NCERT Solutions for Class 11 Chemistry Chapter 3

Q.1 What is the basic theme of organisation in the periodic table?

Ans.

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?

Ans.

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?

Ans.

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.

Ans.

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?

Ans.

Ground state electron configuration: Z=`114 is 5f¹⁴.6d¹⁰.7s².7p²

Shell structure: 2.8.18.32.32.18.4

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 the7^th period and 14^th 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.

Ans.

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?

Ans.

(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?

Ans.

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?

Ans.

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?

Ans.

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.

(i) F–

(ii) Ar

(iii) Mg²⁺

(iv) Rb⁺

Ans.

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 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).

Q.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.

Ans.

(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–

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

Ans.

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.

Ans.

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.

Ans.

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?

Ans.

(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?

Ans.

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?

Ans.

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?

Ans.

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

Ans.

(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.

Ans.

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 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.

O^–_(g) + e ^– → O^2-_(g)

Q.22 What is the basic difference between the terms electron gain enthalpy and electronegativity?

Ans.

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?

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.

Q.24 Describe the theory associated with the radius of an atom as it

(a) Gains an electron

(b) Loses an electron

Ans.

(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.

Ans.

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?

Ans.

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.

Ans.

(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.

Ans.

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.

Ans.

Element General outer electronic configuration

s–block ns^1–2, where n = 2 – 7

p–block ns²np^1–6, where n = 2 – 6

d–block (n–1) d^1–10 ns^0–2, where n = 4 – 7

f–block (n–2)^0–14(n–1)d^0–1ns², 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.

Ans.

(i) It is a p–block element since the last electron occupies the p–orbital. Since n = 3, the element belongs to the 3^rd 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 3^rd period and 16^th group of the periodic table.

The element in 3^rd period and 16^th group is Sulphur.

(ii) For, n = 4, the element belongs to the 4^th 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

=2+2

=4

Therefore, the element is a 4^th period and 4^th group element which is Titanium.

(iii) Since n = 6, the element is present in the 6^th 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

(X=halogen)?

Ans.

(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

Ans.

(a) Li₂O

(b) Mg₃N₂

(c) AlI₃

(d) SiO₂

(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.

Ans.

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.

Ans.

(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.

Ans.

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.

Ans.

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.

Ans.

(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

Ans.

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

Ans.

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

Ans.

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