Class 9 Science Test Paper Chapter Structure of Atom | GSEB Board | Detailed Solutions & Marking Scheme

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 VEDANT CLASSES – A Pathway to Success...

VEDANT IGNITE TEST SERIES 2026–27

Class: IX (GSEB)
Subject: Science
Chapter: Structure of Atom
Topics: Fundamental Particles, Thomson's Model, Rutherford's Model, Bohr's Model, Electronic Configuration (First 20 Elements), Isotopes, Isobars and Average Atomic Mass
Marks: 30
Time: 1 Hour

Instructions:

  1. All questions are compulsory as per internal choice mentioned.
  2. Use only NCERT-based concepts and data.
  3. Calculators are not allowed.
  4. Show necessary calculations wherever required.

Section A (1 Mark Each) – 7 Marks

Multiple Choice Questions (3 × 1 = 3 Marks)

  1. An atom contains 11 protons, 12 neutrons and 11 electrons. Its mass number is:
    (A) 11 (B) 12 (C) 22 (D) 23
  2. The electronic configuration of magnesium (Atomic number 12) is:
    (A) 2, 8, 1 (B) 2, 8, 2 (C) 2, 7, 3 (D) 2, 8, 8
  3. Chlorine occurs as 75% Cl-35 and 25% Cl-37. The average atomic mass of chlorine is:
    (A) 35 u (B) 35.5 u (C) 36 u (D) 36.5 u

Fill in the Blanks (2 × 1 = 2 Marks)

  1. The maximum number of electrons that can be accommodated in the K-shell is ________. (2, 8, 18)
  2. An atom having electronic configuration 2, 8, 7 has atomic number ________. (15, 17, 19)

True or False (2 × 1 = 2 Marks)

  1. The number of neutrons in an atom is equal to the difference between its mass number and atomic number. ________
  2. Isotopes of an element have the same atomic number but different mass numbers. ________

Section B (Write Any 3) – 6 Marks

  1. Calculate the number of protons, neutrons and electrons in a sodium atom having atomic number 11 and mass number 23.
  2. Write the electronic configuration of the following atoms:
    (a) Oxygen (Atomic number 8)
    (b) Calcium (Atomic number 20)
  3. An atom contains 17 electrons and 18 neutrons. Determine:
    (a) Atomic number
    (b) Number of protons
    (c) Mass number
  4. The average atomic mass of boron is 10.8 u. Boron has two isotopes, B-10 and B-11. If the abundance of B-10 is 20%, calculate the abundance of B-11.
  5. An element X has mass number 39 and atomic number 19. Find:
    (a) Number of electrons
    (b) Number of neutrons

Section C (Write Any 3) – 9 Marks

  1. Complete the following table:
ElementAtomic NumberElectronic Configuration
Nitrogen7________
Aluminium13________
Argon18________
  1. An atom has mass number 56 and contains 26 protons.

(a) Find the atomic number.
(b) Find the number of neutrons.
(c) Write the electronic configuration.

  1. The isotopes of neon are present as follows:
  • Ne-20 = 90%
  • Ne-21 = 1%
  • Ne-22 = 9%

Calculate the average atomic mass of neon.

  1. An atom has electronic configuration 2, 8, 8, 1.

(a) Identify the element.
(b) State its atomic number.
(c) Find the total number of valence electrons.

  1. Calculate the average atomic mass of an element having two isotopes:
  • X-63 with abundance 69%
  • X-65 with abundance 31%

Section D (Write Any 2) – 8 Marks

  1. Complete the following table and answer the questions:
ElementAtomic NumberMass Number
A1531
B1735
C2040

For each element, calculate:

(a) Number of protons
(b) Number of electrons
(c) Number of neutrons
(d) Electronic configuration

  1. An element exists naturally as two isotopes:
  • Isotope P: Mass = 24 u, abundance = 78%
  • Isotope Q: Mass = 26 u, abundance = 22%

Calculate:

(a) Average atomic mass of the element
(b) Mass contribution of each isotope
(c) Explain why isotopes have different mass numbers despite belonging to the same element.

  1. The electronic configurations of three elements are given below:
  • X = 2, 8, 1
  • Y = 2, 8, 7
  • Z = 2, 8, 8, 2

Answer the following:

(a) Determine the atomic numbers of X, Y and Z.

(b) Identify the elements.

(c) Calculate the total number of valence electrons in all three elements together.

(d) Arrange the elements in increasing order of number of shells.

End of Question Paper

VEDANT CLASSES – A Pathway to Success...

VEDANT IGNITE TEST SERIES 2026–27

Solved Answer Key with Marking Scheme

Class: IX (GSEB)
Subject: Science
Chapter: Structure of Atom
Marks: 30

Section A (7 Marks)

MCQs (1 Mark Each)

Q1. Mass Number = Protons + Neutrons = 11 + 12 = 23
Answer: (D) 23 [1 Mark]

Q2. Magnesium (Z = 12) = 2, 8, 2
Answer: (B) 2,8,2 [1 Mark]

Q3.
Average Atomic Mass

= (35 × 75 + 37 × 25)/100

= (2625 + 925)/100

= 35.5 u

Answer: (B) 35.5 u [1 Mark]

Fill in the Blanks (1 Mark Each)

Q4. 2 ✔️ [1 Mark]

Q5. 17 ✔️ [1 Mark]

True / False (1 Mark Each)

Q6. True ✔️ [1 Mark]

Q7. True ✔️ [1 Mark]

Section B (Write Any 3)

Q8. (2 Marks)

Given:

Atomic Number = 11

Mass Number = 23

Protons = 11 (½ Mark)

Electrons = 11 (½ Mark)

Neutrons = 23 − 11 = 12 (1 Mark)

Total = 2 Marks

Q9. (2 Marks)

(a) Oxygen (Z = 8)

Electronic configuration = 2,6 (1 Mark)

(b) Calcium (Z = 20)

Electronic configuration = 2,8,8,2 (1 Mark)

Total = 2 Marks

Q10. (2 Marks)

Electrons = 17

Therefore,

Atomic Number = 17 (½ Mark)

Protons = 17 (½ Mark)

Mass Number = 17 +18 = 35 (1 Mark)

Total = 2 Marks

Q11. (2 Marks)

Abundance of B-10 = 20%

Total abundance =100%

Abundance of B-11

=100−20

= 80% (2 Marks)

Q12. (2 Marks)

Atomic Number =19

Mass Number =39

Electrons = 19 (1 Mark)

Neutrons =39−19= 20 (1 Mark)

Section C (Write Any 3)

Q13. (3 Marks)

ElementElectronic Configuration
Nitrogen2,5
Aluminium2,8,3
Argon2,8,8

Marking

Each correct configuration = 1 Mark

Q14. (3 Marks)

Given:

Mass Number =56

Protons =26

(a) Atomic Number = 26 (1 Mark)

(b) Neutrons

=56−26

= 30 (1 Mark)

(c) Electronic Configuration

2,8,14,2 (1 Mark)

Q15. (3 Marks)

Average Atomic Mass

=(20×90 +21×1 +22×9)/100

=(1800+21+198)/100

=2019/100

= 20.19 u

Calculation =2 Marks

Final Answer =1 Mark

Q16. (3 Marks)

Electronic Configuration

2,8,8,1

(a) Element = Potassium (K) (1 Mark)

(b) Atomic Number = 19 (1 Mark)

(c) Valence Electrons = 1 (1 Mark)

Q17. (3 Marks)

Average Atomic Mass

=(63×69 +65×31)/100

=(4347+2015)/100

=6362/100

= 63.62 u

Formula =1 Mark

Calculation =1 Mark

Answer =1 Mark

Section D (Write Any 2)

Q18. (4 Marks)

ElementProtonsElectronsNeutronsElectronic Configuration
A1515162,8,5
B1717182,8,7
C2020202,8,8,2

Marking Scheme

Correct Protons & Electrons (All) – 1 Mark

Correct Neutrons (All) – 1 Mark

Correct Electronic Configuration of A & B – 1 Mark

Correct Electronic Configuration of C – 1 Mark

Q19. (4 Marks)

(a)

Average Atomic Mass

=(24×78 +26×22)/100

=(1872+572)/100

=2444/100

= 24.44 u (2 Marks)

(b)

Mass contribution of P

=24×78/100

= 18.72 u (1 Mark)

Mass contribution of Q

=26×22/100

= 5.72 u

(Award full 1 mark if both contributions are correctly calculated.)

(c)

Explanation:

Isotopes have the same number of protons (same atomic number) but different numbers of neutrons, so their mass numbers are different. (1 Mark)

Q20. (4 Marks)

(a)

Atomic Numbers

X = 11

Y = 17

Z = 20

(1½ Marks)

(b)

X = Sodium

Y = Chlorine

Z = Calcium

(1 Mark)

(c)

Valence Electrons

1 +7 +2

= 10

(½ Mark)

(d)

Increasing order of number of shells

X = Y < Z

(or)

Sodium = Chlorine < Calcium

(1 Mark)

Grand Total = 30 Marks

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