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1. What is the basic theme of organization in the periodic table?
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Solution:
At the beginning of 18th century, only a very few elements were known, it was quite easy to study
and remember their individual properties. Then later, large number of elements were discovered and
scientists ultimately needed a new method to facilitate the study of the properties of various
elements and their compounds. Thus, periodic table, the table giving the arrangement of all the
known elements according to their properties so that similar elements fall within the same vertical
column and dissimilar elements are separated .
2. Which important property did mendeleev use to classify the elements
in his periodic table and did he stick to that?
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Solution:
Atomic mass (weight) of the element was taken by Mendeleev as the fundamental property to
classify the elements in the periodic table.
This classification is based on the fact that the physical and chemical properties of the elements are
periodic functions of their atomic mass.
3. What is the basic difference in approach between the Mendeleev’s periodic law and
the modern Periodic law?
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Solution:
Mendeleev's Periodic Law states that
'The physical and chemical properties of elements are periodic functions of their atomic weights'.
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Modern Periodic Law:
' The physical and chemical properties of elements are periodic functions of their atomic number'.
4. On the basis of quantum numbers, justify that the sixth period of the periodic table
should have 32 elements?
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Solution:
The sixth period of the periodic table must have elements whose electronic configuration starts from
6s and continue filling 4f,5d, and 6p orbitals. As the electron enters 7s orbital such element with
come under 7th period.
The no. of electrons that can be accommodated in 6s,4f,5d and 6p orbitals are 2,14,10 and 6
respectively whose total is 32.
Hence the sixth period of the periodic table should have 32 elements.
5. In terms of period and group where would you locate the element with Z = 114?
•
Solution:
Z = 114
Name of the element: ununquadium; symbol = Uuq Group = 14 Period = 7.
6. Write the atomic number of the element present in the third period and seventeenth
group of the periodic table?
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Solution:
The atomic no of the element present in the period and seventeenth group of the periodic table is 17
and the element is chlorine.
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7. Which element do you think would have been named by (i) Lawrence Berkeley
laboratory? (ii) seaborg’s group?
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Solution:
(i) Berkelium (BK, Atomic number 97 is named by Lawrence Berkely lab.
(ii) Seaborgium (sg), Atomic number 106 is named by seaborg’s group.
8. Why do elements in the same group have similar physical and chemical properties?
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Solution:
Because elements of the same group have same outer electronic configuration. In other words , they
have same number of outermost electrons i.e. same valency. For example alkali metals present in
the I group have ns1 configuration.
9. What does atomic radius and ionic radius really mean to you?
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Solution:
Atomic radius is defined as the distance from the center of nucleus of the atom to the outermost
shell of electron.
Ionic radius is defined as the distance from the center of nucleus of an ion (cation or anion) to the
point in the ionic bond up to which it has influence over the electron cloud. Cationic radius <Atomic
radius < Anionic radius.
10. How do atomic radius vary in a period and in a group ? How do you explain the
variation?
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Solution:
Variation of atomic size down a group: The atomic radius increases as we go from top to bottom
in a group.
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Reason: As we go down the group electrons are added in a new shell. At the same time the nuclear
charge increases down the group. Though the effect of increase in nuclear charge is to reduce the
atomic radius, this effect is offset by the effect of new shell and as a result, the atomic radius
increases down the group.
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Along a period: The atomic radius decreases as we go from left to right along a period.
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Reason: As we go from left to right, the electrons are added to the same shell, while the nuclear
charge increases. As a result the effective nuclearcharge increases and the atomic radius decreases.
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+
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Solution:
Iso electronic species have similar electronic configuration.
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12. Consider the following species:
N3-, O2-, F-, Mg2+ and Al3+
(a) what is common in them?
(b) Arrange them in the order of increasing ionic radii.
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Solution:
(a) All the species given above are isoelectronic and are resembling Ne (inert gas) configuration
(b) Al3+ < Mg2+ < Na+ < F- < O2- < N3-
13. Explain why cation are similar and anions larger in radii than their
parent atoms?
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Solution:
Cationic radius is smaller than that of parent atom:
Reasons:
•€Cations are formed by the loss of one or more electrons from the parent atom.
•€As a result the no of electrons and decreased and no: of protons remains the same.
•€The no of positive charges becomes greater than the negative changes which results in greater
nuclear attraction (increase in effective nuclear charge per electron).
•€The greater the nuclear attraction (electrostatic force of attraction between the nucleus and the
outermost electrons), lesser will be the atomic radius.
For eg:
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The third shell does not have e-s in the case of Na+ ion.
Anionic radius is larger than that of parent atom:
Reason:
•€Anions are formed by the gain of one or more electrons by the gain of one or more electrons by
the gaseous atom.
•€Here nuclear charge remains the same whereas the number of electrons increases by one or more.
•€The nuclear attraction is decreased as the no: of protons are smaller than that of electrons.
•€i.e., the effective nuclear change per electron decreases in the case of anion and hence the
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electrons are less tightly bound by the nucleus which results increased size of the ion.
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14. Energy of an electron in the ground state of the hydrogen atom is –
2.18×10–18J. Calculate the ionization enthalpy of atomic hydrogen in terms of J
mol–1. Hint: Apply the idea of mole concept to derive the answer.
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Solution:
Energy of an electron in the ground state = –2.18
10-18 J.
Energy required for removal of the electron = 0 –(-2.18
10-18 J ) =2.18
10-18 J.
This is the energy required to remove the electron from one atom of hydrogen in ground state.
The energy required to remove electron from one mole (6.022
1023atoms) will be
-18
23
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-1
= 2.18
10
6.022
10 = 13.12
10 J mol .
15. 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 ∆i H than B
(ii) O has lower ∆i H than N and F?
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Solution:
Li < B < Be < C < O < N < F < Ne
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(i) Be has higher ∆i H than B
Electronic configuration of
Be is 1s2 2s2
B is 1s2 2s2 2p1 Be has a completely filled configuration whereas B is ready to lose
one e- to get a completely filled configuration . We also have that the completely filled configuration
are more stable than partially filled configurations Hence B has lower Ionisation enthalpy than Be.
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(ii) O has lower ∆i H than N and F
Electronic configuration of
O - 1s2 2s2 2p4
N - 1s2 2s2 2p3
F - 1s2 2s2 2p5
N has half filled configuration half filled configuration is more stable than partially filled configuration.
F needs only one electron more to get inert gas stable configuration instead
more energy is required to knock out an electron from its outermost shell.
Whereas O is having a partially filled configuration. Hence O needs comparatively lower ionization
energy than N and F.
16. How would you explain the fact that the first ionization enthalpy of
sodium is lower than that of magnesium but its second ionization enthalpy is
higher than that of magnesium?
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Solution:
(i) The 1st ionization enthalpy of Mg is higher than that of Na because the atomic
radii of Mg is smaller and hence its effective nuclear charge is more when
compared to Na.
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(ii) After losing outermost shell electron, Na+ possesses stable octet
configuration,
Na
2,8,1
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Na+
2,8
from which it is very very difficult to remove 2nd electron. Hence 2nd I.E is far
higher than that of Mg.
17. What are the various factors due to which the ionization enthalpy of the
main group elements tends to decrease down a group?
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Solution:
Reasons for the decrease in ionization enthalpies down any group:
(i) There is an increase in the number of the main energy shells (n) in moving from one element to
the other.
(ii) There is also an increase in the magnitude of the screening effect due to the gradual increase in
the number of inner electrons. The valence electrons are well shielded from nuclear attraction.
18. Which of the following pairs of elements would have a more negative
electron gain enthalpy?
(i) O or F
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(ii) F or Cl
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Solution:
(i) F has more negative electron gain enthalpy due to its smaller size and greater effective nuclear
charge.
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(ii) Chlorine (Cl) has more negative electron gain enthalpy than fluorine (F). F has less negative
electron gain enthalpy because in it the added electron goes to the smaller energy level (n=2) and
hence suffers significant repulsion from the electrons already present in this shell. Also fluorine
possesses high charge density.
19. Would you expect the second electron gain enthalpy of O as positive,
more negative or less negative than the first? Justify you answer?
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Solution:
When the fist electron is added to the gaseous atom, it forms a uni-negative ion and the enthalpy
change during the process is called first electron gain enthalpy. Now, if an electron is added to the
uni-negative ion, it experiences a repulsive force from the anion. As a result, the energy has to be
supplied to overcome the repulsive force. Thus, in order to add the second electron, the energy is
required rather than released. Therefore, the value of second electron gain enthalpy is positive.
Similarly, addition to third, fourth electrons, etc., also requires energy. Hence, the values of
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successive electron gain enthalpies are positive. For example, let us study the addition of electrons
to oxygen atom.
O(g) + e- → O- (g) (∆
eg
O-(g) + e- → O2- (g) (∆
H)1 = - 141 kJ
eg
H)2 = +780 kJ.
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20. What is the basic difference between the terms electron gain enthalpy
and electronegativity?
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Solution:
Electronegativity of an element may be defined as the tendency of its atom to attract the shared pair
of electrons towards itself in a covalent bond. whereas electrongain enthalpy is the energy
released when electrons are added to a neutral gaseous atom to form a gaseous anion.
21. How would you react to the statement that the electro negativity of N on Pauling
scale is 3.0 in all the nitrogen compounds?
Solution:
Electro negativity of an element is not constant and varies depending upon the element to which it is
bound.
22. Describe the theory associated with the radius of an atom as it
(a) gains an electron
(b) loses an electron
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Solution:
Cationic radius is smaller than that of parent atom:
Reasons:
•€Cations are formed by the loss of one or more electrons from the parent atom.
•€As a result the no of electrons and decreased and no: of protons remains the same.
•€The no of positive charges becomes greater than the negative changes which results in greater
nuclear attraction (increase in effective nuclear charge per electron).
•€The greater the nuclear attraction (electrostatic force of attraction between the nucleus and the
outermost electrons), lesser will be the atomic radius.
For eg:
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The third shell does not have e-s in the case of Na+ ion.
Anionic radius is larger than that of parent atom:
Reason:
•€Anions are formed by the gain of one or more electrons by the gain of one or more electrons by
the gaseous atom.
•€Here nuclear charge remains the same whereas the number of electrons increases by one or more.
•€The nuclear attraction is decreased as the no: of protons are smaller than that of electrons.
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•€i.e., the effective nuclear change per electron decreases in the case of anion and hence the
electrons are less tightly bound by the nucleus which results increased size of the ion.
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23. Would you expect the first ionization enthalpies for two isotopes of the same
element to be the same or different? Justify your answer.
Solution:
The ionization enthalpies of the two isotopes of an element are expected to be same because the
isotopes have same electronic configuration and same nuclear charge.
24. What are the major differences between metals and non-metals?
Solution:
Metals
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Non-Metals
Differences in Physical Properties
1. Metals are malleable and ductile. That is,
1. Non-metals are brittle. They are neither
metals can be hammered into thin sheets and
malleable nor ductile.
drawn into thin wires.
2.Metals are good conductors of heat and
2. Non-metals are bad conductors of heat and
electricity.
electricity (except graphite which is a good
conductor of electricity).
3. Metals are lustrous (shiny) and can be
3. Non-metals are non-lustrous (dull) and cannot
be polished (except graphite and iodine which are
polished.
lustrous non-metals).
4. Metals are solids at room temperature (except 4. Non-metals may be solid, liquid or gases at the
mercury which is a liquid metal).
room temperature.
5. Metals are strong and tough. They have high
5. Non-metals are not strong. They have low
tensile strength.
tensile strength.
Differences in Chemical Properties
6. Metals form basic oxides.
6. Non-metals form acidic oxides or neutral
oxides.
7. Metals displace hydrogen from dilute acids.
7. Non-metals do not react with dilute acids and
hence do not displace hydrogen from dilute acids.
8. Metals form electrovalent chlorides (ionic
8. Non-metals form covalent chlorides with
chlorides) with chlorine. These electrovalent
chlorine (which are non-electrolytes but volatile).
chlorides are electrolytes but non-volatile.
9. Metals usually do not combine with hydrogen. 9. Non-metals react with hydrogen to form stable,
Only a few reactive metals combine with
covalent hydrides.
hydrogen to form electrovalent metal hydrides.
10. Metals are reducing agents.
10. Non-metals are oxidizing agents (except
carbon which is a reducing agent).
Examples: Cu, Fe, Au, Ag, Al etc.
Examples: P, Si, C, O, N, S etc.
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25. Use the periodic table to answer the following questions.
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(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.
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Solution:
(a) F – 1s2 2s2 2p5 has 5 electrons in the outermost shell.
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(b) 12Mg - 1s2 2s2 2p63s2. The 2 electrons present in the 3s subshell of Mg atom can be lost by it to
get inert gas (Ne) configuration and to form Mg2+ ion.
(c) Oxygen - 1s2 2s2 2p4 needs 2 electrons to get stable configuration and hence it tend to gain 2
electrons.
26. 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
Solution:
(a) atomic number.
27. The increasing order of reactivity among group 1 elements is Li < Na < K < Rb <Cs
whereas that among group 17 elements is F > CI > Br > I. Explain.
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Solution:
The radioactivity of a group of elements depends upon its valency. Alkali metals have one electron in
their outermost shell (ie. ns1 configuration). They tend to give one outermost electron to form a
monopositive ion. Atomic radius is one of the important factor which affects ionization. Atomic radius
and Ionization enthalpy are inversely proportional each other. Hence Cs is having greatest atomic
radius in the I group, it has high reactivity and the order of reactivity can be written as Li < Na < K
< Rb < Cs Li being smaller atom has greater nuclear attraction and hence low reactivity.Among
group 17, F has high reactivity than others and the order of reactivity is F > Cl > Br > I.
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28. 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
I
II
III
IV
V
VI
∆H1
520
419
1681
1008
2372
738
∆H2
7300
3051
3374
1846
5251
1451
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In this group halogens, tend to gain one electron to attain the stable configuration since they have
ns2 np5 configuration. If the atomic radius is less, nuclear attraction will be more. Hence its easy for
the atom to gain an electron. In group 17, the atomic radius increases down the group. The tendency
of accepting electrons also decreases down the group. Hence the order of reactivity also decreases.
∆egH
-60
-48
-328
-295
+48
-40
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Which of the above elements is likely to be :
(a) the least reactive element.
(b) the most reactive metal.
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(c) the most reactive non-metal.
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(d) the least reactive non-metal.
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(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)?
Solution:
(i) The element V having very high ionization enthalpies and positive electron gain enthalpy would be
least reactive.
(ii) The element II would be most reactive metal as it has very low value of ∆iH1.
(iii) The element III would be most reactive non-metal as it has very high negative value of electron
gain enthalpy.
(iv) The element V.
(v) VI, because for this element the first two ionization enthalpies have low values.
(vi) The element II, because it has very low value of first ionization enthalpy.
29. Write the general outer electronic configuration of s-, p-, d- and f- block elements.
Outer Electronic Configuration
ns1-2
ns2 np1-6
(n-1) d1-10 ns1-2
(n-2) f1-14 (n-1) d0-1, ns2
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Types of Element
s-block elements
p-block elements
d-block elements
f-block elements
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Solution:
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30. Predict the formulas of the stable binary compounds that would be formed by the
combination of the following pairs of elements.
(b) Magnesium and nitrogen
(c) Aluminium and iodine
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(a) Lithium and oxygen
(d) Silicon and oxygen
(e) Phosphorus and fluorine
(f) Element 71 and fluorine
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Solution:
(a) Li2O
(b) Mg3N2
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(c) AlI3
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(d) SiO2
(e) PF5
(f) LuF3.
31. Assign the position of the element having outer electronic configuration
(i) ns2np4 for n = 3
(ii) (n-1)d2ns2 for n=4, and
(iii) (n-2) f
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(n-1)d1ns2 for n=6, in the periodic table.
Solution:
(i) ns2np4 for n = 3 - Period - 3, Group - 16.
(ii) (n-1)d2ns2 for n=4 - Period - 4, Group - 4.
(iii) (n-2) f
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(n-1)d1ns2 for n=6 - Period - 6, Group - 3, Lanthanoid.
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