CBSE Class 11 Chemistry Classification of Elements and Periodicity in Properties Worksheet Set 03

Classification of Elements and Periodicity Worksheet

Question 1. What is the basic difference in approach between Mendeleev's periodic law and the modern periodic law?
Answer:
Mendeleev's law: the properties of elements are a periodic function of their atomic masses.
Modern periodic law: the physical and chemical properties of elements are a periodic function of their atomic numbers.
The change came from Moseley's X-ray work. He found that the square root of the frequency of the X-rays emitted by an element varies in a straight line with atomic number, but not with atomic mass.
So atomic number, which is the number of protons and also the nuclear charge, is the more fundamental property of an element.
In simple words: Mendeleev arranged by weight; the modern table arranges by the number of protons.

Teacher's Note:
1. Mention Moseley whenever the change from mass to number is asked.
2. A good extra point: order by atomic number puts Ar before K and Te before I without any exception.

Question 2. Describe two merits of the long form of the periodic table over Mendeleev's periodic table.
Answer:
1. It removes the problem of the position of isotopes. Isotopes have the same atomic number, so they occupy the same position, and no separate place is needed for them.
2. It connects the position of an element with its electronic configuration. The period tells the outermost shell and the group tells the number of valence electrons, so the table has a theoretical basis.
Other merits: elements in a group are really similar (no A and B sub-groups to confuse), and the lanthanoids and actinoids are placed separately at the bottom.
In simple words: Isotopes fit naturally, and the table follows the way electrons are filled.

Teacher's Note:
1. Quote the two points exactly for a 2-mark answer.
2. Do not write 'it is based on atomic number' alone; that is the law, not a merit.

Question 3. What is periodicity in the properties of elements? What is its cause?
Answer:
When the elements are arranged in order of increasing atomic number, elements with similar properties recur after regular intervals. This repetition is called periodicity.
Cause: the same outermost (valence shell) electronic configuration recurs at regular intervals as the atomic number rises.
For example, every alkali metal ends in ns¹ and every halogen in ns²np⁵, so each group shows similar properties.
In simple words: Similar outer electron arrangements come back again and again, so similar properties come back as well.

Teacher's Note:
1. Never say 'because of atomic number alone'; the cause is the repeating electron configuration.
2. Intervals of 2, 8, 8, 18, 18 and 32 elements correspond to the period lengths.

Question 4. (a) State the modern periodic law and list the main features of the long form of the periodic table. (b) Give the general outer electronic configurations of the s-, p-, d- and f-block elements.
Answer:
(a) The physical and chemical properties of the elements are periodic functions of their atomic numbers. Main features:
1. The aufbau principle and the electronic configuration of atoms provide the theoretical basis of the table.
2. There are 7 periods. The period number is the highest principal quantum number (n) of the elements in it.
3. The periods have 2, 8, 8, 18, 18 and 32 elements, and the seventh period is the same as the sixth in capacity (32).
4. There are 18 vertical columns (groups). Elements in a group have similar outer electronic configurations.
5. Groups 1, 2 and 13 to 17 contain the main group (representative) elements; groups 3 to 12 contain the transition elements.
6. The 14 lanthanoids and 14 actinoids are placed in two separate panels at the bottom.
(b) s-block: ns^1–2. p-block: ns²np^1–6. d-block: (n−1)d^1–10ns^0–2. f-block: (n−2)f^1–14(n−1)d^0–1ns².
In simple words: The table is built from electron configurations: periods from n, groups from outer electrons, blocks from the last sub-shell filled.

Teacher's Note:
1. For a 5-mark answer give at least five features and all four configurations.
2. He (1s²) is an s-block element placed in group 18 with the noble gases.

Question 5. Give the general electronic configuration of the f-block elements and describe the characteristic properties of the lanthanoids and actinoids.
Answer:
General configuration: (n−2)f^1–14(n−1)d^0–1ns². The last electron enters the f sub-shell of the pre-penultimate shell (4f for lanthanoids, 5f for actinoids).
They belong to group 3 and are placed in two rows at the bottom of the table.
Characteristic properties:
1. Very small differences in chemical reactivity within each series.
2. Metallic, electropositive, with high melting and boiling points.
3. Most compounds are coloured, and the elements are paramagnetic because of unpaired electrons.
4. The common oxidation state is +3; some also show +2, +4, +5 and +6, which helps catalytic activity.
5. Atomic size falls slightly along each series (lanthanoid contraction and actinoid contraction).
6. Actinoids are radioactive, and the elements after uranium (transuranium elements) are man-made.
In simple words: Lanthanoids and actinoids are the 4f and 5f series; they look alike chemically and show +3 mainly.

Teacher's Note:
1. Remember: lanthanoids 58 to 71 and actinoids 90 to 103.
2. Describe size change as a contraction, not an expansion.

Question 6. Define atomic radius. What are the factors that affect it?
Answer:
Atomic radius is the distance from the centre of the nucleus to the outermost shell of electrons. Practically it is found as half the distance between the nuclei of two bonded identical atoms.
Factors:
1. Number of shells: the radius increases when the number of shells increases.
2. Nuclear charge: for the same number of shells, a larger nuclear charge pulls the electrons in and the radius decreases.
3. Shielding (screening) effect: inner electrons reduce the pull of the nucleus on the valence electrons, so greater shielding gives a larger radius.
In simple words: Size depends on how many shells there are, how strongly the nucleus pulls and how much the inner electrons shield.

Teacher's Note:
1. The size cannot be measured directly because the electron cloud has no sharp edge.
2. Always give the type of radius (covalent, metallic or van der Waals) when quoting a value.

Question 7. Explain why cations are smaller and anions are larger than their parent atoms.
Answer:
Cations: the removal of one or more electrons (often the whole outer shell) leaves the same nuclear charge on fewer electrons. The effective nuclear charge per electron rises, so the electrons are pulled closer and the radius falls. Example: Na = 186 pm and Na⁺ = 95 pm.
Anions: the addition of electrons increases electron–electron repulsion while the nuclear charge is unchanged. The effective nuclear charge per electron falls and the electron cloud expands. Example: F = 64 pm and F⁻ = 136 pm.
In simple words: Fewer electrons means a tighter pull; more electrons means more repulsion and a bigger cloud.

Teacher's Note:
1. Always compare the same element before and after ion formation.
2. Quote one numerical example for each to earn the full marks.

Question 8. Consider the species N³⁻, O²⁻, F⁻, Na⁺, Mg²⁺ and Al³⁺. (a) What is common in them? (b) Arrange them in the order of increasing ionic radii.
Answer:
(a) Each species has 10 electrons: N³⁻ (7 + 3), O²⁻ (8 + 2), F⁻ (9 + 1), Na⁺ (11 − 1), Mg²⁺ (12 − 2) and Al³⁺ (13 − 3). They are isoelectronic.
(b) The nuclear charges are +7, +8, +9, +11, +12 and +13. For the same number of electrons, a higher nuclear charge pulls the electrons in more strongly, so the radius is smaller.
Increasing order: Al³⁺ < Mg²⁺ < Na⁺ < F⁻ < O²⁻ < N³⁻.
In simple words: Same 10 electrons, so the one with the most protons is the smallest.

Teacher's Note:
1. Count electrons as Z minus the positive charge or Z plus the negative charge.
2. Write the nuclear charge under each ion before arranging.

Question 9. Arrange the following in the order of increasing radii: (a) I, I⁺, I⁻ (b) F, Cl, Br.
Answer:
(a) I⁺ has lost an electron, so it is the smallest. I⁻ has gained an electron, so it is the largest: I⁺ < I < I⁻.
(b) F, Cl and Br are in group 17. The size increases down the group because new shells are added: F < Cl < Br.
In simple words: A cation is smaller than its atom, an anion is larger; down a group the atoms grow.

Teacher's Note:
1. Part (a) compares species of the same element; part (b) compares different atoms of one group.
2. Typical values: F 64 pm, Cl 99 pm, Br 114 pm (covalent radii).

Question 10. How does the atomic radius vary (a) in a period and (b) in a group? Explain the variation.
Answer:
(a) In a period the radius decreases from left to right. For period 2: Li 152, Be 111, B 88, C 77, N 74, O 66, F 64 pm. All outer electrons are in the same shell, but the nuclear charge increases by one at every step, so the effective nuclear charge increases and the electrons are drawn closer to the nucleus.
(b) In a group the radius increases from top to bottom. For group 1: Li 152, Na 186, K 231, Rb 244, Cs 262 pm. The principal quantum number n increases, so the valence electrons are farther from the nucleus. The inner electrons also shield them, so the larger nuclear charge cannot make up for the extra shell.
In simple words: Across a period the shell is the same but the pull grows; down a group a new shell is added.

Teacher's Note:
1. Noble gas radii are van der Waals radii, so they appear larger than those of the preceding halogens.
2. Mention the three causes: n, nuclear charge and shielding.

Question 11. Use the periodic table to answer: (a) Identify an element with five electrons in the outer sub-shell. (b) Identify an element that would tend to lose two electrons. (c) Identify an element that would tend to gain two electrons.
Answer:
(a) Five electrons in the outer p sub-shell means ns²np⁵, the configuration of group 17. Examples: F, Cl, Br, I. Chlorine is an answer.
(b) An element that loses two electrons has ns² outside a noble-gas core, so it belongs to group 2 (alkaline earth metals). Example: magnesium (Mg, 3s²).
(c) An element that gains two electrons has ns²np⁴ (six outer electrons) and belongs to group 16. Examples: oxygen or sulphur.
In simple words: Write the outer configuration first; then read the group from it.

Teacher's Note:
1. Group 17 needs one electron, group 16 needs two, group 2 loses two.
2. State the group as well as the element for full marks.

Question 12. Name (i) the only non-metal that is a liquid at room temperature, (ii) the only metal that is a liquid at room temperature, and (iii) three metals with melting points near room temperature.
Answer:
(i) Bromine (Br₂, group 17) is the only liquid non-metal.
(ii) Mercury (Hg, group 12) is the only liquid metal at room temperature.
(iii) Gallium (melting point 29.8 °C), caesium (28.5 °C) and francium (about 27 °C) are metals that melt close to room temperature.
In simple words: Bromine and mercury are the two liquid elements at room temperature; Ga, Cs and Fr melt just above it.

Teacher's Note:
1. Note that Br is in group 17 (p-block) and Hg in group 12 (d-block).
2. Hg is the only d-block element that is a liquid at room temperature.