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Question-1

What is the basic theme of organization in the periodic table?

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 .

Question-2

Which important property did mendeleev use to classify the elements
in his periodic table and did he stick to that?

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.

Question-3

What is the basic difference in approach between the Mendeleev’s periodic law and the modern Periodic law?

Solution:
Mendeleev's Periodic Law states that
'The physical and chemical properties of elements are periodic functions of their atomic weights'.

Modern Periodic Law:

' The physical and chemical properties of elements are periodic functions of their atomic number'.

Question-4

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

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.

Question-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.

Question-6

Write the atomic number of the element present in the third period and seventeenth group of the periodic table?

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.

Question-7

Which element do you think would have been named by (i) Lawrence Berkeley laboratory? (ii) seaborg’s group?

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.

Question-8

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

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.

Question-9

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

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.

Question-10

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

Solution:
Variation of atomic size down a group: The atomic radius increases as we go from top to bottom in a group.

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.

Along a period: The atomic radius decreases as we go from left to right along a period.

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 nuclear charge increases and the atomic radius decreases.

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+

Solution:
Iso electronic species have similar electronic configuration.

Question-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.

Solution:
(a) All the species given above are isoelectronic and are resembling Ne (inert gas) configuration
(b) Al3+ < Mg2+ < Na+ < F- < O2- < N3-

Question-13

Explain why cation are similar and anions larger in radii than their parent atoms?

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:

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 electrons are less tightly bound by the nucleus which results increased size of the ion.

 

Question-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.

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
 
= 2.18 10-18 6.022 1023 = 13.12 105 J mol-1.

Question-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 Δi H than B
(ii) O has lower Δi H than N and F?

Solution:
Li < B < Be < C < O < N < F < Ne

(i) Be has higher Δi H than B
 
Electronic configuration of
Be is 1s2 2s2
B is 1s2 2s2 2p
1 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.


(ii) O has lower Δi H than N and F
Electronic configuration of
O - 1s2 2s2 2p
4 
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.

Question-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?

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.

(ii) After losing outermost shell electron, Na+ possesses stable octet
configuration,

Na     2,8,1

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.

Question-17

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

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.

Question-18

Which of the following pairs of elements would have a more negative electron gain enthalpy? 

(i) O or F

(ii) F or Cl


Solution:
(i) F has more negative electron gain enthalpy due to its smaller size and greater effective nuclear charge.

(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.

 

Question-19

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

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 successive electron gain enthalpies are positive. For example, let us study the addition of electrons to oxygen atom.

O(g) + e- O- (g) (Δ eg H)1 = - 141 kJ

O-(g) + e- O2- (g) (Δ eg H)2 = +780 kJ.

Question-20

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

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.
 

Question-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.

Question-22

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

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:

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 electrons are less tightly bound by the nucleus which results increased size of the ion.

 

Question-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.

Question-24

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

Solution:

Metals

Non-Metals

Differences in Physical Properties

1. Metals are malleable and ductile. That is, metals can be hammered into thin sheets and drawn into thin wires.

1. Non-metals are brittle. They are neither malleable nor ductile.

2.Metals are good conductors of heat and electricity.

2. Non-metals are bad conductors of heat and electricity (except graphite which is a good conductor of electricity).

3. Metals are lustrous (shiny) and can be polished.

3. Non-metals are non-lustrous (dull) and cannot be polished (except graphite and iodine which are lustrous non-metals).

4. Metals are solids at room temperature (except mercury which is a liquid metal).

4. Non-metals may be solid, liquid or gases at the room temperature.

5. Metals are strong and tough. They have high tensile strength.

5. Non-metals are not strong. They have low 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 chlorides) with chlorine. These electrovalent chlorides are electrolytes but non-volatile.

8. Non-metals form covalent chlorides with chlorine (which are non-electrolytes but volatile).

9. Metals usually do not combine with hydrogen. Only a few reactive metals combine with hydrogen to form electrovalent metal hydrides.

9. Non-metals react with hydrogen to form stable, covalent 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.

Question-25

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.

Solution:
(a) F – 1s2 2s2 2p5 has 5 electrons in the outermost shell.

(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.

Question-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.

Question-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.

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.

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.

Question-28

The first (ΔiH1) and the second (ΔiH2) ionization enthalpies (in kJ mol-1) and the egH) electron gain enthalpy (in kJ mol-1) of a few elements are given below:
 
Elements ΔH1 ΔH2 ΔegH
I 520 7300 -60
II 419 3051 -48
III 1681 3374 -328
IV 1008 1846 -295
V 2372 5251 +48
VI 738 1451 -40

Which of the above elements is likely to be :

(a) the least reactive element.

(b) the most reactive metal.

(c) the most reactive non-metal.

(d) the least reactive non-metal.

(e) the metal which can form a stable binary halide of the formula MX2(X=halogen).

(f) the metal which can form a predominantly stable covalent halide of the formula MX (X=halogen)?


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 ΔiH1.

(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.

Question-29

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

Solution:

Types of Element

Outer Electronic Configuration

s-block elements

ns1-2

p-block elements

ns2 np1-6

d-block elements

(n-1) d1-10 ns1-2

f-block elements

(n-2) f1-14 (n-1) d0-1, ns2

Question-30

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

Solution:
(a) Li2O

(b) Mg3N2  

(c) AlI3  

(d) SiO2  

(e) PF5  

(f) LuF3.

Question-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 7 (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 7 (n-1)d1ns2 for n=6 - Period - 6, Group - 3, Lanthanoid.




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