1. Encircle the correct answer.
(i) Number of periods in the periodic table are:
(a) 8
(b) 7
(c) 16
(d) 5
Answer: (b) 7
Explanation: The modern periodic table has 7 horizontal rows called periods. The period number corresponds to the highest principal quantum number (n) of the elements in that period.
(ii) Which of the following groups contain alkaline earth metals?
(a) IA
(b) IIA
(c) VIIA
(d) VIIIA
Answer: (b) IIA
Explanation: Group IIA (Group 2) elements are called alkaline earth metals. They include beryllium, magnesium, calcium, strontium, barium, and radium.
(iii) Which of the following elements belongs to VIIA?
(a) Na
(b) Mg
(c) Br
(d) Xe
Answer: (c) Br
Explanation: Bromine (Br) is a halogen and belongs to Group VIIA (Group 17). Na is in Group IA, Mg in Group IIA, and Xe in Group VIIIA.
(iv) Main group elements are arranged in __ groups.
(a) 6
(b) 7
(c) 8
(d) 10
Answer: (c) 8
Explanation: The main group elements (also called representative elements) are in Groups IA, IIA, IIIA, IVA, VA, VIA, VIIA, and VIIIA, totaling 8 groups.
(v) Period number of ₁₃Al is:
(a) 1
(b) 2
(c) 3
(d) 4
Answer: (c) 3
Explanation: Aluminium (Al) has the electronic configuration 1s², 2s², 2p⁶, 3s², 3p¹. The highest principal quantum number is 3, so it is in Period 3.
(vi) Valence shell electronic configuration of an element M (atomic no. 14) is:
(a) 2s², 2p²
(b) 2s², 2p⁶
(c) 3s², 3p²
(d) 3s², 3p⁶
Answer: (c) 3s², 3p²
Explanation: Element with atomic number 14 is Silicon (Si). Its electronic configuration is 1s², 2s², 2p⁶, 3s², 3p². The valence shell configuration is 3s², 3p².
(vii) Which of the following elements you expect to have greater shielding effect?
(a) Li
(b) Na
(c) Rb
(d) K
Answer: (c) Rb
Explanation: Shielding effect increases down a group as the number of inner electron shells increases. Rubidium (Rb) is the lowest among the options, so it has the most inner shells and the greatest shielding effect.
(viii) As you move from right to left across a period, which of the following does not increase:
(a) electron affinity
(b) ionization energy
(c) nuclear charge
(d) shielding effect
Answer: (d) shielding effect
Explanation: Across a period from right to left, the shielding effect remains relatively constant because electrons are added to the same shell. However, nuclear charge decreases, and atomic radius increases, affecting other periodic trends.
(ix) All the elements of Group IIIA are less reactive than alkali metals. This is because these elements have:
(a) high ionization energies
(b) relatively greater atomic sizes
(c) similar electronic configuration
(d) decreased nuclear charge
Answer: (a) high ionization energies
Explanation: Group IIIA elements have higher ionization energies than alkali metals, meaning they require more energy to lose electrons. This makes them less reactive than the highly reactive alkali metals.
2. Give short answer.
(i) Write the valence shell electronic configuration of an element present in 3rd period and Group IIIA.
Answer: The element in the 3rd period and Group IIIA is Aluminium (Al). Its complete electronic configuration is 1s², 2s², 2p⁶, 3s², 3p¹. The valence shell electronic configuration is 3s², 3p¹. It has 3 valence electrons, which is characteristic of Group IIIA elements.
(ii) Define halogens.
Answer: Halogens are the elements of Group VIIA (Group 17) of the periodic table. They include fluorine (F), chlorine (Cl), bromine (Br), iodine (I), and astatine (At). They are highly reactive non-metals with seven valence electrons (ns² np⁵) and a strong tendency to gain one electron to achieve a stable noble gas configuration. They exist as diatomic molecules (F₂, Cl₂, Br₂, I₂) and are known for forming salts.
(iii) Which atom has higher shielding effect, Li or Na?
Answer: Sodium (Na) has a higher shielding effect than Lithium (Li). Sodium has 11 electrons with the configuration 1s², 2s², 2p⁶, 3s¹, while Lithium has 3 electrons with the configuration 1s², 2s¹. Sodium has more inner electron shells (two inner shells: K and L) compared to Lithium (one inner shell: K). The greater number of inner electrons in sodium shields the valence electron more effectively from the nuclear charge.
(iv) Explain why Na has higher ionization energy than K?
Answer: Sodium (Na) has a higher ionization energy than Potassium (K) because sodium's valence electron is in the 3rd shell (n=3),
while potassium's valence electron is in the 4th shell (n=4). As you go down a group, the atomic size increases, and the valence electron is further from the nucleus. Additionally, the shielding effect increases due to more inner shells in potassium. These factors make it easier to remove the valence electron from potassium, resulting in lower ionization energy compared to sodium.
(v) Alkali metals belong to S-block in the periodic table, why?
Answer: Alkali metals belong to the S-block because their outermost electron enters the s-orbital of the valence shell. Their general valence shell electronic configuration is ns¹ (where n is the period number). For example, Lithium (Li) is 2s¹, Sodium (Na) is 3s¹, and Potassium (K) is 4s¹. Since the last electron goes into the s-orbital, they are classified as S-block elements. They are placed in Group IA of the periodic table.
3. Arrange the elements in each of the following groups in order of increasing ionization energy:
(a) Li, Na, K
Answer: K < Na < Li
Explanation: Ionization energy decreases down a group as atomic size increases and shielding increases. Li is at the top, so it has the highest ionization energy.
(b) Cl, Br, I
Answer: I < Br < Cl
Explanation: Ionization energy decreases down Group VIIA. Cl has the smallest size and highest ionization energy.
4. Arrange the elements in each of the following in order of decreasing shielding effect.
(a) Li, Na, K
Answer: K > Na > Li
Explanation: Shielding effect increases down a group as the number of inner shells increases. K has the most inner shells, so it has the greatest shielding effect.
(b) Cl, Br, I
Answer: I > Br > Cl
Explanation: Shielding effect increases down Group VIIA. I has the most inner shells.
5. Specify which of the following elements you would expect to have the greatest electron affinity. S, P, Cl
Answer: Cl (Chlorine) has the greatest electron affinity among S, P, and Cl.
Explanation: Electron affinity generally increases across a period from left to right. Chlorine is in Group VIIA (Group 17), which is further to the right than Sulfur (Group VIA) and Phosphorus (Group VA). Chlorine has a high effective nuclear charge and a small atomic radius, making it strongly attract an additional electron. Note: Fluorine has a lower electron affinity than Chlorine due to its very small size causing electron-electron repulsion.
6. Electronic configuration of some elements are given below, group elements in pairs that would represent similar chemical properties.
A = 1s², 2s¹
B = 1s², 2s², 2p³
C = 1s², 2s², 2p⁶
D = 1s²
E = 1s², 2s², 2p⁶, 3s¹
F = 1s², 2s²
G = 1s², 2s², 2p⁶, 3s²
H = 1s², 2s², 2p³
Answer:
A (1s², 2s¹) and E (1s², 2s², 2p⁶, 3s¹): Both have one valence electron (ns¹), so they belong to Group IA (Alkali metals) and have similar chemical properties. B (1s², 2s², 2p³) and H (1s², 2s², 2p³): Both have the same valence shell configuration (ns² np³), so they belong to Group VA and have similar chemical properties. C (1s², 2s², 2p⁶) and G (1s², 2s², 2p⁶, 3s²): C is Neon (Noble gas) and G is Magnesium (Alkaline earth metal). They do not have similar properties. D (1s²) and F (1s², 2s²): D is Helium and F is Beryllium. They both have full outer shells (1s² for He, 2s² for Be) but He is a noble gas, and Be is an alkaline earth metal. They are not typically grouped together for similar properties.
7. Arrange the elements in groups and periods in Q. No. 6.
Answer:
A (2s¹): Period 2, Group IA (Li) B (2s², 2p³): Period 2, Group VA (N) C (2s², 2p⁶): Period 2, Group VIIIA (Ne) D (1s²): Period 1, Group VIIIA (He) E (3s¹): Period 3, Group IA (Na) F (2s²): Period 2, Group IIA (Be) G (3s²): Period 3, Group IIA (Mg) H (2s², 2p³): Period 2, Group VA (N) (Same as B)
8. For normal elements, the number of valence electrons of an element is equal to the group number. Find the group number of the following elements. ₁₃Al, ₁₆S, ₁₉K, ₈O
Answer:
₁₃Al: Valence electrons = 3, Group = IIIA (13)
₁₆S: Valence electrons = 6, Group = VIA (16)
₁₉K: Valence electrons = 1, Group = IA (1)
₈O: Valence electrons = 6, Group = VIA (16)
9. Write the valence shell electronic configuration for the following groups:
a. Alkali metals
Answer: ns¹ (where n is the period number)
b. Alkaline earth metals
Answer: ns²
c. Halogens
Answer: ns² np⁵
d. Noble gases
Answer: ns² np⁶ (except Helium: 1s²)
10. Write electron dot symbols for an atom of the following elements:
(a) Be
Answer: Be with two dots (representing 2 valence electrons).
(b) K
Answer: K with one dot (representing 1 valence electron).
(c) N
Answer: N with five dots (representing 5 valence electrons).
(d) I
Answer: I with seven dots (representing 7 valence electrons).
11. Write the valence shell electronic configuration of the atoms of the following elements:
(a) An element present in period 3 of Group VA
Answer: Period 3, Group VA is Phosphorus (P). Valence shell configuration: 3s², 3p³.
(b) An element present in period 2 of Group VIA
Answer: Period 2, Group VIA is Oxygen (O). Valence shell configuration: 2s², 2p⁴.
12. Copy and complete the following table:
| Atomic number | Mass number | No. of protons | No. of neutrons | No. of electrons |
| 11 | 23 | 11 | 12 | 11 |
| 14 | 28 | 14 | 14 | 14 |
| 47 | 108 | 47 | 61 | 47 |
| 27 | 59 | 27 | 32 | 27 |
| 13 | 27 | 13 | 14 | 13 |
Answer:
Row 1: Atomic number = 11, Mass number = 23, Protons = 11, Neutrons = 12, Electrons = 11.
Row 2: Atomic number = 14, Mass number = 28, Protons = 14, Neutrons = 14, Electrons = 14.
Row 3: Atomic number = 47, Mass number = 108, Protons = 47, Neutrons = 61, Electrons = 47.
Row 4: Atomic number = 27, Mass number = 59, Protons = 27, Neutrons = 32, Electrons = 27.
Row 5: Atomic number = 13, Mass number = 27, Protons = 13, Neutrons = 14, Electrons = 13.
13. In which block, group and period in the periodic table where would you place each of the following elements with the following electronic configurations?
(a) 1s², 2s²
Answer: 1s², 2s² = Beryllium (Be). Block: s-block, Group: IIA (2), Period: 2.
(b) 1s², 2s², 2p⁵
Answer: 1s², 2s², 2p⁵ = Fluorine (F). Block: p-block, Group: VIIA (17), Period: 2.
(c) 1s², 2s², 2p⁶, 3s², 3p⁶
Answer: 1s², 2s², 2p⁶, 3s², 3p⁶ = Argon (Ar). Block: p-block, Group: VIIIA (18), Period: 3.
(d) 1s¹
Answer: 1s¹ = Hydrogen (H). Block: s-block, Group: IA (1), Period: 1.
14. What types of elements have the highest ionization energies and what types of elements have the lowest ionization energies. Argue.
Answer:
Highest Ionization Energies: Noble gases (Group VIIIA) have the highest ionization energies. This is because they have a stable, complete valence shell (ns² np⁶) and a high effective nuclear charge. Removing an electron from a stable configuration requires a large amount of energy. Lowest Ionization Energies: Alkali metals (Group IA) have the lowest ionization energies. They have a single valence electron (ns¹) that is shielded from the nucleus by inner electrons and is relatively far from the nucleus. Removing this single electron is easy, requiring little energy. Argument: The ionization energy depends on atomic size, nuclear charge, and shielding. Noble gases have small atomic radii and high nuclear charge, while alkali metals have large atomic radii and low effective nuclear charge.
15. Two atoms have electronic configuration 1s², 2s², 2p⁶ and 1s², 2s², 2p⁶, 3s². The ionization energy of one is 2080 kJ/mole and that of the other is 496 kJ/mole. Match each ionization energy with one of the given electronic configuration. Give reason for your choice.
Answer:
Configuration 1 (1s², 2s², 2p⁶) = Neon (Ne): This is a noble gas with a stable octet. Its ionization energy is very high, so it corresponds to 2080 kJ/mole.
Configuration 2 (1s², 2s², 2p⁶, 3s²) = Magnesium (Mg): This is an alkaline earth metal. It has two valence electrons in the 3s orbital. Its ionization energy is lower, so it corresponds to 496 kJ/mole. Reason: Noble gases have completely filled valence shells and high effective nuclear charge, making it difficult to remove an electron. Alkaline earth metals have a larger atomic size and the valence electrons are more shielded, making it easier to remove an electron.
16. Use the second member of each group from Group IA, IIA and VIIA to judge that the number of valence electron in an atom of the element is the same as its group number.
Answer: The second member of each group is:
Group IA (Li): Electronic configuration = 1s², 2s¹. Valence electrons = 1. Group number = 1.
Group IIA (Be): Electronic configuration = 1s², 2s². Valence electrons = 2. Group number = 2.
Group VIIA (F): Electronic configuration = 1s², 2s², 2p⁵. Valence electrons = 7. Group number = 7. Conclusion: In each case, the number of valence electrons matches the group number (for main group elements).
17. Letter A, B, C, D, E, F indicates elements in the following figure: (Diagram of a periodic table outline with labeled positions A-F)
a. Which elements are in the same periods?
Answer: Elements in the same row (period) are in the same period. Based on the diagram, A and B are in the same period (likely period 2), and C, D, E, F are in the same period (likely period 3).
b. Write valence shell electronic configuration of element D.
Answer: Element D is in Group IVA, Period 3. Its valence shell configuration is 3s², 3p².
c. Which elements are metals?
Answer: Elements on the left and center of the periodic table are metals. Based on the diagram, A, C, and D are likely metals.
d. Which element can lose two electrons?
Answer: Element C is in Group IIA. It can lose two electrons to form a 2+ cation. e. In which group E is present?
Answer: Element E is in Group VIIA (Halogens). f. Which of the element is halogen?
Answer: Element E is a halogen. g. Which element will form dipositive cation?
Answer: Element C (Group IIA) will form a dipositive cation (M²⁺). h. Write electronic configuration of element E.
Answer: Element E is in Group VIIA, Period 3. Its electronic configuration is 1s², 2s², 2p⁶, 3s², 3p⁵.
i. Which two elements can form ionic bond?
Answer: A metal (e.g., C from Group IIA) and a non-metal (e.g., E from Group VIIA) can form an ionic bond.
j. Can element C form C₂ molecule? Interpret.
Answer: No, element C (Group IIA) is a metal and does not form diatomic molecules like C₂. It typically forms ionic compounds. k. Which element can form covalent bonds?
Answer: Non-metals like B (Group IVA), D (Group IVA), and E (Group VIIA) can form covalent bonds.
l. Is element F a metal or non-metal?
Answer: Element F is in Group VIIIA (Noble gas), which is a non-metal.
18. Electronic configurations of four elements are given below:
i) An alkali metal
Answer: (a) 1s², 2s¹ (Lithium) and (c) 1s², 2s², 2p⁶, 3s¹ (Sodium) are alkali metals.
ii) An alkaline earth metal
Answer: None of the given configurations corresponds to an alkaline earth metal (ns²).
iii) A noble gas
Answer: (d) 1s² (Helium) is a noble gas.
iv) A halogen
Answer: None of the given configurations corresponds to a halogen (ns² np⁵).
19. Argue in what region of the periodic table you will find elements with relatively
Answer: High ionization energies are found in the top-right region of the periodic table (excluding noble gases). This includes non- metals like fluorine, chlorine, oxygen, and nitrogen. These elements have small atomic radii and high effective nuclear charges, making it difficult to remove electrons.
Answer: Low ionization energies are found in the bottom-left region of the periodic table. This includes alkali metals like francium, cesium, and rubidium. These elements have large atomic radii and low effective nuclear charges, making it easy to remove electrons.
Identify the group and period of the following elements on the basis of electronic configurations.
(a) ₁₄Si²⁸
Answer: Electronic configuration: 1s², 2s², 2p⁶, 3s², 3p². Period = 3, Group = IVA (14).
(b) ₁₆S³²
Answer: Electronic configuration: 1s², 2s², 2p⁶, 3s², 3p⁴. Period = 3, Group = VIA (16).
(c) ₉F¹⁹
Answer: Electronic configuration: 1s², 2s², 2p⁵. Period = 2, Group = VIIA (17).
(d) ₁₈Ar⁴⁰
Answer: Electronic configuration: 1s², 2s², 2p⁶, 3s², 3p⁶. Period = 3, Group = VIIIA (18).
Electronic configuration of atoms of some elements are given below. Place them into groups and periods.
P = 1s², 2s², 2p³
Answer: Period 2, Group VA (Nitrogen)
Q = 1s², 2s², 3p¹ (Note: The configuration is likely incorrect. Correct configuration for Group IIIA, Period 3 is 1s², 2s², 2p⁶, 3s², 3p¹)
R = 1s¹
Answer: Period 1, Group IA (Hydrogen)
S = 1s², 2s¹
Answer: Period 2, Group IA (Lithium)
T = 1s², 2s²
Answer: Period 2, Group IIA (Beryllium)
W = 1s², 2s², 2p⁵
Answer: Period 2, Group VIIA (Fluorine)
X = 1s², 2s², 2p⁶, 3s², 3p¹
Answer: Period 3, Group IIIA (Aluminium)
Y = 1s², 2s², 2p⁶, 3s², 3p⁶
Answer: Period 3, Group VIIIA (Argon)
Z = 1s², 2s², 2p¹
Answer: Period 2, Group IIIA (Boron)
Obtain the valence shell configuration of Al and S from their position in the periodic table.
Answer:
Al (Aluminium): Group IIIA (13), Period 3. Valence shell configuration = 3s², 3p¹.
S (Sulfur): Group VIA (16), Period 3. Valence shell configuration = 3s², 3p⁴.
Choose the element whose atoms you expect to have smaller shielding effect.
(a) F or Cl
Answer: F (Fluorine) has a smaller shielding effect. It has fewer inner shells (1 inner shell) than Cl (2 inner shells).
(b) Li or Na
Answer: Li (Lithium) has a smaller shielding effect. It has fewer inner shells (1 inner shell) than Na (2 inner shells).
(c) B or Al
Answer: B (Boron) has a smaller shielding effect. It has fewer inner shells (1 inner shell) than Al (2 inner shells).
Using the periodic table but without looking at the figure 4.4, choose the element whose atom you expect to have smaller atomic radius in each of the following pairs.
(a) O or S
Answer: O (Oxygen) has a smaller atomic radius. Atomic radius increases down a group.
(b) O or F
Answer: F (Fluorine) has a smaller atomic radius. Atomic radius decreases across a period from left to right.
Which atom has the smaller ionization energy?
(a) B or N
Answer: B (Boron) has smaller ionization energy. Ionization energy increases across a period.
(b) Be or Mg
Answer: Mg (Magnesium) has smaller ionization energy. Ionization energy decreases down a group.
(c) C or Si
Answer: Si (Silicon) has smaller ionization energy. Ionization energy decreases down a group.
Which element has lower metallic character?
(a) Li or K
Answer: Li (Lithium) has lower metallic character. Metallic character increases down a group.
(b) Mg or Ca
Answer: Mg (Magnesium) has lower metallic character. Metallic character increases down a group.
(c) Compare and contrast ionization energy and electron affinity
Answer:
Ionization Energy: The energy required to remove an electron from a gaseous atom. It is always endothermic (positive). It increases across a period and decreases down a group. Electron Affinity: The energy change when a gaseous atom gains an electron. It is usually exothermic (negative). It increases across a period and decreases down a group. Comparison: Both are measures of an atom's tendency to attract electrons, but ionization energy measures the resistance to losing an electron, while electron affinity measures the attraction for gaining an electron. Both increase across a period and decrease down a group.
Which of the following displacement reactions will occur?
1. Cl₂(g) + 2NaF(aq) [source symbol unavailable] 2NaCl(aq) + F₂(g)
Answer: No reaction. Chlorine cannot displace fluorine because fluorine is more reactive (higher oxidizing power) than chlorine.
2. Br₂(g) + 2KI(aq) [source symbol unavailable] 2KBr(aq) + I₂(g)
Answer: Reaction occurs. Bromine is more reactive than iodine and can displace it from KI.
3. I₂(g) + 2KBr(aq) [source symbol unavailable] 2KI(aq) + Br₂(g)
Answer: No reaction. Iodine is less reactive than bromine and cannot displace it.
4. Cl₂(g) + 2KBr(aq) [source symbol unavailable] 2KCl(aq) + Br₂(g)
Answer: Reaction occurs. Chlorine is more reactive than bromine and can displace it from KBr.
5. Cl₂(g) + 2NaI(aq) [source symbol unavailable] 2NaCl(aq) + I₂(g)
Answer: Reaction occurs. Chlorine is more reactive than iodine and can displace it from NaI.
Compare the general properties of metals and non-metals.
Answer:
| Property | Metals | Non-metals |
| State | Solid (except Hg) | Solid, liquid, or gas |
| Lustre | Shiny (metallic lustre) | Dull (except iodine) |
| Malleability | Malleable and ductile | Brittle (if solid) |
| Conductivity | Good conductors of heat and electricity | Poor conductors (except graphite) |
| Ionization Energy | Low | High |
| Electron Affinity | Low | High |
| Oxides | Basic (e.g., Na₂O) | Acidic (e.g., SO₂) |
| Reaction with Acids | React to produce H₂ | Generally do not react |
| Position in Periodic Table | Left and center | Right |
| Examples | Sodium, Iron, Copper | Oxygen, Chlorine, Sulfur |
Prepare 3D model of the periodic table (Group Activity).
Answer: (Guide for the group activity)
Objective: To create a three-dimensional visual representation of the periodic table. Materials: Cardboard, colored paper, markers, glue, scissors, small boxes or blocks, string, beads or small objects to represent elements. Procedure:
A. Design: Plan the layout of the periodic table on a large cardboard sheet. Mark the positions of periods (rows) and groups (columns).
B. Element Cards: Create small cards or blocks for each element. Write the element's symbol, atomic number, and name.
C. Color Coding: Use different colors for different blocks (s-block, p-block, d-block, f-block) or for metals, non-metals, and metalloids.
D. Assembly: Attach the element cards to the cardboard base in the correct positions. E. 3D Elements: Use small beads or objects to represent electrons in shells around the nucleus for a few selected elements. F. Labels: Add labels for periods, groups, and key regions (e.g., alkali metals, halogens, noble gases). Presentation: Each group presents their model and explains the organization of the periodic table and periodic trends. Learning Outcome: This activity helps students visualize the arrangement of elements, understand periodic trends, and collaborate effectively.
