Chapter 3: Classification of Elements and Periodicity in Properties

Periodic Table Importance

Significance of Periodic Table

• The periodic table is the foundational concept in chemistry, essential for understanding element behavior.
• It organizes vast chemical information into logical, simple patterns.
• Elements are not random; they show trends and group into families with similar properties.

Solved Examples

• Ex 1: Identify the importance of periodic table in predicting element properties.
  Step 1: Recognize element groups.
  Step 2: Use group trends to predict chemical behavior.

Practice Set

• Q1: What does the periodic table primarily organize?
  A. Atomic masses
  B. Elements by properties
  C. Compounds
  D. Isotopes
• Q2: Which statement about elements in the periodic table is true?
  A. Elements are randomly placed
  B. Elements in a group have similar properties
  C. Elements in a period have identical properties
  D. Elements are arranged by color
• Q3: Why is the periodic table considered a great scientific achievement?
  A. It lists all compounds
  B. It groups elements by similar chemical behavior
  C. It shows atomic numbers only
  D. It arranges elements alphabetically

Answer Key

• Q1: B - The periodic table organizes elements based on their chemical properties.
• Q2: B - Elements in the same group share similar chemical properties.
• Q3: B - It groups elements to reveal chemical behavior patterns.

Memory Aids

• Remember: "Groups = Families" with similar properties.
• Periodic table = "Chemical behavior map".

Need for Classification

Why Classify Elements?

• Elements are basic units of matter; their number increased from 31 (1800) to 118 (present).
• Studying each element and its compounds individually is impractical.
• Classification organizes elements systematically by similar properties.
• This organization leads to periodicity and the creation of the periodic table.
• Periodic table groups elements with similar properties together for easy study.

Solved Examples

• Ex 1: Explain why classification of elements is necessary.
  Step 1: Note the increasing number of elements.
  Step 2: Understand difficulty in individual study.
  Step 3: Use classification to group similar elements.

Practice Set

• Q1: What was the number of known elements around 1800?
  A. 31
  B. 63
  C. 118
  D. 50
• Q2: What is the main purpose of classifying elements?
  A. To memorize elements
  B. To organize elements by properties
  C. To increase element count
  D. To separate metals and non-metals only
• Q3: What does classification of elements lead to?
  A. Random grouping
  B. Chemical periodicity
  C. Element isolation
  D. Atomic mass increase

Answer Key

• Q1: A - 31 elements were known in 1800.
• Q2: B - Classification organizes elements by similar properties.
• Q3: B - Classification results in chemical periodicity.

Memory Aids

• Remember: "Classify to simplify" study of elements.
• Periodic table = "Organized element family tree".

Genesis of Periodic Classification

Dobereiner’s Triads (1829)

• Elements grouped in triads with similar properties.
• Atomic weight of middle element ≈ arithmetic mean of other two.
• Example: Cl (35.5), Br (80), I (127); Br ≈ (35.5 + 127)/2 = 81.25

Newland’s Octaves Law (1865)

• Elements arranged by increasing atomic weight.
• Every 8th element has similar properties, like musical notes.
• Example: Li, Na, K show similar properties.

Lothar Meyer’s Atomic Volume Curve (1869)

• Graph plotted: Atomic volume vs atomic weight.
• Elements with similar properties occupy similar graph positions.
• Alkali metals at peaks; alkaline earth metals mid-points; halogens on ascending curve.

Mendeleev’s Periodic Law (1869)

• Properties of elements are periodic functions of atomic weights.
• Elements arranged in increasing atomic weight; similar properties recur periodically.
• Table: 9 vertical columns (groups), 7 horizontal rows (periods).
• Elements with similar properties placed in same group.
• Atomic weight order adjusted to maintain chemical properties.
• Vacant spaces left for undiscovered elements.

Demerits of Mendeleev’s Periodic Table

• Hydrogen’s position anomalous; resembles group 1 and group 17 but placed in group 1.
• Cobalt (58.9) placed before nickel (58.6) despite higher atomic mass.
• Isotopes not given separate places.
• Could not explain cause of periodicity.
• Lanthanoids and actinoids excluded.

Solved Examples

Ex 1

• Find the atomic weight of middle element in triad: Li (7), K (39).
• Calculation: (7 + 39)/2 = 23.
• Middle element: Na (23), confirming Dobereiner’s triad.

Practice Set

• Q1: In Dobereiner’s triads, the atomic weight of the middle element is:
  • A) Geometric mean of other two
  • B) Arithmetic mean of other two
  • C) Harmonic mean of other two
  • D) Median of other two
• Q2: According to Newland’s octaves, elements show similar properties after every:
  • A) 6 elements
  • B) 7 elements
  • C) 8 elements
  • D) 9 elements
• Q3: Which is a limitation of Mendeleev’s periodic table?
  • A) Inclusion of isotopes
  • B) Position of hydrogen
  • C) Use of atomic number
  • D) Explanation of periodicity

Answer Key

• Q1: B - Middle element’s atomic weight is arithmetic mean of other two.
• Q2: C - Every 8th element has similar properties.
• Q3: B - Hydrogen’s position was anomalous in Mendeleev’s table.

Memory Aids

• Dobereiner’s triads: Middle atomic weight = average of outer two.
• Newland’s octaves: Every 8th element repeats properties like musical notes.
• Mendeleev’s table: Left gaps for undiscovered elements; adjusted atomic weight order.

Modern Periodic Law

Explanation

• Henry Moseley (1913) showed atomic number (Z) is fundamental, not atomic mass.
• Plot of √ν (X-ray frequency) vs atomic number is linear.
• Modern periodic law: Properties of elements are periodic functions of atomic numbers.
• Elements arranged by increasing atomic number show repeating properties at regular intervals.

Solved Examples

• Ex 1: Moseley’s experiment showed atomic number is more fundamental than atomic mass by plotting √ν vs Z, resulting in a straight line.

Practice Set

• Q1: What did Moseley’s X-ray spectra study prove?
  • A. Atomic mass is fundamental
  • B. Atomic number is fundamental
  • C. Atomic volume is fundamental
  • D. Atomic radius is fundamental
• Q2: Modern periodic law states properties are periodic functions of?
  • A. Atomic mass
  • B. Atomic number
  • C. Atomic volume
  • D. Atomic radius
• Q3: Which plot gave a straight line in Moseley’s experiment?
  • A. √ν vs atomic mass
  • B. √ν vs atomic number
  • C. Frequency vs atomic mass
  • D. Frequency vs atomic number

Answer Key

• Q1: B - Moseley proved atomic number is fundamental, not atomic mass.
• Q2: B - Modern periodic law is based on atomic number.
• Q3: B - Plot of √ν vs atomic number was linear.

Memory Aids

• "Moseley’s Law": Atomic number defines element properties.
• Modern periodic law = Properties repeat with atomic number, not mass.

Explanation

• Henry Moseley (1913) showed atomic number (Z) is fundamental, not atomic mass.
• Plot of \( \sqrt{\nu} \) (X-ray frequency) vs atomic number is linear.
• Modern periodic law: Properties of elements are periodic functions of atomic numbers.
• Elements arranged by increasing atomic number show repeating properties at regular intervals.

Solved Examples

• Ex 1: Moseley’s experiment showed atomic number is more fundamental than atomic mass by plotting \( \sqrt{\nu} \) vs Z, resulting in a straight line.

Practice Set

• Q1: What did Moseley’s X-ray spectra study prove?
  • A. Atomic mass is fundamental
  • B. Atomic number is fundamental
  • C. Atomic volume is fundamental
  • D. Atomic radius is fundamental
• Q2: Modern periodic law states properties are periodic functions of?
  • A. Atomic mass
  • B. Atomic number
  • C. Atomic volume
  • D. Atomic radius
• Q3: Which plot gave a straight line in Moseley’s experiment?
  • A. \( \sqrt{\nu} \) vs atomic mass
  • B. \( \sqrt{\nu} \) vs atomic number
  • C. Frequency vs atomic mass
  • D. Frequency vs atomic number

Answer Key

• Q1: B - Moseley proved atomic number is fundamental, not atomic mass.
• Q2: B - Modern periodic law is based on atomic number.
• Q3: B - Plot of \( \sqrt{\nu} \) vs atomic number was linear.

Memory Aids

• "Moseley’s Law": Atomic number defines element properties.
• Modern periodic law = Properties repeat with atomic number, not mass.

Periods and Groups

Explanation

• Periodic table has 7 horizontal rows called periods.
• There are 18 vertical columns called groups.
• Period number = highest principal quantum number (n) of elements in that period.
• Elements in a group have similar chemical properties.

Solved Examples

• Ex 1: Period 3 elements have highest principal quantum number n = 3.

Practice Set

• Q1: How many periods are in the long form periodic table?
  • A. 5
  • B. 6
  • C. 7
  • D. 8
• Q2: What does the period number represent?
  • A. Number of valence electrons
  • B. Highest principal quantum number (n)
  • C. Number of groups
  • D. Atomic number
• Q3: How many groups are there in the long form periodic table?
  • A. 16
  • B. 17
  • C. 18
  • D. 19

Answer Key

• Q1: C - There are 7 periods in the long form table.
• Q2: B - Period number equals highest principal quantum number.
• Q3: C - There are 18 groups in the long form table.

Memory Aids

• Periods = Rows, Groups = Columns.
• Period number = highest n (shell) filled.
• Groups have similar properties.

Periodic Table Periods

Explanation

• Periods correspond to electron filling of orbitals.
• Period 1: 2 elements, fills 1s orbital.
• Period 2: 8 elements, fills 2s and 2p orbitals.
• Period 3: 8 elements, fills 3s and 3p orbitals.
• Period 4 & 5: 18 elements, d orbitals start filling.
• Period 6 & 7: 32 elements, include f orbitals (lanthanoids and actinoids).
• Length of periods increases due to more orbitals involved.
• Lanthanides and actinides placed separately at bottom.

Solved Examples

• Ex 1: Period 4 has 18 elements because 4s, 3d, and 4p orbitals fill.

Practice Set

• Q1: How many elements are in period 2?
  • A. 2
  • B. 8
  • C. 18
  • D. 32
• Q2: Which orbitals fill in period 3?
  • A. 1s
  • B. 2s, 2p
  • C. 3s, 3p
  • D. 4s, 3d
• Q3: Why do periods increase in length from period 4 onwards?
  • A. More s orbitals fill
  • B. d and f orbitals start filling
  • C. Less electrons
  • D. Atomic number decreases

Answer Key

• Q1: B - Period 2 has 8 elements filling 2s and 2p orbitals.
• Q2: C - Period 3 fills 3s and 3p orbitals.
• Q3: B - d and f orbitals start filling, increasing period length.

Memory Aids

• Period length = number of orbitals filled.
• Period 1 = 2 elements (1s), Period 2 & 3 = 8 elements (s and p orbitals).
• Period 4 & 5 = 18 elements (d orbitals), Period 6 & 7 = 32 elements (f orbitals).

Classification of Elements

Explanation

• Elements classified as metals, non-metals, and metalloids.
• Zig-zag line separates metals (left/center) and non-metals (right).
• Metals: pink color, include alkali metals, alkaline earth metals, transition metals.
• Non-metals: pink color on right, include noble gases.
• Metalloids: grey color along zig-zag line, have mixed properties.
• Lanthanides and actinides are inner transition elements placed separately at bottom.
• Classification helps predict element properties by position.

Solved Examples

• Ex 1: Alkali metals are in group 1, leftmost column, and are metals.

Practice Set

• Q1: Which side of the periodic table contains metals?
  • A. Right
  • B. Left and center
  • C. Only center
  • D. Bottom
• Q2: Metalloids are located along?
  • A. Left edge
  • B. Right edge
  • C. Zig-zag line
  • D. Bottom rows
• Q3: Which group contains noble gases?
  • A. Group 1
  • B. Group 2
  • C. Group 17
  • D. Group 18

Answer Key

• Q1: B - Metals are on left and center of the table.
• Q2: C - Metalloids lie along the zig-zag line.
• Q3: D - Noble gases are in group 18.

Memory Aids

• Metals = left/center, Non-metals = right, Metalloids = zig-zag line.
• Group 1 = Alkali metals, Group 18 = Noble gases.

Nomenclature of Elements with Atomic Numbers > 100

System Overview

• Elements with atomic number (Z) > 100 are named using the IUPAC systematic method.
• Names are derived from the atomic number using Latin and Greek roots for each digit.
• The suffix -ium is added to the derived name.
• Element symbols are formed from the first letter of each numerical root word.

Solved Examples

• Ex 1: Name element with Z = 102.
  • Digits: 1 (un), 0 (nil), 2 (bi)
  • Name: Unnilbium
  • Symbol: Unb (U from un, n from nil, b from bi)
• Ex 2: Name element with Z = 115.
  • Digits: 1 (un), 1 (un), 5 (pent)
  • Name: Ununpentium
  • Symbol: Uup

Practice Set

• Q1: What is the IUPAC name for element with atomic number 104?
  • A. Unquadium
  • B. Unnilquadium
  • C. Unquadnilium
  • D. Unquadium
• Q2: The symbol for element 113 is:
  • A. Uut
  • B. Uup
  • C. Uus
  • D. Uuo
• Q3: Which suffix is added to the systematic name of elements with Z > 100?
  • A. -ium
  • B. -ine
  • C. -on
  • D. -ide

Answer Key

• Q1: A. Unquadium - Derived from digits 1 (un), 0 (nil), 4 (quad) + ium suffix.
• Q2: A. Uut - Symbol formed from un (U), un (u), tri (t) for 113.
• Q3: A. -ium - Standard suffix for elements with atomic number > 100.

Memory Aids

• Remember digit roots: 0 = nil, 1 = un, 2 = bi, 3 = tri, 4 = quad, 5 = pent, 6 = hex, 7 = sept, 8 = oct, 9 = enn.
• Symbol letters come from first letter of each digit root.
• Always add suffix -ium to complete the element name.

Electronic Configurations and Periodic Table

Electronic Configuration Basics

• Electronic configuration: distribution of electrons in atomic orbitals.
• Element’s position in periodic table reflects quantum numbers of last filled orbital.
• Period number = principal quantum number (n) of valence shell.
• Successive periods fill higher principal energy levels (n=1, 2, 3...).

Solved Examples

• Ex 1: Identify valence shell for period 3 element.
  Solution: Period 3 → n=3, valence shell is 3s and 3p orbitals.
• Ex 2: Write electronic configuration for element in period 2.
  Solution: Period 2 → n=2, fill 2s and 2p orbitals, e.g., Carbon: 1s² 2s² 2p².

Practice Set

• Q1: Which principal quantum number corresponds to period 4?
  A. 2
  B. 3
  C. 4
  D. 5
• Q2: Which orbitals are filled in period 3?
  A. 3s only
  B. 3s and 3p
  C. 3p and 3d
  D. 3d only
• Q3: Element with configuration ending in 4s² 3d⁵ belongs to which period?
  A. 3
  B. 4
  C. 5
  D. 6

Answer Key

• Q1: C. Period number equals principal quantum number n.
• Q2: B. Period 3 fills 3s and 3p orbitals.
• Q3: B. 4s and 3d orbitals filled → period 4.

Memory Aids

• Period number = valence shell principal quantum number (n).
• Orbitals fill in order: s → p → d → f as period increases.

Electronic Configuration Basics

• Electronic configuration: distribution of electrons in atomic orbitals.
• Element’s position in periodic table reflects quantum numbers of last filled orbital.
• Period number = principal quantum number (n) of valence shell.
• Successive periods fill higher principal energy levels (n=1, 2, 3...).

Solved Examples

• Ex 1: Identify valence shell for period 3 element.
  Solution: Period 3 → n=3, valence shell is 3s and 3p orbitals.
• Ex 2: Write electronic configuration for element in period 2.
  Solution: Period 2 → n=2, fill 2s and 2p orbitals, e.g., Carbon: 1s² 2s² 2p².

Practice Set

• Q1: Which principal quantum number corresponds to period 4?
  A. 2
  B. 3
  C. 4
  D. 5
• Q2: Which orbitals are filled in period 3?
  A. 3s only
  B. 3s and 3p
  C. 3p and 3d
  D. 3d only
• Q3: Element with configuration ending in 4s² 3d⁵ belongs to which period?
  A. 3
  B. 4
  C. 5
  D. 6

Answer Key

• Q1: C. Period number equals principal quantum number n.
• Q2: B. Period 3 fills 3s and 3p orbitals.
• Q3: B. 4s and 3d orbitals filled → period 4.

Memory Aids

• Period number = valence shell principal quantum number (n).
• Orbitals fill in order: s → p → d → f as period increases.

Groupwise Electronic Configuration

Group Similarities

• Elements in same group have similar valence shell configurations.
• Same number of electrons in outermost orbitals → similar chemical properties.
• Example: Group 1 (alkali metals) have ns¹ valence configuration.

Solved Examples

• Ex 1: Write valence configuration of sodium (Na).
  Solution: Na atomic number 11 → valence shell 3s¹.
• Ex 2: Identify group of element with valence configuration 4s¹.
  Solution: Group 1 (alkali metals).

Practice Set

• Q1: Which valence configuration corresponds to Group 1?
  A. ns²
  B. ns¹
  C. np⁵
  D. nd¹⁰
• Q2: Lithium (Li) electronic configuration is?
  A. 1s² 2s¹
  B. 1s² 2p¹
  C. 1s² 2s²
  D. 1s² 2p⁶
• Q3: Which element has valence configuration 3s¹?
  A. Sodium (Na)
  B. Magnesium (Mg)
  C. Aluminium (Al)
  D. Neon (Ne)

Answer Key

• Q1: B. Group 1 elements have ns¹ valence configuration.
• Q2: A. Lithium: 1s² 2s¹.
• Q3: A. Sodium has valence 3s¹.

Memory Aids

• Group 1 = ns¹ valence electrons.
• Valence electrons determine group properties.

Periodic Table Groups and Names

Group Names and Properties

• Periodic table columns = groups with similar properties.
• Group 1: Alkali metals (ns¹ valence).
• Group 2: Alkaline earth metals (ns² valence).
• Group 11: Coinage metals.
• Group 15: Pnictogens.
• Group 16: Chalcogens.
• Group 17: Halogens.
• Group 18: Noble gases (aerogens).

Solved Examples

• Ex 1: Name group 17 elements.
  Solution: Halogens.
• Ex 2: Identify group of noble gases.
  Solution: Group 18.

Practice Set

• Q1: Which group is called alkaline earth metals?
  A. Group 1
  B. Group 2
  C. Group 15
  D. Group 18
• Q2: Group 16 elements are known as?
  A. Halogens
  B. Pnictogens
  C. Chalcogens
  D. Noble gases
• Q3: Coinage metals belong to which group?
  A. Group 11
  B. Group 12
  C. Group 13
  D. Group 14

Answer Key

• Q1: B. Group 2 are alkaline earth metals.
• Q2: C. Group 16 are chalcogens.
• Q3: A. Group 11 are coinage metals.

Memory Aids

• Group 1: Alkali metals (ns¹).
• Group 2: Alkaline earth metals (ns²).
• Group 17: Halogens (ns² np⁵).
• Group 18: Noble gases (full valence shell).

Group Names and Properties

• Periodic table columns = groups with similar properties.
• Group 1: Alkali metals (ns¹ valence).
• Group 2: Alkaline earth metals (ns² valence).
• Group 11: Coinage metals.
• Group 15: Pnictogens.
• Group 16: Chalcogens.
• Group 17: Halogens.
• Group 18: Noble gases (aerogens).

Solved Examples

• Ex 1: Name group 17 elements.
  Solution: Halogens.
• Ex 2: Identify group of noble gases.
  Solution: Group 18.

Practice Set

• Q1: Which group is called alkaline earth metals?
  A. Group 1
  B. Group 2
  C. Group 15
  D. Group 18
• Q2: Group 16 elements are known as?
  A. Halogens
  B. Pnictogens
  C. Chalcogens
  D. Noble gases
• Q3: Coinage metals belong to which group?
  A. Group 11
  B. Group 12
  C. Group 13
  D. Group 14

Answer Key

• Q1: B. Group 2 are alkaline earth metals.
• Q2: C. Group 16 are chalcogens.
• Q3: A. Group 11 are coinage metals.

Memory Aids

• Group 1: Alkali metals (ns¹).
• Group 2: Alkaline earth metals (ns²).
• Group 17: Halogens (ns² np⁵).
• Group 18: Noble gases (full valence shell).

Blocks of Elements

Classification by Orbital Type

• Elements classified into s-, p-, d-, and f-blocks based on orbitals being filled.
• s-block: Groups 1 and 2, filling s orbitals.
• p-block: Groups 13 to 18, filling p orbitals.
• d-block: Transition metals, filling d orbitals.
• f-block: Lanthanides and actinides, filling f orbitals.

Solved Examples

• Ex 1: Identify block of element with valence 4s² 3d⁵.
  Solution: d-block (transition metal).
• Ex 2: Which block contains noble gases?
  Solution: p-block.

Practice Set

• Q1: Which block contains alkali metals?
  A. s-block
  B. p-block
  C. d-block
  D. f-block
• Q2: Lanthanides belong to which block?
  A. s-block
  B. p-block
  C. d-block
  D. f-block
• Q3: Elements filling p orbitals are in which groups?
  A. 1-2
  B. 3-12
  C. 13-18
  D. Lanthanides

Answer Key

• Q1: A. Alkali metals are s-block elements.
• Q2: D. Lanthanides are f-block elements.
• Q3: C. p-block elements are in groups 13-18.

Memory Aids

• s-block: Groups 1-2, ns electrons.
• p-block: Groups 13-18, np electrons.
• d-block: Transition metals, (n-1)d electrons.
• f-block: Lanthanides/actinides, (n-2)f electrons.

Classification by Orbital Type

• Elements classified into s-, p-, d-, and f-blocks based on orbitals being filled.
• s-block: Groups 1 and 2, filling s orbitals.
• p-block: Groups 13 to 18, filling p orbitals.
• d-block: Transition metals, filling d orbitals.
• f-block: Lanthanides and actinides, filling f orbitals.

Solved Examples

• Ex 1: Identify block of element with valence 4s² 3d⁵.
  Solution: d-block (transition metal).
• Ex 2: Which block contains noble gases?
  Solution: p-block.

Practice Set

• Q1: Which block contains alkali metals?
  A. s-block
  B. p-block
  C. d-block
  D. f-block
• Q2: Lanthanides belong to which block?
  A. s-block
  B. p-block
  C. d-block
  D. f-block
• Q3: Elements filling p orbitals are in which groups?
  A. 1-2
  B. 3-12
  C. 13-18
  D. Lanthanides

Answer Key

• Q1: A. Alkali metals are s-block elements.
• Q2: D. Lanthanides are f-block elements.
• Q3: C. p-block elements are in groups 13-18.

Memory Aids

• s-block: Groups 1-2, ns electrons.
• p-block: Groups 13-18, np electrons.
• d-block: Transition metals, (n-1)d electrons.
• f-block: Lanthanides/actinides, (n-2)f electrons.

s-Block Elements

Characteristics

• General outer electronic configuration: ns^1-2.
• Soft metals with low melting and boiling points.
• Highly reactive with low ionization enthalpies.
• Form +1 ions (alkali metals) or +2 ions (alkaline earth metals) by losing outermost electrons.
• Highly electropositive and strong reducing agents.
• Metallic character and reactivity increase down the group.
• Never found pure in nature due to high reactivity.
• Compounds are mostly ionic except lithium and beryllium.

Solved Examples

Ex 1: Write the electronic configuration of Calcium (Ca).

• Atomic number = 20
• Configuration: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s²

Practice Set

• Q1: What is the general electronic configuration of s-block elements?
  • A. ns^1-2
  • B. ns² np^1-6
  • C. (n-1)d^1-10 ns^0-2
  • D. (n-2)f^0-14 (n-1)d^0-1 ns²
• Q2: Which ion is formed by alkali metals?
  • A. +2
  • B. +1
  • C. -1
  • D. 0
• Q3: Which of the following is NOT a characteristic of s-block elements?
  • A. Soft metals
  • B. High ionization enthalpy
  • C. Highly electropositive
  • D. Form predominantly ionic compounds

Answer Key

• Q1: A. ns^1-2 - This is the defining electronic configuration of s-block elements.
• Q2: B. +1 - Alkali metals lose one electron to form +1 ions.
• Q3: B. High ionization enthalpy - s-block elements have low ionization enthalpies.

Memory Aids

• s-block = Groups 1 & 2, valence electrons in s-orbital.
• Alkali metals form +1 ions; alkaline earth metals form +2 ions.
• Soft metals, highly reactive, never found free.

Characteristics

• General outer electronic configuration: ns^1-2.
• Soft metals with low melting and boiling points.
• Highly reactive with low ionization enthalpies.
• Form +1 ions (alkali metals) or +2 ions (alkaline earth metals) by losing outermost electrons.
• Highly electropositive and strong reducing agents.
• Metallic character and reactivity increase down the group.
• Never found pure in nature due to high reactivity.
• Compounds are mostly ionic except lithium and beryllium.

Solved Examples

Ex 1: Write the electronic configuration of Calcium (Ca).

• Atomic number = 20
• Configuration: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s²

Practice Set

• Q1: What is the general electronic configuration of s-block elements?
  • A. ns^1-2
  • B. ns² np^1-6
  • C. (n-1)d^1-10 ns^0-2
  • D. (n-2)f^0-14 (n-1)d^0-1 ns²
• Q2: Which ion is formed by alkali metals?
  • A. +2
  • B. +1
  • C. -1
  • D. 0
• Q3: Which of the following is NOT a characteristic of s-block elements?
  • A. Soft metals
  • B. High ionization enthalpy
  • C. Highly electropositive
  • D. Form predominantly ionic compounds

Answer Key

• Q1: A. ns^1-2 - This is the defining electronic configuration of s-block elements.
• Q2: B. +1 - Alkali metals lose one electron to form +1 ions.
• Q3: B. High ionization enthalpy - s-block elements have low ionization enthalpies.

Memory Aids

• s-block = Groups 1 & 2, valence electrons in s-orbital.
• Alkali metals form +1 ions; alkaline earth metals form +2 ions.
• Soft metals, highly reactive, never found free.

p-Block Elements

Characteristics

• General outer electronic configuration: ns² np^1-6.
• Includes non-metals, poor metals, and metalloids; non-metals predominate.
• Exhibit covalency, electrovalency, and coordinate valency.
• Show allotropy (e.g., carbon, silicon, phosphorus, sulfur).
• Noble gases (ns² np⁶) are inert due to full valence shell.
• Halogens (Group 17) and chalcogens (Group 16) have high negative electron gain enthalpies and gain electrons to achieve noble gas configuration.
• Non-metallic character increases left to right; metallic character increases down the group.
• s- and p-block elements are called representative or main group elements.

Solved Examples

Ex 1: Write the electronic configuration of Chlorine (Cl).

• Atomic number = 17
• Configuration: 1s² 2s² 2p⁶ 3s² 3p⁵

Practice Set

• Q1: What is the general electronic configuration of p-block elements?
  • A. ns^1-2
  • B. ns² np^1-6
  • C. (n-1)d^1-10 ns^0-2
  • D. (n-2)f^0-14 (n-1)d^0-1 ns²
• Q2: Which group contains the halogens?
  • A. Group 16
  • B. Group 17
  • C. Group 18
  • D. Group 15
• Q3: Why are noble gases inert?
  • A. They have empty valence shells
  • B. They have completely filled valence shells (ns² np⁶)
  • C. They have low ionization energy
  • D. They are metals

Answer Key

• Q1: B. ns² np^1-6 - Defines p-block elements.
• Q2: B. Group 17 - Halogens belong to group 17.
• Q3: B. Completely filled valence shells - Noble gases are inert due to stable electronic configuration.

Memory Aids

• p-block = Groups 13-18, valence electrons in p-orbitals.
• Halogens (Group 17) gain 1 electron; chalcogens (Group 16) gain 2 electrons.
• Noble gases are inert due to ns² np⁶ configuration.

Characteristics

• General outer electronic configuration: ns² np^1-6.
• Includes non-metals, poor metals, and metalloids; non-metals predominate.
• Exhibit covalency, electrovalency, and coordinate valency.
• Show allotropy (e.g., carbon, silicon, phosphorus, sulfur).
• Noble gases (ns² np⁶) are inert due to full valence shell.
• Halogens (Group 17) and chalcogens (Group 16) have high negative electron gain enthalpies and gain electrons to achieve noble gas configuration.
• Non-metallic character increases left to right; metallic character increases down the group.
• s- and p-block elements are called representative or main group elements.

Solved Examples

Ex 1: Write the electronic configuration of Chlorine (Cl).

• Atomic number = 17
• Configuration: 1s² 2s² 2p⁶ 3s² 3p⁵

Practice Set

• Q1: What is the general electronic configuration of p-block elements?
  • A. ns^1-2
  • B. ns² np^1-6
  • C. (n-1)d^1-10 ns^0-2
  • D. (n-2)f^0-14 (n-1)d^0-1 ns²
• Q2: Which group contains the halogens?
  • A. Group 16
  • B. Group 17
  • C. Group 18
  • D. Group 15
• Q3: Why are noble gases inert?
  • A. They have empty valence shells
  • B. They have completely filled valence shells (ns² np⁶)
  • C. They have low ionization energy
  • D. They are metals

Answer Key

• Q1: B. ns² np^1-6 - Defines p-block elements.
• Q2: B. Group 17 - Halogens belong to group 17.
• Q3: B. Completely filled valence shells - Noble gases are inert due to stable electronic configuration.

Memory Aids

• p-block = Groups 13-18, valence electrons in p-orbitals.
• Halogens (Group 17) gain 1 electron; chalcogens (Group 16) gain 2 electrons.
• Noble gases are inert due to ns² np⁶ configuration.

d-Block Elements

Characteristics

• General outer electronic configuration: (n-1)d^1-10 ns^0-2.
• Elements of groups 3 to 12; called transition elements.
• Classified into 3d, 4d, 5d, and 6d series.
• High melting and boiling points; metals.
• Form coloured salts.
• Paramagnetic due to unpaired electrons.
• Form both ionic and covalent compounds.
• Good catalysts.
• Show variable oxidation states (e.g., Fe²⁺, Fe³⁺, Cu⁺, Cu²⁺).
• Zn, Cd, Hg (d¹⁰ s²) do not show typical transition properties.
• Properties intermediate between s- and p-block elements.

Solved Examples

Ex 1: Write the electronic configuration of Iron (Fe).

• Atomic number = 26
• Configuration: [Ar] 3d⁶ 4s²

Practice Set

• Q1: What is the general electronic configuration of d-block elements?
  • A. ns^1-2
  • B. ns² np^1-6
  • C. (n-1)d^1-10 ns^0-2
  • D. (n-2)f^0-14 (n-1)d^0-1 ns²
• Q2: Which of the following is NOT a property of d-block elements?
  • A. Form coloured salts
  • B. Show variable oxidation states
  • C. Are non-metals
  • D. Are good catalysts
• Q3: Which elements do NOT show typical transition element properties?
  • A. Fe, Cu
  • B. Zn, Cd, Hg
  • C. Sc, Ti
  • D. Mn, Co

Answer Key

• Q1: C. (n-1)d^1-10 ns^0-2 - Defines d-block elements.
• Q2: C. Are non-metals - d-block elements are metals.
• Q3: B. Zn, Cd, Hg - These have d¹⁰ s² configuration and lack typical transition properties.

Memory Aids

• d-block = Groups 3-12, valence electrons in d-orbitals.
• Variable oxidation states and coloured compounds are key features.
• Zn, Cd, Hg are exceptions with filled d-orbitals.

Characteristics

• General outer electronic configuration: (n-1)d^1-10 ns^0-2.
• Elements of groups 3 to 12; called transition elements.
• Classified into 3d, 4d, 5d, and 6d series.
• High melting and boiling points; metals.
• Form coloured salts.
• Paramagnetic due to unpaired electrons.
• Form both ionic and covalent compounds.
• Good catalysts.
• Show variable oxidation states (e.g., Fe²⁺, Fe³⁺, Cu⁺, Cu²⁺).
• Zn, Cd, Hg (d¹⁰ s²) do not show typical transition properties.
• Properties intermediate between s- and p-block elements.

Solved Examples

Ex 1: Write the electronic configuration of Iron (Fe).

• Atomic number = 26
• Configuration: [Ar] 3d⁶ 4s²

Practice Set

• Q1: What is the general electronic configuration of d-block elements?
  • A. ns^1-2
  • B. ns² np^1-6
  • C. (n-1)d^1-10 ns^0-2
  • D. (n-2)f^0-14 (n-1)d^0-1 ns²
• Q2: Which of the following is NOT a property of d-block elements?
  • A. Form coloured salts
  • B. Show variable oxidation states
  • C. Are non-metals
  • D. Are good catalysts
• Q3: Which elements do NOT show typical transition element properties?
  • A. Fe, Cu
  • B. Zn, Cd, Hg
  • C. Sc, Ti
  • D. Mn, Co

Answer Key

• Q1: C. (n-1)d^1-10 ns^0-2 - Defines d-block elements.
• Q2: C. Are non-metals - d-block elements are metals.
• Q3: B. Zn, Cd, Hg - These have d¹⁰ s² configuration and lack typical transition properties.

Memory Aids

• d-block = Groups 3-12, valence electrons in d-orbitals.
• Variable oxidation states and coloured compounds are key features.
• Zn, Cd, Hg are exceptions with filled d-orbitals.

f-Block Elements

Characteristics

• General outer electronic configuration: (n-2)f^0-14 (n-1)d^0-1 ns².
• Called inner-transition elements; divided into Lanthanoids and Actinoids (14 elements each).
• Heavy metals with high melting and boiling points.
• Exhibit variable oxidation states; commonly +3.
• Form coloured salts due to light absorption.
• All actinoids are radioactive.

Solved Examples

Ex 1: Identify the series to which Uranium (U) belongs.

• Atomic number = 92
• Belongs to Actinoids series of f-block elements.

Practice Set

• Q1: What is the general electronic configuration of f-block elements?
  • A. ns^1-2
  • B. ns² np^1-6
  • C. (n-1)d^1-10 ns^0-2
  • D. (n-2)f^0-14 (n-1)d^0-1 ns²
• Q2: Which series are f-block elements divided into?
  • A. Lanthanoids and Actinoids
  • B. Alkali and Alkaline earth metals
  • C. Halogens and Noble gases
  • D. Transition and Representative elements
• Q3: What is a common oxidation state of f-block elements?
  • A. +1
  • B. +2
  • C. +3
  • D. +4

Answer Key

• Q1: D. (n-2)f^0-14 (n-1)d^0-1 ns² - Defines f-block elements.
• Q2: A. Lanthanoids and Actinoids - The two series of f-block elements.
• Q3: C. +3 - Most common oxidation state in f-block.

Memory Aids

• f-block = inner transition metals, valence electrons in f-orbitals.
• Lanthanoids and Actinoids each have 14 elements.
• Actinoids are radioactive; common oxidation state +3.

Characteristics

• General outer electronic configuration: (n-2)f^0-14 (n-1)d^0-1 ns².
• Called inner-transition elements; divided into Lanthanoids and Actinoids (14 elements each).
• Heavy metals with high melting and boiling points.
• Exhibit variable oxidation states; commonly +3.
• Form coloured salts due to light absorption.
• All actinoids are radioactive.

Solved Examples

Ex 1: Identify the series to which Uranium (U) belongs.

• Atomic number = 92
• Belongs to Actinoids series of f-block elements.

Practice Set

• Q1: What is the general electronic configuration of f-block elements?
  • A. ns^1-2
  • B. ns² np^1-6
  • C. (n-1)d^1-10 ns^0-2
  • D. (n-2)f^0-14 (n-1)d^0-1 ns²
• Q2: Which series are f-block elements divided into?
  • A. Lanthanoids and Actinoids
  • B. Alkali and Alkaline earth metals
  • C. Halogens and Noble gases
  • D. Transition and Representative elements
• Q3: What is a common oxidation state of f-block elements?
  • A. +1
  • B. +2
  • C. +3
  • D. +4

Answer Key

• Q1: D. (n-2)f^0-14 (n-1)d^0-1 ns² - Defines f-block elements.
• Q2: A. Lanthanoids and Actinoids - The two series of f-block elements.
• Q3: C. +3 - Most common oxidation state in f-block.

Memory Aids

• f-block = inner transition metals, valence electrons in f-orbitals.
• Lanthanoids and Actinoids each have 14 elements.
• Actinoids are radioactive; common oxidation state +3.

Metals, Non-metals and Metalloids

Metals

• More than 78% of elements are metals, located on the left side of the periodic table.
• Usually solids at room temperature (exceptions: mercury liquid; gallium and caesium low melting points).
• High melting and boiling points.
• Good conductors of heat and electricity.
• Malleable and ductile.

Non-metals

• Located on the top right side of the periodic table.
• Usually solids or gases at room temperature with low melting and boiling points (exceptions: boron, carbon).
• Poor conductors of heat and electricity.
• Most non-metallic solids are brittle, not malleable or ductile.

Metalloids or Semi-metals

• Elements like silicon, germanium, arsenic, antimony, and tellurium show properties of both metals and non-metals.
• Called semi-metals or metalloids.

Periodicity

Concept

• Periodicity: Recurrence of similar properties after definite intervals in increasing atomic number order.
• Caused by repetition of similar valence shell electronic configurations at intervals: 2, 8, 8, 18, 18, 32.
• Observed in properties linked to electronic configuration.

Solved Examples

• Ex 1: Identify periodicity cause in elements: Valence shell electronic configuration repeats every 2, 8, 8, 18, 18, 32 elements.

Practice Set

• Q1: What causes periodicity in elements?
  • A. Atomic mass variation
  • B. Repetition of valence electronic configuration
  • C. Nuclear charge variation
  • D. Isotopic abundance
• Q2: Periodicity intervals in electronic configuration are?
  • A. 2, 8, 8, 18, 18, 32
  • B. 2, 8, 18, 18, 32, 32
  • C. 2, 8, 8, 16, 18, 32
  • D. 2, 8, 8, 18, 16, 32
• Q3: Periodicity is observed in properties linked to?
  • A. Atomic mass
  • B. Electronic configuration
  • C. Isotopes
  • D. Atomic volume

Answer Key

• Q1: B - Periodicity arises due to repetition of valence electronic configuration.
• Q2: A - Correct intervals are 2, 8, 8, 18, 18, 32.
• Q3: B - Properties linked to electronic configuration show periodicity.

Memory Aids

• Periodic intervals: 2, 8, 8, 18, 18, 32 (electron shell capacities).
• Periodicity = repetition of valence shell configuration.

Concept

• Periodicity: Recurrence of similar properties after definite intervals in increasing atomic number order.
• Caused by repetition of similar valence shell electronic configurations at intervals: 2, 8, 8, 18, 18, 32.
• Observed in properties linked to electronic configuration.

Solved Examples

• Ex 1: Identify periodicity cause in elements: Valence shell electronic configuration repeats every 2, 8, 8, 18, 18, 32 elements.

Practice Set

• Q1: What causes periodicity in elements?
  • A. Atomic mass variation
  • B. Repetition of valence electronic configuration
  • C. Nuclear charge variation
  • D. Isotopic abundance
• Q2: Periodicity intervals in electronic configuration are?
  • A. 2, 8, 8, 18, 18, 32
  • B. 2, 8, 18, 18, 32, 32
  • C. 2, 8, 8, 16, 18, 32
  • D. 2, 8, 8, 18, 16, 32
• Q3: Periodicity is observed in properties linked to?
  • A. Atomic mass
  • B. Electronic configuration
  • C. Isotopes
  • D. Atomic volume

Answer Key

• Q1: B - Periodicity arises due to repetition of valence electronic configuration.
• Q2: A - Correct intervals are 2, 8, 8, 18, 18, 32.
• Q3: B - Properties linked to electronic configuration show periodicity.

Memory Aids

• Periodic intervals: 2, 8, 8, 18, 18, 32 (electron shell capacities).
• Periodicity = repetition of valence shell configuration.

Atomic Radius

Types and Trends

• Atomic radius decreases across a period (left to right).
• Atomic radius increases down a group (top to bottom).
• Types of atomic radius:
  • Metallic radius: Half distance between nearest atoms in metallic lattice (metals).
  • Covalent radius: Half distance between nuclei of two atoms joined by single covalent bond (non-metals).
  • van der Waals radius: Half distance between adjacent atoms of neighboring molecules.
• van der Waals radius > metallic radius > covalent radius.
• Noble gases have van der Waals radii as they do not form molecules.

Solved Examples

• Ex 1: Compare radii types for chlorine: Covalent radius < Metallic radius < van der Waals radius.

Practice Set

• Q1: Atomic radius trend across a period is?
  • A. Increases
  • B. Decreases
  • C. Remains constant
  • D. Random
• Q2: Which radius is largest for an atom?
  • A. Covalent radius
  • B. Metallic radius
  • C. van der Waals radius
  • D. Ionic radius
• Q3: Atomic radius increases when moving?
  • A. Left to right in a period
  • B. Top to bottom in a group
  • C. Right to left in a period
  • D. Bottom to top in a group

Answer Key

• Q1: B - Atomic radius decreases across a period due to increasing nuclear charge.
• Q2: C - van der Waals radius is largest.
• Q3: B - Atomic radius increases down a group due to added shells.

Memory Aids

• Radius order: van der Waals > metallic > covalent.
• Across period: radius decreases; down group: radius increases.

Types and Trends

• Atomic radius decreases across a period (left to right).
• Atomic radius increases down a group (top to bottom).
• Types of atomic radius:
  • Metallic radius: Half distance between nearest atoms in metallic lattice (metals).
  • Covalent radius: Half distance between nuclei of two atoms joined by single covalent bond (non-metals).
  • van der Waals radius: Half distance between adjacent atoms of neighboring molecules.
• van der Waals radius > metallic radius > covalent radius.
• Noble gases have van der Waals radii as they do not form molecules.

Solved Examples

• Ex 1: Compare radii types for chlorine: Covalent radius < Metallic radius < van der Waals radius.

Practice Set

• Q1: Atomic radius trend across a period is?
  • A. Increases
  • B. Decreases
  • C. Remains constant
  • D. Random
• Q2: Which radius is largest for an atom?
  • A. Covalent radius
  • B. Metallic radius
  • C. van der Waals radius
  • D. Ionic radius
• Q3: Atomic radius increases when moving?
  • A. Left to right in a period
  • B. Top to bottom in a group
  • C. Right to left in a period
  • D. Bottom to top in a group

Answer Key

• Q1: B - Atomic radius decreases across a period due to increasing nuclear charge.
• Q2: C - van der Waals radius is largest.
• Q3: B - Atomic radius increases down a group due to added shells.

Memory Aids

• Radius order: van der Waals > metallic > covalent.
• Across period: radius decreases; down group: radius increases.

Ionic Radius

Definition and Trends

• Ionic radius: Distance from nucleus to outermost electron shell in an ion.
• Follows same trend as atomic radius.
• Cations are smaller than parent atoms (fewer electrons, same nuclear charge).
• More electrons removed → smaller cation (e.g., Mg > Mg⁺ > Mg²⁺).
• Anions are larger than parent atoms (more electrons, same nuclear charge).
• Size explained by Z/e ratio (Nuclear charge / number of electrons): higher ratio → smaller size.
• Isoelectronic species: Same electron number, different nuclear charge; size decreases with increasing nuclear charge.

Solved Examples

• Ex 1: Arrange isoelectronic ions by size: N^3- > O^2- > F^- > Na⁺ > Mg²⁺ > Al³⁺.

Practice Set

• Q1: Which ion is smallest?
  • A. Mg²⁺
  • B. Na⁺
  • C. F^-
  • D. O^2-
• Q2: Cation size compared to parent atom is?
  • A. Larger
  • B. Smaller
  • C. Same
  • D. Variable
• Q3: Isoelectronic species have?
  • A. Same nuclear charge
  • B. Same number of electrons
  • C. Same size
  • D. Same atomic number

Answer Key

• Q1: A - Mg²⁺ has highest nuclear charge, smallest size.
• Q2: B - Cations are smaller due to electron loss.
• Q3: B - Isoelectronic species have same electron count.

Memory Aids

• Cations smaller than parent atoms; anions larger.
• Size ∝ 1 / (Z/e) ratio.
• Isoelectronic size decreases with increasing nuclear charge.

Definition and Trends

• Ionic radius: Distance from nucleus to outermost electron shell in an ion.
• Follows same trend as atomic radius.
• Cations are smaller than parent atoms (fewer electrons, same nuclear charge).
• More electrons removed → smaller cation (e.g., Mg > Mg⁺ > Mg²⁺).
• Anions are larger than parent atoms (more electrons, same nuclear charge).
• Size explained by Z/e ratio (Nuclear charge / number of electrons): higher ratio → smaller size.
• Isoelectronic species: Same electron number, different nuclear charge; size decreases with increasing nuclear charge.

Solved Examples

• Ex 1: Arrange isoelectronic ions by size: N^3- > O^2- > F^- > Na⁺ > Mg²⁺ > Al³⁺.

Practice Set

• Q1: Which ion is smallest?
  • A. Mg²⁺
  • B. Na⁺
  • C. F^-
  • D. O^2-
• Q2: Cation size compared to parent atom is?
  • A. Larger
  • B. Smaller
  • C. Same
  • D. Variable
• Q3: Isoelectronic species have?
  • A. Same nuclear charge
  • B. Same number of electrons
  • C. Same size
  • D. Same atomic number

Answer Key

• Q1: A - Mg²⁺ has highest nuclear charge, smallest size.
• Q2: B - Cations are smaller due to electron loss.
• Q3: B - Isoelectronic species have same electron count.

Memory Aids

• Cations smaller than parent atoms; anions larger.
• Size ∝ 1 / (Z/e) ratio.
• Isoelectronic size decreases with increasing nuclear charge.

Ionisation Enthalpy

Definition and Factors

• Energy required to remove an electron from a gaseous atom in ground state.
• Reaction: M_g + IE → M⁺_g + e^-
• Successive ionisation energies increase: IE₃ > IE₂ > IE₁.
• Screening effect: Inner electrons repel valence electrons, reducing nuclear attraction.
• Factors affecting IE:
  • Size: IE ∝ 1 / size
  • Nuclear charge: IE ∝ nuclear charge
  • Screening: IE ∝ 1 / screening
  • Stable orbitals (half/full filled) increase IE.
  • Orbital type: IE decreases s > p > d > f.
• IE decreases down group; increases across period.
• Inert gases have exceptionally high IE due to stable configuration.
• Slater’s Rule: Z_eff = Z - σ*; σ* calculated from electron distribution.
• Atomic radius ∝ 1 / Z_eff, IE ∝ Z_eff.

Solved Examples

• Ex 1: Calculate Z_eff for an electron using Slater’s rule.

Practice Set

• Q1: Ionisation enthalpy is the energy to?
  • A. Add an electron
  • B. Remove an electron
  • C. Break a bond
  • D. Form an ion
• Q2: Which factor increases ionisation enthalpy?
  • A. Larger atomic size
  • B. Higher nuclear charge
  • C. More screening
  • D. Filled orbitals
• Q3: Ionisation enthalpy trend down a group is?
  • A. Increases
  • B. Decreases
  • C. Constant
  • D. Random

Answer Key

• Q1: B - IE is energy to remove an electron.
• Q2: B - Higher nuclear charge increases IE.
• Q3: B - IE decreases down group due to size and screening.

Memory Aids

• IE ∝ Z_eff / size.
• Successive IE increases: IE_n+1 > IE_n.
• Stable half/full orbitals raise IE.

Definition and Factors

• Energy required to remove an electron from a gaseous atom in ground state.
• Reaction: M_g + IE → M⁺_g + e^-
• Successive ionisation energies increase: IE₃ > IE₂ > IE₁.
• Screening effect: Inner electrons repel valence electrons, reducing nuclear attraction.
• Factors affecting IE:
  • Size: IE ∝ 1 / size
  • Nuclear charge: IE ∝ nuclear charge
  • Screening: IE ∝ 1 / screening
  • Stable orbitals (half/full filled) increase IE.
  • Orbital type: IE decreases s > p > d > f.
• IE decreases down group; increases across period.
• Inert gases have exceptionally high IE due to stable configuration.
• Slater’s Rule: Z_eff = Z - σ*; σ* calculated from electron distribution.
• Atomic radius ∝ 1 / Z_eff, IE ∝ Z_eff.

Solved Examples

• Ex 1: Calculate Z_eff for an electron using Slater’s rule.

Practice Set

• Q1: Ionisation enthalpy is the energy to?
  • A. Add an electron
  • B. Remove an electron
  • C. Break a bond
  • D. Form an ion
• Q2: Which factor increases ionisation enthalpy?
  • A. Larger atomic size
  • B. Higher nuclear charge
  • C. More screening
  • D. Filled orbitals
• Q3: Ionisation enthalpy trend down a group is?
  • A. Increases
  • B. Decreases
  • C. Constant
  • D. Random

Answer Key

• Q1: B - IE is energy to remove an electron.
• Q2: B - Higher nuclear charge increases IE.
• Q3: B - IE decreases down group due to size and screening.

Memory Aids

• IE ∝ Z_eff / size.
• Successive IE increases: IE_n+1 > IE_n.
• Stable half/full orbitals raise IE.

Electron Gain Enthalpy

Definition and Factors

• Change in enthalpy when an electron is added to a gaseous atom: A_g + e^- → A^-_g.
• Factors affecting electron gain enthalpy (Δ_egH):
  • Size: Δ_egH ∝ 1 / size
  • Nuclear charge: Δ_egH ∝ effective nuclear charge
  • Screening: Δ_egH ∝ 1 / screening
• Δ_egH becomes less negative down group; more negative across period (except noble gases).
• F and O have less negative Δ_egH than Cl and S due to electron-electron repulsion in smaller orbitals.
• Elements with half/full orbitals (Be, Mg, N, inert gases) have positive Δ_egH.

Solved Examples

• Ex 1: Explain why Δ_egH of F is less negative than Cl.

Practice Set

• Q1: Electron gain enthalpy is energy change when?
  • A. Electron is removed
  • B. Electron is added
  • C. Atom ionizes
  • D. Atom forms bond
• Q2: Electron gain enthalpy becomes more negative?
  • A. Down group
  • B. Across period
  • C. Randomly
  • D. For noble gases
• Q3: Why is Δ_egH of O less negative than S?
  • A. Larger size of O
  • B. Electron repulsion in smaller orbitals
  • C. Higher nuclear charge of O
  • D. Screening effect

Answer Key

• Q1: B - Electron gain enthalpy is energy change on adding electron.
• Q2: B - Becomes more negative across period.
• Q3: B - Electron-electron repulsion in smaller n=2 orbitals reduces negativity.

Memory Aids

• Δ_egH ∝ 1/size, ∝ nuclear charge, ∝ 1/screening.
• Less negative down group, more negative across period.
• Repulsion lowers negativity in small orbitals (O, F).

Definition and Factors

• Change in enthalpy when an electron is added to a gaseous atom: A_g + e^- → A^-_g.
• Factors affecting electron gain enthalpy (Δ_egH):
  • Size: Δ_egH ∝ 1 / size
  • Nuclear charge: Δ_egH ∝ effective nuclear charge
  • Screening: Δ_egH ∝ 1 / screening
• Δ_egH becomes less negative down group; more negative across period (except noble gases).
• F and O have less negative Δ_egH than Cl and S due to electron-electron repulsion in smaller orbitals.
• Elements with half/full orbitals (Be, Mg, N, inert gases) have positive Δ_egH.

Solved Examples

• Ex 1: Explain why Δ_egH of F is less negative than Cl.

Practice Set

• Q1: Electron gain enthalpy is energy change when?
  • A. Electron is removed
  • B. Electron is added
  • C. Atom ionizes
  • D. Atom forms bond
• Q2: Electron gain enthalpy becomes more negative?
  • A. Down group
  • B. Across period
  • C. Randomly
  • D. For noble gases
• Q3: Why is Δ_egH of O less negative than S?
  • A. Larger size of O
  • B. Electron repulsion in smaller orbitals
  • C. Higher nuclear charge of O
  • D. Screening effect

Answer Key

• Q1: B - Electron gain enthalpy is energy change on adding electron.
• Q2: B - Becomes more negative across period.
• Q3: B - Electron-electron repulsion in smaller n=2 orbitals reduces negativity.

Memory Aids

• Δ_egH ∝ 1/size, ∝ nuclear charge, ∝ 1/screening.
• Less negative down group, more negative across period.
• Repulsion lowers negativity in small orbitals (O, F).

Electronegativity

Definition and Trends

• Tendency of atom to attract shared electrons in covalent bond.
• Increases across period (left to right).
• Decreases down group (top to bottom).
• Measured by Pauling, Mulliken, and Allred-Rochow scales.
• Bond ionic/covalent character depends on electronegativity difference.

Solved Examples

• Ex 1: Calculate electronegativity difference using Pauling scale from bond energies.

Practice Set

• Q1: Electronegativity trend across period is?
  • A. Decreases
  • B. Increases
  • C. Constant
  • D. Random
• Q2: Which scale uses average of IE and EA?
  • A. Pauling
  • B. Mulliken
  • C. Allred-Rochow
  • D. Born-Haber
• Q3: Bond is mostly ionic if electronegativity difference is?
  • A. 0.5
  • B. 1.0
  • C. 1.9
  • D. >1.9

Answer Key

• Q1: B - Electronegativity increases across period.
• Q2: B - Mulliken scale averages IE and EA.
• Q3: D - Difference >1.9 indicates mostly ionic bond.

Memory Aids

• Electronegativity: Increases across period, decreases down group.
• Pauling scale uses bond energies.
• Mulliken = (IE + EA)/2.

Definition and Trends

• Tendency of atom to attract shared electrons in covalent bond.
• Increases across period (left to right).
• Decreases down group (top to bottom).
• Measured by Pauling, Mulliken, and Allred-Rochow scales.
• Bond ionic/covalent character depends on electronegativity difference.

Solved Examples

• Ex 1: Calculate electronegativity difference using Pauling scale from bond energies.

Practice Set

• Q1: Electronegativity trend across period is?
  • A. Decreases
  • B. Increases
  • C. Constant
  • D. Random
• Q2: Which scale uses average of IE and EA?
  • A. Pauling
  • B. Mulliken
  • C. Allred-Rochow
  • D. Born-Haber
• Q3: Bond is mostly ionic if electronegativity difference is?
  • A. 0.5
  • B. 1.0
  • C. 1.9
  • D. >1.9

Answer Key

• Q1: B - Electronegativity increases across period.
• Q2: B - Mulliken scale averages IE and EA.
• Q3: D - Difference >1.9 indicates mostly ionic bond.

Memory Aids

• Electronegativity: Increases across period, decreases down group.
• Pauling scale uses bond energies.
• Mulliken = (IE + EA)/2.

Chemical Properties

Valence and Oxidation States

• Valence usually equals number of outer electrons or 8 minus outer electrons.
• Oxidation state often used as valence.
• Transition elements and actinoids show variable valencies.
• Valency increases 1 to 4 then decreases 4 to 0 across period.
• Valence electrons same down group; valency same in group.

Anomalous Properties of Second Period

• First elements of groups 1, 2, 13–17 differ from rest (Li, Be, B–F).
• Li and Be show diagonal relationship with Mg and Al.
• Due to small size, high charge/radius ratio, high electronegativity.

Chemical Reactivity Trends

• Reactivity highest at extremes of periodic table, lowest in center.
• Alkali metals reactive due to electron loss; halogens due to electron gain.
• Metallic character decreases left to right; non-metallic increases.
• Metallic character increases down group; non-metallic decreases.
• Oxides: Left extreme form basic oxides (Na₂O), right extreme acidic (Cl₂O₇), center amphoteric/neutral (Al₂O₃, CO).

Solved Examples

• Ex 1: Identify oxide nature of Al₂O₃: Amphoteric.

Practice Set

• Q1: Valency of elements in a group?
  • A. Varies
  • B. Same
  • C. Increases down group
  • D. Decreases down group
• Q2: Which element shows diagonal relationship?
  • A. Na
  • B. Be
  • C. Mg
  • D. Al
• Q3: Oxide of element on extreme right of period is?
  • A. Basic
  • B. Amphoteric
  • C. Acidic
  • D. Neutral

Answer Key

• Q1: B - Valency same down group due to same valence electrons.
• Q2: B - Be shows diagonal relationship with Mg.
• Q3: C - Extreme right oxides are acidic.

Memory Aids

• Valency same in group; varies across period.
• Diagonal relationship: Li-Mg, Be-Al.
• Oxides: Left basic, right acidic, center amphoteric/neutral.

Valence and Oxidation States

• Valence usually equals number of outer electrons or 8 minus outer electrons.
• Oxidation state often used as valence.
• Transition elements and actinoids show variable valencies.
• Valency increases 1 to 4 then decreases 4 to 0 across period.
• Valence electrons same down group; valency same in group.

Anomalous Properties of Second Period

• First elements of groups 1, 2, 13–17 differ from rest (Li, Be, B–F).
• Li and Be show diagonal relationship with Mg and Al.
• Due to small size, high charge/radius ratio, high electronegativity.

Chemical Reactivity Trends

• Reactivity highest at extremes of periodic table, lowest in center.
• Alkali metals reactive due to electron loss; halogens due to electron gain.
• Metallic character decreases left to right; non-metallic increases.
• Metallic character increases down group; non-metallic decreases.
• Oxides: Left extreme form basic oxides (Na₂O), right extreme acidic (Cl₂O₇), center amphoteric/neutral (Al₂O₃, CO).

Solved Examples

• Ex 1: Identify oxide nature of Al₂O₃: Amphoteric.

Practice Set

• Q1: Valency of elements in a group?
  • A. Varies
  • B. Same
  • C. Increases down group
  • D. Decreases down group
• Q2: Which element shows diagonal relationship?
  • A. Na
  • B. Be
  • C. Mg
  • D. Al
• Q3: Oxide of element on extreme right of period is?
  • A. Basic
  • B. Amphoteric
  • C. Acidic
  • D. Neutral

Answer Key

• Q1: B - Valency same down group due to same valence electrons.
• Q2: B - Be shows diagonal relationship with Mg.
• Q3: C - Extreme right oxides are acidic.

Memory Aids

• Valency same in group; varies across period.
• Diagonal relationship: Li-Mg, Be-Al.
• Oxides: Left basic, right acidic, center amphoteric/neutral.

General Trends

Summary of Periodic Properties

• Effective nuclear charge (Z_eff) increases across period.
• Atomic size decreases across period; increases down group.
• Ionisation and electron gain enthalpy increase across period.
• Electronegativity increases across period; decreases down group.
• Screening effect slightly increases down group.
• Metallic character decreases across period; increases down group.
• Oxidising nature increases across period; reducing nature decreases.
• Basic character of oxides decreases across period; acidity increases.
• Thermal stability of carbonates decreases across period.

Solved Examples

• Ex 1: Predict trend of ionisation enthalpy from Na to Ar: Increases.

Practice Set

• Q1: Atomic size trend across period?
  • A. Increases
  • B. Decreases
  • C. Constant
  • D. Random
• Q2: Metallic character trend down group?
  • A. Increases
  • B. Decreases
  • C. Constant
  • D. Random
• Q3: Oxide nature across period?
  • A. Basic to acidic
  • B. Acidic to basic
  • C. Neutral only
  • D. Amphoteric only

Answer Key

• Q1: B - Atomic size decreases across period.
• Q2: A - Metallic character increases down group.
• Q3: A - Oxides change from basic to acidic across period.

Memory Aids

• Across period: size ↓, IE ↑, EN ↑, metallic character ↓.
• Down group: size ↑, metallic character ↑, IE ↓.
• Oxides: basic → amphoteric → acidic across period.