Chapter 3: Classification of Elements and Periodicity in Properties

Periodic Table Importance

Significance of Periodic Table

• The periodic table organizes elements based on their chemical properties and atomic structure.
• It reveals periodic trends and families of elements with similar behavior.
• Enables prediction of element properties and chemical reactions.
• Serves as a fundamental tool for studying chemistry efficiently.

Solved Examples

• Ex 1: Identify the family of an element with atomic number 11.
  Step 1: Atomic number 11 corresponds to Sodium (Na).
  Step 2: Sodium is in Group 1 (alkali metals) with similar properties.

Practice Set

• Q1: Which property is common to elements in the same group?
  A. Atomic mass
  B. Number of valence electrons
  C. Number of neutrons
  D. Atomic number
• Q2: The periodic table arranges elements in order of increasing:
  A. Atomic mass
  B. Atomic number
  C. Density
  D. Melting point
• Q3: Which of the following elements is a noble gas?
  A. Oxygen
  B. Nitrogen
  C. Argon
  D. Chlorine

Answer Key

• Q1: B. Number of valence electrons — Elements in the same group have the same number of valence electrons, determining similar chemical properties.
• Q2: B. Atomic number — Modern periodic table is arranged by increasing atomic number, reflecting periodicity.
• Q3: C. Argon — Argon is a noble gas, chemically inert and located in Group 18.

Memory Aids

• Groups = Columns; same valence electrons.
• Periods = Rows; same number of electron shells.
• "Noble gases" end with "-on" (He, Ne, Ar, Kr, Xe, Rn).

Classification Need

Why Classify Elements?

• Number of known elements increased from 31 (1800) to 118 (present).
• Studying each element and its compounds individually is impractical.
• Classification groups elements with similar properties together.
• Systematic arrangement reveals chemical periodicity and trends.
• Facilitates easier study and prediction of element behavior.

Solved Examples

• Ex 1: Explain the benefit of classifying elements.
  Step 1: Grouping elements reduces complexity.
  Step 2: Similar properties in groups help predict reactions.

Practice Set

• Q1: The main purpose of classifying elements is to:
  A. Increase the number of elements
  B. Group elements with similar properties
  C. Change atomic masses
  D. Separate metals and non-metals only
• Q2: Which scientist is credited with the early classification of elements?
  A. Dalton
  B. Mendeleev
  C. Rutherford
  D. Bohr
• Q3: The periodic table shows chemical periodicity after:
  A. Random intervals
  B. Certain regular intervals
  C. Equal atomic masses
  D. Equal atomic radii

Answer Key

• Q1: B. Group elements with similar properties — Classification organizes elements to simplify study.
• Q2: B. Mendeleev — He developed the first periodic table based on atomic masses and properties.
• Q3: B. Certain regular intervals — Properties repeat periodically with increasing atomic number.

Memory Aids

• Mendeleev’s table arranged elements by increasing atomic mass and properties.
• Periodic law: Properties of elements are periodic functions of their atomic numbers.
• Classification groups elements into metals, non-metals, and metalloids.

Genesis of Periodic Classification

Dobereiner’s Triads (1829)

• Elements grouped in triads with similar properties.
• Atomic weight of middle element ≈ average of other two.
• Example triads: (Ca, Sr, Ba), (Cl, Br, I).

Newland’s Octaves Law (1865)

• Elements arranged by increasing atomic weight.
• Every 8th element has similar properties to the 1st, like musical notes.
• Example: Li to F, then Na to Cl.

Lothar Meyer’s Atomic Volume Curve (1869)

• Graph plotted: Atomic volume vs. Atomic weight.
• Elements with similar properties align at similar graph positions.
• Alkali metals at peaks; 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.
• Order of atomic weights adjusted to fit properties.
• Vacant spaces left for undiscovered elements.

Demerits of Mendeleev’s Periodic Table

• Hydrogen’s position ambiguous (resembles group 1 and 17).
• Cobalt (58.9) placed before nickel (58.6) despite higher atomic weight.
• Isotopes not separately placed.
• Could not explain cause of periodicity.
• Lanthanoids and actinoids excluded.

Solved Examples

Ex 1

Find the atomic weight of the middle element in a triad if the first and third elements have atomic weights 40 and 88.

• Calculate average: (40 + 88) / 2 = 64.
• Middle element atomic weight ≈ 64.

Practice Set

Q1

Dobereiner’s triads are based on which property?

• A. Atomic number
• B. Atomic weight average
• C. Electron configuration
• D. Valency

Q2

According to Newland’s octaves, which element is similar to lithium (Li)?

• A. Sodium (Na)
• B. Magnesium (Mg)
• C. Potassium (K)
• D. Calcium (Ca)

Q3

Which is a limitation of Mendeleev’s periodic table?

• A. Included isotopes separately
• B. Based on atomic number
• C. Could not explain periodicity cause
• D. No vacant spaces for new elements

Answer Key

• Q1: B - Triads are based on the average atomic weight of the first and third elements.
• Q2: A - Sodium is the 8th element after lithium and shares similar properties.
• Q3: C - Mendeleev’s table could not explain the cause of periodicity.

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 order for properties.

Modern Periodic Law

Concept and Significance

• 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 their atomic numbers.
• Elements arranged by increasing atomic number show repeating properties at regular intervals.

Solved Examples

• Ex 1: Why is atomic number preferred over atomic mass in the periodic table?
  Step 1: Moseley’s X-ray studies showed atomic number correlates with element properties.
  Step 2: Atomic mass does not always increase regularly.
  Step 3: Atomic number defines element identity and periodicity.

Practice Set

• Q1: Who established the modern periodic law?
  A. Mendeleev
  B. Henry Moseley
  C. Dobereiner
  D. Newlands
• Q2: According to modern periodic law, element properties depend on:
  A. Atomic mass
  B. Atomic number
  C. Number of neutrons
  D. Isotopic mass
• Q3: Moseley’s experiment involved plotting \( \sqrt{\nu} \) against:
  A. Atomic mass
  B. Atomic number
  C. Number of neutrons
  D. Electron affinity

Answer Key

• Q1: B - Henry Moseley established the modern periodic law through X-ray studies.
• Q2: B - Properties are periodic functions of atomic number, not atomic mass.
• Q3: B - Moseley plotted \( \sqrt{\nu} \) vs atomic number, showing linearity.

Memory Aids

• "Moseley Modernizes" - Moseley linked atomic number to element properties.
• Remember: Atomic Number = Number of Protons = Element Identity.

Periods and Groups

Definitions and Structure

• 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: Identify the period and group of an element with electronic configuration 1s² 2s² 2p⁶ 3s¹.
  Step 1: Highest n = 3 → Period 3.
  Step 2: Last electron in 3s → Group 1 (alkali metals).

Practice Set

• Q1: How many groups are there in the long form periodic table?
  A. 7
  B. 14
  C. 18
  D. 32
• Q2: The period number corresponds to:
  A. Number of valence electrons
  B. Highest principal quantum number (n)
  C. Number of shells filled
  D. Atomic mass
• Q3: Elements in the same group have:
  A. Same atomic number
  B. Similar chemical properties
  C. Same atomic mass
  D. Different valence electrons

Answer Key

• Q1: C - There are 18 groups in the long form periodic table.
• Q2: B - Period number equals highest principal quantum number (n).
• Q3: B - Elements in a group share similar chemical properties.

Memory Aids

• Periods = Rows = Principal Quantum Number (n).
• Groups = Columns = Similar Properties.

Periodic Table Periods Details

Period Length and Orbital Filling

• 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, fills 4s, 3d, 4p and 5s, 4d, 5p orbitals.
• Period 6 & 7: 32 elements, includes filling of f orbitals (lanthanoids and actinoids).
• Length increases due to involvement of d and f orbitals.

Solved Examples

• Ex 1: Why does period 4 have 18 elements?
  Step 1: Period 4 fills 4s, 3d, and 4p orbitals.
  Step 2: 4s (2 electrons) + 3d (10 electrons) + 4p (6 electrons) = 18 elements.

Practice Set

• Q1: How many elements are in period 2?
  A. 2
  B. 8
  C. 18
  D. 32
• Q2: Which orbitals are filled in period 3?
  A. 3s and 3p
  B. 3s, 3p, and 3d
  C. 4s and 3d
  D. 2s and 2p
• Q3: Lanthanoids belong to which period?
  A. 4
  B. 5
  C. 6
  D. 7

Answer Key

• Q1: B - Period 2 has 8 elements filling 2s and 2p orbitals.
• Q2: A - Period 3 fills 3s and 3p orbitals only.
• Q3: C - Lanthanoids are part of period 6.

Memory Aids

• Period length = number of electrons in orbitals filled.
• Remember: s (2), p (6), d (10), f (14) electrons per orbital block.

Classification of Elements

Metals, Non-metals, and Metalloids

• Periodic table divided by zig-zag line separating metals and non-metals.
• Metals: Left and center, pink color; include alkali metals, alkaline earth metals, transition metals.
• Non-metals: Right side, pink color; include noble gases and other non-metals.
• Metalloids: Along zig-zag line, grey color; have properties intermediate between metals and non-metals.
• Lanthanides and actinides placed separately at bottom as inner transition elements.

Solved Examples

• Ex 1: Identify the type of element for silicon (Si).
  Step 1: Si lies along the zig-zag line.
  Step 2: It is a metalloid with mixed properties.

Practice Set

• Q1: Which group contains alkali metals?
  A. Group 1
  B. Group 2
  C. Group 17
  D. Group 18
• Q2: Metalloids are located:
  A. Left side
  B. Right side
  C. Along zig-zag line
  D. Bottom rows
• Q3: Noble gases belong to which group?
  A. 16
  B. 17
  C. 18
  D. 1

Answer Key

• Q1: A - Alkali metals are in Group 1.
• Q2: C - Metalloids lie along the zig-zag line.
• Q3: C - Noble gases are in Group 18.

Memory Aids

• "Zig-zag line" separates metals and non-metals.
• Group 1 = Alkali metals; Group 2 = Alkaline earth 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 digits of the atomic number using Latin and Greek roots.
• The suffix -ium is added at the end of the name.
• Element symbols are formed by taking the first letter of each numerical root word, capitalizing the first letter.

Example of Naming

• For element 115: digits are 1-1-5.
• Roots: 1 = un, 5 = pent.
• Name: ununpentium (now known as moscovium).
• Symbol: Uup (U from un, u from un, p from pent).

Solved Examples

• Ex 1: Name element 102.
  • Digits: 1-0-2
  • Roots: un (1), nil (0), bi (2)
  • Name: unnilbium
  • Symbol: Unb
• Ex 2: Symbol for element 118.
  • Digits: 1-1-8
  • Roots: un (1), un (1), oct (8)
  • Symbol: Uuo

Practice Set

• Q1: What is the IUPAC name for element 113?
  • A) Ununtrium
  • B) Ununpentium
  • C) Ununseptium
  • D) Ununquadium
• Q2: The symbol for element 107 is:
  • A) Uus
  • B) Uup
  • C) Uuo
  • D) Uuq
• Q3: Which suffix is added to the root words in naming elements with Z > 100?
  • A) -ine
  • B) -on
  • C) -ium
  • D) -ide

Answer Key

• Q1: A) Ununtrium - Derived from digits 1-1-3 (un-un-tri) + -ium suffix.
• Q2: A) Uus - For element 107 (1-0-7), roots un-nil-sept, symbol Uus.
• Q3: C) -ium - Standard suffix for elements with atomic number > 100.

Memory Aids

• Remember roots: un (1), bi (2), tri (3), quad (4), pent (5), hex (6), sept (7), oct (8), enn (9), nil (0).
• Symbol letters come from first letter of each root word, first letter capitalized.
• Always add -ium at the end of the name.

Electronic Configurations of Elements and the 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...).

Electronic Configurations in Periods

• Period 1: fills 1s orbital (2 electrons).
• Period 2: fills 2s and 2p orbitals (8 electrons).
• Period 3: fills 3s and 3p orbitals (8 electrons).
• Period 4: fills 4s, 3d, and 4p orbitals (18 electrons).
• Period 5: fills 5s, 4d, and 5p orbitals (18 electrons).
• Period 6: fills 6s, 4f, 5d, and 6p orbitals (32 electrons).
• Period 7: fills 7s, 5f, 6d, and 7p orbitals (32 electrons).

Solved Examples

• Ex 1: Write electronic configuration of Sulfur (Z=16).
  • Period 3 element, valence shell n=3.
  • Configuration: 1s² 2s² 2p⁶ 3s² 3p⁴.

Practice Set

• Q1: Which orbital is being filled in period 4?
  • A) 3p
  • B) 4s
  • C) 2p
  • D) 5s
• Q2: The valence shell principal quantum number of elements in period 5 is:
  • A) 3
  • B) 4
  • C) 5
  • D) 6
• Q3: How many electrons are there in the valence shell of period 6 elements?
  • A) 8
  • B) 18
  • C) 32
  • D) 2

Answer Key

• Q1: B) 4s - Period 4 starts filling 4s orbital.
• Q2: C) 5 - Period number equals valence shell principal quantum number.
• Q3: C) 32 - Period 6 fills up to 32 electrons including 6s, 4f, 5d, 6p orbitals.

Memory Aids

• Period number = valence shell principal quantum number (n).
• Orbitals fill in order: s < p < d < f.
• Period 1: 2 electrons; Period 2 & 3: 8 electrons; Period 4 & 5: 18 electrons; Period 6 & 7: 32 electrons.

Groupwise Electronic Configuration

Group Electronic Configuration

• 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.

Group 1 Alkali Metals Configuration

• Li: [He] 2s¹
• Na: [Ne] 3s¹
• K: [Ar] 4s¹
• Rb: [Kr] 5s¹
• Cs: [Xe] 6s¹
• Fr: [Rn] 7s¹

Solved Examples

• Ex 1: Write valence shell configuration of Potassium (K).
  • Group 1 element, valence shell: 4s¹.

Practice Set

• Q1: What is the valence configuration of Group 2 elements?
  • A) ns¹
  • B) ns²
  • C) ns² np¹
  • D) (n-1)d¹⁰ ns¹
• Q2: Which element has configuration [Ne] 3s¹?
  • A) Sodium (Na)
  • B) Magnesium (Mg)
  • C) Aluminium (Al)
  • D) Chlorine (Cl)
• Q3: Group 1 elements have valence configuration:
  • A) ns²
  • B) ns¹
  • C) (n-1)d¹⁰ ns²
  • D) ns² np⁵

Answer Key

• Q1: B) ns² - Group 2 elements have two electrons in valence s orbital.
• Q2: A) Sodium (Na) - Atomic number 11, configuration [Ne] 3s¹.
• Q3: B) ns¹ - Alkali metals have one electron in valence s orbital.

Memory Aids

• Group 1: ns¹ (alkali metals).
• Group 2: ns² (alkaline earth metals).
• Valence electrons determine group properties.

Periodic Table Groups and Blocks

Group Names and Properties

• Group 1: Alkali metals (ns¹), very reactive metals.
• Group 2: Alkaline earth metals (ns²), less reactive than group 1.
• Group 11: Coinage metals (Cu, Ag, Au), good conductors.
• Group 15: Pnictogens, includes nitrogen family.
• Group 16: Chalcogens, oxygen family.
• Group 17: Halogens, highly reactive nonmetals.
• Group 18: Noble gases (aerogens), inert gases.

Blocks of Elements

• Elements classified into s-, p-, d-, and f-blocks based on last electron orbital filled.
• s-block: Groups 1 and 2, valence electrons in s orbital.
• p-block: Groups 13 to 18, valence electrons in p orbital.
• d-block: Transition metals, valence electrons in d orbital.
• f-block: Lanthanides and actinides, valence electrons in f orbital.

Solved Examples

• Ex 1: Identify the block of element with configuration [Ar] 3d⁵ 4s¹.
  • Last electron enters 3d orbital → d-block element.

Practice Set

• Q1: Which group is called halogens?
  • A) Group 15
  • B) Group 16
  • C) Group 17
  • D) Group 18
• Q2: Elements with valence electrons in f orbital belong to:
  • A) s-block
  • B) p-block
  • C) d-block
  • D) f-block
• Q3: Which group contains coinage metals?
  • A) Group 11
  • B) Group 12
  • C) Group 13
  • D) Group 14

Answer Key

• Q1: C) Group 17 - Halogens.
• Q2: D) f-block - Lanthanides and actinides.
• Q3: A) Group 11 - Coinage metals (Cu, Ag, Au).

Memory Aids

• Group 1: Alkali metals; Group 2: Alkaline earth metals.
• Halogens: Group 17; Noble gases: Group 18.
• Blocks named after orbitals being filled: s, p, d, f.

s-Block Elements

Characteristics

• Valence electron enters s-orbital of outermost shell; general configuration: ns^1-2.
• Includes Group 1 (alkali metals) and Group 2 (alkaline earth metals), except helium.
• Soft metals with low melting and boiling points.
• Highly reactive with low ionization enthalpy.
• Form +1 ions (alkali metals) or +2 ions (alkaline earth metals) by losing valence electrons.
• Strong reducing agents due to high electropositivity.
• Metallic character and reactivity increase down the group.
• Not found pure in nature due to high reactivity.
• Compounds mostly ionic, except lithium and beryllium.

Solved Examples

Ex 1: Write the electronic configuration of calcium and state its block.

• Calcium (Z=20): 1s² 2s² 2p⁶ 3s² 3p⁶ 4s²
• Valence electrons enter 4s orbital → s-block element.

Practice Set

• Q1: Which of the following is an s-block element?
  • A. Carbon
  • B. Sodium
  • C. Chlorine
  • D. Iron
• Q2: The general electronic configuration of s-block elements is:
  • 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
• Q3: Which ion is commonly formed by alkaline earth metals?
  • A. +1
  • B. +2
  • C. -1
  • D. 0

Answer Key

• Q1: B. Sodium is an s-block element with valence electron in s-orbital.
• Q2: A. s-block elements have general configuration ns^1-2.
• Q3: B. Alkaline earth metals form +2 ions by losing two valence electrons.

Memory Aids

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

p-Block Elements

Characteristics

• Valence electrons enter outermost p-orbitals; general configuration: ns² np^1-6.
• Includes Groups 13 to 18; contains non-metals, metalloids, and poor metals.
• Exhibits covalent, electrovalent, and coordinate bonding.
• Show allotropy (e.g., carbon, phosphorus, sulfur).
• Noble gases (Group 18) are inert due to filled ns² np⁶ configuration.
• Halogens (Group 17) and chalcogens (Group 16) have high negative electron gain enthalpy, readily 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 together are called representative or main group elements.

Solved Examples

Ex 1: Identify the block and valence configuration of sulfur (S).

• Sulfur (Z=16): valence electrons in 3s² 3p⁴ → p-block element.

Practice Set

• Q1: The general electronic configuration of p-block elements is:
  • 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: Which of the following is a noble gas?
  • A. Neon
  • B. Nitrogen
  • C. Oxygen
  • D. Fluorine

Answer Key

• Q1: B. p-block elements have ns² np^1-6 configuration.
• Q2: B. Halogens are in Group 17.
• Q3: A. Neon is a noble gas with filled valence shell.

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 full valence shells.

d-Block Elements

Characteristics

• Valence electrons enter (n-1)d orbitals; general configuration: (n-1)d^1-10 ns^0-2.
• Groups 3 to 12; called transition elements.
• Four series: 3d (Sc-Zn), 4d (Y-Cd), 5d (La/Hf-Hg), 6d (Ac-Rf-Cn).
• High melting and boiling points; metals.
• Form coloured compounds and salts.
• Paramagnetic due to unpaired electrons.
• Exhibit variable oxidation states (e.g., Fe²⁺, Fe³⁺).
• Good catalysts.
• 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) and state its common oxidation states.

• Fe (Z=26): [Ar] 3d⁶ 4s²
• Common oxidation states: +2 and +3.

Practice Set

• Q1: Which group contains d-block elements?
  • A. 1-2
  • B. 3-12
  • C. 13-18
  • D. Lanthanoids
• Q2: Which element does NOT show typical transition element properties?
  • A. Fe
  • B. Cu
  • C. Zn
  • D. Ni
• Q3: The general electronic configuration of d-block elements is:
  • 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

Answer Key

• Q1: B. d-block elements are in groups 3 to 12.
• Q2: C. Zn has d¹⁰ s² configuration and lacks typical transition properties.
• Q3: C. General configuration is (n-1)d^1-10 ns^0-2.

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-subshells.

f-Block Elements

Characteristics

• Valence electrons enter (n-2)f orbitals; general configuration: (n-2)f^0-14 (n-1)d^0-1 ns².
• Called inner-transition elements; include Lanthanoids and Actinoids (14 elements each).
• Heavy metals with high melting and boiling points.
• Exhibit variable oxidation states; +3 most common.
• Form coloured salts due to light absorption.
• All actinoids are radioactive.
• Transuranium elements: synthetic elements beyond uranium (Z>92).

Solved Examples

Ex 1: Identify the block and common oxidation state of cerium (Ce).

• Cerium is a lanthanoid (f-block) with common oxidation state +3.

Practice Set

• Q1: The general electronic configuration of f-block elements is:
  • 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 belongs to f-block elements?
  • A. Lanthanoids
  • B. Alkali metals
  • C. Halogens
  • D. Noble gases
• Q3: Which of the following is true about actinoids?
  • A. Non-radioactive
  • B. Radioactive
  • C. Noble gases
  • D. Alkali metals

Answer Key

• Q1: D. f-block elements have configuration (n-2)f^0-14 (n-1)d^0-1 ns².
• Q2: A. Lanthanoids are f-block elements.
• Q3: B. All actinoids are radioactive.

Memory Aids

• f-block = inner transition metals, Lanthanoids & Actinoids.
• Common oxidation state +3; actinoids are radioactive.
• Transuranium elements are synthetic, Z > 92.

Metals, Non-metals and Metalloids

Characteristics

• Metals:
  • Over 78% of elements, located left side of periodic table.
  • Usually solids at room temperature (except Hg, Ga, Cs).
  • High melting and boiling points.
  • Good conductors of heat and electricity.
  • Malleable and ductile.
• Non-metals:
  • Located top right side of periodic table.
  • Usually solids or gases at room temperature (exceptions: B, C).
  • Low melting and boiling points.
  • Poor conductors of heat and electricity.
  • Brittle solids; not malleable or ductile.
• Metalloids (Semi-metals):
  • Elements like Si, Ge, As, Sb, Te.
  • Show properties intermediate between metals and non-metals.

Solved Examples

Ex 1: Classify silicon (Si) as metal, non-metal, or metalloid.

• Si shows properties of both metals and non-metals → metalloid.

Practice Set

• Q1: Which of the following is a metal?
  • A. Oxygen
  • B. Iron
  • C. Sulfur
  • D. Phosphorus
• Q2: Which property is NOT typical of metals?
  • A. Malleability
  • B. High electrical conductivity
  • C. Brittle nature
  • D. High melting point
• Q3: Metalloids are elements that:
  • A. Are good conductors like metals
  • B. Are poor conductors like non-metals
  • C. Show properties intermediate between metals and non-metals
  • D. Are all gases at room temperature

Answer Key

• Q1: B. Iron is a metal.
• Q2: C. Metals are malleable, not brittle.
• Q3: C. Metalloids have intermediate properties.

Memory Aids

• Metals: left side, good conductors, malleable, ductile.
• Non-metals: right side, poor conductors, brittle.
• Metalloids: intermediate properties, e.g., Si, Ge.

Periodicity

Definition and Cause

• 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 the periodicity interval after which properties repeat.
  Answer: After 2, 8, 8, 18, 18, and 32 elements due to valence shell configuration repetition.

Practice Set

• Q1: What causes periodicity in elements?
  A. Atomic mass
  B. Atomic number
  C. Valence electron configuration
  D. Nuclear charge
  
• Q2: After how many elements do similar properties repeat in the periodic table?
  A. 2, 8, 8, 18, 18, 32
  B. 1, 2, 3, 4, 5, 6
  C. 8, 8, 18, 18, 32, 32
  D. 2, 4, 6, 8, 10, 12
  
• Q3: Periodicity is primarily linked to which atomic property?
  A. Nuclear stability
  B. Electron configuration
  C. Atomic mass
  D. Ionization energy
  

Answer Key

• Q1: C - Periodicity arises due to repetition of valence electron configurations.
• Q2: A - Properties repeat after intervals of 2, 8, 8, 18, 18, and 32 elements.
• Q3: B - Periodicity is linked to electronic configuration of atoms.

Memory Aids

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

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 nuclei 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 in neighboring molecules (noble gases).
• Order: van der Waals radius > metallic radius > covalent radius.

Solved Examples

• Ex 1: Compare atomic radius trend in a period.
  Answer: Atomic radius decreases due to increasing nuclear charge pulling electrons closer.

Practice Set

• Q1: Atomic radius generally ______ across a period.
  A. Increases
  B. Decreases
  C. Remains constant
  D. Varies randomly
  
• Q2: Which radius is largest for an atom?
  A. Metallic radius
  B. Covalent radius
  C. van der Waals radius
  D. Ionic radius
  
• Q3: Why does atomic radius increase down a group?
  A. Nuclear charge decreases
  B. Number of electron shells increases
  C. Electron shielding decreases
  D. Ionization energy increases
  

Answer Key

• Q1: B - Atomic radius decreases across a period due to increased nuclear charge.
• Q2: C - van der Waals radius is largest as it measures distance between molecules.
• Q3: B - Atomic radius increases down a group due to addition of electron shells.

Memory Aids

• Atomic radius trend: Left to right ↓, top to bottom ↑.
• van der Waals radius > metallic radius > covalent radius.

Ionic Radius

Definition and Trends

• Distance from nucleus to outermost electron shell in an ion.
• Trend similar to atomic radius: decreases across period, increases down group.
• Cations are smaller than parent atoms (loss of electrons, same nuclear charge).
• More electrons removed → smaller cation (e.g., Mg > Mg⁺ > Mg²⁺).
• Anions are larger than parent atoms (gain of electrons, same nuclear charge).
• Size explained by Z/e ratio (nuclear charge/electrons): higher ratio → smaller size.
• Isoelectronic species: same electron number, different nuclear charge; size decreases with increasing nuclear charge.

Solved Examples

• Ex 1: Arrange Mg, Mg⁺, Mg²⁺ in order of increasing size.
  Answer: Mg²⁺ < Mg⁺ < Mg

Practice Set

• Q1: Which ion is smallest?
  A. O²⁻
  B. O⁻
  C. O
  D. F⁻
  
• Q2: Isoelectronic ions have ______.
  A. Same nuclear charge
  B. Same number of electrons
  C. Same size
  D. Same number of protons
  
• Q3: Why is a cation smaller than its parent atom?
  A. More electrons
  B. Less nuclear charge
  C. Fewer electrons with same nuclear charge
  D. Electron shielding increases
  

Answer Key

• Q1: C - Neutral oxygen atom is smaller than its anions due to less electron repulsion.
• Q2: B - Isoelectronic species have same number of electrons.
• Q3: C - Cations have fewer electrons but same nuclear charge, leading to smaller size.

Memory Aids

• Size order in isoelectronic series: higher positive charge → smaller size.
• Cations < parent atom < anions in size.

Ionisation Enthalpy

Definition and Factors

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

Solved Examples

• Ex 1: Explain why IE increases across a period.
  Answer: Nuclear charge increases, valence shell remains same, increasing effective nuclear charge.

Practice Set

• Q1: Ionisation enthalpy is highest for which element?
  A. Helium
  B. Neon
  C. Argon
  D. Krypton
  
• Q2: Which factor decreases ionisation enthalpy?
  A. Increased nuclear charge
  B. Smaller atomic size
  C. Increased screening effect
  D. Half-filled orbitals
  
• Q3: According to Slater’s rule, effective nuclear charge is calculated by:
  A. Z + σ*
  B. Z - σ*
  C. σ* - Z
  D. Z × σ*
  

Answer Key

• Q1: A - Helium has highest IE due to smallest size and full shell.
• Q2: C - Increased screening reduces effective nuclear charge, lowering IE.
• Q3: B - Z_eff = Z - σ* (screening constant).

Memory Aids

• IE trend: increases across period, decreases down group.
• IE order of orbitals: s > p > d > f.
• Stable half/full orbitals increase IE.

Electron Gain Enthalpy

Definition and Factors

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

Solved Examples

• Ex 1: Why is ΔegH of F less negative than Cl?
  Answer: Smaller size of F causes more electron-electron repulsion when adding electron.

Practice Set

• Q1: Electron gain enthalpy becomes ______ down a group.
  A. More negative
  B. Less negative
  C. Constant
  D. Positive
  
• Q2: Which element has positive electron gain enthalpy?
  A. Oxygen
  B. Nitrogen
  C. Chlorine
  D. Sulfur
  
• Q3: Electron gain enthalpy is influenced by:
  A. Atomic size
  B. Nuclear charge
  C. Screening effect
  D. All of the above
  

Answer Key

• Q1: B - ΔegH becomes less negative down the group.
• Q2: B - Nitrogen has half-filled orbitals, positive ΔegH.
• Q3: D - All factors affect electron gain enthalpy.

Memory Aids

• ΔegH trend: more negative across period, less negative down group.
• Half/full orbital elements have positive ΔegH.

Electronegativity

Definition and Trends

• Tendency of an atom to attract shared electrons in a covalent bond.
• Increases across a period (left to right).
• Decreases down a group (top to bottom).
• Common scales:
  • Pauling scale: Based on bond dissociation energies.
  • Mulliken scale: Average of ionization energy and electron affinity.
  • Allred-Rochow scale: Based on effective nuclear charge and atomic radius.
• Percentage ionic character of bond depends on electronegativity difference.

Solved Examples

• Ex 1: Calculate electronegativity difference using Pauling scale.
  Answer: Use bond energies in formula \(\chi_A - \chi_B = \sqrt{E_{AB} - \frac{E_{AA} + E_{BB}}{2}}\).

Practice Set

• Q1: Electronegativity ______ across a period.
  A. Increases
  B. Decreases
  C. Remains same
  D. Varies randomly
  
• Q2: Which scale uses average of ionization energy and electron affinity?
  A. Pauling
  B. Mulliken
  C. Allred-Rochow
  D. None
  
• Q3: A bond with electronegativity difference of 1.9 is:
  A. Purely covalent
  B. Half covalent, half ionic
  C. Mostly ionic
  D. Purely ionic
  

Answer Key

• Q1: A - Electronegativity increases across a period.
• Q2: B - Mulliken scale uses average of IE and EA.
• Q3: B - 1.9 difference indicates half covalent, half ionic bond.

Memory Aids

• Electronegativity trend: increases across period, decreases down group.
• Pauling scale uses bond energies; Mulliken uses IE and EA.

Chemical Properties

Valence and Reactivity

• 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 a period.
• Elements in a group have same valency (same outer electrons).
• Second period elements show anomalous properties due to small size, high charge/radius ratio, and high electronegativity.
• Diagonal relationship: Li similar to Mg, Be similar to Al.
• Chemical reactivity highest at extremes of periodic table, lowest in center.
• Metallic character decreases left to right; non-metallic increases.
• Metallic character increases down group; non-metallic decreases.
• Oxides:
  • Left extreme: basic oxides (e.g., Na₂O).
  • Right extreme: acidic oxides (e.g., Cl₂O₇).
  • Center: amphoteric (e.g., Al₂O₃) or neutral (e.g., CO, NO).

Solved Examples

• Ex 1: Explain why Li behaves like Mg.
  Answer: Due to diagonal relationship caused by similar size and charge/radius ratio.

Practice Set

• Q1: Valency of elements in a group is generally:
  A. Variable
  B. Same
  C. Increases down group
  D. Decreases down group
  
• Q2: Which oxide is amphoteric?
  A. Na₂O
  B. Cl₂O₇
  C. Al₂O₃
  D. CO₂
  
• Q3: Chemical reactivity is highest for elements:
  A. In the center of periodic table
  B. At the extremes of periodic table
  C. In the middle of groups
  D. None of the above
  

Answer Key

• Q1: B - Elements in a group have same valency.
• Q2: C - Al₂O₃ is amphoteric oxide.
• Q3: B - Reactivity highest at extremes (alkali metals and halogens).

Memory Aids

• Diagonal relationship: Li ~ Mg, Be ~ Al.
• Oxide nature: Left basic, center amphoteric, right acidic.
• Valency trend: increases 1 to 4, then decreases 4 to 0 across period.

Periodic Trends Summary

General Trends

• Effective nuclear charge (Z_eff) increases across period.
• Atomic size decreases across period; increases down group.
• Ionisation enthalpy 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.
• Electropositivity decreases across period; covalent character increases.
• Basic character of oxides, hydroxides, hydrides decreases across period.
• Strength of oxyacids increases; thermal stability of carbonates decreases across period.

Solved Examples

• Ex 1: Explain why atomic size decreases across a period.
  Answer: Increasing Z_eff pulls electrons closer, reducing size.

Practice Set

• Q1: Which property increases across a period?
  A. Atomic size
  B. Metallic character
  C. Ionisation enthalpy
  D. Electropositivity
  
• Q2: Down a group, electronegativity generally:
  A. Increases
  B. Decreases
  C. Remains constant
  D. Varies randomly
  
• Q3: Basic character of oxides across a period:
  A. Increases
  B. Decreases
  C. Remains same
  D. Varies randomly
  

Answer Key

• Q1: C - Ionisation enthalpy increases across period.
• Q2: B - Electronegativity decreases down group.
• Q3: B - Basic character decreases across period.

Memory Aids

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