Periodic Classification of Elements – Numerical Problems with Stepwise Solutions
Class 10
Chemistry — Chapter 14
Board: CBSE
Suitable for board exam practice and revision
Content Bank — Important Formulas & Rules
- Atomic number (Z) = number of protons = number of electrons (in neutral atom).
- Mass number (A) = number of protons + number of neutrons = Z + n.
- Number of neutrons (n) = A − Z.
- Isotopes: atoms with same Z but different A.
- Valency (main-group approx.): For s- and p-block elements, valency ≈ number of electrons lost or gained to achieve noble gas configuration. Often: valency = group number (for metals) or 8 − group number (for non-metals in p-block).
- Electronic configuration order: 1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p ... (use aufbau principle).
- Period = highest principal quantum number (n) of occupied shell.
- Group (main-group) often determined by valence electrons: e.g., ns1 → Group 1, ns2 → Group 2, ns2 npX → Group (12 + X) or (group 13–18).
- Average atomic mass (A_avg): A_avg = Σ(fractional abundance × isotopic mass).
Problem 1. An atom has 11 protons and 12 neutrons. Find its atomic number, mass number and write its chemical symbol.
Step 1: Atomic number Z = number of protons = 11.
Step 2: Mass number A = protons + neutrons = 11 + 12 = 23.
Step 3: Chemical symbol: element with Z=11 is sodium, Na. Write with mass number as superscript: 23Na (often written as ²³Na).
Answer: Atomic number = 11, Mass number = 23, Symbol = ²³Na.
Problem 2. An element X has atomic number 16. (a) How many electrons does X have in its neutral state? (b) Write its electronic configuration. (c) Which group and period does it belong to?
Step 1: Atomic number Z = 16 → number of electrons (neutral atom) = 16.
Step 2: Electronic configuration: fill electrons: 1s² 2s² 2p⁶ 3s² 3p⁴.
Step 3: Highest occupied shell n = 3 → Period 3. Valence electrons are 3s²3p⁴ → 6 valence electrons → Group 16 (chalcogens).
Answer: (a) 16 electrons. (b) 1s² 2s² 2p⁶ 3s² 3p⁴. (c) Group 16, Period 3 (element is sulfur).
Problem 3. Find the number of neutrons in the isotope ³⁵Cl and ³⁷Cl. Explain why both isotopes occupy the same position in the periodic table.
Step 1: For ³⁵Cl: Atomic number of chlorine Z = 17 (number of protons = 17). Neutrons = A − Z = 35 − 17 = 18.
Step 2: For ³⁷Cl: Neutrons = 37 − 17 = 20.
Step 3: Both have same Z (17), so they are isotopes of chlorine. Position in periodic table depends on atomic number (Z), so isotopes occupy same position.
Answer: ³⁵Cl has 18 neutrons; ³⁷Cl has 20 neutrons. Both occupy same place because they have same atomic number (17).
Problem 4. An element Y has isotopes with masses 30 (abundance 20%) and 32 (abundance 80%). Calculate the average atomic mass of element Y.
Step 1: Convert percentages to fractions: 20% = 0.20, 80% = 0.80.
Step 2: Average mass = (0.20 × 30) + (0.80 × 32) = 6 + 25.6 = 31.6.
Step 3: Report to appropriate significant figures (here 31.6 is fine).
Answer: Average atomic mass = 31.6 u.
Problem 5. Determine the valency of an element whose electronic configuration is 1s² 2s² 2p⁶ 3s² 3p³.
Step 1: Highest shell is n=3; valence electrons in 3s² 3p³ total = 5.
Step 2: For p-block non-metals, valency = 8 − number of valence electrons = 8 − 5 = 3.
Step 3: This element is phosphorus (P), Group 15, with common valency 3 (can also show 5 in covalent compounds).
Answer: Valency = 3 (element is phosphorus).
Problem 6. A neutral atom has 20 electrons. (a) What is its atomic number and element? (b) Write its electronic configuration. (c) State its group and period.
Step 1: Atomic number Z = number of electrons = 20.
Step 2: Element with Z=20 is calcium (Ca).
Step 3: Electronic configuration: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² (or [Ar] 4s²).
Step 4: Highest shell n=4 → Period 4. Valence electrons = 4s² → Group 2 (alkaline earth metals).
Answer: (a) Z=20 → Calcium (Ca). (b) 1s²2s²2p⁶3s²3p⁶4s². (c) Group 2, Period 4.
Problem 7. If an element has electronic configuration [Ar] 4s² 3d⁵, identify the element and state one common oxidation state.
Step 1: [Ar]4s²3d⁵ corresponds to element with atomic number 25 → manganese (Mn).
Step 2: Transition metals show variable oxidation states; Mn commonly shows +2, +4, +7 etc. One common oxidation state is +2 (Mn²⁺).
Answer: Element is manganese (Mn). A common oxidation state is +2 (also +7 in permanganate, +4 in MnO₂).
Problem 8. Determine which of the following atoms/ions are isoelectronic: S²⁻, Cl⁻, Ar, K⁺, Ca²⁺. Explain your reasoning.
Step 1: Find electron counts. Argon (Ar) has 18 electrons.
Step 2: S (Z=16) → S²⁻ has 16 + 2 = 18 electrons. Cl (Z=17) → Cl⁻ has 17 + 1 = 18 electrons. K (Z=19) → K⁺ has 19 − 1 = 18 electrons. Ca (Z=20) → Ca²⁺ has 20 − 2 = 18 electrons.
Step 3: All listed species have 18 electrons, so they are isoelectronic with Ar.
Answer: S²⁻, Cl⁻, Ar, K⁺ and Ca²⁺ are all isoelectronic (each has 18 electrons).
Problem 9. An element Z forms an oxide Z₂O₃. Predict the oxidation state of Z and suggest the likely group of the element.
Step 1: Let oxidation state of Z be x. Oxygen is −2. For compound Z₂O₃: 2x + 3(−2) = 0 → 2x − 6 = 0 → x = +3.
Step 2: A +3 oxidation state is typical of Group 13 elements (e.g., Al³⁺, Ga³⁺). So Z likely belongs to Group 13.
Answer: Oxidation state of Z is +3. Z is likely a Group 13 element (e.g., Al, Ga).
Problem 10. Calculate the number of protons, neutrons and electrons in the ion ⁵⁵Fe²⁺.
Step 1: For ⁵⁵Fe, mass number A = 55. Atomic number of Fe (Z) = 26 → protons = 26.
Step 2: Neutrons = A − Z = 55 − 26 = 29.
Step 3: Ion ²⁺ means it has lost 2 electrons, so electrons = Z − 2 = 26 − 2 = 24.
Answer: Protons = 26, Neutrons = 29, Electrons = 24.
Problem 11. Determine the group and period of an element whose valence shell is 4s² 4p³.
Step 1: Valence electrons = 4s²4p³ → total 5 → this corresponds to group 15 (p³ configuration gives group 13+3 = 16? actually group 15).
Step 2: Highest occupied shell n = 4 → Period 4.
Step 3: Element is arsenic (As) which is in Group 15, Period 4.
Answer: Group 15, Period 4 (element = As).
Problem 12. The atomic mass of element X is 40 and atomic number is 20. Is this element likely to be metal or non-metal? Give reasons and name the element.
Step 1: Z=20 → element is calcium (Ca). Mass number 40 (≈ common isotope ⁴⁰Ca).
Step 2: Calcium is located left of periodic table (Group 2) and shows metallic properties (lustrous, conducts electricity, forms basic oxides like CaO).
Answer: Metal (Calcium, Ca). It is an alkaline earth metal with typical metallic properties.
Problem 13. Element A has electronic configuration 1s² 2s² 2p⁶ 3s¹. Predict its group, period and one common compound it forms.
Step 1: Configuration corresponds to Na (Z=11). Highest shell n=3 → Period 3.
Step 2: Valence 3s¹ → Group 1 (alkali metal).
Step 3: Common compound: NaCl (sodium chloride) where Na forms Na⁺ and Cl forms Cl⁻.
Answer: Group 1, Period 3 (element = Na). Common compound: NaCl.
Problem 14. A neutral atom has electronic configuration [Ne] 3s² 3p⁶ 4s². Identify the element and determine how many electrons are in its outermost shell.
Step 1: [Ne]3s²3p⁶4s² corresponds to element with Z = 20 (Ca): because [Ne] = 10, plus 3s²3p⁶ = 8 (total 18) plus 4s² = 2 → total 20.
Step 2: Outermost shell is n=4, electrons in 4s² → 2 electrons in outermost shell.
Answer: Element = Calcium (Ca). Outermost shell electrons = 2.
Problem 15. Calculate the number of electrons, protons and neutrons in the isotope ¹⁴C and explain relevance to dating organic samples.
Step 1: For ¹⁴C, carbon atomic number Z=6 → protons = 6, electrons (neutral) = 6.
Step 2: Neutrons = A − Z = 14 − 6 = 8.
Step 3: Relevance to dating: ¹⁴C is radioactive (half-life ≈ 5730 years) and is used in radiocarbon dating of organic remains by measuring remaining ¹⁴C.
Answer: Protons = 6, Electrons = 6, Neutrons = 8. ¹⁴C is used in radiocarbon dating of organic samples.
Problem 16. An element M forms M₂O₇. Determine the oxidation state of M and suggest a possible family (group) it may belong to.
Step 1: Let oxidation state of M be x. For M₂O₇: 2x + 7(−2) = 0 → 2x − 14 = 0 → x = +7.
Step 2: +7 oxidation state is common for elements like manganese (Mn) in permanganate. Manganese is a transition metal. So M may be a transition metal (e.g., Mn forming Mn₂O₇ is hypothetical; Mn⁷⁺ occurs in compounds like Mn₂O₇ as manganese(VII) oxide exists).
Answer: Oxidation state = +7. M is likely to be a transition element capable of high oxidation states (example Mn).
Problem 17. Determine the element which has 2 electrons in the outermost shell and is in Period 3. Give its symbol and one property.
Step 1: Period 3 elements: Na (3s¹), Mg (3s²), Al (3s²3p¹), ... Outermost shell with 2 electrons → configuration 3s² → element = Mg (magnesium).
Step 2: One property: magnesium is an alkaline earth metal, fairly reactive (reacts with steam to form MgO and H₂), forms +2 ions.
Answer: Magnesium (Mg). Property: forms Mg²⁺, reacts to form basic oxides like MgO.
Problem 18. If an element has atomic number 9, write its electronic configuration, state its group, and indicate whether it is metal or non-metal.
Step 1: Z = 9 → element is fluorine (F).
Step 2: Electronic configuration: 1s² 2s² 2p⁵.
Step 3: Valence electrons = 7 → Group 17 (halogens). Fluorine is a non-metal and the most electronegative element.
Answer: Config = 1s²2s²2p⁵. Group 17 (halogen). Non-metal (very reactive, forms F⁻ ions).
Problem 19. Two isotopes of element Z are ²³Na and ²²Na. Calculate the number of neutrons in each. Which isotope is more stable (qualitatively) and why?
Step 1: For ²³Na: Z (Na) = 11 → neutrons = 23 − 11 = 12.
Step 2: For ²²Na: neutrons = 22 − 11 = 11.
Step 3: Stability: ²³Na is the stable, naturally abundant isotope. ²²Na is radioactive (short-lived) — isotopes with odd neutron-proton ratios or far from valley of stability tend to be less stable.
Answer: ²³Na: 12 neutrons; ²²Na: 11 neutrons. ²³Na is more stable and naturally abundant; ²²Na is radioactive.
Problem 20. Explain using periodic table how you would decide which of Na and Mg has higher first ionisation energy. Provide the expected order and reason.
Step 1: Na and Mg are in same period (Period 3) with Na to the left of Mg.
Step 2: Across a period, ionisation energy generally increases because nuclear charge increases and atomic radius decreases, so electrons are held more strongly.
Step 3: Therefore Mg (to the right) has higher first ionisation energy than Na: IE(Na) < IE(Mg).
Answer: Mg has higher first ionisation energy than Na because Mg’s effective nuclear charge is greater and its valence electrons are held more tightly.