NCERT solutions for class 11 chemistry chapter 3 Classification of Elements and Periodicity in Properties - This chapter starts with, a brief history of the periodic table and historical development of the periodic law and the periodic table and ends with some discussion on physical trends like atomic sizes, ionization enthalpies etc, in the physical and chemical properties of the elements. In this chapter, there are 40 questions in the exercise. The NCERT solutions for class 11 chemistry chapter 3 Classification of Elements and Periodicity in Properties are prepared by respective subject experts. These NCERT solutions will help you in the preparation of your class 11 final exam and also will help in the preparation of various competitive exams like JEE Mains, NEET, VITEEE etc.
In this chapter, you will study the development of the periodic table and the periodic law. Mendeleev arranged elements on the basis of their atomic masses in his periodic table. It is must to go through CBSE NCERT solutions for class 11 chemistry chapter 3 Classification of Elements and Periodicity in Properties to maximise your score in the examination. The modern periodic table arranges the elements in order of their atomic numbers. In the modern periodic table, elements are arranged in eighteen vertical columns( families or groups) and seven horizontal rows (periods). This chapter discusses some periodic trends in the periodic table like ionization enthalpies, atomic sizes, electron gain enthalpies, electronegativity, and valence. After completing solutions of NCERT for class 11 chemistry chapter 3 Classification of Elements and Periodicity in Properties, you will be able to-
1. Understand periodic law
2. Understand the significance of electronic configuration and atomic numbers as a basis for periodic classification
3. Able to classify elements of the periodic table into four different blocks which are s, p, d, f and learn their main characteristics
4. Understand periodic trends in chemical and physical properties of elements and explain the relationship between metallic character and ionization enthalpy.
Important points and terms of Chapter 3 Classification of Elements and Periodicity in Properties-
In the modern periodic table, elements are arranged in the order of their atomic numbers in 7 horizontal rows or periods and 18 vertical columns or groups.
The first period is the smallest and sixth period is the longest.
Elements of periodic table are divided into four s, p, d, f blocks.
Elements of group 1 and 2 are s-block elements and the general valence shell electronic configuration of s-block elements is .
Elements of group 13, 14, 15, 16, 17 and 18 are p-block elements and the general valence shell electronic configuration of p-block elements is .
Elements of group 3, 4, 5, 6, 7, 8, 9, 10, 11 and 12 are d-block elements and the general valence shell electronic configuration of d-block elements is .
Elements of two horizontal rows at the bottom of the periodic table are f-block elements and the general valence shell electronic configuration of f-block elements is
NCERT solutions for class 11 chemistry chapter 3 Classification of Elements and Periodicity in Properties- Solved Exercise Questions
Question 3.1 What is the basic theme of organisation in the periodic table?
The basic theme of organization in the periodic table is to classify all the elements according to similar properties in periods and groups, This arrangement makes the study of elements and their compounds simple and systematic and less confusion can be generated and true information can be regenerated about the different elements present in the periodic table.
Question 3.2 Which important property did Mendeleev use to classify the elements in his periodic table and did he stick to that?
Mendeleev arranged elements in horizontal rows and vertical columns of a table in order of their increasing atomic weights in such a way that the elements with similar properties occupied the same vertical column or group. He fully recognized the significance of periodicity and used a broader range of physical and chemical properties to classify the elements.
In particular, Mendeleev relied on the similarities in the empirical formulas and properties of the compounds formed by the elements. He realized that some of the elements did not fit in with his scheme of classification if the order of atomic weight was strictly followed. He ignored the order of atomic weights, thinking that the atomic measurements might be incorrect, and placed the elements with similar properties together.
Question 3.4 On the basis of quantum numbers, justify that the sixth period of the periodic table should have 32 elements.
The period number corresponds to the highest principal quantum number (n) of the elements in the period. The subsequent periods consists of 8, 8, 18, 18 and 32 elements, respectively. The sixth period (n = 6) contains 32 elements and successive electrons enter 6s, 4f, 5d and 6p orbitals, Now, 6s has 1 orbital, 4f has 7 orbitals, 5d has 5 orbitals, and 6p has 3 orbitals.
Therefore, there is a total of 16 orbitals available in which each orbital carry 2 electrons according to Pauli's exclusion principle. So, we have a total of 32 electrons, Hence the sixth period of the periodic table should have 32 elements.
Question 3.5 In terms of period and group where would you locate the element with Z =114?
Elements with atomic numbers from Z=87 to Z= 114 are present in the 7th period of the periodic table. Thus the element with Z=114 is present in the seventh period.
In the seventh period, the first two elements are s-block elements, the next 14 elements are f-block elements, next 10 elements are d-block elements and the next 6 elements are p-block elements.
Therefore, Z=114 element is the second p-block element in the seventh period. Thus, it is present in the 4th group of the 7th period 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.
Given that it is present in the 3rd period and a 17th group of the periodic table.
So, the first period has 2 elements and the second period has 8 elements. The 3rd period starts with the element with Z=11. And there are 8 elements present in the third period. Thus, the 3rd period ends with the element with Z=18. The element in the 18th group of the 3rd period has Z=18. Hence, the element in the 17th group of the 3rd period has atomic number Z=17.
Question 3.9 What does atomic radius and ionic radius really mean to you?
Atomic radius is the distance between from the centre of an atom to the outer most shell containing the electrons. It is a measure of the size of an atom.
They can be of types like :
(a) Covalent radius, where it is the half of the distance between the line joining the centres of nuclei of two adjacent similar atoms.
(b) Metallic radius is half of the distance between the centres of the nuclei of two adjacent atoms in the metallic crystal.
(c) van der Waal's radius, half of the internuclear distance between 2 similar adjacent atoms.
Whereas, the Ionic radius is the distance from the centre of the nucleus of the ion up to which it exerts its influence on the electron cloud of an ion (cation or anion). Generally, cation has a smaller ionic radius than the parent atom and anions are larger than the parent atom.
Question 3.10 How do atomic radius vary in a period and in a group? How do you explain the variation?
The atomic radius of the elements generally decreases from the left to the right in a period because the nuclear charge gradually increases by one unit and one electron is also added in the electron shell, due to this the electrons get attracted more towards the centre, as a result, the atomic radii decreases.
The atomic radius of the elements increases as we move downwards in a group because there is an increase in the principal quantum number and thus, there is an increase in the number of electrons shells. Therefore, the atomic size is expected to increase.
Question 3.13 Explain why cation are smaller and anions larger in radii than their parent atoms?
A cation is smaller than its parent atom because it has fewer electrons while its nuclear charge remains the same. The size of an anion will be larger than that of the parent atom because the addition of one or more electrons would result in increased repulsion among the electrons and a decrease in effective nuclear charge.
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 (ii) O has lower than N and F?
In oxygen, two of the four 2p-electrons of oxygen occupy the same 2p-orbital and in nitrogen, the three 2p-electrons of nitrogen occupy three different atomic orbitals.
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 sp-electrons from nitrogen.
Hence, oxygen has lower than nitrogen.
In case of fluorine the electron is added to the same shell, the increase in nuclear attraction dominates over the increase in electronic repulsion.
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 a fluorine atom than that required to remove an electron from oxygen atom.
Hence, oxygen has lower than fluorine.
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?
Ionization enthalpy of the main group elements tends to decrease down a group because of the following reasons:
(a) Atomic Size: moving down the group there is an increase in the atomic size, increasing number of electron shells, which results in the weaker binding force with the nucleus. Hence the ionization enthalpy decreases.
(b) Screening or shielding effect: moving down the group increases the new shells, the number of inner electron shells which shield the valence electrons increases. As a result, the force of attraction of the nucleus for the valence electrons further decreases and hence the ionization enthalpy decreases.
Question 3.20(i) Which of the following pairs of elements would have a more negative electron gain enthalpy?
O or F
Oxygen and Fluorine both are of the second group elements and as we move from O to F, there is a decrease in atomic size, as a result, there is an increase in nuclear charge. Further, by a gain of one electron, ion has an inert gas configuration and gives ion which does not have a stable inert gas configuration.
The energy released is much higher in going from fluorine to oxygen or the electron gain enthalpy value of F is much more negative than that of oxygen.
Question 3.22 What is the basic difference between the terms electron gain enthalpy and electronegativity?
The basic differences between the terms electron gain enthalpy and electronegativity are:
Electron gain enthalpy refers to the tendency of an isolated gaseous atom to accept an additional electron from a negative ion, whereas electronegativity refers to the tendency of an atom of an element to attract a shared pair of electrons towards it in a covalent bond.
Question 3.24(a) Describe the theory associated with the radius of an atom as it
gains an electron
When an atom gains an electron it forms an anion. And the size of an anion will be larger than the parent atom because of the addition of one or more electrons would result in increased repulsion among electrons and a decrease in the effective nuclear charge. This results in the increase in the atomic radius of an atom
Question 3.24(b) Describe the theory associated with the radius of an atom as it
loses an electron
When an atom loses an electron it becomes positively charged and now for the same nuclear charge, there are lesser electrons present than the parent atom. Hence there will be an overall increase in the attraction between electrons and nucleus. As a result, there will be a reduction in the atomic radius of an atom and it is smaller than the parent atom.
Question 3.26 What are the major differences between metals and non-metals?
Metals are usually solids at room temperature [mercury is an exception; gallium and caesium also have very low melting points]. Metals usually have high melting and boiling points. They are good conductors of heat and electricity because they have the tendency to lose electrons easily. They are malleable (can be flattened into thin sheets by hammering) and ductile (can be drawn into wires).
Non-metals are usually solids or gases at room temperature with low melting and boiling points (boron and carbon are exceptions). They are poor conductors of heat and electricity. Most nonmetallic solids are brittle and are neither malleable nor ductile hence forming mainly covalent compounds.
Question 3.28 The increasing order of reactivity among group 1 elements is Li < Na < K < Rb <Cs whereas that among group 17 elements is F > CI > Br > I. Explain.
In the case of group 1, we have ordered because there is only one valence electron. and thus have a strong tendency to lose this electron. The tendency to lose electrons in turn also depends upon the ionization enthalpy. Moving down the group there is a decrease in the ionization enthalpy, therefore, increasing reactivity down the group.
In the case of group 17, we have the order because there are seven electrons present in the valence shells and thus have a strong tendency to accept electrons to make it a stable noble gas electronic configuration. So, moving down we have a decrease in both electron gain enthalpy and electronegativity. Hence there is a decrease in the reactivity also.
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