Competency Based Lesson/Unit Plan
1.
Name
of the Teacher: Designation:
2.
Subject CHEMISTRY Class:
11 Section A
3.
Lesson/UnitName:Classification
of elements and their periodicity
4. No. of periods required:6 Duration From: 18/07/2026 To 30/7/2026
A. Curricular Competencies to be developed (As per NCF-SE 2023)
NCF 2023 suggests the Panchaadi (five-step)
framework for lesson planning to achieve these competencies:
1.
Aditi (Introduction/Context Setting):
Need for classification of elements
2.
Bodh (Conceptual Understanding): Classification
of elements using Mendeleev’s periodic law and Modern periodic law
3.
Abhyas (Practice): Active practice
through activities, discussions, and projects.
4.
Prayog (Application): Apply the
knowledge of periodic table for classification of elements and learning their
characteristics
5. Prasar (Extension/Expansion): Expand the knowledge of periodic table for understanding variation of periodic properties and diagonal relationship
B. Learning Outcomes (Specific to the lesson mapped with NCERT learning outcomes):
Students
will be able to learn
1. Understand
a. Need for classification of elements
b. Dobereiner law of triads and Newlands law of octaves
c. Mendeleev’s periodic table
2. Determine
the drawbacks of Mendeleev’s periodic table
3. Recognize
a. The need for using atomic number as a function of periodic properties
b. Properties of s, p, d and f block
4. Define: Modern
periodic law
5. Problem solving:
a. Classify the elements into periods, groups and blocks
b. Variation of atomic radii, ionization energy, electron affinity and electronegativity across a period and down the group
c. Factors affecting Ionization energy
d. Isoelectronic species and variation of atomic radii
C. Details of Pedagogical Strategies/Process (Art integrated /Sports integrated/ Story telling based/Toy based /Any other pedagogy):
1. Art integrated activity: Periodic table
2. Story telling: History of classification of elements
D. Topic of the lesson for presentation by the students (once in a week by rearranging classroom setting suitable for group work):
1. Variation of periodic properties across a period and down the group
E. The 21st Century Skills to be developed:
Critical thinking: Predicting the group and period to which the electron belongs to
Collaboration: Importance of classification of elements to understand their properties
Problem solving: to predict the ionization energy of elements depending on the factors
Technology literacy: Use of Olabs and songs of periodic table
F. Activities/Experiments/Hands-on-learning/Projects:
1. Specify the position of elements in periodic table
2. Table for depicting variation of periodic properties
G. Interdisciplinary linkage and infusion of Life Skills, Values, Gender Sensitivity and Environmental Awareness:
1. Elements present around us and co-relating their properties with periodic table
H. Resources (including ICT):
NCERT textbook, Modern ABC, Physical Chemistry by Atkins, Olabs, Youtube videos related to the experiments, Google forms for assessment, Diksha portal for revision and references, PPT shared by Navodaya teachers, Material provided by Navodayasamiti, Questions available in websites such as Byjus, cbseguide, askiitiansetc to expose the students to different types of questions. Pictures or images available on the internet.
I. Assessment items for measuring the attainment of learning outcomes in the class and as home assignments (Teacher should plan Assessment Items and link the same with learning outcomes mentioned in Para A):
|
Items |
No of Items |
Sl.No. of LO |
Items |
No of Items |
Sl.No. of LO |
|
Oral Quiz |
- |
|
Presentation |
3 |
1-7 |
|
Portfolio |
- |
|
Puzzle |
- |
|
|
Multiple choice Questions |
10 |
1 TO
7 |
Group Project |
- |
|
|
Very Short Answer Questions |
9 |
1 TO
7 |
Individual Project |
1 |
|
|
Short Answer Questions |
10 |
1 TO
7 |
Any other Item |
|
|
|
Long Answer Questions |
7 |
1 TO
7 |
|
|
|
|
Compentancy based Questions |
6 |
1 TO
7 |
|
|
|
Not applicable
Date: 17/7/2026 (Signature of the teacher)
Remarks
of the Principal/Vice Principal:
(Signature of the Principal/Vice Principal)
Assessment
questions
Multiple
Choice questions
1. The vertical columns in the periodic table are termed as ———-.
(a) periods (b) groups (c) series (d) none of these
Ans: (b) Ans: The vertical
columns in the periodic table are termed as groups.
2. The element with atomic number 26 will be found in group :
(a) 2 (b) 8 (c) 6 (d) 10
Ans: (b) Ans: The valence
shell configuration of Z= 26 is [Ar] 3d6 4s2 .
That for the group of element is 6 + 2 =8
3. The elements with atomic numbers 9, 17, 35, 53, 85 are all ———-
(a) halogens (b)
noble gases (c) alkali earth
metals (d) transition metals
Ans: (a) Solution: The
elements of atomic numbers 9, 17, 35, 53, 85 are respectively F, Cl, Br, I,
Atetc are called halogen.
4. Which of the following electronic configurations of an atom has the
lowest ionisation enthalpy?
(a) 1s2 2s2 2p3 (b) 1s2 2s2 2p6 3s1 (c) 1s2 2s2 2p6 (d) 1s2 2s2 2p5
Ans: (b) Solution:
Ionisation enthalpy is the amount of energy required when an electron is
removed from the outermost orbit of an isolated gaseous atom. Electronic
configuration of 1s2 2s2 2p6 3s1 has
lowest ionisation enthalpy.
5. The Ionic radius of cation is always———
(a) Less than the atomic radius (b)
more than the atomic radius
(c) Equal to atomic radius (d)
Cannot be predicted
Ans: (a) Solution: The
Ionic radius of cation is always less than the atomic radius. Cation is formed
by the loss of electrons. So that the effective nuclear charge increases as a
result ionic radius decreases.
6. Which of the following elements has the maximum negative electron gain
enthalpy?
(a) Oxygen (b)
Chlorine (c) Fluorine (d) Nitrogen
Ans: (b) Solution:
Chlorine has the maximum negative electron gain enthalpy.
7. The most electronegative element in the periodic table is———
(a) Nitrogen (b)
Oxygen (c) Chlorine (d) Fluorine
Ans: (d) Solution: The
most electronegative element in the periodic table is Fluorine
8. The elements of group 16 are called———-
(a) noble gases (b)
chalcogens (c) halogens (d) alkali metals
Ans: (b) Solution: The
elements of group 16 are called chalcogens.
9. In a group of the periodic table the Ionization enthalpies of the
elements decreases from top to bottom because of ———-
(a) increase in densities (b)
decrease in chemical reactivities
(c) increase in atomic sizes (d)
decrease in electronegativities
Ans: (c) Solution: In a
group of the periodic table, the Ionization enthalpies of the elements decrease
from top to bottom because of increase in atomic sizes.
10. The smallest ion among the following is
(a) Na+ (b)
Al3+ (c)
Mg2+ (d)
Si4+
Ans: (d)Solution: The isoelectronic ions Si4+, has the
smallest size due to maximum nuclear charge.
Very
Short and Short Answer Type Questions
1. Explain why the electron gain enthalpy of fluorine is less
negative than that of chlorine.
Answer.
This is due to
the small size of the fluorine atom. As a result of the strong interelectronic
repulsions in fluorine’s relatively small 2p orbitals, the incoming electron
does not experience much attraction.
2. All transition elements are d-block elements, but all d-block
elements are not transitionelements. Explain.
Answer.
All transition
metals are d-block elements, but not all d-block elements are transition
elements because all d-block elements that do not have completely filled d-
orbitals are not counted as transition elements, making such elements
exceptional. Zn, Cd, and Hg are a few examples.
3. Identify the group and valency of the element having atomic number
119. Also, predict theoutermost electronic configuration and write the general
formula of its oxide.
Answer.
For element
having atomic number 119:
Group – 1,
Valency – 1, Outermost electronic configuration – 8s1 and the
general formula of the oxide will be M2O.
4. Ionisation enthalpies of elements of second period are given
below:
Ionisation
enthalpy/ kcal mol-1: 520, 899, 801, 1086, 1402, 1314, 1681, 2080.
Match the correct
enthalpy with the elements and complete the graph given in Fig. 3.1.
Also, write
symbols of elements with their atomic number.
Answer.
The order of
ionisation enthalpy of second period elements is: Li < B < Be < C <
O < N < F < Ne
The correct
enthalpy with the elements and the complete graph is given as follows-
5. Among the elements B, Al, C and Si,
(i) Which
element has the highest first ionisation enthalpy?
(ii) Which
element has the most metallic character?
Justify your
answer in each case.
Answer.
Among the
elements, B, Al, C and Si
(i) The element
that has the highest first ionisation enthalpy is C.
(ii) The element
that has the most metallic character is Al.
6. Write four characteristic properties of p-block elements.
Answer.
The four most important
properties of p-block elements are as follows:
(a) Both metals
and nonmetals are present in p-block elements, but the number of nonmetals is
much greater than that of metals. Furthermore, within a group, the metallic
character increases from top to bottom, while the non-metallic character
increases from left to right along a period in this block.
(b) Their
ionisation enthalpies are higher than those of s-block elements.
(c) They mostly
combine to form covalent compounds.
(d) Some of them
have compounds with multiple (variable) oxidation states. In a period, their
oxidising character increases from left to right, while their reducing character
increases from top to bottom.
7. Choose the correct order of atomic radii of fluorine and neon (in
pm) out of the optionsgiven below and justify your answer.
(i) 72, 160 (ii) 160, 160 (iii) 72, 72(iv) 160, 72
Answer.
The atomic
radius of F is usually expressed in terms of covalent radius, whereas the
atomic radius of neon is usually expressed in terms of van der Waals radius. An
element’s van der Waals radius is always greater than its covalent radius. As a
result, the atomic radius of F is less than the atomic radius of Ne (F = 72 pm,
Ne = 160 pm).
8. Illustrate by taking examples of transition elements and
non-transition elements thatoxidation states of elements are largely based on
electronic configuration.
Answer.
An element’s
oxidation state is determined by its electronic configuration.
S-block
elements:
·
Group 1 elements- General electronic configuration of
the valence shell ns1. Oxidation states =+1.
·
Group 2 elements- General electronic configuration of
the valence shell is ns2. Oxidation state =+2
P-block
elements.
·
Group 13 elements: General electronic configuration of
a valence shell is ns2, np1. The oxidation states are +3
and +1.
·
Group 14 elements: General electronic configuration of
the valence shell is ns2, np2. Oxidation states are +4
and +2.
·
Group 15 elements: General valence shell electronic
configuration ns2, np3. Oxidation states are -3,+3, and
+5. Nitrogen exhibits +1,+2,+4 oxidation states as well.
·
Group 16 elements General electronic configuration of
the valence shell ns2, np4. Oxidation states are -2,+2,+4
and +6.
·
Group 17: General electronic configuration of the
valence shell is ns2, np5. Oxidation state is -1. Cl, Br,
and I also have +1,+3,+5, and +7 oxidation states.
·
Group 18: General electronic configuration of the
valence shell is ns2np6 . In bonding, the oxidation
state = zero.
·
A transition metal’s various oxidation states are
caused by the presence of (n-1)d and outer ns electrons. (n-1)d1-10,
ns1-2 is a general electronic configuration. The most common
oxidation states for these elements are +2 and +3.
9. Nitrogen has positive electron gain enthalpy whereas oxygen has
negative. However,oxygen has lower ionisation enthalpy than nitrogen. Explain.
Answer.
The outermost
electronic configuration of nitrogen is 2s2 2px1,
2py1, 2pz1 whereas that of
oxygen is 2s2 2px2, 2py1,
2pz1.
Since oxygen
acquires a stable configuration, i.e., 2p3, by removing one electron
from the 2p-orbital, it has a lower ionisation enthalpy than nitrogen. In the
case of nitrogen, however, due to its stable configuration, it is difficult to
remove one of the three 2p-electrons.
10. First member of each group of representative elements (i.e., s and
p-block elements)shows anomalous behaviour. Illustrate with two examples.
Answer.
The first member
of each group of representative elements (i.e., the s- and p- block elements)
exhibits anomalous behaviour due to:
(i) small size
(ii) higher
ionisation enthalpy
(iii) higher
electronegativity
(iv) the absence
of d- orbitals.
For example, in
s – block elements, lithium behaves differently than the other alkali metals.
(a) Lithium
compounds have a high covalent character. Alkali metal compounds are
predominantly ionic.
(b) Lithium
nitride is formed when lithium reacts with nitrogen, whereas other alkali
metals do not form nitrides.
In p- block
elements, the first member of each group has four orbitals in their valence
shell, one 2s orbital and three 2 p orbitals. As a result, these elements have
a maximum covalency of four, whereas other members of the same or different group
have a maximum covalency that is greater than four due to the availability of
vacant d – orbitals.
11. p-Block elements form acidic, basic and amphoteric oxides. Explain
each property bygiving two examples and also write the reactions of these
oxides with water.
Answer.
Due to their
various properties, p – block elements produce acidic, basic, and amphoteric
oxides:
·
The higher an element’s electronegativity, the more
acidic its oxide.
For example- Boron has an electronegativity of -2, carbon has an electronegativity
of 2.5, and nitrogen has an electronegativity of 3. As a result, the order of
acidic character of B, C, and N oxides is B2O3 <
CO2 < N2O3
·
If the ionisation enthalpy of an element is high, it
will form acidic oxide; if it is intermediate, it will form amphoteric oxide;
and if it is low, it will form basic oxides.
For example, the ionisation enthalpy of boron is 800 while that of carbon is
1086.5, implying that carbon oxide is more acidic than boron oxide.
·
The oxides of the first element in each group in the p
– block are more acidic than the oxides of other elements. As we move down the
group, the acidic character decreases, followed by elements that form
amphoteric oxides and then basic oxides.
For example- In the Boron family, B forms a weak acidic oxide, while Al, Ga,
and In form amphoteric oxides, and Tl forms a strong basic oxide.
Reactions of
some of the oxides with water:
·
Acidic Oxides:
B2O3 + 3H2O → 2H3BO3
·
Basic Oxides:
Ti2O + H2O → 2TiOH
·
Amphoteric Oxides are insoluble in water and thus reacts
with acid and base:
Al2O3 + 2NaOH → 2NaAlO2 + H2O
Al2O3 + 6HCl → 2AlCl3 + 3H2O
12. How would you explain the fact that first ionisation enthalpy of
sodium is lower than thatof magnesium but its second ionisation enthalpy is
higher than that of magnesium?
Answer.
The electronic
configurations of both the atoms are as follows:
Na: [Ne]3s1 Mg: [Ne]3s2
The ion formed
after removing one electron from the sodium atom takes on the configuration of
an inert gas, neon, whereas Mg retains one electron. As a result, the first
ionisation energy of Na is lower than that of Mg. High energy is required to
remove an electron from a noble gas configuration.
As a result, the
second ionisation enthalpy of calcium is greater than that of magnesium.
13. What do you understand by exothermic reaction and endothermic
reaction?Give one example of each type.
Answer.
Exothermic
reactions are those in which heat is produced. The formation of ammonia from
nitrogen and hydrogen, for example, is an exothermic reaction.
N2 +
3H2 → 2NH3, ΔH = – 46.11KJ/mol.
Endothermic
reactions are ones in which heat is absorbed. The dissociation of ammonia into
nitrogen and hydrogen, for example, is an endothermic reaction.
2NH3 →
N2 + 3H2, ΔH = +91.8KJ/mol.
14. Arrange the elements N, P, O and S in the order of-
(i) increasing
first ionisation enthalpy.
(i) increasing
non-metallic character.
Give reason for
the arrangement assigned.
Answer.
(i) Due to the
extra stable exactly half-filled 2p-orbitals, the ionisation enthalpy of
nitrogen (1s2,2s2,2p3) is greater than that of
oxygen (1s2,2s2,2p4). Likewise, the ionisation
enthalpy of phosphorous (1s2,2s2,2p6,3s2,3p3)
is greater than that of sulphur (s2,2s2,2p6,3s2,3p3)
Ionisation
enthalpy decreases with decreasing atomic size as one moves down the group.
As a result, the
increasing order of first ionisation enthalpy is S < P < O < N
(ii) Nonmetallic
character increases across a period (left to right), but decreases as one moves
down the group.
As a result, the
increasing non-metallic order is P < S < N < O.
Answer.
The ionisation
enthalpy of some elements deviates from the general trend, as shown in the
figure. The first ionisation enthalpy of B is lower than that of Be, whereas
the first ionisation enthalpy of nitrogen is higher than that of O.
15. Explain the following:
1. Electronegativity of elements increases on moving from left to right in
the periodic table.
2. Ionisation enthalpy decreases in a group from top to bottom?
Answer.
1. Due to the general decrease in atom size and increase in nuclear charge,
the electronegative of elements increases as one moves from left to right in
the periodic table.
2. The ionisation enthalpy of a group decreases from top to bottom due to
the increase in atomic size caused by the addition of a new shell.
16. How does the metallic and non-metallic character vary on moving
from left to right in aperiod?
Answer.
Moving from left
to right in a period, metallic character decreases and nonmetallic character
increases. This is due to the increase in effective nuclear charge which causes
an increase in ionisation enthalpy and electron gain enthalpy.
17. The radius of Na+ cation is less than that of Na atom.
Give reason.
Answer.
Since Na+ is
formed by losing one energy shell, its radius is smaller than that of Na atom.
Na – 1s2,
2s2, 2p6, 3s1
Na+ –
1s2, 2s2, 2p6
18. Among alkali metals which element do you expect to be least
electronegative and why?
Answer.
Due to the
general increase in size, electronegativity decreases from top to bottom in a
group. As a result, caesium is the least electronegative element.
Competency based questions
Matching
Type Questions
1. Match the correct atomic radius with the element.
|
Element |
Atomic
radius (pm) |
|
Be |
74 |
|
C |
88 |
|
O |
111 |
|
B |
77 |
|
N |
66 |
2. Match the correct ionisation enthalpies and electron gain
enthalpies of the following elements.
|
Elements |
ΔH1 |
ΔH2 |
Δeg H |
|
|
(i)
Most reactive non-metal |
A. |
419 |
3051 |
· 48 |
|
(ii)
Most reactive metal |
B. |
1681 |
3374 |
· 328 |
|
(iii)
Least reactive element |
C. |
738 |
1451 |
· 40 |
|
(iv)Metal
forming binary halide |
D. |
2372 |
5251 |
· 48 |
3. Electronic configuration of some elements is given in Column I and
their electron gain
enthalpies are
given in Column II, Match the electronic configuration with electron gain
enthalpy.
|
Column
I |
Column
II |
|
(i) 1s2 2s2 2p6 |
(A) –
53 |
|
(ii)
1s2 2s2 2p6 3s1 |
(B) –
328 |
|
(iii)
1s2 2s2 2p5 |
(C) –
141 |
|
(iv)
1s2 2s2 2p4 |
(D)
+48 |
Assertion
and Reason Type Questions
In the following
questions a statement of Assertion (A) followed by a statement of reason
(R) is given.
Choose the correct option out of the choices given below each question.
1. Assertion (A): Generally, ionisation enthalpy increases from left to
right in a period.
Reason (R): When
successive electrons are added to the orbitals in the same principal
quantum level,
the shielding effect of the inner core of electrons does not increase very much
to compensate
for the increased attraction of the electron to the nucleus.
(i) Assertion is
correct statement and reason is wrong statement.
(ii) Assertion
and reason both are correct statements and reason is correct explanation of
assertion.
(iii) Assertion
and reason both are wrong statements.
(iv) Assertion
is wrong statement and reason is correct statement.
Answer.
The correct
answer is (ii) Assertion and reason both are correct statements and reason is
correct explanation of assertion.
Ionisation
enthalpy increases over time as effective nuclear charge increases and atomic
size decreases.
2. Assertion (A): Boron has a smaller first ionisation enthalpy than
beryllium.
Reason (R): The
penetration of a 2s electron to the nucleus is more than the 2p electron hence
2p electron is more shielded by the inner core of electrons than the 2s
electrons.
(i) Assertion
and reason both are correct statements but reason is not correct explanation
for assertion.
(ii) Assertion
is correct statement but reason is wrong statement.
(iii) Assertion
and reason both are correct statements and reason is correct explanation for
assertion.
(iv) Assertion
and reason both are wrong statements.
Answer.
The correct
answer is (iii) Assertion and reason both are correct statements and reason is
correct explanation for assertion.
The electron
removed from the beryllium atom during ionisation is from the s-orbital,
whereas the electron removed from the boron atom is from the p-orbital, and the
penetration of the 2s electron to the nucleus is greater than that of the 2p
electron, so the 2p electron of boron is more shielded from the nucleus than
the 2s electron.
3. Assertion (A): Electron gain enthalpy becomes less negative as we
go down a group.
Reason (R): Size
of the atom increases on going down the group and the added electron
would be farther
from the nucleus.
(i) Assertion
and reason both are correct statements but reason is not correct explanation
for assertion.
(ii) Assertion
and reason both are correct statements and reason is correct explanation for
assertion.
(iii) Assertion
and reason both are wrong statements.
(iv) Assertion
is wrong statement but reason is correct statement.
Answer.
The correct
answer is (ii) Assertion and reason both are correct statements and reason is
correct explanation for assertion.
As an atom’s
size increases down the group, its electron gain enthalpy decreases. This is
due to the fact that the screening effect within a group increases as one moves
downward, and the added electron would be further away from the nucleus.
Long
Answer Type Questions
1. Discuss the factors affecting electron gain enthalpy and the trend in
its variation in the
periodic table.
Answer.
Factors
influencing electron gain enthalpy includes-
(i) Nuclear
charge: As the nuclear charge increases, the electron gain enthalpy becomes
more negative. If the nuclear charge is high, there is a greater attraction for
the incoming electron.
(ii) Atomic
size: As the atom’s size increases, so does the distance between the nucleus
and the incoming electron, resulting in less attraction. As a result, as the
size of the element’s atom increases, the electron gain enthalpy becomes less
negative.
(iii) Electronic
configuration: Elements with stable electronic configurations of half-filled
and completely filled valence subshells have a very low tendency to accept
additional electrons, resulting in less negative electron gain enthalpies.
Variations in
electron gain enthalpies in the periodic table
In general,
electron gain enthalpy becomes more negative from left to right in a period and
less negative from top to bottom in a group.
(a) Downward
variation within a group: Moving down a group increases the size and nuclear
charge. However, the effect of increasing atomic size is much more pronounced
than that of increasing nuclear charge, so the additional electron feels less
attraction by the large atom. As a result, the electron gain enthalpy
decreases. This is evident from the decrease in electron gain enthalpy when
transitioning from chlorine to bromine and then to iodine.
(b) Periodic
variation: As one moves across a period, the size of the atom decreases and the
nuclear charge increases. Because both of these factors increase the attraction
for the incoming electron, electron gain enthalpy becomes more negative in a
period from left to right. However, there are some anomalies in the overall
trend. These are primarily due to certain atoms’ stable electronic
configurations.
Important Trends
in Electron Gain Enthalpies
The electron
gain enthalpies of elements have some important characteristics. They are as
follows:
(i) The negative
electron gain enthalpies of halogens are the highest.
(ii) Noble gases
have positive electron gain enthalpy values, whereas Be, Mg, N, and P have
nearly zero.
(iii) Fluorine’s
electron gain enthalpy is unexpectedly less negative than chlorine’s.
2. Define ionisation enthalpy. Discuss the factors affecting
ionisation enthalpy of the elements
and its trends
in the periodic table.
Answer.
Ionisation
Enthalpy: The amount of energy required to remove an e from an isolated gaseous
atom in its gaseous state is defined as an element’s ionisation enthalpy.
The following
factors influence ionisation enthalpy:
1. Atom size: The larger the atomic size, the lower the value of ionisation
enthalpy. The outer e– are far away from the nucleus in large
atoms, so the force of attraction with which they are attracted by the nucleus
is less and thus they can be easily removed.
Ionization enthalpy ∝ 1/atomic size
2. Screening Effect: Because the screening effect reduces the force of
attraction towards the nucleus, the outer e can be easily removed.
Ionization enthalpy ∝ 1/Screening effect
3. Nuclear charge: The Ionisation enthalpy increases as nuclear charge
increases among atoms with the same number of energy shells because of the
force of attraction towards the nucleus increases.
Ionisation enthalpy ∝ nuclear charge
4. Half-filled and fully-filled orbitals: Because atoms with half-filled
and fully-filled orbitals are more stable, it takes more energy to remove an
electron from such atoms. In the case of such an atom, the ionisation enthalpy
is somewhat higher than expected. Ionisation enthalpy ∝ stable
electronic configuration
5. Orbital shape: The s-orbital of the same orbit is closer to the nucleus
than the p-orbital. As a result, removing an electron from a p-orbital is
easier than from an s-orbital. The shape for orbitals: s > p > d > f
Variation of
ionisation enthalpy in the periodic table
In general, as
atomic size increases, the ionisation energy decreases down the group. The
ionisation energy, on the other hand, increases across the period from left to
right, because of a decrease in atomic size from left to right.
3. Justify the given statement with suitable examples— “the
Properties of the elements are a
periodic
function of their atomic numbers”.
Answer.
This statement
means that when elements are arranged in increasing atomic number order, the
similarity of their properties appears at regular intervals.
The electronic
configuration of the elements, for example, repeats itself at regular
intervals, resulting in elements with the same number of valence electrons and
thus similar chemical properties.
·
The elements in the first group are strongly metallic
and electropositive, forming monovalent ions such as Na+, Li +,
K+, and so on. All alkali metals have similar properties due to
their similar outermost shell electronic configuration.
·
Similarly, all elements in the 17th group (halogens)
have the same outermost shell electronic configuration, i.e., ns2 np5,
and thus have similar properties.
4. Write down the outermost electronic configuration of alkali
metals. How will you justify
their placement
in group 1 of the periodic table?
Answer.
Alkali metals’
outermost electronic configuration is ns1.
All elements of
group IA (or I), i.e., alkali metals, have the same outer electronic
configuration, ns1, where n denotes the number of principal shells.
These electronic configurations are shown in the table below.
|
Symbol |
Atomic
Number |
Electronic
Configuration |
|
Li |
3 |
1s2 2s1 |
|
Na |
11 |
1s2 2s2 2p6 3s1 |
|
K |
19 |
1s2 2s2 2p6 3s2 3p6 4s1 |
|
Rb |
37 |
1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 5s1 |
|
Cs |
55 |
1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 4d10 5s2 5p6 6s1 |
|
Fr |
87 |
1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 4d10 4f14 5s2 5p6 5d10 6s2 6p6 7s1 |
As a result of
their similarity in electronic configuration and properties, all of these
elements are placed in group 1 of the periodic table.
5. Write the drawbacks in Mendeleev’s periodic table that led to its
modification.
Answer.
1. Hydrogen’s position: Hydrogen is assigned to group I. It does, however,
resemble elements from Group I (alkali metals) as well as elements from Group
VIIA (halogens). As a result, the position of hydrogen in the periodic table is
incorrect.
2. Anomalous pairs: The increasing order of atomic masses was not followed
in certain pairs of elements. Mendeleev arranged the elements in these cases
based on similarities in their properties rather than the increasing order of
their atomic masses. Argon (Ar, atomic mass 39.9), for example, is placed
before potassium (K, atomic mass 39.1). Likewise, cobalt (Co, atomic mass 58.9)
comes before nickel (Ni, atomic mass 58.6), and tellurium (Te, atomic mass
127.6) comes before iodine (I, atomic mass 126.9). These positions were not
justified.
3. Isotopes are atoms of the same element that have different atomic masses
but the same atomic number. As a result, according to Mendeleev’s
classification, these should be classified differently based on their atomic
masses. For example, hydrogen isotopes with atomic masses 1, 2, and 3 should be
placed in three different locations. Isotopes, on the other hand, do not have
their own spot in the periodic table.
4. Several gaps in the periodic table were left because he believed that
several elements were yet to be discovered, for example, gallium was not
discovered at the time.
5. Position of lanthanoids (or lanthanides) and actinoids (or actinides):
The fourteen elements that follow lanthanum (known as lanthanoids, atomic
numbers 58-71) and the fourteen elements that follow actinium (known as
actinoids, atomic numbers 58-71) are not included separately.
6. In what manner is the long form of periodic table better than
Mendeleev’s periodic table?
Explain with
examples.
Answer.
Due to the
following reasons, the long-form periodic table is considered more letter than
the Mendeleev’s table:
1. All elements in the long-form periodic table are arranged in increasing
order of atomic numbers, whereas the table is arranged in increasing order of
atomic masses.
2. The position of hydrogen in the long-form periodic table has been
justified, whereas there is no such justification in Mendeleev’s periodic
table.
3. The long-form periodic table considers the filling of electrons in
s,p,d, and subshells, whereas the table considers the atomic numbers of the
elements.
4. The periodic table is divided into four blocks: s, p, d, and f, whereas
the periodic table has no blocks.
5. Long-form periodic table groups are not further subdivided into
subgroups, whereas each group in Mendeleev’s periodic table has subgroups A and
B.
6. Long-form periodic tables are simple and easy to reproduce, whereas
Mendeleev’s periodic tables are quite difficult to reproduce.
7. Discuss and compare the trend in ionisation enthalpy of the
elements of group 1 withthose of group 17 elements.
Answer.
1. As the atom grows in size, the distance between the nucleus and the
valence electron grows, resulting in a decrease in the force of attraction
between them. As a result, the amount of energy required to remove an electron is
reduced.
2. Also, as we move down the group, the shielding effect increases, and
thus the force of attraction between the nucleus and the valence electrons
decreases even more.
3. As we move down a group, the combined effect of the two results in a
decrease in ionisation energy.
4. As a result, the group ionisation energy decreases as we move down in
both group 1 and group 17.
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