# NCERT Solutions for Class 11 Chemistry Chapter 3: Classification of Elements and Periodicity in Properties

3.1. What is the basic theme of organisation in the periodic table?

Ans. The basic theme of organisation in the periodic table is classification of the elements as groups and periods according to their properties. The elements with similar properties are placed in the same group of a periodic table. The organisation of elements in the periodic table make the study of these elements and their compound simple and systematic.

3.2. Which important property did Mendeleev use to classify the elements in his periodic table and did he stick to that?

Ans. Mendeleev used atomic mass of elements as the basis of classification. He arranged the elements in the increasing order of their atomic weights. He placed the elements with similar properties in the same group. But while following this principle, he found out that some of the elements did not fit with this scheme of classification. Therefore, he ignored the order of atomic weight in some cases. For example, iodine has atomic weight lower than tellurium (Group VI). But still Mendeleev placed tellurium (in Group VI) before iodine (in group VII) because its properties are similar to fluorine, chlorine and bromine. Thus, he didn’t stick to this classification.

3.3. What is the basic difference in approach between the Mendeleev’s Periodic Law and the Modern Periodic Law?

Ans. According to Mendeleev’s periodic table, the physical and chemical properties of elements are the periodic function of their atomic weights. Thus, elements are grouped according to their atomic weights. While according to modern Periodic Table, the physical and chemical properties of elements are the periodic function of their atomic numbers. Thus, elements are placed in periodic table according to their atomic numbers.

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

Ans. In a modern periodic table, the period is indicative of the principal quantum number (n), For the element belonging to period 6, n = 6. For n = 6, the values of azimuthal quantum number (l) should be 1, 2, 3, 4, 5.

Electrons fill in the different orbitals in order of their increasing energies according to the Aufbau principle, so energy of 6d is higher than that of 7s. And in 6th period, electrons fill in 6s, 4f, 5d, and 6p orbitals. So, there are a total of (1s + 7f + 5d + 3p) 16 orbitals.

According to Pauli’s Exclusion Principle, an orbital can accommodate a maximum of 2 electrons. So, in that case sixth period (16 orbitals) should accommodate a maximum of 32 electrons.

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

Ans. Elements with atomic numbers from Z = 87 to Z = 118 belong to 7th period. Thus, the element with Z = 114 is present in the 7th period of the periodic table. These are a total of 28 elements in the 7th period. First two elements, with Z = 87 and Z = 88 are present in s block, the next 14 elements (excluding Z = 89), i.e., Z = 90 – 103 are present in f-block and 10 elements with Z = 89 and Z = 104 –112 are present in d-block and the elements with Z = 113 – 118 are p-block elements which are present in the 7th period. So, the element with atomic number (Z = 114) is the second p-block element in the 7th period. Thus, the element with Z = 114 belongs to 7th period and 14th group of the periodic table.

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

Ans. There are two elements in the 1st period and eight elements in the 2nd priod. The third period starts with the element with Z = 11. Now, there are eight elements in the third period. Thus, the 3rd period ends with the element with Z = 18 i.e., the element in the 19th group of third period has Z = 18. Hence, theelement in the 17th group of the third period has atomic number Z = 17.

3.7. Which element do you think would have been named by (i) Lawrence Berkeley Laboratory,

(ii) Seaborg’s group?

Ans. (i) The elements named by Lawrence and Berkley Laboratory are Lawrencium (Lr) and Berkelium (Bk).

(ii) The elements named by Seaborg is Seaborgium (Sg).

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

Ans. Elements in the same group have similar physical and chemical properties due to the presence of same number of valence electrons. The physical and chemical properties of an element depend on the number of valence electrons.

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

Ans. Atomic radius is the radius of an atom which measures the size of an atom. It is defined as the distance from the center of the nucleus to the outermost shell containing the electrons. If the atom is a metal, atomic radius states the metallic radius while if atom is a non-metal, atomic radius states the covalent radius.

Metallic radius is half the inter-nuclear 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

$$\text{taken as}\frac{256}{2}\text{pm}=128\text{pm.}$$

Covalent radius is the distance between the two atoms which are bound 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

$$\text{chlorine is taken as}\frac{198}{2}\text{pm}=99\space\text{pm.}$$

Atomic radius can also be measured as Van der waal’s radius. It is defined as one half of the inter nuclear distance between two similar adjacent atoms belonging to the two neighbouring molecules of the same substance in the solid state. Ionic radius is the distance between two ions (cation and anion) in an ionic crystal. It is defined as the distance from the center of the nucleus of the ion upto which it exerts its influence on the electron cloud. The ionic radii can be calculated by measuring the distances between the cations and anions in ionic crystals. For example, the ionic radius of Na+ 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.

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

Ans. In a period, atomic radius generally decreases as we move from left to right. This is because within a period, the outer electrons are present in the same valence shell. As we move from left to right, the atomic number increases and there is gradual increase in their effective nuclear charge. This is why there is increase of attraction of the electrons to the nucleus which results in the decrease of the atomic radius across a period.

On the other hand, atomic radius increases as we move down the group. This is because of the increase in principal quantum number and the number of shells of elements in a group which in turn increases the distance between their nucleus and valence electrons. Thus, electrons are less tightly attracted to nucleus which causes increase in size.

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+

Ans. Isoelectronic species are the atoms and ions having the same number of electrons. They have different magnitude of nuclear charges and belongs to different atoms or ions.

(i) Fion 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), O2– ion (8 + 2 = 10 elcectrons), 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 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), Sr2+ ion (38 – 2 = 36 electrons).

3.12. Consider the following species :

N3– , O2– , F , Na+ , Mg2+ and Al3+

(a) What is common in them?

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

Ans. (a) All of them are isoelectronic in nature and have same number of electrons, i.e., each of the ion have 10 electrons in their valence shell.

(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 followsL

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–

3.13. Explain why cation are smaller and anions larger in radii than their parent atoms?

Ans. A cation is formed by loss of one or more electrons. So, a cation has less electrons than its parent atom while both have the same nuclear charge. So, the attraction of electrons to the nucleus is more in a cation than in an anion. Hence, a cation is smaller in size or radii than its parent atom.

An atom gains electrons and forms an anion. So, an anion has more electrons than its parent atom while both have the same nuclear charge. There is more repulsion among electrons and decrease in the effective nuclear charge in an anion than an atom. So, the distance between the valence electrons and the nucleus is more in an anion than its parent atom. Hence, anion is larger in size or radii than its parent atom.

3.14. What is the significance of the terms — ‘isolated gaseous atom’ and ‘ground state’ while defining the ionisation enthalpy and electron gain enthalpy?

Ans. The energy required to remove an electron from an isolated gaseous atom in its ground state is called its ionisation enthalpy. Even in the gaseous state, there are some amounts of attractive forces among the atoms So, to determine the ionisation enthalpy, a single atom cannot be isolated. But the forces 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.

The most stable state of an atom is its ground state. If an isolated gaseous atom is in this state, than a least amount of energy is required to remove an electron from it.

3.15. Energy of an electron in the ground state of the hydrogen atom is –2.18 × 10–18 J. Calculate the ionisation enthalpy of atomic hydrogen in terms of J mol–1.

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

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

Ionisation 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

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 Di H than B

(ii) O has lower Di H than N and F

Ans. (i) Electronic configuration of Be is 1s22s2.

Electronic configuration of B is 1s22s22p1.

As it is clear from above electronic configuration, in case of boron valence electron is present in 2p orbitals that needs to be removed whereas in case of boron atom valence electron is present in 2s orbital. s orbital is more penetrating than p. Hence, 2s-electrons are more strongly attached to the nucleus than 2p-electrons. As a result, more energy is required to remove an electron from 2s orbital than that required to remove an electron from 2p-orbital of boron. Hence, Be has higher ΔiH than B.

(ii) Electronic configuration of N is 1s22s22px1 2py12pz1.

Electronic configuration of O is 1s22s22px2 2py12pz1.

Electronic configuration of F is 1s22s22px2 2py22pz1.

As it is clear from above electronic configuration, in nitrogen atom, the three 2p-electrons of nitrogen occupy three different atomic orbitals forming a half filled p-oribtal. Half filled and fully filled orbitals are more stable. In oxygen, two of the four 2p-electrons of oxygen occupy the same 2px-orbital. This causes increased electron-electron repulsion in oxygen atom. Therefore, it is easy to remove the fourth 2p-electron from oxygen as compared to one of the three 2p-electrons from nitrogen. So, the energy required to remove the fourth 2p-electron from oxygen is also less. Hence, O has lower Di H than N.

Fluorine has more electrons and protons than oxygen. This increases effective nuclear charge in fluorine than that experienced by the electrons present in oxygen. As a result, it is difficult to remove an electron from fluorine atom than to remove an electron from oxygen atom. So, more energy is required to remove an electron from fluorine atom as compared to oxygen atom. Hence, O has lower ΔiH than N.

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?

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

Firstly, the atomic size of sodium is greater than that of magnesium.

Secondly, 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 when sodium loses one valence electron, it attains a noble electronic configuration while when magnesium loses one valence electron, it still possess one electron in the 3s orbital Thus, in order to attain stable noble electronic configuration, it still has to lose one more electron. This can be represented as:

After the loss of first electron, the electronic configuration of :

After the loss of first electron, the electronic configuration of :

Na+ = 1s22s22p6

Mg+ = 1s22s22p63s1

So, the energy required to remove the second electron in case of sodium is much higher than that required in case of magnesium. Therefore, the second ionisation enthalpy of sodium is greater than that of magnesium.

3.18. What are the various factors due to which the ionization enthalpy of the main group elements tends to decrease down a group?

Ans. The various factors due to which the ionization enthalpy of the main group elements tends to decrease down a group are as follows:

(i) Increase in the atomic size of elements: As we move down a group, the number of shells increases, So, the atomic size also increases gradually on moving down a group due to which the distance of the valence electrons from the nucleus increases and 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: There are more number of inner shells of electrons in the atoms as we move 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.

3.19. The first ionization enthalpy values (in kJ mol–1) of group 13 elements are :
 B Al Ga In Tl 801 577 579 558 589

How would you explain this deviation from the general trend?

Ans. In general, as we move down a group, the ionization enthalpy of atoms decrease due to increase in their atomic size and shielding effect. Therefore, on moving down group 13, ionization enthalpy decreases as we move from B to Al. But when we move down from Al to Ga, it increases. This is because in the periodic table, Al is placed immediately after s-block elements while Ga is placed after d-block elements. The shielding effect provided by electrons of d-block elements is not very effective. These electrons do not shield the valence electrons effectively. This is why the valence electrons of Ga experience a lot of effective nuclear charge than those of Al. Hence, the ionisation enthalpy of Ga is more than Al. Now, when we move from Ga to In, the ionisation enthalpy increases due to increase in their atomic size and shielding effect. And as we move from In to Tl, the ionisation enthalpy again increases. This is because in the periodic table, TI 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 ionisation energy of Tl is on the higher side.

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

Ans. (i) Electron gain enthalpy provides a measure of the ease with which an atom adds an electron to form anion. The atom which accepts an electron easily will have more electron gain enthalpy. Both O and F are present in the same period of the periodic table but F has one proton and one electron more than O.
As F contain one proton more than O, the electron being added will be more strongly attracted by the nucleus of F as compared to the nucleus of O. Also, since F needs only one more electron to attain stable electronic gas configuration,therefore, F has more negative electron gain enthalpy than O.

(ii) Both F and Cl are present in seventeen group of the periodic table. In general, the electron gain enthalpy becomes less negative on moving down a group. But in this case, electron gain enthalpy of Cl is more negative than that of F . This is because when an electron is added to F, it will be added to quantum level n = 2 or 2nd shell in F while in Cl, the electron will be added to quantum level n = 3 or 3rd shell. Also, Cl is bigger in size than F, therefore, it has less electron-electron repulsions, so, it will easily accommodate the incoming electron. This is why Cl has more negative electron gain enthalpy than F.

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.

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

O(g) + e → O (g)

But when second electron is added to O ion to form O2– ion, in order to overcome the force of repulsion between monovalent anion and second incoming electron, energy is needed. Hence, the second electron gain enthalpy is positive.

3.22. What is the basic difference between the terms electron gain enthalpy and electronegativity?
Ans. (i) Electron gain enthalpy provides a measure
of the ease with which an atom adds an electron to form anion while electronegativity is the measure of the tendency of an atom in a chemical compound to attract a shared pair of electrons.

(ii) Electron gain enthalpy is the property of isolated atoms whereas electronegativity is the property of atoms in molecules.

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?

Ans. 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.

3.24. Describe the theory associated with the radius of an atom as it (a) Gains an electron (b) Loses an electron.

Ans (i) An anion is formed when an atom gains electrons. So, an anion has more electrons than its parent atom while both have the same nuclear charge. There is more repulsion among electrons and decrease in the effective nuclear charge in an anion than an atom. So, the distance between the valence electrons and the nucleus is more in an anion than its parent atom. Hence, anion has larger radius than its parent atom.

(ii) A cation is formed when an atom loses electron(s). So, a cation has less electrons than its parent atom while both have the same nuclear charge. So, the attraction of electrons to the nucleus is more in a cation than in an anion. Hence, a cation is smaller in size than its parent atom or cation have smaller radius than its parent atom.

3.25. Would you expect the first ionization enthalpies for two isotopes of the same element to be the same or different? Justify your answer.

Ans. The ionisation enthalpy of an atom depends upon the number of electrons and protons (nuclear charge) present in it. Since the isotopes of an element have the same number of electrons and protons, therefore, the first ionisation enthalpies for the two isotopes of same element should be same.

3.26. What are the major differences between metals and non-metals?

Ans.

 Metals Non-Metals 1. Metals generally lose electrons. 1. Non-metals generally gain electrons. 2. Metallic oxides are basic in nature. 2. Non-metallic oxides are acidic in nature. 3. They generally form ionic compounds. 3. They generally form covalent compounds. 4. They have low ionization enthalpies. 4. They have high ionization enthalpies. 5. Metals have less negative electron gain enthalpies 5. Non metals have high positive electron gain enthalpies. 6. They have a high reducing power. 6. They have a low reducing power. 7. Metals are electropositive elements 7. Non metals are electronegative elements.

3.27. 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.

Ans. (a) The elements present in halogen group have 5 electrons in its outermost or valence shell , i.e., ns2np5 Thus, the elements F, Cl, Br, I have 5 electrons in outer shell.

(b) An element would tend to lose two electrons in order to attain stable noble gas configuration. Thus, the general electronic configuration of such an element will be ns2 which is the electronic configuration of group 2 elements. Among group 2 elements, Be, Sr, Mg, Ca and Ba can lose two electrons to attain noble gas configuration.

(c) An element would tend to gain two electrons if it needs two electrons to attain stable noble gas configuration. Thus, the general electronic configuration of such an element will be ns2np4 which is the electronic configuration of oxygen family. Hence, among oxygen family, O can gain two electrons to attain stable configuration.

(d) Group 17 has metal (Iodine) , non-metal (Chlorine) , liquid (Bromine) as well as gas (Fluorine) at the room temperature.

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

Ans. The elements of group 1 have only 1 valence electron, so they tend to lose it. The energy required to lose an electron keeps on decreasing as we move down in group 1 as their atomic radii also increases. This means that on moving down group 1, the ionisation enthalpies decrease. Thus, reactivity increases on moving down the group. Hence, the increasing order of reactivity among group 1 elements is as follows: Li < Na < K < Rb < Cs.

Elements of group 17 have 7 valence electrons so, they tend to gain one more electron to attain noble gas configuration. As we move down group 17 from Cl to I, electron gain enthalpy becomes less negative which means the tendency of atoms to gain electrons decreases down group . The electron gain enthalpy of F is less negative than Cl but still it is most reactive halogen because of its low bond dissociation energy. Hence, the decreasing order of reactivity among group 17 elements is F > CI > Br > I.

3.29. Write the general outer electronic configuration of s-, p-, d- and f- block elements.

Ans. General outer electronic configuration of s-block elements: ns1-2 where, n = 2 – 7

General outer electronic configuration of p-block elements: ns2 np1-6 where, n = 2 – 6.

General outer electronic configuration of d-block elements: (n – 1)d1–10 ns0–2 where, n = 4 – 7

General outer electronic configuration of f-block elements: (n – 2)f1–14 (n–1)d0-10 ns2where, n = 6 – 7

3.30. Assign the position of the element having outer electronic configuration :

(i) ns2np4 for n = 3

(ii) (n – 1)d2ns2 for n = 4, and

(iii) (n – 2)f7(n – 1)d1ns2 for n = 6, in the periodic table.

Ans. (i) As n = 3, the element belongs to the 3rd period. It is a p-block element because the last electron occupies the p-orbital. Since it has 4 electrons in its p-orbital, so the corresponding group of the element is:

= 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. Hence, the element is Sulphur.

(ii) As n = 4, the element belongs to the 4th period. Its d-orbitals are incompletely filled, so the element belongs to d-block. It has 2 electrons in the d-orbital so, the corresponding group of the element is:

= Number of s-block groups + number of d-block groups

= 2 + 2

= 4

Therefore, it is a 4th period and 4th group element. Hence, the element is Titanium.

(iii) As n = 6, the element belongs to the 6th period. Its last electrons are filled in f-orbitals, so it is an f-block element. It belongs to group 3 since all f-block elements belong to group 3. Element present in 6th period and 3rd group is Gadolinium.

Its electronic configuration is [Xe]

4f75d16s2. Thus, its atomic number is 54 + 7 + 2 + 1 = 64. Hence, the element is Gadolinium.

3.31. The first (ΔiH1) and the second (ΔiH2) ionization enthalpies (in kJ mol–1) and the (ΔegH) electron gain enthalpy (in kJ mol–1) of a few elements are given below:

 Elements Δ1H1 ΔiH2 ΔegH 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 (X = halo)

Ans. (a) Element V is likely to be the least reactive element as it has the highest first ionisation enthalpy (ΔiH1) and a positive electron gain enthalpy (ΔegH).

(b) Element II is likely to be the most reactive metal as it has the lowest first ionization enthalpy (ΔiH1) and a low negative electron gain enthalpy (ΔegH).

(c) Element III is likely to be the most reactive non-metal as it has high first ionization enthalpy (lesser than Element 1) (ΔiH1) and high negative electron gain enthalpy (ΔegH).

(d) Element IV, because its electron gain enthalpy is highly negative and first ionisation enthalpy is very low. It should be the least reactive nonmetal. Hence it is a less reactive halogen (l).

(e) Element VI has a low negative electron gain enthalpy (ΔegH). Thus, it is a metal. Further, it has the lowest second ionisation enthalpy (Δ1H2). Hence, it can form a stable binary halide of the formula MX2 (X = halogen).

(f) The value of first ionisation enthalpy (ΔiH1) is low but the value of second ionization enthalpy (ΔiH2) of element is high. So, it can form a stable binary halide of the formula MX. It is an alkaline metal which must be Li because it forms stable covalent halide of the formula MX.

3.32. 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

Ans. (a) Li2O (Lithium oxide)

(b) Mg3N2 (Magnesium nitrite)

(c) AlI3 (Aluminium triiodide)

(d) SiO2 (Silicon dioxide)

(e) PF3/PF5 (Phosphorous trifluoride or pentafluoride)

(f) LuF3 (Lutetium trifluoride)

3.33. 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.

Ans. (c) Principal quantum number

Explanation:

In the modern periodic table, each period begins with the filling of a new shell. Thus, the value of the principal quantum number (n) for the outermost shell or the valence shell indicates the period in the Modern periodic table.

3.34. 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.

Ans. (b) The d-block has 8 columns, because a maximum of 8 electrons can occupy all the orbitals in a d-subshell.

Explanation:

The d-block has 10 columns, because a maximum of 10 electrons can occupy all the orbitals in a d-subshell.

3.35. 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.

Ans. (c) Nuclear mass

Explanation:

Nuclear mass does not affect the valence electrons because the nuclear mass is so small, it is considered as negligible.

3.36. The size of isoelectronic species — F, Ne and Na+ is affected by:

(a) Nuclear charge (Zeff)

(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

Ans. (a) Nuclear charge (Zeff).

Explanation:

The size of an isoelectronic species is inversely proportional to nuclear charge. It increases with a decrease in Zeff.

The order of the decreasing nuclear charge of the given species is as follows:

Na+ > Ne > F

Therefore, the order of the decreasing size of the given species is as follows:

F– > Na+ > Ne

Thus, the correct option is (a).

3.37. 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.

Ans. (d) Removal of electron from orbitals bearing lower n value is easier than from orbital having higher n value.

Explanation:

Electrons in orbitals bearing a lower n value are more attracted to the nucleus than electrons in orbitals bearing a higher n value. Hence, the removal of electrons from orbitals bearing a higher n value is easier than the removal of electrons from orbitals having a lower n value.

3.38. 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

Ans. (d) K > Mg > Al > B

Explanation:

The metallic character of elements decreases from left to right across a period. Thus, the metallic character of Mg is more than that of Al. The metallic character of elements increases down a group. Thus, the metallic character of Al is more than that of B. Hence, the correct order of metallic character is K > Mg > Al > B.

3.39. 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

Ans. (c) F > N > C > B > Si

Explanation:

The non-metallic character of elements increases from left to right across a period. Thus, the decreasing order of non-metallic character is F > N > C > B. Also, the non-metallic character of elements decreases down a group. Thus, the decreasing order of non-metallic characters of C and Si are C > Si. However, Si is less non- metallic than B i.e., B > Si. Hence, the correct order of their non-metallic character is F > N > C > B > Si.

3.40. 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

Ans. (b) F > O > Cl > N

Explanation:

The oxidizing character of elements increases from left to right across a period. Thus, we get the decreasing order of oxidizing property as
F > O > N.

Again, the oxidizing character of elements decreases down 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 chemical reactivity of F, Cl, O and N in terms of their oxidizing property is F > O > Cl > N.