic order).
Correct Answer: Across Period 3, oxides progress from strongly basic (Na₂O) → basic (MgO) → amphoteric (Al₂O₃) → acidic (SO₃) → strongly acidic (Cl₂O₇).
Incorrect: MgO is more basic (less acidic) than amphoteric Al₂O₃.
Incorrect: Na₂O is strongly basic and least acidic.