Practise 22 real Federal Board questions on Atomic Structure from 2023 to 2026. See the answer and explanation as soon as you answer.
Atomic Structure is one of the 19 main chapters in our Federal Board FSC Chemistry question bank (11th and 12th class), and about 4% of its real questions come from it. Tayyari Karo has 22 real questions on this chapter from 2023 to 2026; 8 are multiple choice and marked instantly, and the rest are short and long questions with a model answer. By difficulty: 10 easy, 8 medium, 4 hard.
Work through the questions below, then open the explanation for each one. Redo the ones you missed a few days later.
These are real past-paper questions. Think of your answer first, then open the explanation.
Br has higher electron affinity than K due to:
Answer: B. Atomic size
Br's much smaller atomic size (compared to K) means an added electron feels stronger nuclear attraction, giving it much higher electron affinity.
If an element 'X' has 1st ionization energy 495 kJ/mol and 2nd ionization energy 4560 kJ/mol, this indicates that is/are valence electrons.
Answer: A. 1
A huge jump between IE1 and IE2 (495→4560) shows the 2nd electron is removed from a stable inner shell, so the element has only 1 valence electron.
The number of valence electrons in an atom can be deduced from its:
Answer: C. Group number
For main-group elements, the group number directly gives the number of valence electrons.
If ²⁹₆₃Cu and ⁷⁹₁₉₇Au are used as anodes for production of X-rays, then "Cu" will produce the X-rays of:
Answer: C. Longer wavelength
By Moseley's law, characteristic X-ray energy increases with atomic number Z. Cu (Z=29) has lower Z than Au (Z=79), so Cu produces lower-energy, longer-wavelength X-rays.
The third line of Balmer series appears due to the transition of electron from:
Answer: C. n₅ → n₂
Balmer series lines (transitions to n=2): 1st = n3→n2, 2nd = n4→n2, 3rd = n5→n2.
The wave number (ṽ) of a radiation having λ = 2×10⁸ nm will be:
Answer: C. 0.5×10⁸ nm⁻¹
Wave number ṽ = 1/λ. Using the numbers as given, ṽ = 1/2 × 10⁸ = 0.5×10⁸ nm⁻¹.
Compare s and p orbitals. (Three points)
1. s orbitals are spherical in shape, while p orbitals are dumbbell-shaped.
2. s orbitals have no directional/angular nodes, while p orbitals have one angular node and are directional (px, py, pz).
3. Each shell has only one s orbital, while each shell (from n=2) has three degenerate p orbitals.
What is mass spectrometry? Explain its working and its applications in determining the relative atomic mass.
Mass spectrometry is an analytical technique that ionizes a sample and separates the resulting ions according to their mass-to-charge (m/z) ratio.
Working: the vaporized sample is bombarded by high-energy electrons in the ionization chamber, forming positive ions; these are accelerated by an electric field into a beam, which is deflected by a magnetic field into a circular path whose radius depends on m/z; a detector records the abundance of ions at each m/z value, producing the mass spectrum.
Determining relative atomic mass: for an element with several isotopes, each isotope's peak gives its exact mass and its relative abundance (from peak height/intensity). …
The full model answer is in the app.
| Session | Questions |
|---|---|
| 2026 Annual | 5 |
| 2025 1st Annual | 4 |
| 2025 2nd Annual | 6 |
| 2024 Annual | 3 |
| 2023 Annual | 4 |
22 real Federal Board (FBISE) Chemistry questions on Atomic Structure from 2023 to 2026, for 11th and 12th class. Each one shows the session it came from.
No. We could not find an official FBISE marking scheme for these papers, so the answers and explanations are written by Tayyari Karo for practice. Check anything important with your teacher or textbook.
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