CG Board Class 10 Science Chapter 4 Solutions: Periodic Classification of Elements (तत्वों का आवर्ती वर्गीकरण) (2026-27)

(i) The position of which of the following elements is not fixed in the periodic table:**

• (a) Sodium
• (b) Chlorine
• (c) Helium
• (d) Hydrogen

(d) Hydrogen

Explanation: Hydrogen exhibits properties similar to both alkali metals (Group 1) and halogens (Group 17), making its position anomalous.

Answer: (b) Increasing order of atomic weight

Explanation: Mendeleev’s periodic law states that physical and chemical properties of elements are periodic functions of their atomic weights.

Answer: (b) Moseley

Explanation: Henry Moseley showed in 1913 that atomic number is the fundamental property of elements and gave the modern periodic law.

Answer: (c) Increases

Explanation: As atomic size increases down a group, nuclear attraction on outer electrons weakens, making it easier to lose electrons and form positive ions.

Answer: (a) Li < Na < K

Explanation: Li, Na, and K belong to Group 1. Atomic size increases down the group as new shells are added (Li = 2,1; Na = 2,8,1; K = 2,8,8,1).

Elements belonging to the same group have the same number of valence electrons in their outermost shell. Since chemical reactivity and bonding depend directly on valence electrons, elements of the same group exhibit similar physical and chemical properties.

Position:** Period 1, Group 1 (or Group 17).

• Reasons:

1. Resemblance to Group 1 (Alkali Metals): Has 1 valence electron (1s¹), forms a positive ion (H^+), and forms oxides (H_2O) and halides (HCl) similar to Na_2O and NaCl.
2. Resemblance to Group 17 (Halogens): Needs 1 electron to complete its duplet shell, exists as a diatomic gas (H₂), and forms a hydride ion (H^-).

(a) Mg is bigger than Cl:**

• Reason: Mg (2,8,2) and Cl (2,8,7) belong to the same period (Period 3). Moving left to right across a period, nuclear charge increases while electron shells remain the same, pulling electrons closer and decreasing atomic size.
• (b) K is bigger than Na:
• Reason: Na (2,8,1, Period 3) and K (2,8,8,1, Period 4) belong to Group 1. Moving down a group, a new electron shell is added, increasing atomic size.

Electronic Configurations:** A(5) = 2, 3; B(7) = 2, 5; C(10) = 2, 8.

• Analysis: Elements A (Boron) and B (Nitrogen) belong to the same period (Period 2) and are p-block elements that form covalent compounds. In contrast, C (Neon) is an extremely unreactive noble gas with a complete octet. Thus, A and B show closer non-metallic/covalent bonding properties compared to the inert noble gas C.

(a) Group 18 (Noble gas): Element with atomic number 10 (Neon, configuration 2,8).

• (b) Group 15 (5 valence electrons): Element with atomic number 7 (Nitrogen, configuration 2,5; Group = 10 + 5 = 15).
• (c) Group 13 (3 valence electrons): Element with atomic number 5 (Boron, configuration 2,3; Group = 10 + 3 = 13).

Given:**

• Atomic weight of A = 7
• Atomic weight of C = 39
• Döbereiner’s Triad Law:

Atomic weight of B = (Atomic weight of A + Atomic weight of C)/(2)

• Calculation:

Atomic weight of B = (7 + 39)/(2) = 46/2 = 23

• Result: The atomic weight of element ‘B’ is 23 (Sodium, Na).

(i) How does the metallic character of the elements change?**

(ii) How does electronegativity change?

(iii) How does electron affinity change?

(i) ?

(iii) ?

• (i) Metallic character: Decreases across a period (transitions from electropositive metals to electronegative non-metals).
• (ii) Electronegativity: Increases across a period due to increased nuclear attraction for shared electrons.
• (iii) Electron Affinity: Increases across a period as elements readily accept electrons to achieve a stable octet.

Reason:** Mendeleev predicted that certain elements were yet to be discovered. Instead of treating gaps as defects, he left blank spaces and predicted their atomic masses and chemical properties based on group trends.

• Example: Mendeleev left a gap below aluminium and named the undiscovered element Eka-aluminium. When Gallium (Ga) was discovered later, its atomic weight (69.72) and properties matched Mendeleev’s predictions perfectly.

Yes, Döbereiner’s triads exist within Newlands’ octaves.

• Example: Consider Döbereiner’s triad of Alkali Metals: Lithium (Li), Sodium (Na), and Potassium (K).
• In Newlands’ octaves, if we start at Lithium (Li), Sodium (Na) is the 8th element after Li, and Potassium (K) is the 8th element after Na. Thus, they form the same column in Newlands’ octaves as well.

| ** | | 1 4 , 0 |

Valency of B:** Element B (2,8,7) has 7 valence electrons. Its valency is 8 – 7 = 1.

• Chemical Formula: The compound formed is AB₂. By criss-cross valency rule

A^{Valency of A} B¹ ⇒ AB₂

• Therefore, the valency of element A must be 2.

| Feature | Mendeleev’s Periodic Table | Modern Periodic Table |