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

NCERT solutions for class 11 chemistry chapter 3 Classification of Elements and Periodicity in Properties - This chapter starts with, a brief history of the periodic table and historical development of the periodic law and the periodic table and ends with some discussion on physical trends like atomic sizes, ionization enthalpies etc, in the physical and chemical properties of the elements. In this chapter, there are 40 questions in the exercise. The NCERT solutions for class 11 chemistry chapter 3 Classification of Elements and Periodicity in Properties are prepared by respective subject experts. These NCERT solutions will help you in the preparation of your class 11 final exam and also will help in the preparation of various competitive exams like JEE Mains, NEET, VITEEE etc.

In this chapter, you will study the development of the periodic table and the periodic law. Mendeleev arranged elements on the basis of their atomic masses in his periodic table. It is must to go through CBSE NCERT solutions for class 11 chemistry chapter 3 Classification of Elements and Periodicity in Properties to maximise your score in the examination. The modern periodic table arranges the elements in order of their atomic numbers. In the modern periodic table, elements are arranged in eighteen vertical columns( families or groups) and seven horizontal rows (periods). This chapter discusses some periodic trends in the periodic table like ionization enthalpies, atomic sizes, electron gain enthalpies, electronegativity, and valence. After completing solutions of NCERT for class 11 chemistry chapter 3 Classification of Elements and Periodicity in Properties, you will be able to-

1. Understand periodic law

2. Understand the significance of electronic configuration and atomic numbers as a basis for periodic classification

3. Able to classify elements of the periodic table into four different blocks which are s, p, d, f and learn their main characteristics

4. Understand periodic trends in chemical and physical properties of elements and explain the relationship between metallic character and ionization enthalpy.

## Important points and terms of Chapter 3 Classification of Elements and Periodicity in Properties-

1. In the modern periodic table, elements are arranged in the order of their atomic numbers in 7 horizontal rows or periods and 18 vertical columns or groups.
2. The first period is the smallest and sixth period is the longest.
3. Elements of periodic table are divided into four s, p, d, f blocks.
4. Elements of group 1 and 2 are s-block elements and the general valence shell electronic configuration of s-block elements is $ns^{1-2}$.
5. Elements of group 13, 14, 15, 16, 17 and 18 are p-block elements and the general valence shell electronic configuration of p-block elements is $ns^{2}np^{1-6}$.
6. Elements of group 3, 4, 5, 6, 7, 8, 9, 10, 11 and 12 are d-block elements and the general valence shell electronic configuration of d-block elements is $(n-1)d^{1-10}ns^{1-2}$.
7. Elements of two horizontal rows at the bottom of the periodic table are f-block elements and the general valence shell electronic configuration of f-block elements is $\dpi{100} (n-1)f^{1-14}(n-1)d^{0-1}ns^{2}$

NCERT solutions for class 11 chemistry chapter 3 Classification of Elements and Periodicity in Properties- Solved Exercise Questions

The basic theme of organization in the periodic table is to classify all the elements according to similar properties in periods and groups, This arrangement makes the study of elements and their compounds simple and systematic and less confusion can be generated and true information can be regenerated about the different elements present in the periodic table.

Mendeleev arranged elements in horizontal rows and vertical columns of a table in order of their increasing atomic weights in such a way that the elements with similar properties occupied the same vertical column or group. He fully recognized the significance of periodicity and used a broader range of physical and chemical properties to classify the elements.

In particular, Mendeleev relied on the similarities in the empirical formulas and properties of the compounds formed by the elements. He realized that some of the elements did not fit in with his scheme of classification if the order of atomic weight was strictly followed. He ignored the order of atomic weights, thinking that the atomic measurements might be incorrect, and placed the elements with similar properties together.

Mendeleev's Periodic Law and can be stated as : The physical and chemical properties of the elements are periodic functions of their atomic weights.

But Modern Periodic Law states that, The physical and chemical properties of the elements are periodic functions of their atomic numbers.

The period number corresponds to the highest principal quantum number (n) of the elements in the period. The subsequent periods consists of 8, 8, 18, 18 and 32 elements, respectively. The sixth period (n = 6) contains 32 elements and successive electrons enter 6s, 4f, 5d and 6p orbitals, Now, 6s has 1 orbital, 4f has 7 orbitals, 5d has 5 orbitals, and 6p has 3 orbitals.

Therefore, there is a total of 16 orbitals available in which each orbital carry 2 electrons according to Pauli's exclusion principle. So, we have a total of 32 electrons, Hence the sixth period of the periodic table should have 32 elements.

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

In the seventh period, the first two elements are s-block elements, the next 14 elements are f-block elements, next 10 elements are d-block elements and the next 6 elements are p-block elements.

Therefore, Z=114 element is the second p-block element in the seventh period. Thus, it is present in the 4th group of the 7th period of the periodic table.

Given that it is present in the 3rd period and a 17th group of the periodic table.

So, the first period has 2 elements and the second period has 8 elements. The 3rd period starts with the element with Z=11. And there are 8 elements present in the third period. Thus, the 3rd period ends with the element with Z=18. The element in the 18th group of the 3rd period has Z=18. Hence, the element in the 17th group of the 3rd period has atomic number Z=17.

Lawrence Berkeley Laboratory named the elements Lawrencium (Lr) with $Z =103$ and Berkelium (Bk) with $Z=97$.

Question

Seaborg’s group named the elements Seaborgium (Sg) with $Z =106$.

Elements in the same group have similar physical and chemical properties because they have the same number of valence electrons. So, their most properties are similar, as they react in the same manner as the group has the same number of valence electrons.

Atomic radius is the distance between from the centre of an atom to the outer most shell containing the electrons. It is a measure of the size of an atom.

They can be of types like :

(a) Covalent radius, where it is the half of the distance between the line joining the centres of nuclei of two adjacent similar atoms.

(b) Metallic radius is half of the distance between the centres of the nuclei of two adjacent atoms in the metallic crystal.

(c) van der Waal's radius, half of the internuclear distance between 2 similar adjacent atoms.

Whereas, the Ionic radius is the distance from the centre of the nucleus of the ion up to which it exerts its influence on the electron cloud of an ion (cation or anion). Generally, cation has a smaller ionic radius than the parent atom and anions are larger than the parent atom.

The atomic radius of the elements generally decreases from the left to the right in a period because the nuclear charge gradually increases by one unit and one electron is also added in the electron shell, due to this the electrons get attracted more towards the centre, as a result, the atomic radii decreases.

The atomic radius of the elements increases as we move downwards in a group because there is an increase in the principal quantum number and thus, there is an increase in the number of electrons shells. Therefore, the atomic size is expected to increase.

$F^-$

Atoms and ions which contains the same number of electrons are isoelectronic species.

GIven $F^-$ ion which has total of $9+1 =10$ electrons.

Thus, the species having the same number of electrons are $Na^+$ ion having $(11-1=10)$ electrons, or $Ne$ having 10 electrons, or $O_{2}^-$ ion $(8+2 =10)$ electrons, and $Al^{3+}$ ion having $13-3 =10$ electrons.

Ar

Atoms and ions which contains the same number of electrons are isoelectronic species.

GIven $Ar$  which has a total of 18 electrons.

Thus, the species having the same number of electrons are $S_{2}^-$ ion having $(16+2=18)$electrons, or $Cl^-$ having $(17+1 =18)$ electrons, or $K^+$ ion $(19-1=18)$ electrons, and $Ca^{2+}$ ion having $(20-2=18)$ electrons.

$Mg ^{2+}$

Atoms and ions which contains the same number of electrons are isoelectronic species.

GIven $Mg ^{2+}$  which has total of $(12-2=10)$ electrons.

Thus, the species having the same number of electrons are

$F^-$ ion having $(9+1=10)$electrons,  or $O_{2}^-$ having $(8+2 =10)$ electrons,

or $Ne$  $10$ electrons, and $Al^{3+}$ ion having $(13-3=10)$ electrons.

$Rb^+$

Atoms and ions which contains the same number of electrons, are isoelectronic species.

GIven $Rb^+$  which has total of $(37-1=36)$ electrons.

Thus, the species having the same number of electrons are

$Br^-$ ion having $(35+1=36)$electrons,  or $Kr$ having $36$ electrons,

or $Sr^{2+}$ ion  $(38-2 =36)$ electrons.

Given species $N^{3-} , O ^{2- } , F^- , Na ^+ , Mg ^{2+ }, Al ^{3+}$ are isoelectronic species as they have the same number of electrons i.e., 10 electrons.

$N^{3-}has\ 7+3 =10\ electrons.$

$O^{2-}has\ 8+2 =10\ electrons.$

$F^{-}has\ 9+1 =10\ electrons.$

$Na^{+}has\ 11-1 =10\ electrons.$

$Mg^{2+}has\ 12-2 =10\ electrons.$

$Al^{3+}has\ 13-3 =10\ electrons.$

Arrange them in the order of increasing ionic radii.

Given species $N^{3-} , O ^{2- } , F^- , Na ^+ , Mg ^{2+ }, Al ^{3+}$.

As we know that the ionic radii of isoelectric species increases with a decrease in the magnitude of nuclear charge.

So, here is the increasing ionic radii arrangement :

$Al^{3+}

A cation is smaller than its parent atom because it has fewer electrons while its nuclear charge remains the same. The size of an anion will be larger than that of the parent atom because the addition of one or more electrons would result in increased repulsion among the electrons and a decrease in effective nuclear charge.

The significance of the term 'isolated gaseous atom' indicates that the atoms in the gaseous phase are much far separated that there does not have any mutual attraction or repulsion interactions present which is an isolated state. Here the value of ionization enthalpy and electron gain enthalpy are not influenced by the presence of the other atoms.

The significance of the term 'ground state' means that in an atom, electrons are present in the lowest energy state where they neither lose nor gain an electron. Ionization enthalpy and electron gain enthalpy are generally expressed with respect to the ground state of an atom only.

Given that Energy of an electron in the ground state of the hydrogen atom is $-2.18 \times 10 ^{-18 }J$.

So, the ionization enthalpy is for 1 mole of atoms.

Therefore, the ground state energy of the atoms may be expressed as :

E (ground state) = $-2.18 \times 10 ^{-18 }J \times (6.022\times 10^{23} mol^{-1})$

$= -1.312\times 10^{6} J\ mol^{-1}$

Whereas, ionization enthalpy is,

$=E_{\infty } - E_{ground\ state}$

$=0-(-1.312\times10^6 mol^{-1}) =1.312\times10^{6} J\ mol^{-1}$

The electronic configuration of Be is $(1s^22s^2)$. The outermost electron is present in the s-orbital. Whereas, in B the electronic configuration is $(1s^22s^22p^1)$ here outermost electron is present in p-orbital.

The electrons in s-orbital are more tightly attracted by the nucleus than the p-orbital, as a result, more amount of energy is required to knock out a s-orbital electron.

Hence Be has higher $\Delta _ i$ H than B.

In oxygen, two of the four 2p-electrons of oxygen occupy the same 2p-orbital and in nitrogen, the three 2p-electrons of nitrogen occupy three different atomic orbitals.

This results in increased electron-electron repulsion in oxygen atom as a result, the energy required to remove the fourth 2p-electron from oxygen is less as compared to the energy required to remove one of the three sp-electrons from nitrogen.

Hence, oxygen has lower $\Delta _ i H$ than nitrogen.

In case of fluorine the electron is added to the same shell, the increase in nuclear attraction dominates over the increase in electronic repulsion.

Therefore, the valence electrons in fluorine atom experience a more effective nuclear charge than that experienced by the electrons present in oxygen. As a result, more energy is required to remove an electron from a fluorine atom than that required to remove an electron from oxygen atom.

Hence, oxygen has lower $\Delta _ i H$ than fluorine.

The electronic configuration of Na and Mg are:

$Na = 1s^22s^22p^63s^1$   and  $Mg = 1s^22s^22p^63s^2$

So, the first electron in case of both has to be removed from 3s-orbital, but the nuclear charge of Sodium has +11 charge and Magnesium has +12 charge, which causes electrons to held more tightly in case of Mg, therefore, the first ionization energy of sodium is lower than that of magnesium.

After the first ionization happens, the second electron has to be removed from the p-orbital in case of sodium which has already attained its stable noble gas configuration and from the s-orbital in case of magnesium.

Therefore, 2nd ionization enthalpy of sodium is higher than that of magnesium.

Ionization enthalpy of the main group elements tends to decrease down a group because of the following reasons:

(a) Atomic Size: moving down the group there is an increase in the atomic size, increasing number of electron shells, which results in the weaker binding force with the nucleus. Hence the ionization enthalpy decreases.

(b) Screening or shielding effect: moving down the group increases the new shells, the number of inner electron shells which shield the valence electrons increases. As a result, the force of attraction of the nucleus for the valence electrons further decreases and hence the ionization enthalpy decreases.

The given trend can be explained by the following steps:

(i) Moving from B to Al, there is an increase in the size of the atom as a result decrease in the value of ionization enthalpy.

(ii) Moving from Al to Ga, there are 10 electrons in Ga which do not screen as is done by Sulphur and Phosphorus. Therefore, there is an unexpected increase in the value of effective nuclear charge resulting in increased ionization energy value.

(iii) Moving from Ga to In and Tl, there are 14 electrons in Tl with very poor shielding effect, which increases the effective nuclear charge thus the value of ionization energy increases.

Oxygen and Fluorine both are of the second group elements and as we move from O to F, there is a decrease in atomic size, as a result, there is an increase in nuclear charge. Further, by a gain of one electron, $F \rightarrow F^-$ ion has an inert gas configuration and $O \rightarrow O^-$ gives ion which does not have a stable inert gas configuration.

The energy released is much higher in going from fluorine to oxygen or the electron gain enthalpy value of F is much more negative than that of oxygen.

F or Cl

The electron gain enthalpy value of Cl  $\triangle_{eg}H = -349kJ\ mol^{-1}$ is more negative than that of F  $\triangle_{eg}H = -328kJ\ mol^{-1}$.

This is because of the reason for the smaller size of F due to its small size, the electron repulsions in the relatively compact 2p-subshell are comparatively large and hence the attraction for an incoming electron is less as in the case of Cl.

When an electron is added to neutral O atom, a monovalent anion $(O^-)$ is formed and there is release in energy, i.e., the first electron gain enthalpy is negative. After that when the second electron is added to anion $(O^-)$ to form $(O^{2-})$ anion, there is a lot of electrostatic repulsions is there as both $(O^-)$ anion and electron have a negative charge, more difficult to add the second electron. To overcome the force of repulsion, we have to give some energy. Thus, the second electron gain enthalpy of Oxygen is positive. i.e., $O_{g} \rightarrow O^- \rightarrow O^{2-}$

The basic differences between the terms electron gain enthalpy and electronegativity are:

Electron gain enthalpy refers to the tendency of an isolated gaseous atom to accept an additional electron from a negative ion, whereas electronegativity refers to the tendency of an atom of an element to attract a shared pair of electrons towards it in a covalent bond.

The Given statement "the electronegativity of N on Pauling scale is 3.0 in all the nitrogen compounds" is wrong, because the electronegativity is a variable property of any element and is different for different kinds of compounds. For example electronegativity in $NO_{2}$ compound is different from $NH_{3}$ compound.

Question

gains an electron

When an atom gains an electron it forms an anion. And the size of an anion will be larger than the parent atom because of the addition of one or more electrons would result in increased repulsion among electrons and a decrease in the effective nuclear charge. This results in the increase in the atomic radius of an atom

Question

loses an electron

When an atom loses an electron it becomes positively charged and now for the same nuclear charge, there are lesser electrons present than the parent atom. Hence there will be an overall increase in the attraction between electrons and nucleus. As a result, there will be a reduction in the atomic radius of an atom and it is smaller than the parent atom.

As the ionization enthalpy, depends upon the number of electrons (electronic configuration) and nuclear charge (number of protons) in the nucleus, And there are the same electronic configuration of isotopes and also the nuclear charge.

Hence they will have the same ionization enthalpy.

Metals are usually solids at room temperature [mercury is an exception; gallium and caesium also have very low melting points]. Metals usually have high melting and boiling points. They are good conductors of heat and electricity because they have the tendency to lose electrons easily. They are malleable (can be flattened into thin sheets by hammering) and ductile (can be drawn into wires).

Non-metals are usually solids or gases at room temperature with low melting and boiling points (boron and carbon are exceptions). They are poor conductors of heat and electricity. Most nonmetallic solids are brittle and are neither malleable nor ductile hence forming mainly covalent compounds.

Question

Identify an element with five electrons in the outer subshell.

The element having five electrons in the outer subshell belongs to the nitrogen family or group 15, Example Nitrogen.

Question

Identify an element that would tend to lose two electrons.

The element that would tend to lose two electrons should belong to alkaline earth family or group 2. For example Magnesium.

Question

Identify an element that would tend to gain two electrons.

The element that would tend to gain two electrons should belong to the oxygen family or group 16. For example Oxygen.

Question

Identify the group having metal, non-metal, liquid as well as gas at the room temperature.

Group 17 has metals, non-metals, liquid as well as gas at room temperature, i.e., chlorine, bromine are non-metals while iodine is a metal.

Fluorides are liquids in the state while chlorine, Br and Iodine are gases at room temperature.

In the case of group 1, we have ordered$Li < Na < K < Rb  because there is only one valence electron. and thus have a strong tendency to lose this electron. The tendency to lose electrons in turn also depends upon the ionization enthalpy. Moving down the group there is a decrease in the ionization enthalpy, therefore, increasing reactivity down the group.

In the case of group 17, we have the order $F > CI > Br > I$ because there are seven electrons present in the valence shells and thus have a strong tendency to accept electrons to make it a stable noble gas electronic configuration. So, moving down we have a decrease in both electron gain enthalpy and electronegativity. Hence there is a decrease in the reactivity also.

The general electronic configuration of s-,p-,d- and f- block elements are shown below:

For the s-block elements: $ns^{1-2}$, where $n=2-7$.

For the p-block elements: $ns^2np^{1-6}$, where $n =2-6$.

For the d-block elements, $(n-1)d^{1-10}ns^{0-2}$, where $n = 4-7$.

For the f- block elements: $(n-2)f^{0-14}(n-1)d^{0-1}ns^2$, where $n= 6-7$.

Given the outer electronic configuration of the element:

$ns^2 np ^4 \: \:for \: \: n = 3$

For n=3 hence the element belongs to the third period, p-block element.

Since the valence shell contains 6 electrons and Group number = 10+6 =16 configuration $= 1s^22s^22p^63s^23p^4$ element name is Sulphur.

Given the outer electronic configuration of the element:

$( n-1) d^ 2 ns^2\ for\ n=4$.

For n=4 hence the element belongs to the fourth period.

Since the valence shell contains (2+2) electrons and Group number = 4 configuration $= 1s^22s^22p^63s^23p^63d^24s^2$ element name is Titanium (Ti).

Question

$( n-2 ) f^ 7 ( n-1 ) d^ 1 ns^2 \: \: for\: \: n = 6$ , in the periodic table.

Given the outer electronic configuration of the element:

$( n-2 ) f^ 7 ( n-1 ) d^ 1 ns^2 \: \: for\: \: n = 6$.

For n=6 hence the element belongs to the sixth period and the last electron goes to the f-orbital, the element is from f-block.

We have the group number =3. Electronic configuration: $[X_{e}]4f^75d^16s^2.$

Hence the element is Gadolinium with Z=64.

the least reactive element.

The element which has highest first ionization enthalpy $(\triangle_{i}H_{1})$ and positive electron gain enthalpy $(\triangle_{eg}H)$ is $element\ V$.

And also element V shows similar behaviour like inert gases because of positive electron gain enthalpy.

the most reactive metal.

The element II which has the least first ionization enthalpy value $(\triangle_{i}H_{1})$ and a low negative electron gain enthalpy value $(\triangle_{eg}H)$ is the most reactive metal.

the most reactive non-metal.

The element III which has high first ionization enthalpy $(\triangle_{i}H_{1})$ and a very high negative electron gain enthalpy $(\triangle_{eg}H)$ is the most reactive non-metal.

the least reactive non-metal.

The element IV has a high negative electron gain enthalpy $(\triangle_{eg}H)$ but not so high that the first ionization enthalpy $(\triangle_{i}H_{1})$ and is the least reactive non-metal.

the metal which can form a stable binary halide of the formula $MX_2$ (X=halogen).

The metal which can form a stable binary halide of the formula: $MX_2$ (X=halogen) is the element VI which has low first ionization enthalpy $(\triangle_{i}H_{1})$ but higher than that of alkali metals.Hence it is an alkaline earth metal and form binary halide.

Lithium and oxygen

For lithium and oxygen, the formula of the stable binary compound is: $LiO_{2}$ (Lithium oxide)

Magnesium and nitrogen

For Magnesium and nitrogen, the formula of the stable binary compound is: $Mg_{3}N_{2}$ (Magnesium nitride) .

Aluminium and iodine

For Aluminium and iodine, the formula of the stable binary compound is: $AlI_{3}$ (Aluminium iodide) .

Silicon and oxygen

The stable binary compound that would be formed by the combination of Silicon and oxygen is: Silicon dioxide $SiO_{2}$ .

Phosphorus and fluorine

The stable binary compound that would be formed by the combination of Phosphorus and fluorine are: Phosphorus trifluoride $PF_{3}$ or Phosphorus pentafluoride $PF_{5}$.

Element 71 and fluorine

The element with the atomic number 71 is Luteteium (Lu) having valency of 3.

Hence, the formula of the compound is $LuF_{3}$ (Lutetium trifluoride).

(a) Atomic number
(b) Atomic mass
(c) Principal quantum number
(d) Azimuthal quantum number.

Answer - (C) Principal quantum number

In the modern periodic table, the period indicates the value of, principal quantum number (n) for the outermost shell or the valence shell.

(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.

Answer - (b) because d-block has a maximum of 10 electrons in all orbitals in a d-subshell and there are 10 columns. As each subshell can have a maximum of 5-orbitals and 10 electrons, therefore, there are 10 vertical columns in a d-block.

(a) Valence principal quantum number (n)

(b) Nuclear charge (Z )

(c) Nuclear mass

(d) Number of core electrons.

Valence shell electrons are present in the outermost shell of an atom and nuclear mass does not affect the valence shell because the nuclear mass has so small value that it is considered as negligible.

(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.

Answer - (a) Nuclear charge (Z)

Because as the nuclear charge increases the size of an isoelectronic species decreases.

Here we have the isoelectronic species: $F^-$, Ne and $Na^+$

$F^-$  has  $Z=9$

$Ne$  has  $Z=10$

$Na^{+}$  has  $Z=11$

Therefore, the order of the increasing size is as follows:

$Na^{+}

(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.

Because 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.

(a) B > Al > Mg > K

(b) Al > Mg > B > K

(c) Mg > Al > K > B

(d) K > Mg > Al > B

Answer - (d) K > Mg > Al > B

Because the metallic character of an element is the tendency of an element to lose electrons easily. So, moving from left to right across the period metallic character of element decreases.

Here, K and Mg are s-block elements while B and Al belong to p-block elements.

Hence the order is as follows:

K > Mg > Al > B

(a) B > C > Si > N > F

(b) Si > C > B > N > F

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

(d) F > N > C > Si > B

Answer -  (c) F > N > C > B > Si

Because non-metallic character of an elements increases from left to right across the period. And here given position of elements:

Boron (B)  -  $2^{nd}$ period and $13^{th}$ group.

Carbon (C)  -  $2^{nd}$ period and $14^{th}$ group.

Nitrogen (N)  -  $2^{nd}$ period and $15^{th}$ group.

Fluorine (F)  -  $2^{nd}$ period and $17^{th}$ group.

And the Silicon (Si) is present in the $3^{rd}$ period and  $14^{th}$ group.

Hence the correct order of non-metallic character is as follows:

F > N > C > B > S

(a) F > Cl > O > N

(b) F > O > Cl > N

(c) Cl > F > O > N

(d) O > F > N > Cl

Answer - (b) $F>O>Cl>N$

The oxidizing property of elements increases from left to right across a period because of the presence of vacant d-orbitals in their valence shells.

Thus, we get the decreasing order of oxidizing property as $F>O>N$.

And the oxidizing character of elements decreases down a group. Thus we get $F>Cl$.

However, the oxidizing character of Oxygen is more than that of Chlorine.

So, $O>Cl$.

Hence the correct order their chemical reactivity in terms of oxidizing property is :

$F>O>Cl>N$.

## NCERT solutions for class 11 chemistry

 Chapter 1 NCERT solutions for class 11 chemistry chapter 1 Some Basic Concepts of Chemistry Chapter-2 CBSE NCERT solutions for class 11 chemistry chapter 2 Structure of Atom Chapter-3 NCERT solutions for class 11 chemistry chapter 3 Classification of Elements and Periodicity in Properties Chapter-4 NCERT solutions for class 11 chemistry chapter 4 Chemical Bonding and Molecular Structure Chapter-5 CBSE NCERT solutions for class 11 chemistry chapter 5 States of Matter Chapter-6 Solutions of NCERT class 11 chemistry chapter 6 Thermodynamics Chapter-7 NCERT solutions for class 11 chemistry chapter 7 Equilibrium Chapter-8 CBSE NCERT solutions for class 11 chemistry chapter 8 Redox Reaction Chapter-9 Solutions of NCERT class 11 chemistry chapter 9 Hydrogen Chapter-10 NCERT solutions for class 11 chemistry chapter 10 The S-Block Elements Chapter-11 CBSE NCERT solutions for class 11 chemistry chapter 11 The P-Block Elements Chapter-12 Solutions of NCERT class 11 chemistry chapter 12 Organic chemistry- some basic principles and techniques Chapter-13 NCERT solutions for class 11 chemistry chapter 13 Hydrocarbons Chapter-14 CBSE NCERT solutions for class 11 chemistry chapter 14 Environmental Chemistry

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