Page No 93: - Chapter 3 Classifications Of Elements & Periodicity In Properties class 11 ncert solutions Chemistry - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Question 3.5:

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

Answer:

Elements with atomic numbers from Z = 87 to Z = 114 are present in the 7^th period of the periodic table. Thus, the element with Z = 114 is present in the 7^th period of the periodic table.

In the 7^th period, first two elements with Z = 87 and Z= 88 are s-block elements, the next 14 elements excluding Z = 89 i.e., those with Z = 90 – 103 are f – block elements, ten elements with Z = 89 and Z = 104 – 112 are d – block elements, and the elements with Z = 113 – 118 are p – block elements. Therefore, the element with Z = 114 is the second p – block element in the 7^th period. Thus, the element with Z = 114 is present in the 7^th period and 14^th group of the periodic table.

Question 3.6:

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

Answer:

There are two elements in the 1^st period and eight elements in the 2^nd period., The third period starts with the element with Z = 11. Now, there are eight elements in the third period. Thus, the 3^rd period ends with the element with Z = 18 i.e., the element in the 18^th group of the third period has Z = 18. Hence, the element in the 17^th group of the third period has atomic number Z = 17.

Question 3.7:

Which element do you think would have been named by

(i) Lawrence Berkeley Laboratory

(ii) Seaborg’s group?

Answer:

(i) Lawrencium (Lr) with Z = 103 and Berkelium (Bk) with Z = 97

(ii) Seaborgium (Sg) with Z = 106

Question 3.8:

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

Answer:

The physical and chemical properties of elements depend on the number of valence electrons. Elements present in the same group have the same number of valence electrons. Therefore, elements present in the same group have similar physical and chemical properties.

Question 3.9:

What does atomic radius and ionic radius really mean to you?

Answer:

Atomic radius is the radius of an atom. It measures the size of an atom. If the element is a metal, then the atomic radius refers to the metallic radius, and if the element is a non-metal, then it refers to the covalent radius. Metallic radius is calculated as half the internuclear distance separating the metal cores in the metallic crystal. For example, the internuclear distance between two adjacent copper atoms in solid copper is 256 pm. Thus, the metallic radius of copper is taken as .

Covalent radius is measured as the distance between two atoms when they are found together by a single bond in a covalent molecule. For example, the distance between two chlorine atoms in chlorine molecule is 198 pm. Thus, the covalent radius of chlorine is taken as .

Ionic radius means the radius of an ion (cation or anion). The ionic radii can be calculated by measuring the distances between the cations and anions in ionic crystals.

Since a cation is formed by removing an electron from an atom, the cation has fewer electrons than the parent atom resulting in an increase in the effective nuclear charge. Thus, a cation is smaller than the parent atom. For example, the ionic radius of  ion is 95 pm, whereas the atomic radius of Na atom is 186 pm. On the other hand, an anion is larger in size than its parent atom. This is because an anion has the same nuclear charge, but more electrons than the parent atom resulting in an increased repulsion among the electrons and a decrease in the effective nuclear charge. For example, the ionic radius of F^– ion is 136 pm, whereas the atomic radius of F atom is 64 pm.

Question 3.10:

How does atomic radius vary in a period and in a group? How do you explain the variation?

Answer:

Atomic radius generally decreases from left to right across a period. This is because within a period, the outer electrons are present in the same valence shell and the atomic number increases from left to right across a period, resulting in an increased effective nuclear charge. As a result, the attraction of electrons to the nucleus increases.

On the other hand, the atomic radius generally increases down a group. This is because down a group, the principal quantum number (n) increases which results in an increase of the distance between the nucleus and valence electrons.

Question 3.11:

What do you understand by isoelectronic species? Name a species that will be

isoelectronic with each of the following atoms or ions.

(i) F^–

(ii) Ar

(iii) Mg²⁺

(iv) Rb⁺

Answer:

Atoms and ions having the same number of electrons are called isoelectronic species.

(i) F^– ion has 9 + 1 = 10 electrons. Thus, the species isoelectronic with it will also have 10 electrons. Some of its isoelectronic species are Na⁺ ion (11 – 1 = 10 electrons), Ne (10 electrons), O^2– ion (8 + 2 = 10 electrons), and Al³⁺ ion (13 – 3 = 10 electrons).

(ii) Ar has 18 electrons. Thus, the species isoelectronic with it will also have 18 electrons. Some of its isoelectronic species are S^2– ion (16 + 2 = 18 electrons), Cl^– ion (17 + 1 = 18 electrons), K⁺ ion (19 – 1 = 18 electrons), and Ca²⁺ ion (20 – 2 = 18 electrons).

(iii) Mg²⁺ ion has 12 – 2 = 10 electrons. Thus, the species isoelectronic with it will also have 10 electrons. Some of its isoelectronic species are F^– ion (9 + 1 = 10 electrons), Ne (10 electrons), O^2– ion (8 + 2 = 10 electrons), and Al³⁺ ion (13 – 3 = 10 electrons).

(iv) Rb⁺ ion has 37 – 1 = 36 electrons. Thus, the species isoelectronic with it will also have 36 electrons. Some of its isoelectronic species are Br^– ion (35 + 1 = 36 electrons), Kr (36 electrons), and Sr²⁺ ion (38 – 2 = 36 electrons).

Question 3.12:

Consider the following species:

N^3–, O^2–, F^–, Na⁺, Mg²⁺ and Al³⁺

(a) What is common in them?

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

Answer:

(a) Each of the given species (ions) has the same number of electrons (10 electrons). Hence, the given species are isoelectronic.

(b) The ionic radii of isoelectronic species increases with a decrease in the magnitudes of nuclear charge.

The arrangement of the given species in order of their increasing nuclear charge is as follows:

N^3– < O^2– < F^– < Na⁺ < Mg²⁺ < Al³⁺

Nuclear charge = +7 +8 +9 +11 +12 +13

Therefore, the arrangement of the given species in order of their increasing ionic radii is as follows:

Al³⁺ < Mg²⁺ < Na⁺ < F^– < O^2– < N^3–

Question 3.13:

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

Answer:

A cation has a fewer number of electrons than its parent atom, while its nuclear charge remains the same. As a result, the attraction of electrons to the nucleus is more in a cation than in its parent atom. Therefore, a cation is smaller in size than its parent atom.

On the other hand, an anion has one or more electrons than its parent atom, resulting in an increased repulsion among the electrons and a decrease in the effective nuclear charge. As a result, the distance between the valence electrons and the nucleus is more in anions than in it’s the parent atom. Hence, an anion is larger in radius than its parent atom.

Question 3.14:

What is the significance of the terms – ‘isolated gaseous atom’ and ‘ground state’ while defining the ionization enthalpy and electron gain enthalpy?

Hint: Requirements for comparison purposes.

Answer:

Ionization enthalpy is the energy required to remove an electron from an isolated gaseous atom in its ground state. Although the atoms are widely separated in the gaseous state, there are some amounts of attractive forces among the atoms. To determine the ionization enthalpy, it is impossible to isolate a single atom. But, the force of attraction can be further reduced by lowering the pressure. For this reason, the term ‘isolated gaseous atom’ is used in the definition of ionization enthalpy.

Ground state of an atom refers to the most stable state of an atom. If an isolated gaseous atom is in its ground state, then less amount energy would be required to remove an electron from it. Therefore, for comparison purposes, ionization enthalpy and electron gain enthalpy must be determined for an ‘isolated gaseous atom’ and its ‘ground state’.

Question 3.15:

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.

Answer:

It is given that the energy of an electron in the ground state of the hydrogen atom is –2.18 × 10^–18 J.

Therefore, the energy required to remove that electron from the ground state of hydrogen atom is 2.18 × 10^–18 J.

Ionization enthalpy of atomic hydrogen = 2.18 × 10^–18 J

Hence, ionization enthalpy of atomic hydrogen in terms of J mol^–1 = 2.18 × 10^–18 × 6.02 × 10²³ J mol^–1 = 1.31 × 10⁶ J mol^–1

Question 3.16:

Among the second period elements the actual ionization enthalpies are in the

order Li < B < Be < C < O < N < F < Ne.

Explain why

(i) Be has higher Δ_iH than B

(ii) O has lower Δ_iH than N and F?

Answer:

(i) During the process of ionization, the electron to be removed from beryllium atom is a 2s-electron, whereas the electron to be removed from boron atom is a 2p­-electron. Now, 2s-electrons are more strongly attached to the nucleus than 2p-electrons. Therefore, more energy is required to remove a 2s­-electron of beryllium than that required to remove a 2p-electron of boron. Hence, beryllium has higher Δ_iH than boron.

(ii) In nitrogen, the three 2p-electrons of nitrogen occupy three different atomic orbitals. However, in oxygen, two of the four 2p-electrons of oxygen occupy the same 2p-orbital. This results in increased electron-electron repulsion in oxygen atom. As a result, the energy required to remove the fourth 2p-electron from oxygen is less as compared to the energy required to remove one of the three 2p-electrons from nitrogen. Hence, oxygen has lower ΔiH than nitrogen.

Fluorine contains one electron and one proton more than oxygen. As the electron is being added to the same shell, the increase in nuclear attraction (due to the addition of a proton) is more than the increase in electronic repulsion (due to the addition of an electron). Therefore, the valence electrons in fluorine atom experience a more effective nuclear charge than that experienced by the electrons present in oxygen. As a result, more energy is required to remove an electron from fluorine atom than that required to remove an electron from oxygen atom. Hence, oxygen has lower ΔiH than fluorine.

Question 3.17:

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:

The first ionization enthalpy of sodium is lower than that of magnesium. This is primarily because of two reasons:

1. The atomic size of sodium is greater than that of magnesium

2. The effective nuclear charge of magnesium is higher than that of sodium

For these reasons, the energy required to remove an electron from magnesium is more than the energy required in sodium. Hence, the first ionization enthalpy of sodium is lower than that of magnesium.

However, the second ionization enthalpy of sodium is higher than that of magnesium. This is because after losing an electron, sodium attains the stable noble gas configuration. On the other hand, magnesium, after losing an electron still has one electron in the 3s-orbital. In order to attain the stable noble gas configuration, it still has to lose one more electron. Thus, the energy required to remove the second electron in case of sodium is much higher than that required in case of magnesium. Hence, the second ionization enthalpy of sodium is higher than that of magnesium.

Question 3.18:

What are the various factors due to which the ionization enthalpy of the main

group elements tends to decrease down a group?

Answer:

The factors responsible for the ionization enthalpy of the main group elements to decrease down a group are listed below:

(i) Increase in the atomic size of elements: As we move down a group, the number of shells increases. As a result, the atomic size also increases gradually on moving down a group. As the distance of the valence electrons from the nucleus increases, the electrons are not held very strongly. Thus, they can be removed easily. Hence, on moving down a group, ionization energy decreases.

(ii) Increase in the shielding effect: The number of inner shells of electrons increases on moving down a group. Therefore, the shielding of the valence electrons from the nucleus by the inner core electrons increases down a group. As a result, the valence electrons are not held very tightly by the nucleus. Hence, the energy required to remove a valence electron decreases down a group.

Question 3.19:

The first ionization enthalpy values (in kJmol^–1) of group 13 elements are :

How would you explain this deviation from the general trend?

Answer:

On moving down a group, ionization enthalpy generally decreases due to an increase in the atomic size and shielding. Thus, on moving down group 13, ionization enthalpy decreases from B to Al. But, Ga has higher ionization enthalpy than Al. Al follows immediately after s–block elements, whereas Ga follows after d–block elements. The shielding provided by d-electrons is not very effective. These electrons do not shield the valence electrons very effectively. As a result, the valence electrons of Ga experience a greater effective nuclear charge than those of Al. Further, moving from Ga to In, the ionization enthalpy decreases due to an increase in the atomic size and shielding. But, on moving from In to Tl, the ionization enthalpy again increases. In the periodic table, Tl follows after 4f and 5d electrons. The shielding provided by the electrons in both these orbitals is not very effective. Therefore, the valence electron is held quite strongly by the nucleus. Hence, the ionization energy of Tl is on the higher side.

Question 3.20:

Which of the following pairs of elements would have a more negative electron gain enthalpy?

(i) O or F (ii) F or Cl

Answer:

(i) O and F are present in the same period of the periodic table. An F atom has one proton and one electron more than O and as an electron is being added to the same shell, the atomic size of F is smaller than that of O. As F contains one proton more than O, its nucleus can attract the incoming electron more strongly in comparison to the nucleus of O atom. Also, F needs only one more electron to attain the stable noble gas configuration. Hence, the electron gain enthalpy of F is more negative than that of O.

(ii) F and Cl belong to the same group of the periodic table. The electron gain enthalpy usually becomes less negative on moving down a group. However, in this case, the value of the electron gain enthalpy of Cl is more negative than that of F. This is because the atomic size of F is smaller than that of Cl. In F, the electron will be added to quantum level n = 2, but in Cl, the electron is added to quantum level n = 3. Therefore, there are less electron- electron repulsions in Cl and an additional electron can be accommodated easily. Hence, the electron gain enthalpy of Cl is more negative than that of F.

Question 3.21:

Would you expect the second electron gain enthalpy of O as positive, more negative or less negative than the first? Justify your Answer.

Answer:

When an electron is added to O atom to form O^– ion, energy is released. Thus, the first electron gain enthalpy of O is negative.

On the other hand, when an electron is added to O^– ion to form O^2– ion, energy has to be given out in order to overcome the strong electronic repulsions. Thus, the second electron gain enthalpy of O is positive.

Question 3.22:

What is the basic difference between the terms electron gain enthalpy and electronegativity?

Answer:

Electron gain enthalpy is the measure of the tendency of an isolated gaseous atom to accept an electron, whereas electronegativity is the measure of the tendency of an atom in a chemical compound to attract a shared pair of electrons.

Question 3.23:

How would you react to the statement that the electronegativity of N on Pauling scale is 3.0 in all the nitrogen compounds?

Answer:

Electronegativity of an element is a variable property. It is different in different compounds. Hence, the statement which says that the electronegativity of N on Pauling scale is 3.0 in all nitrogen compounds is incorrect. The electronegativity of N is different in NH₃ and NO₂.