JEE Main & Advanced

Classification of Elements and Periodicity

Classification of Elements and Periodicity for JEE Main & Advanced

1
Module 1

Periodic Table and Electronic Configurations

Historical Development and Modern Periodic TableTopic 1

Historical Development:

YearScientistContribution
1817DöbereinerTriads — atomic mass of middle element = mean of other two
1864NewlandsLaw of Octaves — every 8th element similar (failed beyond Ca)
1869MendeleevPeriodic Law — properties are periodic functions of atomic masses
1913MoseleyAtomic number ($Z$) more fundamental than atomic mass

Mendeleev's Periodic Law: Properties of elements are periodic function of their atomic masses.

Modern Periodic Law (Moseley): Properties of elements are periodic function of their atomic numbers.

Mendeleev's Achievements:

  • Predicted properties of undiscovered elements (eka-Al = Ga; eka-Si = Ge; eka-B = Sc)
  • Arranged elements based on similarities

Mendeleev's Defects:

  • Position of hydrogen ambiguous
  • Anomalous pairs: Ar before K, Co before Ni, Te before I (mass order violated)
  • Isotopes had no separate place
  • Did not explain lanthanide/actinide series

Modern Periodic Table:

  • $18$ groups (vertical columns)
  • $7$ periods (horizontal rows)
  • $4$ blocks: $s, p, d, f$
  • Total elements known up to $Z = 118$ (Oganesson, Og)

IUPAC Naming for $Z > 100$:

DigitRootSymbol
0niln
1unu
2bib
3trit
4quadq
5pentp
6hexh
7septs
8octo
9enne

Suffix '-ium'. Example: $Z = 119$ = unununium (Uue); $Z = 120$ = unbinilium (Ubn).

Worked Examples
1

Predict IUPAC symbol of element with $Z = 116$.

Show solution

Digits: 1, 1, 6 → un + un + hex + ium = ununhexium (Uuh). (Officially named Livermorium, Lv.)

Final Answer: Uuh (Livermorium).

2

Mendeleev placed Te (atomic mass $128$) before I ($127$). Why?

Show solution

Although Te is heavier, Mendeleev placed it in Group VI (with O, S, Se) based on similar chemical properties, and I in Group VII (halogens). Modern periodic law (based on $Z$) resolves this: Te = 52, I = 53 — natural order.

Final Answer: Based on chemical property similarity; modern law (using $Z$) fixed this issue.

✎ Self-Check — 5 questions0 / 5
Q1.

Modern periodic law arranges elements by:

Q2.

Number of periods in modern periodic table:

Q3.

Mendeleev's table failed to place:

Q4.

Eka-silicon predicted by Mendeleev turned out to be:

Q5.

Newlands' law was called:

Electronic Configuration, Blocks, Periods and GroupsTopic 2

Period number = principal quantum number ($n$) of outermost shell.

Periods and Number of Elements:

Period$n$ElementsNumber
1$1$H, He$2$
2$2$Li → Ne$8$
3$3$Na → Ar$8$
4$4$K → Kr$18$
5$5$Rb → Xe$18$
6$6$Cs → Rn$32$
7$7$Fr → Og$32$ (still incomplete in some respects)

Blocks (based on subshell of last electron entering):

BlockLast subshellGroupsProperties
s-block$ns^{1-2}$1, 2Most reactive metals (alkali, alkaline earth)
p-block$np^{1-6}$13–18Mixed: metals, non-metals, noble gases
d-block$(n-1)d^{1-10}\,ns^{0-2}$3–12Transition metals; variable oxidation states, colored
f-block$(n-2)f^{1-14}$3 (sub)Lanthanoids ($4f$), actinoids ($5f$)

Group Determination:

  • For $s$-block: group = number of valence electrons (1 or 2)
  • For $p$-block: group = $10 + $ number of valence electrons (13 to 18)
  • For $d$-block: group = number of $(n-1)d + ns$ electrons (3 to 12)

Examples:

  • Na ($Z = 11$): $[Ne]\,3s^1$ → period 3, group 1, s-block
  • Cl ($Z = 17$): $[Ne]\,3s^2\,3p^5$ → period 3, group 17, p-block
  • Fe ($Z = 26$): $[Ar]\,3d^6\,4s^2$ → period 4, group 8, d-block
  • Ce ($Z = 58$): $[Xe]\,4f^1\,5d^1\,6s^2$ → period 6, f-block (lanthanide)

Metals, Non-metals, and Metalloids:

  • Metals: Left and centre — about $78\%$ of elements
  • Non-metals: Right (p-block top), e.g., H, C, N, O, halogens, noble gases
  • Metalloids: B, Si, Ge, As, Sb, Te, Po (diagonal line)
Worked Examples
1

Find period and group of element with $Z = 33$.

Show solution

Config: $[Ar]\,3d^{10}\,4s^2\,4p^3$. Outermost $n = 4$ → Period 4. Valence electrons in $s+p$ = $2 + 3 = 5$. p-block. Group = $10 + 5 = 15$.

Final Answer: Period 4, Group 15 (As — Arsenic).

2

Identify block, period, group of $Z = 64$.

Show solution

Configuration: $[Xe]\,4f^7\,5d^1\,6s^2$ (Gd, gadolinium). Block: f. Period: 6 (highest $n$). Group: officially Group 3 (f-block placed under Group 3).

Final Answer: f-block, Period 6, Group 3 (Gd).

✎ Self-Check — 5 questions0 / 5
Q1.

The element with $Z = 17$ belongs to:

Q2.

Number of elements in 6th period:

Q3.

Configuration $[Ar]\,3d^5\,4s^1$ corresponds to:

Q4.

Group 18 elements have configuration:

Q5.

f-block elements include:

2
Module 2

Periodic Trends

Atomic Radius, Ionic Radius, Ionization EnthalpyTopic 1

Effective Nuclear Charge ($Z_{\text{eff}}$): $$Z_{\text{eff}} = Z - \sigma$$ where $\sigma$ = screening (shielding) constant calculated by Slater's rules.

Slater's Rules (for $\sigma$ on a given electron):

  • Other electrons in same group (or higher): $0$
  • Each other electron in same $(ns, np)$ group: $0.35$ (but $1s$: $0.30$)
  • Each electron in $(n-1)$ shell: $0.85$
  • Each electron in $(n-2)$ or deeper: $1.00$
  • For $d, f$ electrons: each electron in same group $0.35$; all underlying $= 1.00$

Atomic Radius: Half the distance between two bonded atoms.

  • Covalent radius (for non-metals): half of bond length in covalent molecule
  • Metallic radius (for metals): half the distance between two adjacent metal atoms in crystal
  • van der Waals radius: half the distance between two non-bonded atoms

Trends:

  • Down a group: Atomic radius increases (new shell added)
  • Across a period (L→R): Atomic radius decreases ($Z_{\text{eff}}$ increases for same shell)

Ionic Radius:

  • Cations are smaller than parent atoms (loss of outer shell, reduced repulsions)
  • Anions are larger than parent atoms (added electrons, more repulsion)
  • For isoelectronic species (same e⁻): radius decreases as $Z$ increases. e.g., O²⁻ > F⁻ > Na⁺ > Mg²⁺ > Al³⁺

Ionization Enthalpy ($\Delta_i H$): Minimum energy to remove an electron from an isolated gaseous atom. $$M(g) \to M^+(g) + e^-; \quad \Delta_i H_1 > 0$$

Trends:

  • Across period: Increases (smaller atom, more $Z_{\text{eff}}$)
  • Down group: Decreases (larger atom)

Successive IEs: $\Delta_i H_1 < \Delta_i H_2 < \Delta_i H_3 < \ldots$

Big jump occurs when noble gas configuration is reached.

Anomalies in IE:

  • B < Be (filled $2s^2$ extra stable; removing from $2p^1$ easier than $2s^2$)
  • O < N ($2p^3$ half-filled extra stable; removing from $2p^4$ easier than $2p^3$)
  • Similar: Al < Mg; S < P
Worked Examples
1

Arrange in increasing order of ionic radius: Mg²⁺, Al³⁺, Na⁺, O²⁻, F⁻.

Show solution

All have $10$ electrons (isoelectronic). Radius decreases as $Z$ increases: $Z$: O²⁻ (8) < F⁻ (9) < Na⁺ (11) < Mg²⁺ (12) < Al³⁺ (13). Order of radii: Al³⁺ < Mg²⁺ < Na⁺ < F⁻ < O²⁻.

Final Answer: Al³⁺ < Mg²⁺ < Na⁺ < F⁻ < O²⁻.

2

Why is the first ionization enthalpy of N greater than that of O?

Show solution

N: $2p^3$ (half-filled, extra stable). Removing one electron disrupts stable arrangement → requires more energy. O: $2p^4$ — removal of one electron gives stable $2p^3$. Hence, lower IE.

Final Answer: N has stable half-filled $2p^3$; harder to remove an electron.

✎ Self-Check — 5 questions0 / 5
Q1.

Across a period, atomic radius:

Q2.

Among Na, Na⁺, K, K⁺:

Q3.

Which has highest first ionization energy?

Q4.

Down group 1, ionization enthalpy:

Q5.

Isoelectronic species N³⁻, O²⁻, F⁻, Ne. Radius order:

Electron Gain Enthalpy, Electronegativity, AnomaliesTopic 2

Electron Gain Enthalpy ($\Delta_{eg} H$): Enthalpy change when an electron is added to a neutral gaseous atom. $$X(g) + e^- \to X^-(g); \quad \Delta_{eg}H$$

Trends:

  • Across period: Becomes more negative (electron readily accepted by smaller, more electronegative atoms)
  • Down group: Less negative (atom larger; added electron experiences less attraction)

Highest electron affinity: Cl (more than F because F is small → electron-electron repulsion in compact $2p$ subshell makes F's $\Delta_{eg}H$ less negative than Cl's).

Halogen$\Delta_{eg}H$ (kJ/mol)
F$-328$
Cl$-349$
Br$-325$
I$-295$

Electronegativity (EN): Tendency of an atom to attract shared electron pair in a covalent bond. Dimensionless.

Pauling scale: F = $4.0$ (highest); Cs (or Fr) = $0.7$ (lowest among non-radioactive). H = $2.1$.

Other scales: Mulliken (avg of IE and EA), Allred-Rochow.

Trends in EN:

  • Across period: Increases (smaller atom, higher $Z_{\text{eff}}$)
  • Down group: Decreases (larger atom)

Diagonal Relationship: Diagonally adjacent elements (Li-Mg, Be-Al, B-Si) show similar properties because of similar charge density (charge/size ratio).

Anomalous Properties of 2nd Period Elements:

  • Small size, high EN, no d-orbitals → limit covalency to 4
  • Examples: Li (resembles Mg), Be (resembles Al)

Periodic Trends Summary:

PropertyAcross period (L→R)Down group
Atomic radius
Ionic radius (cation)
IE
Electron gain enthalpy (magnitude)More negativeLess negative
Electronegativity
Metallic character
Non-metallic character
Oxidising power
Reducing power
Worked Examples
1

Why is the electron affinity of fluorine less than chlorine?

Show solution

F is very small ($2p$ subshell compact). Added electron experiences significant electron-electron repulsion → less energy released. In Cl ($3p$), the orbital is larger → less repulsion. Hence, Cl has more negative $\Delta_{eg}H$.

Final Answer: Electron-electron repulsion in compact $2p$ of F is the cause.

2

Arrange in increasing electronegativity: O, F, N, C.

Show solution

Period 2 elements; EN increases left to right: C < N < O < F.

Final Answer: C < N < O < F.

✎ Self-Check — 5 questions0 / 5
Q1.

The most electronegative element:

Q2.

Electron gain enthalpy of Cl is more negative than F because:

Q3.

Electronegativity scale by Pauling: $F$ value:

Q4.

Across a period, metallic character:

Q5.

Diagonal relationship is shown by:

Ready to test yourself?

Attempt the full timed mock tests — Main & Advanced level.

Start Mock Test 1 →