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State Board · Class 11 · Chemistry

Lesson 7: Modern Periodic Table

7 concepts · 56 practice questions with answers · Free to read

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Classification of elements: from triads to the modern periodic law

Link to Std 9–10 : In Std 9–10 you saw Mendeleev's table; here we see how it developed and what Moseley changed.

From daily life

Line up classmates by height and groups appear; line up elements in the right order and properties repeat.

Simple definition

Döbereiner's triads: the middle element's mass is about the mean of the outer two. Newlands' law of octaves: the eighth element resembles the first. Mendeleev: arranged by atomic mass, leaving gaps. Moseley: atomic number Z is the basis. Modern periodic law: properties are periodic functions of atomic number.

Classification1817Döbereinertriads1865Newlandsoctaves1869Mendeleevatomic mass1913Moseleyatomic number ZModern periodic law: properties are a periodic function of ZLi 7, Na 23, K 39: (7 + 39) ÷ 2 = 23The journey of classification

In the figure : Four stages: triads → octaves → Mendeleev → Moseley.

Rule 1: Döbereiner's triads — Middle mass ≈ (first + third) ÷ 2.

  • Easy : Li 7, Na 23, K 39: (7 + 39) ÷ 2 = 23.
  • Medium : Ca 40, Sr 88, Ba 137: (40 + 137) ÷ 2 = 88.5 ≈ 88.
  • Exam-type : Cl 35.5, Br 80, I 127: mean 81.25 ≈ 80.

Rule 2: Mendeleev's success and limits — He left gaps and predicted undiscovered elements (eka-aluminium = Ga, eka-silicon = Ge). Limits: position of H, isotopes, a few reversed pairs (Ar before K).

  • Easy : Ar (39.9) must come before K (39.1) because Z = 18 < 19.
  • Medium : Co (58.9) and Ni (58.7) are reversed in the same way.
  • Exam-type : Isotopes fit one place because Z is the same.

Rule 3: Modern periodic law — Arranged by increasing atomic number Z (Moseley, 1913), properties recur at regular intervals.

  • Easy : Z = 3 (Li), 11 (Na), 19 (K) have similar properties.
  • Medium : Ne (10), Ar (18), Kr (36) are noble gases.
  • Exam-type : So the table is based on Z, not mass.

Derivation: how Moseley used Z

  • Step 1: He measured the X-ray frequency ν emitted by each element.
  • Step 2: The graph of √ν against Z was a straight line (√ν ∝ Z).
  • Step 3: So the true identity of an element is Z, not mass — the basis of the modern law.

CET level (MCQ)

  • CET 1) The modern periodic law is based on? 1) atomic mass 2) atomic number 3) valency 4) density — Answer: 2).
  • CET 2) Li 7, Na 23, K 39 form? 1) octave 2) Döbereiner's triad 3) isotopes 4) isobars — Answer: 2).

Real-life use: When a new element (e.g. Z = 118) is made its properties are predicted from its place in the table; the same logic helps choose elements for batteries and semiconductors.

Remember

Basis is Z; Mendeleev left gaps; Moseley (1913) ⇒ the modern law.

Common mistake

Writing that the modern law is based on atomic mass; that was Mendeleev's law.

CET trap

Ar–K, Co–Ni and Te–I are reversed pairs quoted as Mendeleev's limits; Z resolves them in the modern table.

Shortcut

Memory clue: "Döbereiner 3, Newlands 8, Mendeleev mass, Moseley Z".

Spot the difference

ScientistBasis
DöbereinerGroups of 3
Newlands8th = 1st
MendeleevAtomic mass
MoseleyAtomic number

Practice questions

Easy
  1. 1State the modern periodic law.
  2. 2What is a triad?
  3. 3Who gave the basis of the modern table?
Medium
  1. 4If Ca is 40 and Ba is 137 estimate Sr.
  2. 5How does the modern table solve the Ar–K problem?
  3. 6State one success of Mendeleev.
Challenging
  1. 7If Cl is 35.5 and I 127 estimate Br.
  2. 8What was Moseley's experiment?
Show answers (try them yourself first)
  1. 1Properties are periodic functions of atomic number.
  2. 2A set of three elements whose middle mass equals the mean.
  3. 3Moseley.
  4. 4(40 + 137) ÷ 2 = 88.5.
  5. 5By Z, Ar (18) comes before K (19).
  6. 6Predicting unknown elements like Ga and Ge.
  7. 7(35.5 + 127) ÷ 2 = 81.25.
  8. 8Relating X-ray frequency to Z.

Structure of the modern periodic table: periods, groups and blocks

From daily life

A city map has rows (streets) and columns (lanes); the periodic table has 7 periods (rows) and 18 groups (columns).

Simple definition

A period is a horizontal row (7 in all). A group is a vertical column (18). A block is s, p, d or f according to the subshell receiving the last electron.

Periodic tablessdpfElements per period: 2, 8, 8, 18, 18, 32, 32 = 118

In the figure : s (groups 1–2), d (3–12), p (13–18), f (two rows below).

Rule 1: Length of a period — Elements per period: 2, 8, 8, 18, 18, 32, 32 — 118 in all.

  • Easy : Period 1: H, He (2).
  • Medium : Period 2: Li to Ne (8).
  • Exam-type : Period 4: K to Kr (18); period 6 has 32 (including La to Lu).

Rule 2: Groups and blocks — s-block: groups 1, 2; p-block: 13–18; d-block: 3–12; f-block: inner transition (lanthanoids, actinoids).

  • Easy : Na — s; Cl — p; Fe — d; Ce — f.
  • Medium : Group 1: alkali metals; group 2: alkaline earth metals; group 17: halogens; group 18: noble gases.
  • Exam-type : The d-block has 10 columns; p has 6; s has 2.

Rule 3: Number of elements by block — Each subshell holds orbitals × 2 electrons: s 2, p 6, d 10, f 14.

  • Easy : Period 4: 4s (2) + 3d (10) + 4p (6) = 18.
  • Medium : Period 6: 6s (2) + 4f (14) + 5d (10) + 6p (6) = 32.
  • Exam-type : Period 2: 2s (2) + 2p (6) = 8.

Derivation: period length from the filling order

  • Step 1: Each period starts with the ns subshell and ends at np.
  • Step 2: Add the electrons of the subshells in between (period 4: 3d).
  • Step 3: The sum is the number of elements: 2 + 10 + 6 = 18.

CET level (MCQ)

  • CET 1) How many elements in period 5? 1) 8 2) 18 3) 32 4) 10 — Answer: 2) 18.
  • CET 2) Which block is groups 3–12? 1) s 2) p 3) d 4) f — Answer: 3).

Real-life use: Li (group 1) in batteries and Nd (f-block) in magnets: the table position hints at the use.

Remember

7 periods, 18 groups; s 2 + p 6 + d 10 + f 14 columns; lengths 2, 8, 8, 18, 18, 32, 32.

Common mistake

Forgetting that He is 1s² (s-block electronically) yet sits in group 18 as a noble gas.

CET trap

Zn, Cd, Hg (group 12) belong to the d-block but are not strict transition elements.

Shortcut

Memory clue: "2-8-8-18-18-32-32" and "s 2, p 6, d 10, f 14".

Spot the difference

BlockGroupsLast e⁻
s1, 2ns¹⁻²
p13–18np¹⁻⁶
d3–12(n−1)d¹⁻¹⁰
f—(n−2)f¹⁻¹⁴

Practice questions

Easy
  1. 1What are a period and a group?
  2. 2What is group 17 called?
  3. 3How many elements in period 6?
Medium
  1. 4Which block is Fe in?
  2. 5Why 18 in period 4?
  3. 6How many columns in the d-block?
Challenging
  1. 7Total elements (2 + 8 + 8 + 18 + 18 + 32 + 32)?
  2. 8How many groups in the p-block?
Show answers (try them yourself first)
  1. 1Horizontal row and vertical column.
  2. 2Halogens.
  3. 332.
  4. 4d-block.
  5. 54s (2) + 3d (10) + 4p (6).
  6. 610.
  7. 7118.
  8. 86 (groups 13–18).

Electronic configuration and position in the table

From daily life

A street and a house number locate a house; the configuration locates the period and group.

Simple definition

Period number = highest principal quantum number n. Group: s-block = ns electrons; p-block = 10 + (s + p) electrons; d-block = (n−1)d + ns electrons.

Finding the positionZ = 26Fe[Ar] 3d⁶ 4s²configurationperiod = highest n = 4(n of 4s)group = 6 + 2 = 8(d + s electrons)s: group = ns e⁻  p: group = 10 + (s + p) e⁻  d: group = (n−1)d + ns e⁻

In the figure : Fe: [Ar] 3d⁶ 4s² ⇒ period 4, group 8.

Live figure

Electronic configuration and position in the table

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Rule 1: s and p blocks — s: group = number of ns e⁻ (1 or 2). p: group = 10 + ns + np.

  • Easy : Ca (Z = 20): [Ar] 4s² ⇒ period 4, group 2.
  • Medium : Cl (Z = 17): [Ne] 3s² 3p⁵ ⇒ period 3, group 10 + 7 = 17.
  • Exam-type : Br (Z = 35): [Ar] 3d¹⁰ 4s² 4p⁵ ⇒ period 4, group 17.

Rule 2: d-block — Group = (n−1)d electrons + ns electrons.

  • Easy : Fe: 3d⁶ 4s² ⇒ group 8.
  • Medium : Zn: 3d¹⁰ 4s² ⇒ group 12.
  • Exam-type : Cr: 3d⁵ 4s¹ ⇒ group 6; Cu: 3d¹⁰ 4s¹ ⇒ group 11.

Rule 3: Reverse: Z from the position — Add the electrons of earlier periods then count by group.

  • Easy : Period 3, group 18: 2 + 8 + 8 = 18 (Ar).
  • Medium : Period 4, group 2: 18 + 2 = 20 (Ca).
  • Exam-type : Period 3, group 13: 10 + 3 = 13 (Al).

Derivation: position of Cl

  • Step 1: Z = 17 ⇒ 1s² 2s² 2p⁶ 3s² 3p⁵.
  • Step 2: Highest n = 3 ⇒ period 3; the last electron is in p ⇒ p-block.
  • Step 3: Group = 10 + (2 + 5) = 17 — a halogen.

CET level (MCQ)

  • CET 1) Position of Z = 26? 1) period 4, group 8 2) period 3, group 8 3) period 4, group 6 4) period 5, group 8 — Answer: 1).
  • CET 2) [Ne] 3s² 3p⁴ belongs to group? 1) 14 2) 15 3) 16 4) 17 — Answer: 3). 10 + 6.

Real-life use: Scientists place a new element by its configuration and predict its properties.

Remember

Period = highest n; group: s = ns, p = 10 + s + p, d = (n−1)d + ns.

Common mistake

Forgetting to add 10 for the p-block (writing 7 instead of 17).

CET trap

With Cr and Cu (4s¹) groups are 6 and 11; Zn is group 12.

Shortcut

Memory clue: "period = biggest n; p-group = 10 + outer e⁻".

Spot the difference

BlockGroup rule
sns e⁻
p10 + (ns + np)
d(n−1)d + ns

Practice questions

Easy
  1. 1How is the period number fixed?
  2. 2Position of Na (Z = 11)?
  3. 3Period and group of Ca?
Medium
  1. 4Write the position of Br.
  2. 5How is Fe's group found?
  3. 6Group of Zn?
Challenging
  1. 7Element in period 3, group 13?
  2. 8Element and group of [Ne] 3s² 3p⁴?
Show answers (try them yourself first)
  1. 1From the highest n.
  2. 2Period 3, group 1.
  3. 3Period 4, group 2.
  4. 4Period 4, group 17.
  5. 56 + 2 = 8.
  6. 612.
  7. 7Al.
  8. 8S; group 16.

Blockwise characteristics of elements

From daily life

A phone battery (Li), chlorine in water (Cl) and an iron bridge (Fe): three blocks, three uses.

Simple definition

s-block: soft reactive metals that lose 1 or 2 e⁻. p-block: metals, non-metals and metalloids; non-metals gain e⁻. d-block: hard metals, coloured ions and variable oxidation states. f-block: lanthanoids (mainly +3).

Periodic tablemetalsmetalloidsnon-metals

In the figure : Blue: metals; orange: metalloids; red: non-metals (staircase line).

Rule 1: s-block — Group 1: M → M⁺; group 2: M → M²⁺. Reactivity increases down the group.

  • Easy : Na → Na⁺ + e⁻.
  • Medium : Mg → Mg²⁺ + 2e⁻.
  • Exam-type : K reacts with water more vigorously than Na.

Rule 2: p-block — Group 17 (halogens): X + e⁻ → X⁻; group 18: inert (filled shell). The metal/non-metal boundary is the staircase line.

  • Easy : Cl + e⁻ → Cl⁻.
  • Medium : Si and Ge are metalloids (semiconductors).
  • Exam-type : He, Ne, Ar do not react.

Rule 3: d- and f-blocks — d: many oxidation states (ns and (n−1)d energies are close), coloured compounds, catalysts, paramagnetic. f: mostly +3.

  • Easy : Mn: +2 to +7.
  • Medium : Cu²⁺ (3d⁹, 1 unpaired e⁻) is blue; Zn²⁺ (3d¹⁰) is colourless.
  • Exam-type : Fe is a catalyst (Haber); V₂O₅ (contact process).

Derivation: why variable oxidation states in the d-block?

  • Step 1: The (n−1)d and ns subshells have nearly equal energies.
  • Step 2: So some d electrons can join the ns electrons in reactions.
  • Step 3: Example Mn: 3d⁵ 4s² ⇒ using all 7 electrons gives +7.

CET level (MCQ)

  • CET 1) Which ion is colourless? 1) Cu²⁺ 2) Zn²⁺ 3) Fe³⁺ 4) Mn²⁺ — Answer: 2). Filled 3d¹⁰.
  • CET 2) Variable valency is prominent in which block? 1) s 2) p 3) d 4) halogens — Answer: 3).

Real-life use: Iron, copper and nickel (d) in construction and coins; silicon (p) in chips; sodium (s) in lamps.

Remember

s: reactive metals; p: mixed; d: coloured, catalytic, variable valency; f: +3.

Common mistake

Saying the p-block has only non-metals; it also has metals like Al and Pb.

CET trap

In 'coloured ion' questions d⁰ and d¹⁰ ions are colourless — Sc³⁺ and Zn²⁺ appear as distractors.

Shortcut

Memory clue: "s = super-reactive, d = dazzling colours, p = a mix".

Spot the difference

BlockFeatureExample
sReactive metalsNa, Ca
pMetals/non-metals/metalloidsAl, Cl, Si
dColoured, catalyticFe, Cu
f+3 stateCe, U

Practice questions

Easy
  1. 1What is group 1 called?
  2. 2Which are the noble gases?
  3. 3Example of a metalloid?
Medium
  1. 4Why is Cu²⁺ blue?
  2. 5Maximum oxidation state of Mn?
  3. 6Why is Zn²⁺ colourless?
Challenging
  1. 7Why variable valency in d-block?
  2. 8Which catalyst is used in the Haber process?
Show answers (try them yourself first)
  1. 1Alkali metals.
  2. 2He, Ne, Ar, Kr, Xe, Rn.
  3. 3Si or Ge.
  4. 4One unpaired 3d electron.
  5. 5+7.
  6. 6Filled 3d¹⁰, no unpaired electron.
  7. 7ns and (n−1)d energies are close.
  8. 8Fe.

Effective nuclear charge, atomic radius and ionic radius

From daily life

A player in the middle is pulled strongly, but with others standing in between the pull feels weaker; the outer electron feels the nucleus less because of inner electrons.

Simple definition

Z_eff = Z − σ (σ = shielding constant). Atomic radius decreases across a period and increases down a group. A cation is smaller than its atom; an anion is larger.

Atomic radiusAtomic radius (pm): increases down a groupLi152Na186K231Left to right in a period: Zeff rises ⇒ radius fallsNa (186) → Cl (99) pm

In the figure : Li 152, Na 186, K 231 pm — radius grows down the group.

Rule 1: Slater's rules (for s, p) — σ = 0.35 × (other e⁻ in the same shell) + 0.85 × (e⁻ in shell n−1) + 1.00 × (e⁻ in n−2 and lower).

  • Easy : Na: σ = 0.85 × 8 + 1 × 2 = 8.8 ⇒ Z_eff = 2.2.
  • Medium : Mg: σ = 0.35 + 6.8 + 2 = 9.15 ⇒ 2.85.
  • Exam-type : Cl: σ = 0.35 × 6 + 6.8 + 2 = 10.9 ⇒ Z_eff = 6.1.

Rule 2: Trends in atomic radius — Across a period Z_eff rises with the same shell ⇒ radius falls. Down a group more shells ⇒ radius rises.

  • Easy : Na (186) > Mg (160) > Al (143) pm.
  • Medium : Li < Na < K.
  • Exam-type : Cl (99 pm) is much smaller than Na (186 pm).

Rule 3: Ionic radius — Cation: loses e⁻ ⇒ smaller. Anion: gains e⁻ ⇒ larger. In an isoelectronic series radius falls as Z rises.

  • Easy : Na 186 pm → Na⁺ 102 pm.
  • Medium : Cl 99 pm → Cl⁻ 181 pm.
  • Exam-type : O²⁻ (140) > F⁻ (133) > Na⁺ (102) > Mg²⁺ (72).

Derivation: why does radius fall across a period?

  • Step 1: From Na to Cl, Z goes 11 → 17 while electrons enter the same shell (n = 3).
  • Step 2: Electrons in the same shell shield each other little (0.35) ⇒ Z_eff 2.2 → 6.1.
  • Step 3: Stronger pull ⇒ the shell contracts ⇒ radius 186 → 99 pm.

Formulas, units and dimensional formulae

QuantityFormula / trendUnit
Effective chargeZ_eff = Z − σnone
Atomic radius↓ across, ↑ downpm (10⁻¹² m)
Ionic radiuscation < atom < anionpm

CET level (MCQ)

  • CET 1) Z_eff of Na (Slater)? 1) 1 2) 2.2 3) 8.8 4) 11 — Answer: 2) 2.2. 11 − 8.8.
  • CET 2) Largest to smallest: 1) Mg²⁺ > Na⁺ > F⁻ > O²⁻ 2) O²⁻ > F⁻ > Na⁺ > Mg²⁺ 3) F⁻ > O²⁻ > Na⁺ > Mg²⁺ 4) Na⁺ > Mg²⁺ > O²⁻ > F⁻ — Answer: 2).

Real-life use: Crystal structure and melting point depend on ionic radius; the small Li⁺ ion moves easily in batteries.

Remember

Z_eff = Z − σ; r falls across a period, rises down a group; cation < atom < anion.

Common mistake

Believing radius increases with Z across a period; Z_eff rises and radius falls.

CET trap

In isoelectronic series 'higher Z = smaller radius'; options give the reverse.

Shortcut

Memory clue: "tighter across, looser down" (radius).

Spot the difference

PointAcross →Down ↓
Z_effIncreasesAlmost constant
RadiusDecreasesIncreases

Practice questions

Easy
  1. 1What is Z_eff?
  2. 2Order of radius in Li, Na, K.
  3. 3Why is a cation smaller?
Medium
  1. 4Find Z_eff of Mg.
  2. 5Find Z_eff of Cl.
  3. 6Arrange Na⁺, Mg²⁺, F⁻ from large to small.
Challenging
  1. 7Why does radius fall in period 3?
  2. 8O²⁻, F⁻, Na⁺ all have 10 e⁻; why do radii differ?
Show answers (try them yourself first)
  1. 1The nuclear charge actually felt by an outer electron, Z − σ.
  2. 2Li < Na < K.
  3. 3Fewer electrons/shells and less repulsion.
  4. 42.85.
  5. 56.1.
  6. 6F⁻ > Na⁺ > Mg²⁺.
  7. 7Z_eff rises with the same shell.
  8. 8Different Z: higher Z pulls harder ⇒ smaller radius.

Ionisation enthalpy and electron gain enthalpy

From daily life

Plucking a fruit needs effort, more if it is tight. Removing an electron from an atom also needs energy.

Simple definition

Ionisation enthalpy (ΔᵢH) is the energy needed to remove the outermost electron of a gaseous atom, in kJ mol⁻¹. Electron gain enthalpy (Δ_egH) is the energy change when a gaseous atom gains an electron.

Ionisation enthalpyIE₁Li520Be899B801C1086N1402O1314F1681Ne2080Na496First ionisation enthalpy (kJ/mol)Be > B and N > O: filled / half-filled subshells are stable

In the figure : Li to Na: general rise; Be > B and N > O are exceptions.

Live figure

Ionisation enthalpy and electron gain enthalpy

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Rule 1: Trends in ionisation enthalpy — Rises across a period (Z_eff ↑, r ↓); falls down a group. IE₁ < IE₂ < IE₃ (increasing positive charge).

  • Easy : Li 520 < Be 899 < … < Ne 2080 kJ/mol.
  • Medium : Na 496 < Li 520 (decrease down the group).
  • Exam-type : Na: IE₁ = 496, IE₂ = 4562 — a huge jump.

Rule 2: Exceptions and successive IE — Be (2s²) > B (2p¹); N (2p³, half-filled) > O (2p⁴). The big jump tells how many valence electrons there are.

  • Easy : Mg: 737, 1450, 7733 — big jump at the third IE ⇒ 2 valence electrons.
  • Medium : IE₁ of Na per atom = 496 × 10³ ÷ 6.022 × 10²³ ÷ 1.602 × 10⁻¹⁹ = 5.14 eV.
  • Exam-type : Ionising 0.5 mol of Na vapour: 0.5 × 496 = 248 kJ.

Rule 3: Electron gain enthalpy — Halogens have the most negative Δ_egH; Cl (−349) is more negative than F (−328) because the small F has strong electron repulsion. Noble gases are positive.

  • Easy : Cl: −349 kJ/mol; F: −328; Br: −325.
  • Medium : O: −141; O⁻ → O²⁻: +780 (positive because a second electron is added to an anion).
  • Exam-type : Ne, Ar: positive (filled shell).

Derivation: why IE rises across a period

  • Step 1: Z rises while the outer e⁻ stays in the same shell ⇒ Z_eff rises.
  • Step 2: Radius falls ⇒ stronger pull.
  • Step 3: More energy is needed ⇒ IE rises.

Formulas, units and dimensional formulae

QuantityProcessUnit
IE₁M(g) → M⁺(g) + e⁻kJ mol⁻¹
Δ_egHX(g) + e⁻ → X⁻(g)kJ mol⁻¹
1 eV1.602 × 10⁻¹⁹ JJ

CET level (MCQ)

  • CET 1) Highest IE₁? 1) Li 2) Be 3) Ne 4) Na — Answer: 3) Ne.
  • CET 2) Which order of IE is correct? 1) B > Be 2) N > O 3) O > N 4) Li > Be — Answer: 2).

Real-life use: The yellow glow of sodium lamps (low IE, easy excitation) and chlorine's disinfecting power (very negative Δ_egH).

Remember

IE: ↑ across, ↓ down; exceptions Be > B, N > O; Δ_egH: Cl most negative.

Common mistake

Calling electron gain enthalpy positive; energy is released, so it is negative.

CET trap

Reversed Be/B and N/O orders appear in options; between F and Cl, Cl has the more negative Δ_egH.

Shortcut

Memory clue: "Be above B, N above O" — the two exceptions.

Spot the difference

QuantityAcrossDown
IEIncreasesDecreases
Δ_egHMore negativeLess negative

Practice questions

Easy
  1. 1What is ionisation enthalpy?
  2. 2Which has the highest IE₁?
  3. 3Why is Δ_egH negative?
Medium
  1. 4Why is IE₂ of Na far above IE₁?
  2. 5Why is Be > B?
  3. 6Energy to ionise 0.5 mol Na (IE₁ = 496) in kJ.
Challenging
  1. 7From Mg: 737, 1450, 7733 find the valency.
  2. 8Why is Cl's Δ_egH more negative than F's?
Show answers (try them yourself first)
  1. 1Energy to remove the outer electron of a gaseous atom.
  2. 2Ne (period 2).
  3. 3Energy is released.
  4. 4Na⁺ has the stable Ne configuration.
  5. 5Filled 2s².
  6. 6248 kJ.
  7. 72 (big jump at the third).
  8. 8The small F has stronger electron repulsion.

Electronegativity, valency, metallic character and reactivity

From daily life

In a tug of war the stronger team pulls the rope; in a bond the shared electrons are pulled to the more electronegative atom.

Simple definition

Electronegativity (EN) is an atom's ability to attract the shared electrons in a bond (Pauling scale). Valency is its combining capacity. Metallic character is the tendency to lose electrons.

ElectronegativityLi1.0Be1.5B2.0C2.5N3.0O3.5F4.0Electronegativity (Pauling): rises across a periodΔEN = EN(A) − EN(B): HF = 1.9, HCl = 0.9, NaCl = 2.1falls down a group; F is highest (4.0)

In the figure : Li 1.0 to F 4.0: electronegativity rises across the period.

Rule 1: Trends in electronegativity — Rises across a period and falls down a group; F is highest (4.0). Large ΔEN (> 1.7) ⇒ ionic bond; small ⇒ covalent.

  • Easy : F 4.0 > O 3.5 > N 3.0 > C 2.5.
  • Medium : HCl: ΔEN = 3.0 − 2.1 = 0.9 ⇒ polar covalent.
  • Exam-type : NaCl: 3.0 − 0.9 = 2.1 ⇒ ionic; HF: 1.9.

Rule 2: Valency and the nature of oxides — Valency = outer electrons (or 8 − outer electrons). Across a period oxides go basic → amphoteric → acidic.

  • Easy : Na 1, Mg 2, Al 3, Si 4, P 3 (or 5), S 2, Cl 1.
  • Medium : Na₂O, MgO basic; Al₂O₃ amphoteric.
  • Exam-type : SiO₂ weakly acidic; P₄O₁₀, SO₃, Cl₂O₇ acidic.

Rule 3: Metallic character and reactivity — Metallic character falls across a period and rises down a group. Reactivity of metals rises down a group (Li < Na < K); that of non-metals falls (F > Cl > Br > I).

  • Easy : Metallic character Na > Mg > Al.
  • Medium : Cl₂ displaces bromide: Cl₂ + 2Br⁻ → 2Cl⁻ + Br₂.
  • Exam-type : F₂ is the most reactive non-metal; Cs the most reactive metal.

Derivation: bond type from ΔEN

  • Step 1: Look up the EN of both atoms (H 2.1, Cl 3.0).
  • Step 2: ΔEN = larger − smaller = 0.9.
  • Step 3: 0 ⇒ pure covalent; 0 to 1.7 ⇒ polar; > 1.7 ⇒ ionic. So HCl is polar covalent.

Formulas, units and dimensional formulae

QuantityTrendUnit
Electronegativity↑ across, ↓ downnone (Pauling)
Metallic character↓ across, ↑ down—
Valency1, 2, 3, 4, 3, 2, 1, 0—

CET level (MCQ)

  • CET 1) Most electronegative? 1) O 2) N 3) F 4) Cl — Answer: 3) F.
  • CET 2) Which oxide is amphoteric? 1) Na₂O 2) Al₂O₃ 3) SO₃ 4) MgO — Answer: 2).

Real-life use: Chlorine disinfects water (reactive non-metal); aluminium oxide reacts with both acids and bases.

Remember

EN: F = 4.0 is highest; ΔEN > 1.7 ionic; oxides: basic → amphoteric → acidic.

Common mistake

Treating EN and electron gain enthalpy as the same; EN is attraction in a bond, Δ_egH is for an isolated gaseous atom.

CET trap

ΔEN > 1.7 is only a guide; HF (1.9) is treated as polar covalent. Amphoteric: Al₂O₃, ZnO.

Shortcut

Memory clue: "F-O-N-Cl": most electronegative — 4.0, 3.5, 3.0, 3.0".

Spot the difference

PropertyMetalNon-metal
e⁻LosesGains
OxideBasicAcidic
ENLowHigh

Practice questions

Easy
  1. 1What is electronegativity?
  2. 2Most electronegative element?
  3. 3What type of oxide is Al₂O₃?
Medium
  1. 4Find ΔEN of HCl.
  2. 5Why is NaCl ionic?
  3. 6Order of metallic character of Na, Mg, Al.
Challenging
  1. 7Write the reaction of Cl₂ with KBr.
  2. 8Order of reactivity of F, Cl, Br, I with reason.
Show answers (try them yourself first)
  1. 1Ability to attract the shared electron pair.
  2. 2F.
  3. 3Amphoteric.
  4. 43.0 − 2.1 = 0.9.
  5. 5ΔEN = 2.1 > 1.7.
  6. 6Na > Mg > Al.
  7. 7Cl₂ + 2KBr → 2KCl + Br₂.
  8. 8F > Cl > Br > I; bigger size lowers the tendency to gain e⁻.

Quick revision of the whole lesson

ConceptKey formula / fact
BasisModern law: atomic number Z
Period/group7 / 18; 2, 8, 8, 18, 18, 32, 32
Group (p)10 + (ns + np)
Z_effZ − σ
Radius↓ across, ↑ down
IE↑ across; Be > B, N > O
ENF = 4.0 highest; ΔEN > 1.7 ionic
Oxidesbasic → amphoteric → acidic
Live figure

🚀 Quick revision of the whole lesson

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🎯 Most important points for the exam

  • Moseley and the modern law; Mendeleev's limits.
  • Period, group and block from configuration.
  • d-block features: colour, variable valency.
  • Slater's Z_eff and radius orders.
  • Trends and exceptions of IE and Δ_egH.
  • EN, ΔEN and nature of oxides.
Study mantra

understand the idea → remember the formula → solve the example → solve problems yourself.