Class 11: Classification of Elements and Periodicity in Properties
MCQs from Unit – 3
Q1. Generally, IE increases from left to right in a period, but there are some exceptions. In which of the following cases do no such exceptions occur?
(A) \( \text{N and O} \) (B) \( \text{Na and Mg} \) (C) \( \text{Mg and Al} \) (D) \( \text{Be and B} \)
(B) Na and Mg where no such exception occurs.
N and O: Nitrogen has a higher ionization energy than oxygen due to half-filled stability of the p-orbital in nitrogen.
Na and Mg: Magnesium has a higher ionization energy than sodium, following the trend.
Mg and Al: Magnesium has a higher ionization energy than aluminum because of the filled s-orbital being more stable.
Be and B: Beryllium has a higher ionization energy than boron due to the stable filled s-orbital.
The acidic character of oxides increases with the increase in oxidation state of the central atom and as we move from left to right across a period.
– \( \text{CaO} \) (basic)
– \( \text{PbO}_2 \) (amphoteric)
– \( \text{SO}_3 \) (acidic)
– \( \text{Cl}_2\text{O}_7 \) (highly acidic)
The oxidizing power of halogens decreases as we move down the group in the periodic table.
– Fluorine is the strongest oxidizing agent.
– Followed by chlorine, bromine, and iodine in that order.
Q4. \( \text{K}^+ \) and \( \text{Ca}^{2+} \) contain the same number of:
(A) Neutrons (B) Protons (C) Electrons (D) Both (A) and (B)
(C) Electrons.
– \( \text{K}^+ \) (potassium ion) has 19 protons and 18 electrons.
– \( \text{Ca}^{2+} \) (calcium ion) has 20 protons and 18 electrons. They contain the same number of electrons.
Q5. In aqueous solution, ionic mobility is least for:
Ionic mobility decreases with the increase in charge and size of the ion. Higher charge and smaller size lead to higher hydration energy and lower mobility.
– \( \text{Al}^{3+} \) has the highest charge and smallest size.
Hydration energy decreases down the group in the periodic table.
– \( \text{Ca}^{2+} \) has the highest hydration energy.
– Followed by \( \text{Sr}^{2+} \) and \( \text{Ba}^{2+} \).
Q7. In which of the following pairs do the species have nearly the same size?
– \( \text{Zr}^{4+} \) and \( \text{Hf}^{4+} \) have nearly the same size due to the lanthanide contraction, which makes them similar in size despite being in different periods.
Q8. Which of the following shows the highest oxidation state?
The highest oxidation state is often determined by the number of unpaired electrons. Configuration \( 3d^5 4s^1 \) will likely have the highest oxidation state because \( 3d^5 \) (half-filled d-subshell) and one \( 4s \) electron give a higher number of valence electrons available for oxidation.
Q9. The correct sequence of stability of +1 oxidation state is:
Stability of +1 oxidation state increases down the group in group 13 due to the inert pair effect.
– \( \text{Tl} \) has the most stable +1 oxidation state.
Q10. The correct sequence of ionisation enthalpy of \( \text{Ti}(22), \text{V}(23), \text{Cr}(24), \text{Mn}(25) \) is:
Generally, ionisation enthalpy increases across a period. Exceptions occur due to electron configuration stability:
– \( \text{Cr} \) has higher ionisation energy due to its half-filled \( 3d^5 \) stability.
– \( \text{Mn} \) has a higher ionization enthalpy than Ti but less than Cr.
Q11. The correct sequence of thermal stability of the hydrides of alkali metals is:
– Bond dissociation energy: \( \text{Cl}_2 > \text{Br}_2 > \text{F}_2 > \text{I}_2 \) (due to small size, \( \text{F}_2 \) has lower bond dissociation energy).
Q13. The number of naturally occurring p-block elements that are diamagnetic is:
(A) 18 (B) 6 (C) 7 (D) 8
(C) 7.
Diamagnetic elements have all their electrons paired. Among naturally occurring p-block elements:
– Group 18 elements (Noble gases) are diamagnetic.
– \( \text{N}_2 \), \( \text{O}_2 \), \( \text{P}_2 \), \( \text{S}_2 \), \( \text{Cl}_2 \), \( \text{Se}_2 \), \( \text{Br}_2 \) are diamagnetic.
Q14. The increasing order of the density of alkali metals is:
The densities of alkali metals generally increase from lithium to cesium, with potassium being an exception due to its unique crystal structure.
Q15. Ease of formation of the cation is favoured by:
(A) Lower value of ionisation potential (B) Lower value of electron affinity (C) Higher value of electron affinity (D) Higher value of electronegativity
(A) Lower value of ionisation potential.
Formation of cations is easier if the ionisation potential is lower.
Q16. The element with the electronic configuration as [Ar] \(3d^4 4s^2 \) represents a:
(A) Non-metal (B) Metal (C) Metalloid (D) Transition element
(D) Transition element.
Configuration [Ar] \(3d^4 4s^2\) corresponds to chromium (\( \text{Cr} \)), which is a transition metal.
Q17. Which one of the following is correct order of the size:
Anion (\( \text{I}^- \)) is larger than the neutral atom (\( \text{I} \)), which is larger than the cation (\( \text{I}^+ \)).
Q18. The first ionisation energies of the elements of the first transition series (Ti → Cu):
(A) increase as the atomic number increases (B) decrease as the atomic number increases (C) do not show any change as the addition of electrons take place in the inner (n-1)d-orbitals (D) increase from Ti to Mn and then decrease from Mn to Cu
(D) increase from Ti to Mn and then decrease from Mn to Cu.
The first ionisation energies of transition elements generally increase slightly across a period due to increasing nuclear charge, with some irregularities. Specifically, there is a slight increase from Ti to Mn and then a more irregular pattern.
Q19. Which represents the correct order of first ionisation potential of third period elements:
The ionisation potential increases across a period from left to right. The correct order for the third period elements is:
– \( \text{Na} < \text{Mg} < \text{Al} < \text{Si} \)
Q20. The elements X, Y and Z form oxides which are acidic, basic, and amphoteric respectively. The correct order of their electronegativity is:
(A) \( X > Y > Z \) (B) \( Z > Y > X \) (C) \( Y > Z > X \) (D) \( X > Z > Y \)
(D) X > Z > Y.
Acidic oxides are formed by elements with high electronegativity, basic oxides by elements with low electronegativity, and amphoteric oxides by elements with intermediate electronegativity.
– Hence, \( X \) (acidic) > \( Z \) (amphoteric) > \( Y \) (basic).
Q21. The atomic numbers of vanadium (V), chromium (Cr), manganese (Mn) and iron (Fe) are respectively 23, 24, 25 and 26. Which one of these may be expected to have the highest second ionisation enthalpy:
(A) V (B) Cr (C) Mn (D) Fe
(B) Cr.
The second ionisation enthalpy is highest when removing the second electron disrupts a stable configuration. Chromium has a half-filled \( 3d^5 \) configuration after losing one electron.
– Cr (removal of second electron is from \( 3d^5 \)).
Q22. In which of the following sets, the IE of the first member is less than that of the second:
– \( \text{IE of N} < \text{IE of P} \)
– \( \text{IE of Be}^{+} < \text{IE of Be} \)
– \( \text{IE of S} > \text{IE of P} \)
– \( \text{IE of N} < \text{IE of N}^- \)
Q23. Which one of the following is correct in respect of ionic radii:
– \( \text{Sc}^{3+} > \text{Cr}^{6+} \)
– \( \text{I}^- > \text{Cl}^- \)
– \( \text{N}^{3-} \) (larger due to gain of electrons) > \( \text{N}^{5+} \) (smaller due to loss of electrons).
– \( \text{Na}^+ < \text{P}^{3-} \)
Q24. Select the correct statement:
(A) on moving down a group electronegativity increases and non-metallic character decreases (B) electronegativity increases as metallic character of an element increases (C) non-metals are reluctant to accept electrons (D) none of the above are correct
(D) none of the above are correct.
– Electronegativity generally decreases as you move down a group, and non-metallic character decreases.
– Electronegativity increases with decreasing metallic character.
– Non-metals are more eager to accept electrons.
Q25. Which is the correct sequence of metallic character of the given elements with the following electronic configurations:
(A) III < I < II < IV (B) IV < II < I < III (C) I < III < II < IV (D) IV < III < II < I
(A) III < I < II < IV.
The metallic character decreases as we move from left to right across a period. Configuration III is most metallic as it has fewer p-electrons (more metallic nature) and configuration II has a complete p-subshell (least metallic).
Q26. The property of attracting electrons by the halogen atoms in a molecule is called:
(A) Ionisation potential (B) Electron affinity (C) Electronegativity (D) Electron gain enthalpy
(C) Electronegativity.
Electronegativity is the property of attracting electrons within a molecule.
Q27. On moving from top to bottom in a group, the electropositive character of elements:
(A) Increases (B) Decreases (C) Does not change (D) Changes irregularly
(A) Increases.
Electropositive character increases down a group due to the increase in atomic size and decrease in ionisation energy.
Q28. The correct sequence of electronegative character is:
Higher positive oxidation states attract electrons more strongly.
\( \text{M}^+3 \) has higher electronegative character than \( \text{M}^+2 \) and \( \text{M}^+ \).
Q29. The correct sequence of electronegativity is:
– \( \text{Al}_2\text{O}_3 \) is amphoteric (can act as both acidic and basic).
– \( \text{CO}_2 \), \( \text{N}_2\text{O}_5 \), and \( \text{SO}_3 \) are acidic oxides.
Q34. Which of the following oxides is most acidic:
\( \text{SO}_3 \) is the most acidic oxide among the given options.
Q35. Metallic sodium forms \( \text{Na}^+ \) ion but not \( \text{Na}^{2+} \) ion because:
(A) Both \( \text{IE}_1 \) and \( \text{IE}_2 \) are low (B) Both \( \text{IE}_1 \) and \( \text{IE}_2 \) are high (C) \( \text{IE}_1 \) is low, but \( \text{IE}_2 \) is high (D) \( \text{IE}_1 \) is high, but \( \text{IE}_2 \) is low
(C) \( \text{IE}_1 \) is low, but \( \text{IE}_2 \) is high
The first ionisation energy (\( \text{IE}_1 \)) is low, allowing \( \text{Na} \) to form \( \text{Na}^+ \). However, the second ionisation energy (\( \text{IE}_2 \)) is very high, making it difficult to remove another electron and form \( \text{Na}^{2+} \).
Thermal stability of hydrides decreases down the group in the periodic table. This is because the bond strength between hydrogen and the central atom (N, P, As) decreases as the size of the central atom increases. Therefore, the given sequence is correct as \( \text{NH}_3 \) is more thermally stable than \( \text{PH}_3 \), which in turn is more stable than \( \text{AsH}_3 \).
Basicity of oxides increases as you move from left to right across a period and down a group in the periodic table. Therefore, the basicity should decrease from \( \text{K}_2\text{O} \) to \( \text{Al}_2\text{O}_3 \), because \( \text{K}_2\text{O} \) (potassium oxide) is more basic than \( \text{Na}_2\text{O} \) (sodium oxide), which is more basic than \( \text{MgO} \) (magnesium oxide), and \( \text{Al}_2\text{O}_3 \) (aluminum oxide) is amphoteric, not basic. The given sequence is reversed, so it is incorrect.
First ionization energy (IE1) generally increases across a period from left to right. Lithium (Li) has the lowest ionization energy and nitrogen (N) has the highest in the given sequence. Therefore, this sequence is correct.
Ionic radius increases as you move down a group in the periodic table. Lithium ion (\( \text{Li}^+ \)) is the smallest, and cesium ion (\( \text{Cs}^+ \)) is the largest. Therefore, this sequence is correct.
Ionisation enthalpy decreases with increasing negative charge.
Q44. Five successive IE of an element (\( \text{IE}_1 \) to \( \text{IE}_5 \)) are 800, 2427, 3658, 25024, and 36823 kJ/mol respectively. So the number of valence electrons in the element is:
(A) 1 (B) 2 (C) 3 (D) 4
(C) 3
Large jump between \( \text{IE}_3 \) and \( \text{IE}_4 \) indicates three valence electrons.
Q45. Which one of the following has the lowest electron affinity:
(A) B (B) F (C) N (D) C
(C) N.
Nitrogen (N) has half-filled p-orbital, hence lower affinity.
Q46. Which one of the following are isoelectronic: \( \stackrel{\oplus}{\mathrm{C}} \mathrm{H}_{3}(\mathrm{I}) \), \( \stackrel{\ominus}{\mathrm{N}} \mathrm{H}_{2} \) (II), \( \stackrel{\oplus}{\mathrm{N}} \mathrm{H}_{4} \) (III), \( \mathrm{NH}_{3} \) (IV) –
(A) II, III, IV
(B) I, II, III
(c) I, II, IV
(D) II, I
(A) II, III, IV
1. \( \stackrel{\oplus}{\mathrm{C}} \mathrm{H}_{3}(\mathrm{I}) \): – Carbon (C) normally has 6 electrons. – Each hydrogen (H) atom contributes 1 electron, and there are 3 hydrogen atoms, adding 3 electrons. – The positive charge (\( \oplus \)) indicates the loss of 1 electron. – Total electrons = \( 6 + 3 – 1 = 8 \) electrons.
2. \( \stackrel{\ominus}{\mathrm{N}} \mathrm{H}_{2} \) (II): – Nitrogen (N) normally has 7 electrons. – Each hydrogen (H) atom contributes 1 electron, and there are 2 hydrogen atoms, adding 2 electrons. – The negative charge (\( \ominus \)) indicates the gain of 1 electron. – Total electrons = \( 7 + 2 + 1 = 10 \) electrons.
3. \( \stackrel{\oplus}{\mathrm{N}} \mathrm{H}_{4} \) (III): – Nitrogen (N) normally has 7 electrons. – Each hydrogen (H) atom contributes 1 electron, and there are 4 hydrogen atoms, adding 4 electrons. – The positive charge (\( \oplus \)) indicates the loss of 1 electron. – Total electrons = \( 7 + 4 – 1 = 10 \) electrons.
4. \( \mathrm{NH}_{3} \) (IV): – Nitrogen (N) normally has 7 electrons. – Each hydrogen (H) atom contributes 1 electron, and there are 3 hydrogen atoms, adding 3 electrons. – No charge, so no electrons are added or removed. – Total electrons = \( 7 + 3 = 10 \) electrons.
Now, comparing the total number of electrons in each species: – \( \stackrel{\oplus}{\mathrm{C}} \mathrm{H}_{3}(\mathrm{I}) \) has 8 electrons. – \( \stackrel{\ominus}{\mathrm{N}} \mathrm{H}_{2} \) (II) has 10 electrons. – \( \stackrel{\oplus}{\mathrm{N}} \mathrm{H}_{4} \) (III) has 10 electrons. – \( \mathrm{NH}_{3} \) (IV) also has 10 electrons.
Species II, III, and IV all have 10 electrons and are therefore isoelectronic with each other.
Q47. Which one of the following elements has the highest ionisation energy:
Ne[3s\(^2\)3p\(^3\)] has highest IE due to half-filled p-orbital stability.
Q48. Which of the following statements regarding ionisation enthalpy is wrong:
(A) \( \text{IE}_1 < \text{IE}_2 < \text{IE}_3 \) (usually) (B) Sudden increase of one of the successive ionisation energies indicates the removal of the last valence electron (C) IE increases with the value of principal quantum number (D) IE of species having noble gas like electronic configuration is very high
(C) IE increases with the value of principal quantum number.
– IE decreases with increasing principal quantum number.
Q49. Second electron-gain enthalpy:
(A) Always negative (exothermic) (B) Always positive (endothermic) (C) May be positive or negative (D) Always zero
(B) Always positive (endothermic).
– Second electron-gain enthalpy is always endothermic due to repulsion.
Q50. Non-metallic character decreases in the sequence:
As you move from left to right across a period in the periodic table, the elements change from metallic to non-metallic in nature (i.e. non-mettallic character increases). This is due to the increase in nuclear charge (number of protons) which attracts the electrons more strongly, resulting in a higher tendency to gain electrons (electronegativity).