Class 11 Chemistry

Periodic Table

NCERT Exercise

Part 1

Question 1: What is the basic theme of organization in the periodic table?

Answer: The periodic table has rows and columns. Rows are called periods and columns are called groups. There are 18 groups, viz. IA, IIA, IIIA, IVA, VA, VIA, VIIA, VIII, IB, IIB, IIIB, IVB, VB, VIB, VIIB, 18. There are seven periods in the periodic table. Metals are on the right hand side and take most of the space in the periodic table. Non-metals are on the LHS, preceded by metalloids.

Question 2: Which important property did Mendeleev use to classify the elements in his periodic table and did he stick to that?

Answer: Mendeleev used atomic mass to classify elements in his periodic table. But in order to group some elements with similar properties, he put certain elements with greater atomic mass before those with lower atomic mass.

Question 3: What is the basic difference in approach between the Mendeleev’s Periodic Law and the Modern Periodic Law?

Answer: Mendeleev’s Periodic Law is based on atomic mass, while Modern Periodic Law is based on Atomic number.

Question 4: On the basis of quantum numbers, justify that the sixth period of the periodic table should have 32 elements.

Answer: The principal quantum number for sixth period is 6, or n = 6

For n = 6, the value of l = 0, 1, 2, 3, 4 which are named as s, p, d, f

Number of orbitals in s, p, d, f = 1 + 3 + 5 + 7 = 16

So, no. of elements = 2 × 16 = 32

Question 5: In terms of period and group where would you locate the element with Z = 114?

Answer: Electronic configuration of 114 = (Rn) + 5f14 6d10 7s2 7p2

Period number = n = 7

Block = p

Group number for p-block element = 10 + no. of electrons in valence shell

= 10 + 4 = 14

So, it is a p-block element of Group 14 and Period 7

Question 6: Write the atomic number of the element present in the third period and seventeenth group of the periodic table.

Answer: Cl (17)

Question 7: Which element do you think would have been named by

  1. Lawrence Berkeley Laboratory
  2. Seaborg’s group

Answer: (a) Lawrencium (Lr [103]), (b) Seaborgium (Sg [106])

Question 8: Why do elements in the same group have similar physical and chemical properties?

Answer: Elements in the same group have same number of valence or oxidation number. Due to this, elements in the same group have similar physical and chemical properties.

Question 9: What does atomic radius and ionic radius really mean to you?

Answer: Atomic radius is the distance between nucleus and outermost electron in an orbit. Ionic radius is the distance between nucleus and outermost electron in an ion.

Question 10: How do atomic radii vary in a period and in a group? How do you explain the variation?

Answer: Atomic radii decrease when we move left to right in a period. This happens because addition of electron in the same shell increases nuclear attraction. On the other hand, addition of new shell when we go down a group results in reduction in nuclear attraction. Due to this, atomic radii increase when we go down a group.

Question 11: What do you understand by isoelectronic species? Name a species that will be isoelectric with each of the following atoms or ions: (a) F- (b) Ar (c) Mg2+ (d) Rb+

Answer: Atoms or ions with same number of electrons are called isoelectronic species. For example, O2-, F-, Na+ and Mg2+ have the same number of electrons (10).

Answer: (a) Mg2+, (b) K+, (c) F-

Question 12: Consider the following species: N3-, O2-, F-, Na+, Mg2+ and Al3+

(a) What is common in them?

Answer: All of them are isoelectronic species

(b) Arrange them in the order of increasing ionic radii.

Answer: Al3+ < Mg2+ < Na+ F- < O2- N3-

Question 13: Explain why cations are smaller and anions larger in radii than their parent atoms?

Answer: When an atom loses an electron it becomes a cation. Due to less number of electrons there will be greater attraction of electrons towards nucleus. So, a cation has smaller radius than its parent atom. Opposite to this happens in case of anion.

Question 14: What is the significance of the terms – ‘isolated gaseous atom’ and ‘ground state’ while defining ionization enthalpy and electron gain enthalpy? (Hint: Requirements for comparison purposes.)

Answer: In gaseous state, atoms are quite far from each other so one atom does not influence another atom’s ionization enthalpy or electron gain enthalpy. So, example of isolated gaseous atom is taken in order to study ionization enthalpy and electron gain enthalpy.

Ground state refers to the electrically neutral state of an atom. So, ground state is ideal for defining ionization enthalpy and electron gain enthalpy.

Question 15: Energy of an electron in the ground state of the hydrogen atom is -2.18 × 10-18 J. Calculate the ionization enthalpy of atomic hydrogen in terms of J mol-1. (Hint: Apply the idea of mole concept to derive the answer)

Answer: Energy of an atom is given by following:

-2.18 × 10-18 J × (6.022 × 1023

= -13.13 × 105 J = -1.313 × 106 J

Electron enthalpy = E - E

= 0 + 1.313 × 106 J mol-1

Question 16: Among the second period elements the actual ionization enthalpies are in the order Li < B < Be < C < O &let; N < F < Ne.

Explain why

(a) Be has higher Δi H than B

Answer: Electronic configuration of Beryllium and Boron are as follows:

Be (1s2 2s2)

B (1s2 2s2 2p1)

Be has last electrons in 2s orbital, while B has last electron in 2p orbital. Electrons in s orbital are strongly attracted to nucleus than those in p orbital. Due to this, Be has higher Δi H than B

(b) O has lower Δi than N and F?

Answer: Electronic configuration of oxygen, nitrogen and fluorine are as follows:

O: 1s2 22 2p4

N: 1s2 22 2p3

F: 1s2 22 2p5

N has half filled 2p orbital which is not the case with N and O. Due to this, O has lower Δi than N and F.

Question 17: How would you explain the fact that the first ionization enthalpy of sodium is lower than that of magnesium but its second ionization enthalpy is higher than that of magnesium?

Answer: Electronic configuration of Na and Mg are as follows:

Na: 1s2 2s2 2p6 3s1

Mg: 1s2 2s2 2p6 3s2

The 3s orbital of Na is half filled while it is fully filled in Mg. It is more difficult to remove electrons from a fully filled orbital than from a half filled orbital. Due to this, first ionization enthalpy of sodium is lower than that of magnesium. In case of Na+, the outermost orbital is 2p which is fully filled. But in case of Mg+ the outermost orbital 2s is half filled. So, second ionization enthalpy of sodium is greater than that of magnesium.

Question 18: What are the various factors due to which the ionization enthalpy of the main group elements tends to decrease down a group?

Answer: When we go down a group, atomic radii decrease. It results in reduction in nuclear attraction on electrons in outermost shell. Due to this, ionization enthalpy decreases when we go down a group.

Question 19: The first ionization enthalpy values (in kJ mol-1) of group 13 elements are:


How would you explain this deviation from the general trend?

Answer: Ionisation enthalpy decreases from B to Al due to addition of a shell. But in case of Ga, addition of 3d orbital (with 10 electrons in it) increases ionization enthalpy. Same thing happens in case of Tl with addition of 4f orbitals (with 14 electrons). 3d or 4f orbitals do not provide shielding effect the way s or p orbitals do.

Question 20: Which of the following pairs of elements would have a more negative electron gain enthalpy? (a) O or F (b) F or Cl

Answer: Generally, electron gain enthalpy becomes more negative from left to right in a period. This happens because increase in effective nuclear charge makes it easier to add an electron to a smaller atom. Electron enthalpy generally becomes less negative when we move down a group.

Oxygen is towards left of Fluorine so Oxygen has less negative electron gain enthalpy than Fluorine. Chlorine is below fluorine in the group of halogens. So, Fluorine has higher negative electron gain enthalpy than chlorine.