NCERT Solutions for Class 11 Chemistry Chapter 3
NCERT Solutions for Class 11 Chemistry Chapter 3 – Classification of Elements and Periodicity in Properties
Do you want to understand the concepts of Chemistry Class 11 Chapter 3 thoroughly? NCERT Solutions for Class 11 Chemistry Chapter 3 will assure just that. You can get an insight into the crucial topics of Classification of Elements and Periodicity with Class 11 Chemistry Chapter 3 NCERT Solutions. These NCERT Solutions, curated by subject experts, help students understand how to answer questions and assure that students score good marks in their exams.
Access NCERT Solutions for Science Chapter 3 – Classifications of Element and Periodicity
NCERT Solutions For Class 11 Chemistry Chapter 3 Classification of Elements and Periodicity
Chemistry Class 11 Chapter 3 NCERT Solutions help students prepare well for the exams. This chapter will let students know about how the periodic table was developed.
NCERT Solutions Class 11 Chemistry Chapter 3 – Classification of Elements and Periodicity in Properties
Chapter 3- Classification of Elements and Periodicity in Properties discusses the history of the development of the periodic table and the current form of the periodic table. It also includes a part about the current trends in the properties of elements. This unit covers multiple topics like properties of elements, their nomenclature in the periodic table, etc. as well.
Students will know the reason for classifying elements, preparing the periodic table, and grouping elements having similar properties together. The difference between metals, non-metals, and semimetals is also explained to the students in this unit.
Mendeleev’s Periodic Table along with its advantages and disadvantages, the Modern Periodic Table, the Periodic Table’s structural features, Long Form of Periodic Table’s advantages, etc. are some of the topics that are a part of this chapter. The chapter aids in teaching students about ionisation enthalpy, electron gain enthalpy, electronegativity, trends in physical properties, anomalous properties of second-period elements, etc.
NCERT Solutions for Class 11 Chemistry Chapter 3 helps in learning concepts like electronic configuration, atomic and ionic radii, valency, etc. very well. Students can test their knowledge by solving questions and checking the answers with the Extramarks’ solutions for this chapter.
Class 11 Chemistry Chapter 3 Classification of Elements and Periodicity in Properties – Marks Distribution
The periodic classification of elements in Chemistry CBSE Class 11 is based on the fundamental features of the periodic table and the importance of periodicity. This unit carries a weightage of 4 marks in the annual exam. NCERT Solutions Class 11 includes the answers to all the problems of this chapter in the NCERT book.
Ex 3.1 Why do we Need to Classify Elements?
Ex 3.2 Genesis of Periodic Classification.
Ex 3.3 Modern Periodic Law and The Present Form of the Periodic Table.
Ex 3.4 Nomenclature of Elements with Atomic Numbers >100.
Ex 3.5 Electronic Configurations of Elements and The Periodic Table.
Ex 3.6 Electronic Configurations and Types of Elements: s, p, d, f-blocks.
Ex 3.7 Periodic Trends in Properties of Elements.
Benefits of Classification of Elements and Periodicity in Properties Chapter 3 NCERT Solutions
The Chemistry Chapter 3 – Classification of Elements and Periodicity in Properties of Class 11 is a relatively easier unit. Students can score well in questions from this chapter, if they have a conceptual understanding of the topics. The benefits of NCERT Solutions for Class 11 Chemistry Chapter 3 are;
- The Class 11 Chemistry Chapter 3 Solutions are framed by teachers who are experts in their fields and have extensive experience.
- Students can refer to NCERT Class 11 Solutions to solve subjective and objective questions from Chapter 3.
- The solutions are easy to comprehend, hence revising the syllabus becomes easy for the students.
- Extramarks Solutions cover all there is to know about Chemistry Chapter 3. Students do not have to refer to other books to get answers to NCERT textbook questions.
Q.1 What is the basic theme of organisation in the periodic table?
The periodic table is based on the classification of elements in periods and groups according to their properties which helps in systematic studies of elements and their compound in simpler way. In periodic table elements placed in a group show similar properties.
Q.2 Which important property did Mendeleev use to classify the elements in his periodic table and did he stick to that?
Mendeleev uses the property of atomic weight or mass to classify the elements in his periodic table. In his periodic table the elements are arranged in periods and group in order of their increasing atomic weight and the elements with similar properties are placed in the same group.
Later, he found that the some of the elements did not fit within this type of classification. So, he ignored the increasing order of atomic weight for some elements. For example tellurium having higher atomic weight is placed in group VI whereas; iodine having lower atomic weight than tellurium is placed in group VII because iodine shows similar properties to fluorine, chlorine and bromine (Halogens).
Q.3 What is the basic difference in approach between the Mendeleev’s Periodic Law and the Modern Periodic Law?
Mendeleev’s Periodic Law states that the physical and chemical properties of elements are periodic functions of their atomic weights whereas; the Modern periodic Law states that the physical and chemical properties of elements are periodic functions of their atomic numbers.
Q.4 On the basis of quantum numbers, justify that the sixth period of the periodic table should have 32 elements.
The value of the principal quantum number (n) for the outermost shells is always equals to the number of period. i.e: n=value of period.
For, sixth period n=6
For, n =6,
According to Aufbau’s principle, energy of 6d >7s sub-shell.In the 6th period, electrons can be filled in the order 6s, 4f, 5d, and 6 p sub shells. As, 6s has only one orbital, 4f has seven, 5d has five, and 6p has three orbitals. Therefore, total 1 + 7 + 5 + 3 = 16 orbitals. According to Pauli’s exclusion principle, maximum two electrons can be accommodated in an orbital. Hence, 16 orbitals can have a maximum of 32 electrons. Hence, the sixth period contains 32 elements.
Q.5 In terms of period and group where would you locate the element with Z =114?
Ground state electron configuration: Z=`114 is 5f14.6d10.7s2.7p2
Shell structure: 18.104.22.168.32.18.4
Hence, it is present in the 7th period of the periodic table.
Therefore, the element with Z = 114 is the second p – block element in the 7th period. Thus, the element with Z = 114 is present in the7th period and 14th group of the periodic table.
Q.6 Write the atomic number of the element present in the third period and seventeenth group of the periodic table.
First period of the periodic table contains two elements and second period contains eight elements. The third period contains eight elements from Z=11-18. The element in the 18th group of the third period has Z = 18. Hence, the element in the 17th group of the third period has atomic number Z =17.
Q.7 Which element do you think would have been named by
(i) Lawrence Berkeley Laboratory
(ii) Seaborg’s group?
(i) Lawrencium (Lr) with Z =103 and Berkelium (Bk) with Z = 97
(ii) Seaborgium (Sg) with Z = 106
Q.8 Why do elements in the same group have similar physical and chemical properties?
The physical and chemical properties of elements depend on the number of valence electrons. Elements of the same group have the same number of valence electrons. Therefore, elements present in the same group show similar physical and chemical properties.
Q.9 What does atomic radius and ionic radius really mean to you?
The radius of an atom is termed as atomic radius. It gives 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.
Covalent radius is measured as the distance between two atoms when they are found together by a single bond in a covalent molecule.
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 less electrons than the parent atom which results an increase in the effective nuclear charge.
Hence, a cation is smaller than the parent atom. 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.
Q.10 How does atomic radius vary in a period and in a group? How do you explain the variation?
On moving from left to right across a period, the atomic radius decreases. 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.
Q.11 What do you understand by isoelectronic species? Name a species that will be isoelectronic with each of the following atoms or ions.
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. Isoelectronic species is Na+ ion (11 – 1 = 10 electrons).
(ii) Ar has 18 electrons. Thus, the species isoelectronic with it will also have 18 electrons. Some of its isoelectronic species are S2– 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), O2– 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).
Q.12 Consider the following species:
N3–, O2–, F–, Na+, Mg2+ and Al3+
(a) What is common in them?
(b) Arrange them in the order of increasing ionic radii.
(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:
N3– < O2– < 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– < O2– < N3–
Q.13 Explain why cations are smaller and anions larger in radii than their parent atoms?
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.
Q.14 What is the significance of the terms- ‘isolated gaseous atom’ and ‘ground state’ while defining the ionisation enthalpy and electron gain enthalpy?
Hint: Requirements for comparison purposes.
What is the significance of the terms- ‘isolated gaseous atom’ and ‘ground state’ while defining the ionisation enthalpy and electron gain enthalpy?
Hint: Requirements for comparison purposes.
Ionisation 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 ionisation 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 ionisation enthalpy.
Ground state of an atom refers to the most stable state of an atom, having least energy.
Q.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.
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
Q.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 ∆iH than B
(ii) O has lower ∆iH than N and F?
(i) During the process of ionization, the electron to be removed from beryllium atom is a 2s-electron which, 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 ∆iH than boron.
(ii) In nitrogen, the three 2p-electrons of nitrogen occupy three different atomic orbitals. Nitrogen has greater ionization enthalpy as compared to oxygen, as it has half filled subshells which are more stable. In general the ionization enthalpy increase across a period due to decrease in size and increased nuclear charge. Fluorine has therefore greater ionization enthalpy.
Q.17 How would you explain the fact that the first ionisation enthalpy of sodium is lower than that of magnesium but its second ionisation enthalpy is higher than that of magnesium?
The first ionisation 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. Mg which has electronic configuration with two electrons in 3s is more stable as it is fully filled than Na as it has one electron in 3s. It is therefore easier to remove the first electron from Na than Mg.
However, the second ionisation 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.
Q.18 What are the various factors due to which the ionization enthalpy of the main group elements tends to decrease down a group?
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.
Q.19 The first ionization enthalpy values (in kJmol-1) of group 13 elements are :
How would you explain this deviation from the general trend?
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.
Q.20 Which of the following pairs of elements would have a more negative electron gain enthalpy?
(i) O or F (ii) F or Cl
(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.
Q.21 Would you expect the second electron gain enthalpy of O as positive, more negative or less negative than the first? Justify your 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.
O (g) + e– → O–(g)
On the other hand, when an electron is added to O– ion to form O2– 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.
O–(g) + e – → O2-(g)
Q.22 What is the basic difference between the terms electron gain enthalpy and electronegativity?
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.
Q.23 How would you react to the statement that the electronegativity of N on Pauling scale is 3.0 in all the nitrogen compounds?
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.
Q.24 Describe the theory associated with the radius of an atom as it
(a) Gains an electron
(b) Loses an electron
(a) When an atom gains an electron, its size increases. When an electron is added, the number of electrons goes up by one. This results in an increase in repulsion among the electrons. However, the number of protons remains the same. As a result, the effective nuclear charge of the atom decreases and the radius of the atom increases.
(b) When an atom loses an electron, the number of electrons decreases by one while the nuclear charge remains the same. Therefore, the interelectronic repulsions in the atom decrease. As a result, the effective nuclear charge increases. Hence, the radius of the atom decreases.
Q.25 Would you expect the first ionisation enthalpies for two isotopes of the same element to be the same or different? Justify your answer.
The ionisation enthalpy of an atom depends on the number of electrons and protons (nuclear charge) of that atom. Now, the isotopes of an element have the same number of protons and electrons. Therefore, the first ionisation enthalpy for two isotopes of the same element should be the same.
Q.26 What are the major differences between metals and non-metals?
|1.||Metals can lose electrons easily.||Non-metals cannot lose electrons easily.|
|2.||Metals cannot gain electrons easily.||Non-metals can gain electrons easily.|
|3.||Metals generally form ionic compounds.||Non–metals generally form covalent compounds.|
|4.||Oxides of metals are basic in nature.||Oxides of non–metals are acidic in nature.|
|5.||The ionization enthalpies are very low in metals.||The ionization enthalpies are very high in non metals.|
|6.||Metals have less negative electron gain enthalpies.||Non–metals have high negative electron gain enthalpies.|
|7.||Metals are less electronegative. They are rather electropositive elements.||Non–metals are electronegative.|
|8.||Metals have a high reducing power.||Non–metals have a low reducing
Q.27 Use the periodic table to answer the following questions.
(a) Identify an element with five electrons in the outer subshell.
(b) Identify an element that would tend to lose two electrons.
(c) Identify an element that would tend to gain two electrons.
(d) Identify the group having metal, non-metal, liquid as well as gas at the room temperature.
(a) The outer most electronic configuration of element having five electrons in its outermost subshell should be ns2 np3. This is the electronic configuration of the nitrogen group.
Thus, the element can be N, P, As, Sb, or Bi.
(b) An element having two valence electrons will lose two electrons easily to attain the stable noble gas configuration. The general electronic configuration of an element having two valence electrons will be ns2 as it can lose the two valence electrons easily to attain the stable noble gas configuration. This is the electronic configuration of group 2 elements. The elements present in group 2 are Be, Mg, Ca, Sr, Ba.
(c) An element is likely to gain two electrons if it needs only two electrons to attain the stable noble gas configuration. Thus, the general electronic configuration of such an element should be ns2 np4. This is the electronic configuration of the oxygen family.
(d) Group 17, halogens, has metal, non–metal, liquid as well as gas at room temperature.
Q.28 The increasing order of reactivity among group 1 elements is Li < Na < K < Rb <cs< p=””>< Cs</cs<>
whereas that among group 17 elements is F > CI > Br > I. Explain.
The number of valence electron in group 1 elements is one. The elements tend to lose this single valence electron.
Whereas Group 17 elements, lack only one electron to attain the noble gas configuration. On moving down group 1, the ionization enthalpies decrease. This means that the nergy required to lose the valence electron decreases.Thus, reactivity increases on moving down a group. Thus, the increasing order of reactivity among group
Group 1 elements are as follows:
Li < Na < K < Rb < Cs
In group 17, as we move down the group from Cl to I, the electron gain enthalpy becomes less negative i.e., its tendency to gain electrons decreases down group 17.
Thus, reactivity decreases down a group. The electron gain enthalpy of F is less negative than Cl. Still, it is the most reactive halogen. This is because of its low bond dissociation energy. Thus, the decreasing order of reactivity among group 17 elements is as follows:
F > Cl > Br > I
Q.29 Write the general outer electronic configuration of s-, p-, d- and f- block elements.
Element General outer electronic configuration
s–block ns1–2, where n = 2 – 7
p–block ns2np1–6, where n = 2 – 6
d–block (n–1) d1–10 ns0–2, where n = 4 – 7
f–block (n–2)0–14(n–1)d0–1ns2, where n = 6 – 7
Q.30 Assign the position of the element having outer electronic configuration
(i) ns2 np4 for n = 3 (ii) (n – 1)d2 ns2 for n = 4, and (iii) (n – 2) f7 (n – 1)d1 ns2 for n = 6, in the periodic table.
(i) It is a p–block element since the last electron occupies the p–orbital. Since n = 3, the element belongs to the 3rd period.
The p orbital contains four element, hence the group for this element = Number of s–block groups + number of d–block groups + number of p–electrons = 2 + 10 + 4 = 16
Therefore, the element belongs to the 3rd period and 16th group of the periodic table.
The element in 3rd period and 16th group is Sulphur.
(ii) For, n = 4, the element belongs to the 4th period. It is a d–block element as d–orbitals are incompletely filled.
There are 2 electrons in the d–orbital.
Thus, the corresponding group of the element
= Number of s–block groups + number of d–block groups
Therefore, the element is a 4th period and 4th group element which is Titanium.
(iii) Since n = 6, the element is present in the 6th period. It is an f –block element as the last electron occupies the f–orbital. It belongs to group 3 of the periodic table since all f-block elements belong to group 3. Its electronic configuration is [Xe] 4f7 5d1 6s2.
Thus, its atomic number is 54 + 7 + 2 + 1 = 64. Thus, the element is Gadolinium.
Q.31 The first (∆iH1) and the second (∆iH2) ionization enthalpies (in kJ mol–1) and the (∆egH)
electron gain enthalpy (in kJ mol–1) of a few elements are given below:
Elements ∆iH1 ∆iH2 ∆egH
I 520 7300 –60
II 419 3051 –48
III 1681 3374 –328
IV 1008 1846 –295
V 2372 5251 +48
VI 738 1451 –40
Which of the above elements is likely to be:
(a) the least reactive element.
(b) the most reactive metal.
(c) the most reactive non-metal.
(d) the least reactive non-metal.
(e) the metal which can form a stable binary halide of the formula MX2, (X=halogen).
(f) the metal which can form a predominantly stable covalent halide of the formula MX
(a) The first ionization enthalpy (∆iH1) and a positive electron gain enthalpy (∆egH) is highest for element V. Hence, the least reactive element is Element V.
(b) The first ionization enthalpy (∆iH1) is lowest and negative electron gain enthalpy (∆egH) is also very low for element II. Hence, element II is likely to be the most reactive metal
(c) The first ionization enthalpy (∆iH1)is very high and the highest negative electron gain enthalpy(∆egH) for element III. Hence, it is the most reactive non metal.
(d) Element IV is likely to be the least reactive non–metal since it has the least negative electron gain enthalpy (∆egH) among the non-metals.
(e) Element VI has a low negative electron gain enthalpy (∆egH). Thus, it is a metal. Also, it has the lowest second ionization enthalpy (∆iH2). Hence, it can form a stable binary halide of the formula MX2 (X=halogen).
(f) Element I has low first ionization energy and high second ionization energy. Therefore, it can form a predominantly stable covalent halide of the formula MX (X=halogen).
Q.32 Predict the formulas of the stable binary compounds that would be formed by the combination of the following pairs of elements.
(a) Lithium and oxygen (b) Magnesium and nitrogen
(c) Aluminium and iodine (d) Silicon and oxygen
(e) Phosphorus and fluorine (f) Element 71 and fluorine
(e) PF3 or PF5
(f) The element with the atomic number 71 is Lutetium (Lu). It has valency 3. Hence, the formula of the compound is LuF3.
Q.33 In the modern periodic table, the period indicates the value of:
(a) Atomic number
(b) Atomic mass
(c) Principal quantum number
(d) Azimuthal quantum number.
The value of the principal quantum number (n) for the outermost shell or the valence shell indicates a period in the Modern periodic table. Thus, the correct answer is option (c) Principal quantum number.
Q.34 Which of the following statements related to the modern periodic table is incorrect?
(a) The p-block has 6 columns, because a maximum of 6 electrons can occupy all the orbitals in a p-shell.
(b) The d-block has 8 columns, because a maximum of 8 electrons can occupy all the orbitals in a d-subshell.
(c) Each block contains a number of columns equal to the number of electrons that can occupy that subshell.
(d) The block indicates value of azimuthal quantum number (l) for the last subshell that received electrons in building up the electronic configuration.
(b) The d-block has 10 columns because a maximum of 10 electrons can occupy all the orbitals in a d-subshell.
Q.35 Anything that influences the valence electrons will affect the chemistry of the element.
Which one of the following factors does not affect the valence shell?
(a) Valence principal quantum number (n)
(b) Nuclear charge (Z)
(c) Nuclear mass
(d) Number of core electrons.
Nuclear mass does not affect the valence electrons. Thus, the correct option is (c) Nuclear mass.
Q.36 The size of isoelectronic species — F–, Ne and Na+ is affected by
(a) Nuclear charge (Z)
(b) Valence principal quantum number (n)
(c) Electron-electron interaction in the outer orbitals
(d) None of the factors because their size is the same.
As the nuclear charge(Z) decreases the size of an isoelectronic species increases. Thus, the correct option is (a) Nuclear charge (Z).
Q.37 Which one of the following statements is incorrect in relation to ionization enthalpy?
(a) Ionization enthalpy increases for each successive electron.
(b) The greatest increase in ionization enthalpy is experienced on removal of electron from core noble gas configuration.
(c) End of valence electrons is marked by a big jump in ionization enthalpy.
(d) Removal of electron from orbitals bearing lower n value is easier than from orbital having higher n value.
(d) Removal of electron from orbitals bearing lower n value is easier than from orbital having higher n value.
Q.38 Considering the elements B, Al, Mg, and K, the correct order of their metallic character is:
(a) B > Al > Mg > K (b) Al > Mg > B > K
(c) Mg > Al > K > B (d) K > Mg > Al > B
The metallic character of Mg is more than that of Al as the metallic character of elements decreases from left to right across a period.
The metallic character of Al is more than that of B as the metallic character of elements increases down a group.
Hence, K > Mg and the correct order of metallic character is
(d) K > Mg > Al > B
Q.39 Considering the elements B, C, N, F, and Si, the correct order of their non-metallic character is:
(a) B > C > Si > N > F b) Si > C > B > N > F
(c) F > N > C > B > Si d) F > N > C > Si > B
In a periodic table from left to right across a period the non-metallic character of elements. Thus, the decreasing order of non-metallic character is F > N > C > B.
The decreasing order of non-metallic characters of C and Si are C > Si as the non metallic character decreases on moving down in a group. B is more non-metallic than Si i.e., B > Si.
So, the correct order of their non-metallic characters is
( c ) F > N > C > B > Si
Q.40 Considering the elements F, Cl, O and N, the correct order of their chemical reactivity in terms of oxidizing property is:
(a) F > Cl > O > N
(b) F > O > Cl > N
(c) Cl > F > O > N
(d) O > F > N > Cl
In a periodic table on moving from left to right the oxidizing character of elements increases. Thus, nitrogen to fluorine we get the decreasing order of oxidizing property as F > O > N.
Also, the oxidizing character of elements decreases from top to bottom in a group. Thus, we get F > Cl.
However, the oxidizing character of O is more than that of Cl i.e., O > Cl.
Hence, the correct order of oxidizing property is
( b ) F > O > Cl > N
FAQs (Frequently Asked Questions)
1. Why is Atomic Number a better option for classifying elements than Atomic Mass?
Even though atomic mass is related to the nucleus that is situated in the center of the atom, it is not ideal to classify elements based on that. This is because the properties of the elements are dependent upon the electronic configuration related to the atomic number. Change in the electronic configuration of all elements leads to changes in their properties. It is a more easy and efficient way to classify elements based on their chemical and physical properties. Atomic number is a better option for classifying elements than atomic mass.
2. Explain Mendeleev's Periodic Table.
The Mendeleev’s Periodic Table was framed by Dimitri Ivanovich Mendeleev. The table has elements arranged in vertical columns called groups and horizontal rows called periods.
- The table has a total of nine rows which are named in Roman numerals (I, II, III, IV, V, VI, VII, VIII, zero). The eighth group has nine elements in it. These elements are arranged in three triads. The group name zero consists of gases and elements that have zero valency.
- There are a total of 7 periods in Mendeleev’s Periodic Table.
3. What is the difference between Mendeleev's Periodic Law and Modern Periodic Law?
Basis of periodicity is the major difference between Mendeleev’s Periodic Law and the Modern Periodic Law. Mendeleev’s Periodic Law says that the basis of periodicity is the atomic mass of the elements. Mendeleev had put elements having the same properties into a group. According to the Modern Periodic Law, the basis of periodicity is the atomic number of elements. Arranging elements based on atomic numbers in the periodic table was the right thing to do as per Henry Moseley. The properties of elements were considered a periodic function of their atomic masses by Mendeleev while as per Modern Periodic Law properties of elements were considered a periodic function of their atomic numbers.
4. How are NCERT Solutions for Class 11 Chemistry Chapter 3 helpful for Class 11 students?
Class 11 Chemistry Chapter 3 NCERT Solutions are one of the best guides out there. Incorporated with all types of exercises, the Class 11 Chemistry Chapter 3 Solutions help students learn all the important concepts. Solving questions will assure that students have thoroughly studied everything in this chapter. This will assure that they score well in the exam.
5. What are the limitations of Newlands' Law of Octaves?
British Chemist John Newlands arranged elements in ascending order based on atomic masses in the year 1864 in such a way that every eighth element had similar properties. This observation led to the formulation of Newlands’ Law of Octaves.
This law said that when elements are arranged in the increasing order of their atomic masses, every eighth element has similar properties. Newlands compared this similarity of every eighth element to octaves of music where every eighth note is comparable to the first. However, Newlands’ Law of Octaves had the following limitations as well;
- Elements that had dissimilar properties were grouped together.
- Classification under the Newlands’ Law of Octaves was successful only until the calcium element. After calcium, every eighth element lacked the same properties that the element placed above it in the same group possessed.
- The discovery of noble gases disrupted the entire table as including the noble gases elements disturbed the entire arrangement.
- Multiple elements were fit into the same slot. Eg: cobalt and nickel were placed together.
- This method of classification was not ideal since it did not leave room for new elements as other elements that were later discovered could not fit into the octave pattern. At the time of Newlands’ discovery, only 56 elements were known. However, 118 elements are known to us today.
6. What is electronegativity?
The ability of an atom that is a part of a given chemical compound to attract shared electrons is called electronegativity. Measuring electronegativity is not possible quantitatively like ionization enthalpy and electron gain enthalpy, hence it is a qualitative measure. No given element’s electronegativity is constant. It varies depending on the element it is bound to. To get more in-depth knowledge students can refer to the NCERT Solutions for Chapter 3 by downloading the solutions for free from Extramarks.
7. What is the present form of the Periodic Table?
The current form of the Periodic Table is based on the Modern Periodic Law. Elements are arranged in the increasing order of their atomic numbers in the present form of the periodic table. Therefore, the modern periodic table with 18 vertical columns (called groups) and 7 horizontal rows (called periods) is the present form of the periodic table.
8. How to prepare for Chemistry Class 11 Chapter 3?
Students need to have a thorough understanding of the periodic tables that have evolved throughout the ages. Hence, students should be able to differentiate the multiple periodic tables that are a part of this chapter. They need to be prepared well enough to list down all the properties of the elements that each periodic table includes. They need to practise well for this. NCERT Solutions for Class 11 Chemistry Chapter 3 will cover exercises that will help students score high marks in their exams.