CBSE Class 11 Chemistry HOTs Classification of Elements and Periodicity in Properties Set 02

Higher Order Thinking Questions

Question 1. Assertion (A): The fifth period of the periodic table contains 18 elements and not 32.
Reason (R): In the fifth period the orbitals 5s, 4d and 5p are filled; there are 9 of them, and they hold 18 electrons.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Answer:
Ans. (a) The fifth period starts with the filling of 5s (1 orbital), then 4d (5 orbitals) and ends with 5p (3 orbitals). This is 1 + 5 + 3 = 9 orbitals, so a maximum of 18 electrons and therefore 18 elements.
The 4f orbitals (needed to reach 32) are filled only from the sixth period onwards, because 4f lies above 6s in energy. Hence A is true and R explains it.
In simple words: Count the orbitals filled in the period and double the number to get the number of elements.

Teacher's Note:
1. Period 6 fills 6s, 4f, 5d and 6p: 1 + 7 + 5 + 3 = 16 orbitals, which gives 32 elements.
2. Write the filling order before answering any question on period length.

Question 2. Assertion (A): The 4f and 5f series of inner transition elements are placed separately at the bottom of the periodic table.
Reason (R): This arrangement prevents the table from becoming too wide and also keeps elements with similar properties together in one column.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Answer:
Ans. (a) If the 28 f-block elements were placed in the main body, the table would be 32 columns wide, which is clumsy. The lanthanoids and actinoids have very similar properties within each series, so they are shown as separate panels below, linked to group 3.
Both statements are true and the Reason explains the Assertion.
In simple words: The f-block would stretch the table, so it is moved below to keep the table compact.

Teacher's Note:
1. The f-block has 14 elements in each of the two rows, 28 in all.
2. The 'Reason' contains two points; both must be right for the answer to be (a).

Question 3. Assertion (A): The third ionisation enthalpy of phosphorus is larger than that of sulphur.
Reason (R): There is greater stability associated with filled s and p sub-shells (a noble-gas configuration with eight electrons in the valence shell).
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Answer:
Ans. (d) P²⁺ is 3s² 3p¹, so the third electron of P is the lone 3p electron; removing it leaves the stable, filled 3s² configuration. This is relatively easy (about 2914 kJ mol⁻¹).
S²⁺ is 3s² 3p²; removing an electron from it gives 3s² 3p¹, which has no special stability, and the nuclear charge is higher. The third ionisation enthalpy of S is therefore larger (about 3357 kJ mol⁻¹).
So A is false. R is a true statement about stability.
In simple words: Phosphorus loses its third electron more easily because it ends up with a filled 3s sub-shell.

Teacher's Note:
1. Write the configuration of M²⁺ before judging the third ionisation enthalpy.
2. A true Reason does not make a false Assertion true.

Question 4. Which of the following sets contain only isoelectronic ions? (more than one may be correct)
(a) Zn²⁺, Ca²⁺, Ga³⁺, Al³⁺
(b) K⁺, Ca²⁺, Sc³⁺, Cl⁻
(c) P³⁻, S²⁻, Cl⁻, K⁺
(d) Ti⁴⁺, Ar, Cr³⁺, V⁵⁺
Answer:
Ans. (b) and (c) Count the electrons as (Z − charge).
Set (a): Zn²⁺ = 28, Ca²⁺ = 18, Ga³⁺ = 28, Al³⁺ = 10, which are not all equal.
Set (b): K⁺ = 18, Ca²⁺ = 18, Sc³⁺ = 18, Cl⁻ = 18, all with 18 electrons.
Set (c): P³⁻ = 18, S²⁻ = 18, Cl⁻ = 18, K⁺ = 18, all with 18 electrons.
Set (d): Ti⁴⁺ = 18, Ar = 18, Cr³⁺ = 21, V⁵⁺ = 18; Cr³⁺ breaks the set.
In simple words: Isoelectronic means every member has the same number of electrons.

Teacher's Note:
1. Atoms or ions with the same number of electrons need not come from the same period.
2. One odd member is enough to rule out a set.

Question 5. The first ionisation enthalpy of the first atom is greater than that of the second atom, whereas the order is reversed for their second ionisation enthalpies. Which pairs of elements fit this statement? (more than one may be correct)
(a) C, B
(b) P, S
(c) Be, B
(d) Mg, Na
Answer:
Ans. (a), (b), (c) and (d) All four pairs fit. Values in kJ mol⁻¹ (first and second ionisation enthalpies):
C (1086, 2353) and B (801, 2427): C > B first, B > C second.
P (1012, 1907) and S (1000, 2252): P > S first, S > P second.
Be (899, 1757) and B (801, 2427): Be > B first, B > Be second.
Mg (738, 1451) and Na (496, 4562): Mg > Na first, Na > Mg second.
In each pair the second atom forms an ion (B⁺, S⁺, B⁺, Na⁺) with a filled or half-filled sub-shell, which makes the second electron harder to remove.
In simple words: The ion left after the first ionisation decides the second value.

Teacher's Note:
1. Write the configuration of M⁺ for both atoms in the pair.
2. Na⁺ has the neon configuration and has by far the highest second value.

Question 6. Which of the following elements gain one electron more readily than the other elements of their own group? (more than one may be correct)
(a) S
(b) Na
(c) O
(d) Cl
Answer:
Ans. (a) and (d) In groups 16 and 17, the second-period elements O and F have smaller electron gain enthalpies (in magnitude) than S and Cl, because the added electron enters the compact 2p sub-shell and is repelled strongly.
S (−200 kJ mol⁻¹) is more negative than O (−141), and Cl (−349) is more negative than F (−328).
Na is a metal in group 1 and does not gain electrons readily; in group 1 the first member Li (−60) is only slightly different from Na (−53).
In simple words: Sulphur and chlorine take up an electron more readily than oxygen and fluorine.

Teacher's Note:
1. The first member of a group is out of order in groups 16 and 17.
2. Group 15 and group 2 have their own exceptions because of stable sub-shells.

Question 7. Which of the following statements are correct? (more than one may be correct)
(a) Helium has the highest first ionisation enthalpy in the periodic table.
(b) Chlorine has a less negative electron gain enthalpy than fluorine.
(c) Mercury and bromine are liquids at room temperature.
(d) In any period, the atom of the alkali metal has the largest radius (excluding the noble gases).
Answer:
Ans. (a), (c) and (d)
(a) is correct: He has ΔiH₁ = 2372 kJ mol⁻¹, the highest known.
(b) is wrong: Cl (−349 kJ mol⁻¹) is more negative than F (−328 kJ mol⁻¹).
(c) is correct: Hg is the only liquid metal and Br the only liquid non-metal at room temperature.
(d) is correct: atomic radius falls across a period, so the group 1 atom is the largest.
In simple words: Only the statement about Cl and F is false.

Teacher's Note:
1. The F and Cl exception is a favourite in exams.
2. Noble gas radii are van der Waals radii, so they are left out of such comparisons.

Question 8. The first and second ionisation enthalpies (kJ mol⁻¹) of four elements A, B, C and D are: A 2372 and 5251; B 520 and 7300; C 900 and 1760; D 1680 and 3380. (i) Which is a reactive metal? (ii) Which is a reactive non-metal? (iii) Which is a noble gas? (iv) Which is a metal that forms a stable M²⁺ ion?
Answer:
(i) B. It has a low first value (520) and a huge jump to the second (7300), so it readily loses one electron and then reaches a stable core. It behaves like an alkali metal (such as Li).
(ii) D. Both values are very high (1680 and 3380), so it does not lose electrons easily, but it would readily gain one; it resembles fluorine.
(iii) A. It has the highest first and second values (2372, 5251), like helium, a complete shell.
(iv) C. The values 900 and 1760 are low and close, so two electrons are lost easily, as in beryllium (899 and 1757); it forms M²⁺.
In simple words: A big gap between the first and second values means one valence electron; two low values mean two.

Teacher's Note:
1. Compare the ratio of the second to the first value for each element.
2. Very high values of both mean a non-metal or a noble gas.

Question 9. Three hydroxy-compounds A, B and C belong to elements X, Y and Z of the same period. A gives an aqueous solution with pH less than 7, B reacts with both strong acids and strong alkalis, and C gives a strongly alkaline solution. Which statements are true?
I. The three elements are metals.
II. The electronegativities decrease from X to Y to Z.
III. The atomic radii decrease in the order X, Y, Z.
IV. X, Y and Z could be phosphorus, aluminium and sodium respectively.
(a) I, II and III only
(b) I only
(c) II and IV only
(d) II, III and IV only
Answer:
Ans. (c) A is acidic, B is amphoteric and C is strongly basic. Across a period the nature goes from acidic to amphoteric to basic as we move from right to left, so X is on the right and Z on the left.
I is false because the element giving an acidic solution (X) is a non-metal.
II is true because electronegativity increases from left to right, so X > Y > Z.
III is false because the radius increases from right to left, so the order X, Y, Z is increasing and not decreasing.
IV is true: H₃PO₄ (acidic), Al(OH)₃ (amphoteric) and NaOH (strongly basic) fit exactly.
In simple words: Move across the period: acidic hydroxide, then amphoteric, then basic means the elements go from non-metal to metal.

Teacher's Note:
1. Draw the period 3 sequence Na, Mg, Al, Si, P and place X, Y, Z on it.
2. Radius and electronegativity move in opposite directions across a period.

Question 10. Consider the statements: (I) The radius of an anion is larger than that of its parent atom. (II) The ionisation enthalpy generally increases with atomic number in a period. (III) The electronegativity of an element is the tendency of an isolated atom to attract an electron. Which are correct?
(a) I alone
(b) II alone
(c) I and II
(d) II and III
Answer:
Ans. (c)
I is correct: an added electron raises repulsion and lowers the effective nuclear charge per electron.
II is correct: the nuclear charge rises and the size falls across a period.
III is wrong: the tendency of an isolated gaseous atom to gain an electron is its electron gain enthalpy. Electronegativity refers to an atom in a molecule attracting the shared pair of electrons.
In simple words: Statement III mixes up electronegativity with electron gain enthalpy.

Teacher's Note:
1. Look for the phrases 'isolated atom' and 'in a molecule'.
2. Electronegativity has no unit.

Question 11. Which of the following statements is wrong? (more than one may be wrong)
(a) The van der Waals radius of iodine is more than its covalent radius.
(b) All isoelectronic ions belong to the same period of the periodic table.
(c) The first ionisation enthalpy of N is higher than that of O, while the second ionisation enthalpy of O is higher than that of N.
(d) Nitrogen has an electron gain enthalpy close to zero, whereas phosphorus has a clearly negative one (about −72 kJ mol⁻¹).
Answer:
Ans. (b) Isoelectronic ions need not be in the same period; for example, N³⁻ and Al³⁺ (both 10 electrons) come from periods 2 and 3, and K⁺ (period 4) is isoelectronic with Cl⁻ (period 3).
(a) is correct because the van der Waals radius is always larger than the covalent radius of the same atom (for Cl, 180 pm against 99 pm).
(c) is correct: N (2p³) is higher than O in the first ionisation, but O⁺ (2p³) is half-filled, so the second ionisation enthalpy of O is higher.
(d) is correct because N has a stable half-filled 2p³ sub-shell.
In simple words: Only the statement about isoelectronic ions being in the same period is wrong.

Teacher's Note:
1. Isoelectronic species are linked by their electron count, not by their position.
2. Check every option when the question asks for the wrong statement.

Question 12. The formation of the oxide ion O²⁻(g) from an oxygen atom takes two steps: O(g) + e⁻ → O⁻(g), ΔH = −141 kJ mol⁻¹, and O⁻(g) + e⁻ → O²⁻(g), ΔH = +780 kJ mol⁻¹. Find the overall enthalpy change for O(g) + 2e⁻ → O²⁻(g) and explain why the formation of O²⁻ in the gas phase is unfavourable even though O²⁻ has the neon configuration.
Answer:
Overall change: ΔH = (−141) + (+780) = +639 kJ mol⁻¹. The process is endothermic.
Reason: the first electron is added to a neutral atom and energy is released. The second electron has to be pushed onto an ion that is already negative, and the strong electron–electron repulsion outweighs the gain in stability from the noble-gas (neon) configuration.
Oxides such as MgO and Na₂O are nevertheless stable because the oxide ions are held in a crystal lattice, and the large lattice energy released compensates for the +639 kJ mol⁻¹.
In simple words: Adding a second electron to a negative ion costs more energy than the first one gives out.

Teacher's Note:
1. Second and later electron gain enthalpies are always positive.
2. The O⁻ to O²⁻ step is a good example of repulsion beating stability.

Question 13. Halogens other than fluorine show positive oxidation states of +1, +3, +5 and +7 in addition to −1. Explain why fluorine shows only −1.
Answer:
All halogens have the configuration ns²np⁵, so each can gain one electron and show −1.
Chlorine, bromine and iodine can take part in extra bonding in their compounds with more electronegative atoms (for example in ClO₄⁻, BrO₃⁻ and IO₃⁻), and they can show +1, +3, +5 and +7 because they have vacant d orbitals in their valence shell to make this possible.
Fluorine has no d orbitals in its valence shell (n = 2) and it is the most electronegative element, so it always attracts the shared electrons. It shows only −1 (and 0 in F₂).
In simple words: Cl, Br and I can use more of their valence electrons in bonding; F cannot and always pulls electrons.

Teacher's Note:
1. In OF₂ oxygen is +2 because fluorine is more electronegative than oxygen, but fluorine itself stays −1.
2. Always state both reasons: no d orbitals and highest electronegativity.

Question 14. Among the halogens, the amount of energy released when an electron is added is in the order F < Cl > Br > I. Why is fluorine not at the top, although the trend down the group is a decrease?
Answer:
Electron gain enthalpy values (kJ mol⁻¹) are F −328, Cl −349, Br −325 and I −295. In magnitude, the usual fall down the group is seen from Cl to I.
Fluorine has a very small 2p sub-shell. When an electron is added, it is crowded into a small volume and repelled strongly by the seven electrons already present (the electron density is high).
In chlorine the incoming electron enters the larger 3p sub-shell, so the repulsion is much less and more energy is released. Hence Cl is more negative than F, and the expected order starts only after chlorine.
In simple words: Fluorine's small size crowds the extra electron, so it releases less energy than chlorine.

Teacher's Note:
1. The same effect makes S more negative than O.
2. Fluorine is still more electronegative than chlorine; electron gain enthalpy and electronegativity are different properties.