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

(a) Order of Increasing Atomic Weights:**

H (1) < He (4) < Li (6.9) < Be (9) < B (10.8) < C (12) < N (14) < O (16) < F (19) < Ne (22.2) < Na (23) < Mg (24.3) < Al (27) < Si (28.08) < P (31) < S (32.1) < Cl (35.5) < K (39.1) < Ar (39.9) < Ca (40.1)

• (b) Order of Increasing Atomic Numbers:

H (1) < He (2) < Li (3) < Be (4) < B (5) < C (6) < N (7) < O (8) < F (9) < Ne (10) < Na (11) < Mg (12) < Al (13) < Si (14) < P (15) < S (16) < Cl (17) < Ar (18) < K (19) < Ca (20)

No, the order is slightly different between Argon (Ar) and Potassium (K). By atomic weight, Ar (39.9) comes after K (39.1), whereas by atomic number, Ar (18) comes before K (19).

H (1): 1

• He (2): 2
• Li (3): 2, 1
• Be (4): 2, 2
• B (5): 2, 3
• C (6): 2, 4
• N (7): 2, 5
• O (8): 2, 6
• F (9): 2, 7
• Ne (10): 2, 8
• Na (11): 2, 8, 1
• Mg (12): 2, 8, 2
• Al (13): 2, 8, 3
• Si (14): 2, 8, 4
• P (15): 2, 8, 5
• S (16): 2, 8, 6
• Cl (17): 2, 8, 7
• Ar (18): 2, 8, 8
• K (19): 2, 8, 8, 1
• Ca (20): 2, 8, 8, 2

Yes, elements having the same number of valence electrons in their outermost shell exhibit similar chemical properties and can be grouped together (e.g., Li, Na, K all have 1 valence electron).

(i) Which element will lie in the middle of group ‘A’?**

(ii) What will be the atomic weight of the middle element in group ‘B’?

(iii) What will be the atomic weight of the middle element in group ‘C’?

(i) ‘A’ ?

(iii) ‘C’ ?

• (i) Group ‘A’ Middle Element: Sodium (Na).
• (ii) Group ‘B’ Middle Element Atomic Weight (Sr):

Average = (40.1 + 137.3)/(2) = (177.4)/(2) = 88.7

• (iii) Group ‘C’ Middle Element Atomic Weight (Br):

Average = (35.5 + 126.9)/(2) = (162.4)/(2) = 81.2

Döbereiner could identify only three triads from the elements known at that time. His law of triads was not applicable to all known elements, making this classification system unsuccessful.

In Newlands’ octaves, Sodium (Na) resembles Lithium (Li) placed before it and Potassium (K) placed eighth after it.

The general formula for oxides of group three elements in Mendeleev’s periodic table is R_2O₃.

1. Lothar Meyer’s Basis: Atomic volume plotted against atomic weight.
2. First four elements:

• Group I: Hydrogen (H), Lithium (Li), Sodium (Na), Potassium (K).
• Group II: Beryllium (Be), Magnesium (Mg), Calcium (Ca), Zinc (Zn).

1. Elements placed in empty gaps: Scandium (Sc, Eka-boron), Gallium (Ga, Eka-aluminium), and Germanium (Ge, Eka-silicon).
2. Atomic weights corrected: Beryllium (Be) and Indium (In).
3. Reason for separate group for inert gases: Inert gases (He, Ne, Ar) are unreactive (zero valency) and were discovered later. Adding them in a separate ‘Zero Group’ did not disturb the existing arrangement.

1. Modern Periodic Law: Atomic number.
2. Number of elements in second period: 8 elements.
3. Position of isotopes: Isotopes have the same atomic number (same number of protons), so all isotopes of an element occupy the exact same single position in the modern periodic table.
4. Argon and Potassium positions: Argon has atomic number 18 (placed in Group 18, Period 3), while Potassium has atomic number 19 (placed in Group 1, Period 4). Their positions are strictly based on increasing atomic numbers.
5. Electronic configurations: x(6) = 2, 4; y(10) = 2, 8; z(18) = 2, 8, 8.

• (a) Same group: y(10) and z(18) belong to Group 18 (noble gases, 8 valence electrons).
• (b) Same period: x(6) and y(10) belong to Period 2 (both have 2 electron shells).

(a) Valency Calculation: If valence electrons (n) ≤ 4, Valency = n. If n > 4, Valency = 8 – n.

• (b) Valency down a group: Remains constant because the number of valence electrons is identical for all group members.
• (c) Group 17 Atomic Radii: Smallest = Fluorine (F, 64 pm); Largest = Astatine (At, 144 pm). Atomic radius increases down a group due to the addition of new electron shells.
• (d) Period 3 Atomic Radii: Atomic radius decreases from left to right across a period due to increased nuclear charge pulling electrons closer.
• (e) Ionization Energy trend: Decreases down a group (atomic size increases, nuclear pull weakens) and increases across a period from left to right (nuclear charge increases).
• (f) Difference:
• Electron Affinity: Energy released when an electron is added to an isolated gaseous atom (Unit: eV).
• Electronegativity: Relative tendency of an atom in a covalent molecule to attract shared pair of electrons towards itself (Unitless).
• Highest Electron Affinity Group: Group 17 (Halogens).

(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.

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 |