CLASSIFICATION OF ELEMENTS AND PERIODICITY

The periodic table is one of the most important tools in chemistry. It organizes the myriad of elements into a logical framework based on their electronic configurations. This chapter explores the historical attempts at classification, the modern periodic law, and the fundamental trends that dictate chemical behavior. It is heavily weighted in both Board Exams and NEET/JEE.

1. Genesis of Periodic Classification

Before the modern periodic table, scientists made several attempts to classify elements based on their atomic weights.

Dobereiner's Triads (1829): Grouped elements in threes (triads) where the atomic weight of the middle element was roughly the average of the other two (e.g., Li, Na, K). Failed because it didn't apply to all elements.
Newlands' Law of Octaves (1865): Arranged elements in increasing atomic weight and noted that every eighth element had properties similar to the first (like musical notes). Failed beyond Calcium.
Mendeleev's Periodic Law (1869): "The properties of the elements are a periodic function of their atomic weights." He successfully predicted the existence of undiscovered elements (e.g., Eka-Aluminium $\rightarrow$ Gallium, Eka-Silicon $\rightarrow$ Germanium).

2. Modern Periodic Law (Henry Moseley, 1913)

Moseley conducted X-ray experiments on various elements and observed a regular pattern when he plotted the square root of the X-ray frequency ($\sqrt{\nu}$) against the atomic number ($Z$).

Concept Modern Periodic Law: The physical and chemical properties of the elements are periodic functions of their atomic numbers (not atomic weights).

Moseley's Equation: $$ \sqrt{\nu} = a(Z - b) $$ Where $\nu$ is the frequency of X-rays, $Z$ is the atomic number, and $a, b$ are constants.

Practice Problems 1 — Historical & Modern Laws Q1 (School Focus): What was the basic theme of organization in the periodic table proposed by Mendeleev, and how did Moseley change it?

Solution:
Mendeleev organized elements based on their atomic weights. He stated that physical and chemical properties are a periodic function of atomic weights.
Moseley changed it by organizing elements based on their atomic numbers, establishing the Modern Periodic Law.

Q2 (NCERT Focus): Explain why Mendeleev left gaps in his periodic table.

Solution:
Mendeleev was bold enough to leave gaps in his periodic table for elements that had not yet been discovered. He predicted their existence and properties based on the properties of other elements in the same group (e.g., Eka-Aluminium and Eka-Silicon).

3. Nomenclature of Elements with Z > 100

To avoid competing claims for the discovery of new elements, IUPAC established a systematic naming process based on the digits of the atomic number.

Digit	Root	Abbreviation	Digit	Root	Abbreviation
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

Example: Element 104 $\rightarrow$ un + nil + quad + ium = Unnilquadium (Unq). Its accepted IUPAC name is Rutherfordium (Rf).

Practice Problems 2 — Nomenclature Z > 100 Q1 (NCERT Focus): What would be the IUPAC name and symbol for the element with atomic number 120?

Solution:
The roots for 1, 2, and 0 are un, bi, and nil respectively.
Name: un + bi + nil + ium = Unbinilium.
Symbol: Ubn.

Q2 (School Focus): Write the IUPAC name and symbol for the element with atomic number 115.

Solution:
The roots for 1, 1, and 5 are un, un, and pent respectively.
Name: un + un + pent + ium = Ununpentium.
Symbol: Uup.

4. Electronic Configurations and Types of Elements

The periodic table is divided into four main blocks based on the subshell in which the last electron enters.

s-Block Elements (Groups 1 and 2): Last electron enters the s-orbital.
General Configuration: $ns^{1-2}$
Highly reactive metals, low ionization enthalpies.
p-Block Elements (Groups 13 to 18): Last electron enters the p-orbital.
General Configuration: $ns^2 np^{1-6}$
Contains metals, non-metals, and metalloids. Group 18 contains Noble Gases ($ns^2 np^6$).
d-Block Elements (Groups 3 to 12): Last electron enters the (n-1)d orbital. Known as Transition Elements.
General Configuration: $(n-1)d^{1-10} ns^{0-2}$
Form colored ions, exhibit variable valency, and act as catalysts.
f-Block Elements (Lanthanoids and Actinoids): Last electron enters the (n-2)f orbital. Known as Inner Transition Elements.
General Configuration: $(n-2)f^{1-14} (n-1)d^{0-1} ns^2$

Metals, Non-metals, and Metalloids

Metals: Comprise more than 78% of all known elements and appear on the left side of the Periodic Table. They are usually solids at room temperature (except mercury), have high melting and boiling points, and are good conductors of heat and electricity.
Non-metals: Located at the top right hand side of the Periodic Table. They are usually solids or gases at room temperature with low melting and boiling points (except boron and carbon). They are poor conductors of heat and electricity.
Metalloids (Semi-metals): Elements that show properties of both metals and non-metals. They border the zig-zag line that separates metals from non-metals (e.g., Silicon, Germanium, Arsenic, Antimony, and Tellurium).

Practice Problems 3 — Electronic Configurations & Blocks Q1 (NCERT Exercise 3.6): How would you justify the presence of 18 elements in the $5^{\text{th}}$ period of the Periodic Table?

Solution:
The 5th period corresponds to the filling of the 5th energy level ($n=5$). According to the Aufbau principle, the order of filling is $5s, 4d,$ and $5p$.
- $5s$ orbital holds 2 electrons.
- $4d$ subshell holds 10 electrons.
- $5p$ subshell holds 6 electrons.
Total electrons = $2 + 10 + 6 = 18$. Therefore, the 5th period has 18 elements.

Q2 (NEET Focus): An element has the electronic configuration $[\text{Kr}] 5s^2 4d^{10} 5p^4$. Predict its block, group, and period.

Solution:
- Period: Highest principal quantum number is 5, so it belongs to the 5th period.
- Block: The last electron enters the $p$-orbital, so it is a p-block element.
- Group: For p-block elements, Group Number = $12 + \text{number of p electrons} = 12 + 4 = $ 16.

5. Periodic Trends in Properties of Elements

This is the most critical section for competitive exams. Trends are governed primarily by the Effective Nuclear Charge ($Z_{eff}$).

Effective Nuclear Charge & Slater's Rules (Advanced)

Inner electrons shield the outer electrons from the full pull of the nucleus. This is the screening or shielding effect. The actual charge felt by the outermost electron is $Z_{eff}$.

Formula for Effective Nuclear Charge: $$ Z_{eff} = Z - \sigma $$ Where $Z$ = Actual atomic number (nuclear charge) and $\sigma$ = Shielding/Screening constant.

Slater's Rules for calculating $\sigma$ (For an $ns$ or $np$ electron):

Electrons in the same shell ($n$) contribute $0.35$ each (except $1s$ which is $0.30$).
Electrons in the $(n-1)$ shell contribute $0.85$ each.
Electrons in $(n-2)$ or lower shells contribute $1.00$ each.

A. Atomic Radius

Cannot be measured directly. Types include Covalent Radius, Metallic Radius, and van der Waals Radius. van der Waals > Metallic > Covalent.

Across a Period (Left to Right): Atomic radius decreases. Reason: Electrons are added to the same shell, but nuclear charge ($Z$) increases, pulling electrons closer.
Down a Group (Top to Bottom): Atomic radius increases. Reason: New shells are added, which outweighs the increase in nuclear charge.

B. Ionic Radius

Cations are smaller than their parent atoms (loss of electron(s) increases $Z_{eff}$). Anions are larger than their parent atoms (gain of electron(s) decreases $Z_{eff}$ and increases electron-electron repulsion).

JEE Shortcut Isoelectronic Species: Atoms and ions with the same number of electrons.
Example: $\text{N}^{3-}, \text{O}^{2-}, \text{F}^-, \text{Ne}, \text{Na}^+, \text{Mg}^{2+}, \text{Al}^{3+}$ (All have 10 electrons).

Rule: For isoelectronic species, as the positive nuclear charge ($Z$) increases, the size decreases.
Size order: $\text{N}^{3-} > \text{O}^{2-} > \text{F}^- > \text{Na}^+ > \text{Mg}^{2+} > \text{Al}^{3+}$

C. Ionization Enthalpy ($\Delta_i H$)

The minimum amount of energy required to remove the most loosely bound electron from an isolated gaseous atom in its ground state.

$$ M_{(g)} + \text{Energy} \rightarrow M^+_{(g)} + e^- $$

Across a Period: Generally increases due to decreasing size and increasing $Z_{eff}$.
Down a Group: Generally decreases due to increasing size and shielding effect.

Crucial Exceptions 1. Be vs B: $\Delta_i H$ of Beryllium ($899 \text{ kJ/mol}$) is greater than Boron ($801 \text{ kJ/mol}$).
Reason: Removing an electron from fully-filled, more penetrating $2s$ orbital in Be ($1s^2 2s^2$) requires more energy than removing a $2p$ electron in B ($1s^2 2s^2 2p^1$).

2. N vs O: $\Delta_i H$ of Nitrogen is greater than Oxygen.
Reason: N ($1s^2 2s^2 2p^3$) has a highly stable exactly half-filled p-orbital. O ($1s^2 2s^2 2p^4$) wants to lose an electron to achieve that half-filled stability.

Practice Problems 4 — Radius & Ionization Enthalpy Q1 (JEE Main Focus): Arrange the following in decreasing order of their size: $\text{O}^{2-}, \text{F}^-, \text{Na}^+, \text{Mg}^{2+}$.

Solution:
All these are isoelectronic species (10 electrons each).
Their nuclear charges ($Z$) are: $\text{O}=8, \text{F}=9, \text{Na}=11, \text{Mg}=12$.
As nuclear charge increases, the electrons are pulled closer to the nucleus, decreasing the size.
Decreasing order: $\text{O}^{2-} > \text{F}^- > \text{Na}^+ > \text{Mg}^{2+}$.

Q2 (NCERT Focus): The first ionization enthalpy of Sodium ($\text{Na}$) is lower than that of Magnesium ($\text{Mg}$), but its second ionization enthalpy is higher. Explain.

Solution:
- 1st IE: $\text{Na}$ ($3s^1$) loses an electron easily to achieve a stable noble gas core ($[\text{Ne}]$). $\text{Mg}$ ($3s^2$) has a stable fully-filled $3s$ subshell, requiring more energy. Thus, $\text{Mg} > \text{Na}$.
- 2nd IE: Removing a second electron from $\text{Na}^+$ involves breaking a highly stable noble gas configuration ($2p^6$). Removing a second electron from $\text{Mg}^+$ ($3s^1$) is easier. Thus, $\text{Na}^+ > \text{Mg}^+$.

D. Electron Gain Enthalpy ($\Delta_{eg} H$)

The enthalpy change accompanying the addition of an electron to a neutral gaseous atom. It can be negative (energy released) or positive (energy required).

Across a Period: Becomes more negative (releases more energy) as we move toward halogens.
Down a Group: Becomes less negative.

Crucial Exceptions 1. F vs Cl: Chlorine has a more negative electron gain enthalpy than Fluorine.
Reason: F is exceptionally small, leading to high inter-electronic repulsions in its compact $2p$ subshell. Incoming electrons face resistance. Cl has a larger $3p$ orbital, accommodating the electron easily.

2. O vs S: Similarly, Sulfur has a more negative $\Delta_{eg} H$ than Oxygen.

E. Electronegativity (EN)

The qualitative ability of an atom in a chemical compound to attract shared electrons to itself. Measured on Pauling scale (Fluorine = 4.0).

Across a Period: Increases (size decreases, $Z_{eff}$ increases).
Down a Group: Decreases (size increases).

Practice Problems 5 — Electron Gain & Electronegativity Q1 (NCERT Exercise 3.19): Which element has the most negative electron gain enthalpy, and which has the highest electronegativity in the periodic table?

Solution:
- Most negative electron gain enthalpy: Chlorine ($\text{Cl}$). (Fluorine has a less negative value due to strong electron-electron repulsion in its small $2p$ subshell).
- Highest electronegativity: Fluorine ($\text{F}$).

Q2 (NEET Focus): Would you expect the first electron gain enthalpy of $\text{O}$ to be more negative or less negative than that of $\text{S}$? Justify.

Solution:
The first electron gain enthalpy of Oxygen is less negative than that of Sulfur.
Because of the small size of the oxygen atom, adding an electron to the compact $2p$ subshell causes significant electron-electron repulsion. Sulfur has a larger $3p$ orbital, which accommodates the incoming electron more easily, releasing more energy.

6. Periodic Trends in Chemical Properties

Periodicity of Valence or Oxidation States

The valence of representative elements is usually (though not necessarily) equal to the number of electrons in the outermost orbitals and/or equal to eight minus the number of outermost electrons.

Across a period: The number of valence electrons increases from 1 to 8. The valence of the elements with respect to hydrogen and oxygen first increases from 1 to 4 and then decreases to zero.
Down a group: The number of valence electrons remains the same, hence elements in a group exhibit similar valence/oxidation states. For example, all Group 1 elements show an oxidation state of +1.

However, transition elements (d-block) and inner transition elements (f-block) exhibit variable valencies due to the involvement of (n-1)d and (n-2)f electrons.

Anomalous Properties of Second Period Elements

The first element of each group in the s- and p-blocks (Li, Be, B, C, N, O, F) differs significantly from the rest of their respective groups due to:

Exceptionally small size.
High electronegativity and high ionization enthalpy.
Absence of vacant d-orbitals in their valence shell (limiting their maximum covalency to 4).

Diagonal Relationship

Certain elements of the second period show similarities with elements of the third period positioned diagonally to them (e.g., Li-Mg, Be-Al, B-Si). This is due to similar polarizing power (Ionic charge / Ionic radius squared).

Chemical Reactivity

Metals (Left side): Low IE, tend to lose electrons, form Basic Oxides (e.g., $\text{Na}_2\text{O}$).
Non-metals (Right side): High EN, tend to gain electrons, form Acidic Oxides (e.g., $\text{Cl}_2\text{O}_7$, $\text{SO}_3$).
Middle Elements: Form amphoteric oxides (e.g., $\text{Al}_2\text{O}_3$, $\text{ZnO}$) or neutral oxides (e.g., $\text{CO}, \text{NO}, \text{N}_2\text{O}$).

Practice Problems 6 — Chemical Reactivity Q1 (School Focus): Predict whether the following oxides are acidic, basic, or amphoteric: $\text{Na}_2\text{O}, \text{Al}_2\text{O}_3, \text{Cl}_2\text{O}_7$.

Solution:
- $\text{Na}_2\text{O}$: Metallic oxide on the extreme left. It is highly basic.
- $\text{Al}_2\text{O}_3$: Oxide of an element in the middle of the period. It acts as both acid and base, so it is amphoteric.
- $\text{Cl}_2\text{O}_7$: Non-metallic oxide on the extreme right. It is highly acidic.

Q2 (NCERT Focus): Why do elements in the same group show similar chemical properties?

Solution:
Elements in the same group have the same number of valence electrons and identical outer electronic configurations. Since chemical properties are determined by valence electrons, elements in a group behave similarly.