10.3 Position of an Element in the Periodic Table

The electronic configuration of an element provides basic information about its position in the periodic table. By examining the valence electron configuration, n-value, and sub-shell type, you can identify an element’s group, period, and block.

Key Concept

Group Determination: For s-block elements, group number = number of valence electrons. For p-block elements, group number = number of valence electrons + 10.

Period Determination: Period number = n value of the valence shell (the largest principal quantum number).

Memorization Trick

G-P-B Method: Remember “Group = Period + Block connection”.

  • Group: Count valence electrons (s-block) or add 10 for p-block
  • Period: Look at the highest n value
  • Block: Check the last electron’s subshell (s, p, d, f)

Example 10.1: Nitrogen (Atomic Number 7)

Step 1: Electronic configuration: 1s² 2s² 2p³

Step 2: Valence shell = L shell (n=2) → Period 2

Step 3: Valence electrons = 2s² 2p³ = 5 electrons

Step 4: Since it’s p-block, group = 5 + 10 = 15 (Group VA)

Result: Nitrogen is in Period 2, Group 15 (VA)

H
Li
Be
B
C
N
O
F
Ne

10.5 Periodicity of Properties

Elements with similar valence shell electronic configurations are placed in the same group. Because all elements in a given group have a similar valence shell electronic configuration, they have similar chemical properties.

10.5.1 Electron Affinity

Electron affinity is defined as the amount of energy released when an electron adds up in the valence shell of an isolated atom to form a uni-negative gaseous ion.

X(g) + e⁻ → X⁻(g) + Electron Affinity

Trends in Electron Affinity

Across a period (left to right): Increases due to increasing nuclear charge and decreasing atomic size.

Down a group (top to bottom): Decreases due to increasing atomic size and shielding effect.

Memorization Trick

Affinity Acronym: “LARD” – Left to Right Affinity Increases, Downward Affinity Decreases

Exception Trick: Remember “Fluorine is Funny” – Fluorine has lower electron affinity than chlorine due to its small size causing electron-electron repulsion.

10.4 Metals, Non-Metals and Metalloids

The periodic table provides a general framework for organizing elements. The elements are generally classified in three categories based on their properties:

Classification

1. Metals: Left side of periodic table (grey). Good conductors, malleable, ductile.

2. Non-Metals: Right side of periodic table (yellow). Poor conductors, non-malleable, non-ductile.

3. Metalloids: Adjacent to the dividing line (blue). Properties of both metals and non-metals.

10.6 Trends in Metallic and Non-Metallic Behaviours

Memorization Trick

Metal Mania: “Metals Lose Electrons, Non-Metals Gain”

Group Trend: “Down the group, metals get more metallic, non-metals get less non-metallic”

Example: Identifying Unknown Elements

Element 1 (Atomic Number 19): Electronic configuration = 1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹

Valence shell: n=4 → Period 4. Valence electrons: 1 → Group 1 (Alkali metal)

Conclusion: Potassium (K) – reacts violently with water

Element 2 (Atomic Number 17): Electronic configuration = 1s² 2s² 2p⁶ 3s² 3p⁵

Valence shell: n=3 → Period 3. Valence electrons: 7 → Group 17 (Halogen)

Conclusion: Chlorine (Cl) – exists as Cl₂, forms salts with metals

MCQ Exercises

1. In the periodic table the number of period of an element is same as:

(a) atomic number
(b) atomic mass
(c) group number
(d) principal quantum number

Correct Answer: (d) principal quantum number

The period number equals the principal quantum number (n) of the valence shell. For example, if valence electrons are in the 3rd shell (n=3), the element is in period 3.

2. Which element possesses the highest electronegativity?

(a) F
(b) Cl
(c) Br
(d) I

Correct Answer: (a) F (Fluorine)

Fluorine has the highest electronegativity (3.98 on Pauling scale) due to its small atomic size and high nuclear charge. Electronegativity decreases down the halogen group.

3. Which of the following will produce a base in water?

(a) Na₂O
(b) CO₂
(c) SO₃
(d) SiO₂

Correct Answer: (a) Na₂O

Metal oxides (like Na₂O) are basic and produce bases in water: Na₂O + H₂O → 2NaOH. Non-metal oxides (CO₂, SO₃) produce acids in water.

4. Which is correct for ₁₂Mg?

(a) period 1 element
(b) group 1 element
(c) period 2 element
(d) alkaline earth metal

Correct Answer: (d) alkaline earth metal

Magnesium (atomic number 12) has electronic configuration: 1s² 2s² 2p⁶ 3s². It has 2 valence electrons in the 3rd shell, so it’s in period 3, group 2 (alkaline earth metals).

5. An element has electron configuration 1s², 2s², 2p⁶, 3s², 3p², it belongs to:

(a) period 2
(b) group 14
(c) group IIIA
(d) period 1

Correct Answer: (b) group 14

The highest n value is 3 → period 3. Valence electrons: 3s² 3p² = 4 electrons. Since it’s p-block, group = 4 + 10 = 14 (Group IVA). This element is Silicon (Si).

6. Which of the following element has highest ionization energy?

(a) Li
(b) Na
(c) K
(d) Rb

Correct Answer: Actually, the question seems reversed. Among alkali metals, ionization energy decreases down the group, so Li has the highest, Rb the lowest.

Correcting the question: Li has highest ionization energy among alkali metals. Ionization energy decreases down a group due to increasing atomic size and shielding effect.

7. Which of the following element has lowest electron affinity?

(a) F
(b) Cl
(c) Br
(d) I

Correct Answer: (d) I (Iodine)

Electron affinity decreases down the halogen group. Iodine has the lowest electron affinity among halogens due to its large atomic size and strong shielding effect.

8. Which of the following oxide is amphoteric?

(a) MgO
(b) CO₂
(c) Al₂O₃
(d) SO₃

Correct Answer: (c) Al₂O₃

Aluminum oxide (Al₂O₃) is amphoteric – it reacts with both acids and bases. MgO is basic, CO₂ and SO₃ are acidic oxides.

9. Element of which group have strong non-metallic character?

(a) G-IA
(b) G-IIIA
(c) G-IVA
(d) G-VIIA

Correct Answer: (d) G-VIIA (Group 17 – Halogens)

Group 17 elements (halogens) have the strongest non-metallic character because they have high electronegativity and tend to gain electrons to achieve stable octet.

10. Which atom has smaller ionization energy?

(a) B
(b) Al
(c) C
(d) N

Correct Answer: (b) Al

Aluminum has smaller ionization energy than boron because ionization energy decreases down a group. Al is below B in group 13, so it has larger atomic size and more shielding, making electron removal easier.

Short Questions

2. (i) What is a period?

A period is a horizontal row in the periodic table. Elements in the same period have the same number of electron shells. The period number equals the principal quantum number (n) of the valence shell.

2. (ii) Differentiate between s and p-block elements.

s-block elements: Last electron enters s-orbital. Groups 1 & 2. Generally metals, low ionization energy, form cations.

p-block elements: Last electron enters p-orbital. Groups 13 to 18. Include metals, non-metals, and metalloids. More varied properties.

2. (iii) Find out the position of sulphur (atomic number 16) in the periodic table.

Solution:

Electronic configuration: 1s² 2s² 2p⁶ 3s² 3p⁴

Valence shell: n=3 → Period 3

Valence electrons: 3s² 3p⁴ = 6 electrons

Since it’s p-block, group = 6 + 10 = 16 (Group VIA)

Result: Sulphur is in Period 3, Group 16

2. (iv) Find out the position of an element with electronic configuration: 1s², 2s², 2p⁶, 3s², 3p³

Solution:

Electronic configuration: 1s² 2s² 2p⁶ 3s² 3p³

Valence shell: n=3 → Period 3

Valence electrons: 3s² 3p³ = 5 electrons

Since it’s p-block, group = 5 + 10 = 15 (Group VA)

Result: The element is Phosphorus (P) in Period 3, Group 15

2. (v) In what respect metals differ from metalloids?

Metals: Good conductors of heat and electricity, malleable, ductile, lose electrons to form cations, generally have 1-3 valence electrons.

Metalloids: Have properties intermediate between metals and non-metals, semiconductors (conductivity between metals and non-metals), can gain or lose electrons depending on conditions.

2. (vi) A cation is smaller than its atom. Justify

When an atom loses electrons to form a cation:

  1. The electron-electron repulsion decreases
  2. The effective nuclear charge (Zeff) on remaining electrons increases
  3. The remaining electrons are pulled closer to the nucleus
  4. Sometimes an entire electron shell is lost

All these factors cause the cation to be smaller than its parent atom.