Chemistry · Classification of elements and periodicity · chapter 3 of 9

Class 11 Chemistry Chapter 3: Classification of Elements and Periodicity in Properties worksheet with answer key

A 20-mark worksheet on Classification of Elements and Periodicity in Properties for about 40 minutes of class or homework: 5 multiple-choice, 3 very short answer and 3 short answer questions, written from the NCERT chapter and checked by a second reviewer. Print the PDF for the class; the answer key carries the marking scheme. Teachers may photocopy it freely.

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Question 1
An element X has atomic number 38. The element whose chemical properties would be most like those of X has the atomic number
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Answer: (b) 20

Z = 38 is [Kr]5s²: period 5, group 2. Z = 20 is [Ar]4s², also group 2, with the same ns² valence configuration and so similar properties. 37 and 39 are its neighbours in the period (groups 1 and 3), and 36 is a noble gas.

Question 2
Which of the following elements is a metalloid?
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Answer: (a) germanium

Germanium borders the zig-zag line and shows properties of both metals and non-metals. Gallium and tin are metals; selenium is a non-metal.

Question 3
Which of the following pairs is isoelectronic?
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Answer: (c) N³⁻ and Ne

N³⁻ has 7 + 3 = 10 electrons and Ne has 10. Na⁺ (10) and Mg (12), O²⁻ (10) and Cl⁻ (18), K⁺ (18) and Ne (10) differ.

Question 4
Moseley plotted the square root of the frequency of the characteristic X-rays of elements against another property and obtained a straight line. That property is
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Answer: (c) atomic number

The plot of √ν against atomic number (Z) is a straight line; against atomic mass it is not. This showed atomic number to be the more fundamental property.

Question 5
Sulphur (atomic weight 32), selenium and tellurium (atomic weight 127.6) behave as a Dobereiner triad. The atomic weight the Law of Triads predicts for selenium is
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Answer: (b) 79.8

In a triad the middle element's atomic weight is about half way between the other two: (32 + 127.6)/2 = 79.8. 95.6 is the difference, 47.8 half of it, 159.6 the sum.

Question 6 · 2 marks
Why are the lanthanoids and actinoids placed in separate rows at the foot of the periodic table?
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Fourteen elements in each of periods 6 and 7 fill the 4f and 5f orbitals and have very similar properties. Placing them in the main body would break the table's structure (it would need 32 columns), and keeping them apart preserves the principle that elements with similar properties stand in a single column.

  • to maintain the structure of the table (1)
  • to keep elements of similar properties in one column / they resemble one another (1)
Question 7 · 2 marks
Why is the chemistry of the early actinoids more complicated than that of the corresponding lanthanoids? What are transuranium elements?
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The early actinoids can show a large number of oxidation states, and they are radioactive, many made only in nanogram quantities, so their chemistry is complicated and not fully studied. The elements after uranium (Z > 92) are called transuranium elements.

  • many possible oxidation states (and radioactivity) of actinoids (1)
  • transuranium: elements after uranium (1)
Question 8 · 2 marks
Lithium and beryllium form compounds with pronounced covalent character, while the other members of their groups form mainly ionic compounds. Give two reasons for this behaviour of the first members.
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The first member of a group is anomalous because of (i) its small size and large charge/radius ratio, which lets the cation distort the anion's electron cloud, and (ii) its high electronegativity compared with the rest of the group. (It also has only four valence orbitals, 2s and 2p.)

  • small size / large charge to radius ratio (1)
  • high electronegativity (or only four valence orbitals) (1)
Question 9 · 3 marks
Classify MgO, Al₂O₃ and SO₃ as basic, acidic or amphoteric, and write one equation for each to show its nature.
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MgO is basic: MgO + 2HCl → MgCl₂ + H₂O. SO₃ is acidic: SO₃ + H₂O → H₂SO₄ (or SO₃ + 2NaOH → Na₂SO₄ + H₂O). Al₂O₃ is amphoteric, reacting with both acids and bases: Al₂O₃ + 6HCl → 2AlCl₃ + 3H₂O and Al₂O₃ + 2NaOH + 3H₂O → 2Na[Al(OH)₄]. The trend follows the period: basic on the left, amphoteric in the centre, acidic on the right.

  • MgO basic, with equation (1)
  • SO₃ acidic, with equation (1)
  • Al₂O₃ amphoteric, with an equation with acid or base (1)
Question 10 · 3 marks
What did Moseley observe in the X-ray spectra of the elements? How did this change the Periodic Law, and what became of Mendeleev's placing of tellurium before iodine?
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Moseley found that a plot of √ν (ν, the frequency of the characteristic X-rays) against atomic number is a straight line, while the plot against atomic mass is not; atomic number is the more fundamental property. The law became: the physical and chemical properties of the elements are periodic functions of their atomic numbers. Te (Z = 52) has a lower atomic number than I (Z = 53), so ordering by Z places Te before I naturally; the anomaly disappears.

  • √ν against Z is a straight line, not against atomic mass (1)
  • Modern Periodic Law in terms of atomic number (1)
  • Te (52) before I (53) follows from atomic number (1)
Question 11 · 3 marks
Mendeleev left a gap below silicon in his table and called the missing element eka-silicon. (a) Explain how he could predict the formulas of its oxide and chloride. (b) He predicted an atomic weight of 72 and a density of 5.5 g cm⁻³; germanium, discovered later, has 72.6 and 5.36 g cm⁻³. What do these figures show about his gaps? (c) What did the success of such predictions do for his Periodic Table?
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(a) Mendeleev relied on the similarities in the empirical formulas of the compounds of elements in a group. Silicon forms SiO₂ and SiCl₄, so eka-silicon would form EO₂ and ECl₄; germanium was indeed found to form GeO₂ and GeCl₄. (b) The predicted and found values agree closely (72 against 72.6; 5.5 against 5.36 g cm⁻³). The gap was a real, then undiscovered element, and its properties could be foretold from its place among elements of similar properties. (c) The boldness of these quantitative predictions and their eventual success made Mendeleev and his Periodic Table famous.

  • (a) EO₂ and ECl₄, from similar empirical formulas within silicon's group (1)
  • (b) close agreement: the gap was an undiscovered element whose properties followed from its place in the table (1)
  • (c) the success made Mendeleev and his table famous / gained it acceptance (1)
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