Class 11 | Chapter 3 (Classification of Elements and Periodicity in Properties) NCERT Chemistry Solutions

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

Similarity in properties is the basic theme for organisation of elements in the periodic table.

According to their characteristics, the elements in the periodic table are organised into groups (shown by vertical columns) and periods (represented by horizontal rows).

On the periodic chart, elements are grouped together in the same category if they have similar chemical and physical characteristics.

There are now 118 elements listed in the periodic table. With such a huge number of elements, it is very challenging to study the chemistry of each of these elements and the uncountable compounds that may be formed from them.

Honestly quite difficult to study properties of so many elements and compounds, sometimes continuously studying chemistry for just an hour can be tiring.

Number of scientists in the past looked for methodical approaches to organising their knowledge by categorising the elements. 

Systematic organization of  information obtained from a variety of experts resulted in the classification of elements into groups and periods, which resulted as periodic table.


Question 2
Which important property did Mendeleev use to classify the elements in his periodic table and did he stick to it?

Atomic mass was an essential feature that Mendeleev used to classify elements.

He organised all known elements according to their atomic masses. Later, he grouped elements with similar properties together rather than in order of their atomic masses.

However, he did not stick to this arrangement for long. He found out that if the elements were arranged strictly in order of their increasing atomic weights, then some elements did not fit within this scheme of classification.

Therefore, he ignored the order of atomic weights in some cases. For example, the atomic weight of iodine is lower than that of Tellurium. Still, Mendeleev placed tellurium (in Group VI) before iodine (in Group VII) simply because iodine’s properties are so similar to Fluorine, Chlorine, and Bromine.


Question 3
What is the basic difference in approach between the Mendeleev’s Periodic Law and the Modern Periodic Law?

Mendeleev periodic law states that elements physical and chemical qualities are a periodic function of their atomic weights, while Modern periodic law states that properties of elements are a periodic function of their atomic numbers.

Thus, the fundamental difference in approach between Mendeleev’s periodic Law and Modern Periodic Law is the shift in the basis of element classification from atomic weight to atomic number.

Below are some key difference between Mendeleev’s Periodic Law and Modern Periodic Law

Mendeleev’s Periodic TableModern Periodic Law
This periodic table is based on atomic massThis periodic table is based on atomic numbers
There are a total of 7 groups and 6 periodsThere are a total of 18 groups and 7 periods
Elements are arranged based on their increasing in atomic massElements are arranged based on the increasing order of their atomic numbers
There is no place allotted for isotopesIsotopes are kept in the same place due to their atomic numbers being the same

Question 4
On the basis of quantum numbers, justify that the sixth period of the periodic table should have 32 elements

In the periodic table of the elements, a period indicates the value of the principal quantum number (n) for the outermost shells. The value of n for the sixth period is 6. For n = 6, azimuthal quantum number (l) can have values of 0, 1, 2, 3, 4.

According to Aufbau’s principle, electrons are added to different orbitals in order of their increasing energies. The energy of the 6d subshell is even higher than that of the 7s subshell.

In the 6th period, electrons can be filled in only 6s, 4f, 5d, and 6 p subshells. Now, 6s has one orbital, 4f has seven orbitals, 5d has five orbitals, and 6p has three orbitals.

Therefore, there are a total of sixteen (1 + 7 + 5 + 3 = 16) orbitals available. According to Pauli’s exclusion principle, each orbital can accommodate a maximum of 2 electrons.

Thus, 16 orbitals can accommodate a maximum of 32 electrons. Hence, the sixth period of the periodic table should have 32 elements.


Question 5
In terms of period and group where would you locate the element with Z = 114?

Electron configuration of element with atomic number Z = 114 is [Rn] 5f14  6d10 7s2 7p2
From the electron configuration its clear that this element have n = 7 for it’s outermost shell.

Therefore this element belongs to p-block in the periodic table

Group number for any p-block element = 10 + Number of Valence Electrons

Z = 114 have electron configuration [Rn] 5f14  6d10 7s2 7p2

Therefore number of valence electrons in it = 2 + 2 = 4

Hence
Group number of Z = 114 is 10 + 4 = 14


Question 6
Write the atomic number of the element present in the third period and seventeenth group of the periodic table.

Chlorine


Question 7
Which element do you think would have been named by
(i) Lawrence Berkeley Laboratory
(ii) Seaborg’s group?

(i) Lawrencium
(ii) Seaborgium


Question 8
Why do elements in the same group have similar physical and chemical properties?

Elements in the group have similar chemical properties. This is because their atoms have the same number of electron in the highest occupied energy level.


Question 9
What does atomic radius and ionic radius really mean to you?

Atomic radius is the distance between nucleus and outermost shell where electrons are present.

Ionic radius is the distance between nucleus and outermost shell of an ion.


Question 10
How do atomic radius vary in a period and in a group? How do you explain the variation?

In a period, from left to right, with increase in the atomic number, the atomic radius decreases due to increase in the nuclear charge which increases the attraction of the nucleus for the valence electrons. 

In a group, on moving from top to bottom, the ionic radius increases as a new energy level is added at each succeeding element but the number of valence electrons remains same.


Question 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+

Isoelectronic species are those which have same number of electrons.

F have 10 electrons, species isoelectronic to this are N3+, O2-, Ne, Na+, Mg2+

Ar have 18 electrons, species isoelectronic to this are P3-, S2-, Cl, K+, Ca2+

Mg2+ have 10 electrons, species isoelectronic to this are N3+, O2-, Ne, Na+, Mg2+

Rb+ have 36 electrons, specifies isoelectronic to this are Br, Kr, Sr2+


Question 12
Consider the following species :
N3-, O2-, F, Na+, Mg2+ and Al3+

(a) What is common in them?
(b) Arrange them in the order of increasing ionic radii.

(a) N3-, O2-, F, Na+, Mg2+ and Al3+ all of these have same number of electrons (10) and therefore are isoelectronic.

(b) N3- > O2- > F  > Na+ > Mg2+ > Al3+


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

A gaseous atom looses one or more electrons to form cation.
The nuclear charge remains same but the number of electrons decreases.

Same nuclear charge acts on lesser number of electrons. The effective nuclear charge per electron increases.

The electrons are more strongly attracted and pulled towards the nucleus. Thus, the size decreases.

A gaseous atom gains one or more electrons to form anion.
The nuclear charge remains same but the number of electrons increases.

Same nuclear charge acts on greater number of electrons. The effective nuclear charge per electron increases.

The electrons are less strongly attracted and pulled away from the nucleus. Thus, the size decreases.


Question 14
What is the significance of the terms – ‘isolated gaseous atom’ and ‘ground state’ while defining the ionisation enthalpy and electron gain enthalpy?

Ionization 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.
If an isolated gaseous atom is in its ground state, then less amount energy would be required to remove an electron from it.

Therefore, for comparison purposes, ionisation enthalpy and electron gain enthalpy must be determined for an ‘isolated gaseous atom’ and its ‘ground state’.


Question 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.

For 1 mole of hydrogen atoms, energy will be – 2.18 × 10-18 × 6.023 × 1023 = 13.13 × 105 = 1.31 × 106 J

Therefore ionisation enthalpy of atomic hydrogen in terms of J mol-1 is 1.31 × 106 J


Question 16
Among the second period elements the actual ionisation enthalpies are in the order Li < B < Be < C < O < N < F < Ne. Explain why?

(i) Be has higher ∆iH than B
(ii) O has lower ∆iH than N and F?

(i) Be has higher ∆iH than B
The electronic configuration of Be has higher stability than the electronic configuration of B due to completely filled 2s orbital.
Electronic configuration of Be is 1s2 2s2 (fully filled)
Electronic configuration of B is 1s2 2s2 2p1 (Not fully filled)

Also during ionisation of Be, s electron is removed.  During ionisation of B, p electron is removed.
2s electron penetrates to the nucleus to greater extent than 2p electron. Thus, 2p electron is more shielded than 2s electron.

The attraction of nucleus for 2s electron is higher than the attraction of nucleus for 2p electron. Thus, removal of 2s electron requires higher energy than the removal of 2p electron.

Therefore, the ionisation enthalpy of Be is higher than the ionisation enthalpy of B.

(ii) O has lower ∆iH than N and F?
Electron configuration of N is 1s2 2s2 2p3
Electron configuration of O is 1s2 2s2 2p4
Electron configuration of F is 1s2 2s2 2p5
The 2p orbital contains 4 electrons out of which 2 are present in the same 2p-orbital.

Due to this, the electron repulsion increases.

N has stable half filled configuration. F has greater nuclear charge.

Hence, the ionisation enthalpy of O is lower than that of N and F


Question 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?

Ionization enthalpy is defined as the energy required to remove an electron from an isolated atom in its ground state.

Electronic configuration of sodium (Na) is 1s2 2s2 2p6 3s1

Electronic configuration of magnesium (Mg) is 1s2 2s2 2p6 3s2

First ionisation enthalpy is the energy required to remove the first most loosely bound electron.

Here, the first electron in both sodium and magnesium has to be removed from 3s orbital but the effective nuclear charge of sodium is less than that of magnesium.

Therefore, first ionisation enthalpy of sodium is less than that of magnesium. 

Now, after the removal of first electron, electronic configuration of Na and Mg is:
Electronic configuration of Na+ is 1s2 2s2 2p6
Electronic configuration of Mg+ is 1s2 2s2 2p6 3s1

Sodium after losing one electron acquires the stable noble gas configuration of neon 1s2 2s2 2p6.

Therefore in case of sodium, the second electron has to be removed from the stable noble gas configuration.

Consequently, removal of a second electron from sodium requires more energy in comparison to that required in magnesium.

Therefore, the second ionisation enthalpy of sodium is higher than that of magnesium.


Question 18
What are the various factors due to which the ionisation enthalpy of the main group elements tends to decrease down a group?

Size of Atom
As atomic size of the elements becomes larger as one moves down the periodic table. This is because the number of main energy shells increases (n).

As a result of the growth in size, the distance that separates the electron from the nucleus also increases, which results in a decrease in the attractive force that exists between the nucleus and the valence electrons.

Therefore, the amount of energy that is required to remove valence electrons will decrease, and it will be much easier to remove the valence electrons. As a result, the ionisation energy drops down the group.

Shielding Effect
As one moves down the group, the number of electrons in the inner shells also grows, and as a result, the shielding impact on the valence electron, will increase.

Therefore the force that the nucleus exerts on the electrons farther from it will weaken, and the electrons will no longer be held with strong force.

As a consequence of this, the amount of energy necessary to remove a valence electron will decrease down the group. Therefore ionisation energy decreases down the group due to shielding effect.

Nuclear charge
On moving from top to down in a group, the nuclear charge increases.

The effect of increase in the atomic size and the shielding effect is much more significant than the nuclear charge.

The electron becomes less lightly held to the nucleus on moving down the group. 


Question 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?

(i) Al has lower ionisation enthalpy than B due to larger size.

(ii) Ga has slightly higher ionisation enthalpy than Al due to ineffective shielding by 3d electrons.

(iii) In has lower ionisation enthalpy than Ga due to larger size.

(iv) Tl has higher ionisation enthalpy than In due to ineffective shielding by 4f electrons.


Question 20
Which of the following pairs of elements would have a more negative electron gain enthalpy?
(i) O or F
(ii) F or Cl

F will have more negative electron gain enthalpy because in going from O to F.

The atomic size decreases and the nuclear charge increases.

Both these factors make the value of electron gain enthalpy of fluorine more negative because they tend to increase the nuclear pull for the incoming electron.

Cl has more negative electron gain enthalpy than fluorine.

This is because the incoming electron repel each other in fluorine due to small size but in chlorine, the incoming electrons due to not face the electronic repulsion due to large size of chlorine than fluorine.


Question 21
Would you expect the second electron gain enthalpy of O as positive, more negative or less negative than the first? Justify your answer.

Because the oxygen atom becomes negatively charged after gaining one electron, the second electron gain enthalpy of O is positive.

This is due to the fact that after gaining one electron atom of oxygen becomes O

Now as the second electron approaches the negatively charged oxygen ion, there is repulsion between negatively charged oxygen ion and incoming electron.

Therefore energy is required to overcome this repulsion and add second electron to oxygen atom, thus second electron gain enthalpy of oxygen is positive.


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

Electronegativity is the tendency of atom of an element to draw a shared pair of electrons towards it in a covalent bond. This tendency is characterised as the ability of an element to be electronegative.

Whereas the term “electron gain enthalpy” refers to the change in enthalpy that occurs when a neutral gaseous atom takes on an additional electron in order to form an anion.


Question 23
How would you react to the statement that the electronegativity of N on Pauling scale is 3.0 in all the nitrogen compounds?

The term “electronegativity” was coined by Linus Pauling, who defined it as “the capacity of an atom in a molecule to attract electrons towards itself”.

An atom’s electronegativity may be thought of as a relative number that describes how well that atom is able to pull election density towards itself when it forms a bond with another atom. For example on Linus Pauling scale electronegativity of fluorine is 4.0 (maximum amongst all elements in periodic table).

When the electronegativity of an element is high, the atoms of that element will try to attract electrons towards themselves in a bond.

The primary characteristics of an atom that are responsible for determining its electronegativity are the atomic number and the atomic radius of the atom.

Which are same even if nitrogen is forming compounds with different atoms of different elements. That’s why electronegativity of N on Pauling Scale is 3.0 in all the nitrogen compounds.


Question 24
Describe the theory associated with the radius of an atom as it
(a) gains an electron
(b) loses an electron

(a) gains an electron
An anion is formed whenever an atom acquires an additional electron. The radius of the anion is more than the atomic radius.

This is due to the fact that the number of electrons has increased while the charge on the nucleus has remained same.

Because of this, the effective nuclear charge acting on attracting each electron in the anion is less as compared to that in atom, therefore force of attraction between nucleus and electrons become lesser as an atom gains electron.

As less force is acting on electrons, therefore size of anion is bigger than that of atom.

(b) loses an electron
Cations are formed whenever an atom loses electrons.

The radius of an atom is much bigger than the radius of cation. Although there is no change in the nuclear charge, there is a one-to-one decrease in the total number of electrons.

As a result, there is an increase in the effective nuclear charge per electron. Therefore, t he nucleus and the valence electrons are becoming more and more attracted to one another.

As a result of this, the radius becomes smaller. That’s why cation is smaller as compared to atom.


Question 25
Would you expect the first ionisation enthalpies for two isotopes of the same element to be the same or different? Justify your answer.

Isotopes are variations of elements that have the same number of protons and electrons but different number of neutrons than the parent element.

This difference in neutron count is what differentiates isotopes from one another.

For example – hydrogen have 3 isotopes which are hydrogen, deuterium, and tritium.

Ionisation energy primarily depends upon force between nucleus and electrons and number of electrons in an atom.

As all isotopes of a given element have same number of protons and electrons therefore first ionisation enthalpies of two isotopes of an element are same.


Question 26
What are the major differences between metals and non-metals?

MetalsNon-Metals
These are solids at room temperature except mercuryThese exist in all three states (solid, liquid and gas)
These are very hard except sodium
(Sodium can be cut with a knife at room temperature)
These are soft except diamond
(Diamond which is an allotrope of carbon is quite solid)
These are malleable and ductileThese are brittle and can break down into pieces
These are shinyThese are non-lustrous except iodine
Electropositive in natureElectronegative in nature
Have high densitiesHave low densities

Question 27
Use the periodic table to answer the following questions.
(a) Identify an element with five electrons in the outer subshell.
(b) Identify an element that would tend to lose two electrons.
(c) Identify an element that would tend to gain two electrons.
(d) Identify the group having metal, non-metal, liquid as well as gas at the room temperature.

(a) Fluorine
(b) Magnesium
(c) Oxygen
(d) Group 17 (Halogens)


Question 28
The increasing order of reactivity among group 1 elements is Li < Na < K < Rb < Cs whereas that among group 17 elements is F > Cl > Br > I.
Explain.

Elements in Group 1 of periodic table have a valence shell that have just one electron.
General electron configuration of elements in first group of periodic table is ns1 where n is period of element.

They have a high tendency for losing this one valence electron and  and achieve electron configuration of nearest noble gas.

This tendency of an element to loose electron is measured using Ionisation Enthalpy, as we go down the first group of periodic table size of atoms of elements increase therefore force of attraction between nucleus and valence electrons become lesser and lesser, which means Ionisation Enthalpy is decreasing down the group.

Therefore reactivity is increase, thus order of reactivity of elements in group 1 is Li < Na < K < Rb < Cs.

Group 1 ElementFirst Ionisation Enthalpy
Li520.2 kJ/mol
Na495.8 kJ/mol
K418.8 kJ/mol
Rb403.0 kJ/mol
Cs375.7 kJ/mol

From the above table, its clear that ionisation enthalpy is decreasing as we’re going down the group 1 in the periodic table.

Elements in group 17 of periodic table have 7 valence electrons, so just gaining one electron would make them achieve electron configuration of nearest noble gas and therefore become more stable.

So reactivity of group 17 elements can be measured in terms of tendency to gain electrons. As we go down the group 17, size of atom is increasing thus repulsion on incoming electron is decreasing therefore as we go down the group 17, it becomes more and more easier to gain the electron and achieve more stable electron configuration of nearest noble gas.

Therefore reactivity of elements in group 17 increases down the group as it become easier to gain electrons.


Question 29
Write the general outer electronic configuration of s-, p-, d- and f- block elements

Block of ElementElectronic Configuration
s – blockns1
p – blockns2 np(1 – 6)
d – block(n – 1)d(1 – 10) ns(0 – 2)
f – block(n – 2)f(0 – 14) (n – 1)d(0 – 1) ns2

Where n is period of element in the Periodic Table.


Question 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)f 7 (n-1)d1 ns2 for n = 6, in the periodic table.

(i) For n=3, the period in which the element belongs is third. The electronic configuration is 3s2 3p4 and the element belongs to p block. The group number of the element is 10 + number of electrons in the valence shell.
=10 + 6 = 16

Thus, the element belongs to third period and sixteenth group.

(ii) For n=4, the period in which the element belongs is fourth. The electronic configuration is 3d2 4s2 and the element belongs to d block. The group number of the element is = number of electrons in the (n−1)d subshell + number of electrons in ns subshell = 2+2 = 4

Thus, the element belongs to fourth period and fourth group.

(iii) For n=6, the period in which the element belongs is sixth. The electronic configuration is 4f7 5d1 6s2  and the element belongs to f block. All f block elements belong to third group.

Thus, the element belongs to sixth period and third group.


Question 31
The first (∆iH1) and the second (∆iH1) ionisation enthalpies (in kJ mol-1) and the (∆H1) electron gain enthalpy (in kJ mol-1) of a few elements are given below :

Elements∆H1∆H2egH
I5207300– 60
II4193051– 48
III16813374– 328
IV10081846– 295
V23725251+ 48
VI7381451– 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)?

(a) V
The element V has highest first ionisation enthalpy and positive electron gain enthalpy hence, it is least reactive.

(b) II
The element II has the least first ionisation enthalpy hence, it is most reactive metal.

(c) III
The element III has very high negative electron gain enthalpy hence, it is most reactive non-metal.

(d) IV
Element IV has high negative electron gain enthalpy but ionisation energy is not that high hence, it is least reactive non-metal.

(e) VI
The first and second ionisation energies of element VI indicate that it can form a stable binary halide.

(f) I
The element 1 has very low value of first ionisation energy but very high second ionisation energy. Hence, it will form a stable covalent halide of the formula MX.


Question 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

(a) Li2O

(b) Mg3N2

(c) AlI3

(d) SiO2

(e) PF5

(f) LuF3


Question 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

(c) Principal Quantum Number


Question 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 (/) for the last subshell that received electrons in building up the electronic configuration.

(b) is incorrect


Question 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.

(c) Nuclear mass


Question 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

(a) Nuclear charge (Z)


Question 37
Which one of the following statements is incorrect in relation to ionisation enthalpy?

(a) Ionization enthalpy increases for each successive electron.

(b) The greatest increase in ionisation enthalpy is experienced on removal of electron from core noble gas configuration.

(c) End of valence electrons is marked by a big jump in ionisation enthalpy.

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

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


Question 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

(d) K > Mg > Al > B


Question 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

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


Question 40
Considering the elements F, Cl, O and N, the correct order of their chemical reactivity in terms of oxidising property is
(a) F > Cl > O > N
(b) F > O > Cl > N
(c) CI > F > O > N
(d) O > F > N > Cl

(b) F > O > Cl > N


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