Classification of Elements and Periodicity
Classification of Elements and Periodicity for JEE Main & Advanced
Periodic Table and Electronic Configurations
Historical Development and Modern Periodic TableTopic 1
Historical Development:
| Year | Scientist | Contribution |
|---|---|---|
| 1817 | Döbereiner | Triads — atomic mass of middle element = mean of other two |
| 1864 | Newlands | Law of Octaves — every 8th element similar (failed beyond Ca) |
| 1869 | Mendeleev | Periodic Law — properties are periodic functions of atomic masses |
| 1913 | Moseley | Atomic 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 $Z > 100$:
| Digit | Root | Symbol |
|---|---|---|
| 0 | nil | n |
| 1 | un | u |
| 2 | bi | b |
| 3 | tri | t |
| 4 | quad | q |
| 5 | pent | p |
| 6 | hex | h |
| 7 | sept | s |
| 8 | oct | o |
| 9 | enn | e |
Suffix '-ium'. Example: $Z = 119$ = unununium (Uue); $Z = 120$ = unbinilium (Ubn).
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).
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.
Modern periodic law arranges elements by:
Number of periods in modern periodic table:
Mendeleev's table failed to place:
Eka-silicon predicted by Mendeleev turned out to be:
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 | $n$ | Elements | Number |
|---|---|---|---|
| 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):
| Block | Last subshell | Groups | Properties |
|---|---|---|---|
| s-block | $ns^{1-2}$ | 1, 2 | Most reactive metals (alkali, alkaline earth) |
| p-block | $np^{1-6}$ | 13–18 | Mixed: metals, non-metals, noble gases |
| d-block | $(n-1)d^{1-10}\,ns^{0-2}$ | 3–12 | Transition 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)
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).
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).
The element with $Z = 17$ belongs to:
Number of elements in 6th period:
Configuration $[Ar]\,3d^5\,4s^1$ corresponds to:
Group 18 elements have configuration:
f-block elements include:
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
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⁻.
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.
Across a period, atomic radius:
Among Na, Na⁺, K, K⁺:
Which has highest first ionization energy?
Down group 1, ionization enthalpy:
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:
| Property | Across period (L→R) | Down group |
|---|---|---|
| Atomic radius | ↓ | ↑ |
| Ionic radius (cation) | ↓ | ↑ |
| IE | ↑ | ↓ |
| Electron gain enthalpy (magnitude) | More negative | Less negative |
| Electronegativity | ↑ | ↓ |
| Metallic character | ↓ | ↑ |
| Non-metallic character | ↑ | ↓ |
| Oxidising power | ↑ | ↓ |
| Reducing power | ↓ | ↑ |
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.
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.
The most electronegative element:
Electron gain enthalpy of Cl is more negative than F because:
Electronegativity scale by Pauling: $F$ value:
Across a period, metallic character:
Diagonal relationship is shown by:
Ready to test yourself?
Attempt the full timed mock tests — Main & Advanced level.
Start Mock Test 1 →