CBSE Class 11 Chemistry Classification of Elements and Periodicity in Properties MCQs Set 05

Answer: (b) p-block elements

Explanation:
1. The last electron enters a p orbital in ns²np¹ to ns²np⁶.
2. These elements lie in groups 13 to 18 (with ns²np⁶ for the noble gases).
3. Hence they form the p-block.

Teacher's Note:
1. A block is named after the sub-shell that receives the last electron.
2. Learn the four general configurations together: s, p, d and f.

Answer: (b) atomic number

Explanation:
1. Moseley showed that the square root of the X-ray frequency varies linearly with atomic number, not with atomic mass.
2. So the modern periodic law says that the properties of elements are periodic functions of their atomic numbers.

Teacher's Note:
1. Mendeleev used atomic mass; the modern table uses atomic number.
2. Moseley's X-ray work is the evidence for this change.

Answer: (c) 0

Explanation:
1. The fourth period fills 4s, 3d and 4p only.
2. 4d orbitals begin to fill in the fifth period (after 5s).
3. So no element of the fourth period has a 4d electron.

Teacher's Note:
1. Period number is the highest principal quantum number n of the valence shell.
2. Use the filling order 4s, 3d, 4p, 5s, 4d, 5p.

Answer: (c) d-block

Explanation:
1. By the aufbau principle Ca is 1s² 2s² 2p⁶ 3s² 3p⁶ 4s², an s-block element.
2. Without it, the last two electrons would go into 3d: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d².
3. An element whose last electron enters a d orbital belongs to the d-block.

Teacher's Note:
1. The block follows the sub-shell of the last electron.
2. Write the alternative configuration first and then read off the block.

Answer: (c) d-block elements

Explanation:
1. The configuration of La is [Xe] 5d¹ 6s².
2. The last electron enters a 5d orbital, so La is placed in group 3 as a d-block element.
3. The 14 lanthanoids that follow it (58 to 71) fill the 4f sub-shell.

Teacher's Note:
1. La is the first member of the 5d series; Ce to Lu form the 4f series.
2. Do not confuse 'lanthanoid' with 'f-block element' for La itself.

Answer: (a) metal

Explanation:
1. The configuration has 12 electrons, so Z = 12 and the element is magnesium.
2. It has two valence electrons in 3s, which it loses easily to form Mg²⁺.
3. So it is a metal (an s-block, group 2 element).

Teacher's Note:
1. Count the electrons to get Z.
2. Elements that lose their ns electrons readily are metals.

Answer: (d) f-block

Explanation:
1. Elements from Z = 58 to 71 are filling the 4f sub-shell and are called lanthanoids.
2. Their general configuration is (n−2)f^1–14(n−1)d^0–1ns².
3. So Ce belongs to the f-block.

Teacher's Note:
1. Lanthanoids: Z = 58 to 71. Actinoids: Z = 90 to 103.
2. Both series are placed at the bottom of the table.

Answer: (c) 33

Explanation:
1. The configuration has 15 electrons, so the element is phosphorus (group 15).
2. The next element in group 15 is arsenic. Period 4 has 18 elements, so the element below P has Z = 15 + 18 = 33 (As).

Teacher's Note:
1. Going from period 3 to period 4 in a group, add 18 for groups 13 to 18 (add 8 for groups 1 and 2).
2. Check: As is 4s² 3d¹⁰ 4p³, which has the same ns²np³ outer pattern.

Answer: (c) For transition elements, the 3d orbitals are filled with electrons after the 3p orbitals and before the 4s orbitals.

Explanation:
1. The 4s orbital is lower in energy than 3d, so 4s is filled before 3d in the first transition series.
2. The 3d orbitals are therefore filled after 4s (and after 3p), not before 4s.
3. The other three statements are correct.

Teacher's Note:
1. Learn the filling order 3p, 4s, 3d, 4p.
2. In a question with 'not correct', test each statement on its own.

Answer: (a) maximum principal quantum number of any element in the period

Explanation:
1. The period number is the number of the outermost (valence) shell.
2. That is the highest principal quantum number n of the elements in the period.
3. Example: Na (3s¹) is in period 3 because its highest n is 3.

Teacher's Note:
1. Number of shells = period number.
2. Example: Fe, [Ar]3d⁶ 4s², is in period 4 because the highest n is 4.

Answer: (d) lanthanoids

Explanation:
1. Filling of 4f takes place from cerium (Z = 58) to lutetium (Z = 71).
2. These 14 elements follow lanthanum and are called lanthanoids.
3. Actinoids fill 5f, and transition elements fill d orbitals.

Teacher's Note:
1. 4f means lanthanoids; 5f means actinoids.
2. Lanthanoids are also called rare earth elements.

Answer: (a) Mg, Ba

Explanation:
1. Mg and Ba both have the outer configuration ns² and are alkaline earth metals of group 2.
2. Na is in group 1, Cu in group 11 and Cl in group 17.

Teacher's Note:
1. Same group means same number of valence electrons.
2. Write the outer configuration of each element and compare.

Answer: (a) Cs

Explanation:
1. Atomic size increases down a group, so Cs is larger than K (both are in group 1).
2. Kr and Xe lie at the right end of periods 4 and 5, where atoms are much smaller than the group 1 atoms of those periods.
3. Cs is placed at the extreme left of period 6, where atoms are largest.

Teacher's Note:
1. Size decreases across a period and increases down a group.
2. Cs is the largest of the stable elements.

Answer: (b) 1.54 Å

Explanation:
1. Covalent radius = half the distance between the nuclei of two bonded atoms of the same element.
2. So bond length = 2 × covalent radius = 2 × 0.77 Å = 1.54 Å.

Teacher's Note:
1. Bond length d(A–A) = 2 r(A).
2. 1 Å = 100 pm, so 1.54 Å = 154 pm.

Answer: (a) cation is smaller but the anion is larger

Explanation:
1. A cation has fewer electrons but the same nuclear charge, so the nucleus pulls the remaining electrons closer.
2. An anion has more electrons, so electron–electron repulsion rises and the effective nuclear charge per electron falls.
3. Hence a cation is smaller and an anion is larger than the atom.

Teacher's Note:
1. Example: Na = 186 pm and Na⁺ = 95 pm; F = 64 pm and F⁻ = 136 pm.
2. This is true for every element.

Answer: (d) 1 and 2; 13 to 18; 3 to 12

Explanation:
1. The s-block consists of groups 1 and 2 (including He, whose configuration is 1s²).
2. The p-block consists of groups 13 to 18, where the np sub-shell is being filled.
3. The d-block (transition elements) consists of groups 3 to 12.

Teacher's Note:
1. Count the groups: 2 + 6 + 10 = 18 columns.
2. Helium is an s-block element that sits in group 18 because of its filled valence shell.

Answer: (b) Be²⁺

Explanation:
1. Li⁺ and Be²⁺ both have 2 electrons, but Be²⁺ has the larger nuclear charge (+4), so it is smaller.
2. Na⁺ and K⁺ have more shells and are larger.
3. So Be²⁺ has the smallest radius.

Teacher's Note:
1. Compare ions with the same number of electrons by the nuclear charge.
2. The cation with the higher charge is smaller.

Answer: (b) increases

Explanation:
1. Cl⁻ is formed by adding one electron to Cl, so there are 18 electrons for 17 protons.
2. Electron–electron repulsion increases and the effective nuclear charge per electron falls.
3. The electron cloud expands, so the ion is larger than the atom.

Teacher's Note:
1. Anions are always larger than their parent atoms.
2. Cl atom is about 99 pm (covalent radius) and Cl⁻ is about 181 pm.

Answer: (d) Mg²⁺ < Na⁺ < F⁻ < O²⁻ < N³⁻

Explanation:
1. All five ions have 10 electrons.
2. The nuclear charges are N +7, O +8, F +9, Na +11 and Mg +12.
3. The ionic radius decreases as nuclear charge increases, so the increasing order is Mg²⁺ < Na⁺ < F⁻ < O²⁻ < N³⁻.

Teacher's Note:
1. Write the nuclear charge below each ion and arrange from high to low.
2. Check that you have an increasing order and not a decreasing one.

Answer: (d) Cl⁻ and Ca²⁺

Explanation:
1. Number of electrons: Ne = 10, Cl⁻ = 17 + 1 = 18, Ca²⁺ = 20 − 2 = 18 and Rb⁺ = 37 − 1 = 36.
2. Cl⁻ and Ca²⁺ both have 18 electrons, so they are isoelectronic.

Teacher's Note:
1. Isoelectronic means the same number of electrons, not the same number of protons.
2. Always count electrons as (Z − charge).