CBSE Class 10 Chemistry Periodic Classification of Elements MCQs Set 28

Answer: (a) P, Co

Explanation:
1. Cobalt (Z = 27) is a d-block transition element.
2. P, Sr, Cs, S, K and Cl are all s-block or p-block elements, so they are representative.

Teacher's Note:
1. Representative elements are those of the s-block and p-block.
2. Look for the odd element that is in the d-block.

Answer: (c) N → N^–

Explanation:
1. Approximate electron affinities in kJ mol^–1: B 27, C 122, N about 0 (energy must be supplied), O 141.
2. Nitrogen, with its half-filled 2p³ subshell, releases no energy, so the energy released is a minimum for N.

Teacher's Note:
1. Half-filled orbitals are extra stable and resist addition of an electron.
2. This is why N does not follow the general trend in the second period.

Answer: (c) transition element

Explanation:
1. The configuration is [Ar]3d¹⁰4s¹, which is copper.
2. The penultimate shell (n = 3) holds 18 electrons because the 3d subshell is filled, so this is a d-block transition element.

Teacher's Note:
1. A normal element has 8 electrons in the penultimate shell (except period 2).
2. Inner transition elements fill f subshells.

Answer: (a) almost equal size (charge to radius ratio)

Explanation:
1. Diagonal pairs such as Li–Mg have similar ionic size and polarising power.
2. Their electronic configurations are different, so only (a) is correct.

Teacher's Note:
1. The shift of group raises the charge, while the shift of period raises the size.
2. The two effects roughly cancel, giving similar properties.

Answer: (b) N > O < F

Explanation:
1. The general trend across period 2 is an increase, so F is highest.
2. N has a half-filled 2p³ configuration, so its ionisation energy (1402) is higher than that of O (1314).
3. Thus N > O < F (1681).

Teacher's Note:
1. Half-filled and filled subshells give higher ionisation energies than expected.
2. Remember the exceptions Be > B and N > O.

Answer: (c) P, H

Explanation:
1. Pauling values: F 3.98, N 3.04 (difference 0.94); Na 0.93, Cs 0.79 (0.14); P 2.19, H 2.20 (0.01); C 2.55, H 2.20 (0.35).
2. The smallest difference is for P and H.

Teacher's Note:
1. A small electronegativity difference means a nearly non-polar bond.
2. P–H bonds are essentially non-polar.

Answer: (a) Br^– > S^2– > N^3–

Explanation:
1. Br^– has 4 shells (radius about 196 pm), S^2– has 3 shells (184 pm) and N^3– has 2 shells (171 pm).
2. More shells mean a larger ion, so Br^– > S^2– > N^3–.

Teacher's Note:
1. Number of shells is the first thing to compare.
2. Within the same shell number, higher negative charge gives a larger ion.

Answer: (a) (ii), (iii) and (iv) only

Explanation:
1. Pr is a lanthanide and Co is a transition element, so pair (i) is not representative.
2. Sn, As, S, K, P and Cl are s-block or p-block elements, so pairs (ii), (iii) and (iv) are correct.

Teacher's Note:
1. Pr (Z = 59) is an f-block element.
2. Representative = s-block plus p-block.

Answer: (a) (i), (ii) and (iii) only

Explanation:
1. Li–Mg, Be–Al and B–Si are the classical diagonal pairs.
2. Mg and Sc do not show a diagonal relationship (Sc is a transition element in period 4).

Teacher's Note:
1. The diagonal pairs lie between periods 2 and 3.
2. Learn them as Li–Mg, Be–Al, B–Si.

Answer: (a) (ii), (iii) and (iv) only

Explanation:
1. Radium occurs naturally, so statement (i) is wrong.
2. Elements beyond uranium are man-made, so most actinides are man-made; alkali metals (Li, Na, K) have very low densities; alkaline earth metals lose electrons easily and are good reducing agents.
3. So (ii), (iii) and (iv) are correct.

Teacher's Note:
1. Technetium was the first man-made element.
2. Lithium is the lightest metal of all.

Answer: (a) (i), (ii) and (iv) only

Explanation:
1. Atomic size increases down a group, so (i) is incorrect.
2. A cation is smaller than its atom and an anion is larger than its atom, so (ii) and (iv) are incorrect.
3. Atomic size decreases along a period, so (iii) is correct.

Teacher's Note:
1. Cation: fewer electrons and the same nuclear charge, so it is smaller.
2. Anion: more electrons repel each other, so it is larger.

Answer: (a) smaller size

Explanation:
1. Fluorine is very small, so the incoming electron is repelled strongly by the electrons already in the compact 2p subshell.
2. This reduces the energy released and makes its electron gain enthalpy lower than that of chlorine.

Teacher's Note:
1. Small size is the cause of several second-period anomalies.
2. The passage itself states this reason.

Answer: (b) chlorine

Explanation:
1. Chlorine releases the most energy on gaining an electron (about 349 kJ mol^–1), more than fluorine, oxygen or nitrogen.

Teacher's Note:
1. Among all elements chlorine has the highest electron gain enthalpy.
2. Use the passage to see why fluorine is lower.

Answer: (d) almost zero

Explanation:
1. As the passage states, elements with completely filled orbitals have zero electron gain enthalpy.
2. Noble gases have filled valence shells, so their value is almost zero.

Teacher's Note:
1. Noble gases do not want to add an electron to a stable octet.
2. The word 'almost' allows for the small positive or negative values seen in practice.

Answer: (a) gain of electrons

Explanation:
1. Highly electronegative atoms attract electrons strongly.
2. They complete their octet by gaining electrons and form negative ions (anions).

Teacher's Note:
1. Metals lose electrons, non-metals gain electrons.
2. Gaining electrons makes the atom an oxidising agent, not a reducing agent.

Answer: (d) 7s¹

Explanation:
1. Electronegativity is lowest at the bottom-left of the table, which is francium (and caesium).
2. Francium has the outer configuration 7s¹.

Teacher's Note:
1. Alkali metals at the bottom of group 1 are the least electronegative.
2. Fluorine, at the top right, is the most electronegative.

Answer: (b) high ionisation enthalpy

Explanation:
1. A strongly electronegative atom is small and holds its electrons firmly, so its ionisation enthalpy is high.
2. It forms negative ions and has a high electron gain enthalpy.

Teacher's Note:
1. Electronegativity, ionisation enthalpy and electron gain enthalpy usually move in the same direction.
2. Large size reduces all three.

Answer: (a) A-S, B-P, C-Q, D-R

Explanation:
1. The Law of Octaves was proposed by Newlands (A-S).
2. Boron shows a diagonal relationship with silicon (B-P).
3. Removing an electron from M²⁺ to form M³⁺ is the third ionisation energy (C-Q), and chlorine is a halogen (D-R).

Teacher's Note:
1. Learn the diagonal pairs and the scientist names.
2. Each ionisation step removes one electron from a gaseous species.

Answer: (a) A-Q, B-S, C-R, D-P

Explanation:
1. Li, Na and K form a Dobereiner triad (A-Q).
2. Mendeleev's law arranged elements by atomic weight (B-S).
3. Lanthanum (Z = 57) is placed with the d-block transition elements (C-R), and the alkaline earth metal barium has the configuration [Xe]6s² (D-P).

Teacher's Note:
1. Mendeleev used atomic weight; the modern law uses atomic number.
2. Barium is an alkaline earth metal with Z = 56.

Answer: (a) A-Q, B-P, C-P and R, D-Q and S

Explanation:
1. Cu, Ag and Au are the coinage metals (A-Q).
2. Tin has the largest number of stable isotopes among the metals (B-P).
3. Tin, bromine and krypton are p-block elements (C-P, R), and Cu, Ag, Au, Sc, Ti and V are all transition elements (D-Q, S).

Teacher's Note:
1. Tin has 10 stable isotopes.
2. Coinage metals are the group 11 elements.

Answer: (a) A-Q, B-P and R, C-S, D-P, Q, R and S

Explanation:
1. Chlorine is a halogen (A-Q).
2. Oxygen and sulphur are chalcogens (B-P, R).
3. Neon is a noble gas (C-S).
4. All four elements are s-block or p-block elements, so each is a representative element (D-P, Q, R, S).

Teacher's Note:
1. Chalcogens are the group 16 elements, halogens group 17.
2. Noble gases belong to group 18.

Answer: (b) Na, K, Rb

Explanation:
1. In a triad, the atomic mass of the middle element is about the average of the other two.
2. Na, K, Rb: average of 23 and 85 is 54, but K is 39, so they do not follow the law.
3. The other three sets follow it: for example (35.5 + 127)/2 = 81.2 for Br = 80.

Teacher's Note:
1. Check the average of the first and third masses.
2. Na, K and Rb show a strong mismatch.