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Free Flip-Book Chemistry Class 11th by Study Innovations. 534 Pages

Free Flip-Book Chemistry Class 11th by Study Innovations. 534 Pages

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It is observed that magnitude of effective nuclear charge increases in a period when we move from
left to right.

In a subgroup of normal elements the magnitude of effective nuclear charge remains almost the
same.

Atomic, Ionic and Van der Waals' Radii.

The radius of an atom is the distance between the centre of its nucleus and electrons in the last
orbit. However, according to quantum mechanics, there is no certainty about the exact position of electrons at any
time. Theoretically, an electron, at one time, may be very close to the nucleus while at other time it may be far away
from the nucleus. In spite of these limitations we need to have some operational definition of the term atomic
radius. There are three operational concepts of atomic radius.

(1) If the bonding is covalent, the radius is called a covalent radius.

(2) If the two atoms are not bonded by a chemical bond (as in noble gases) the radius is called van der Waal's radius.

(3) If the bonding is ionic, the radius is called ionic radius.

(1) Covalent radius : It is half of the distance between the nuclei of two like atoms bonded
together by a single bond. Thus covalent radius of carbon in a compound having C – C single bond can be
determined by dividing the bond length by 2, i.e.,

rc = C−C ∴ C − C = 2rc or rc + rc
2

where, rc is the single bond covalent radii (SBCR) of carbon. However, if atoms forming the covalent bond
are different i.e., one is more electronegative than the other then the atomic radius is determined by the relation,
A − B = rA + rB − 0.009(χ A − χ B ) , where χ A and χ B are electronegativities of the atoms A and B respectively. This
relation was given by Stevenson in 1941.

In a given period, atomic radius generally decreases from left to right and thus in any period, alkali
metal is the largest and halogen is the smallest atom. For example, in second period elements the covalent radii
decrease from Li to F.

3Li 4Be 5B 6C 7N 8O 9F 10Ne
1.23 0.89 0.80 0.77 0.74 0.74 0.72 1.6

The decrease in size along a period is due to the effect of successive increasing nuclear charge without
addition of a new shell, i.e., in each element of a given period a new electron is added in the same principal
quantum number. For example, in the second period the nuclear charge increases from + 3 in Li to + 9 in F. The
increased nuclear charge attracts the electrons more strongly to the nucleus and thus decreases the size of the atom.
In case of noble gases, the atomic radii are only the van der Waal's radii which are naturally higher than the
covalent radii of other elements.

In a given group, Atomic radius generally increases as one moves from top to bottom, e.g., in group
1 atomic size increases steadily from lithium to cesium, i.e. rCs > rRb > rK > rNa > rLi


The increase in size on descending a group is due to addition of extra shell (i.e., the number of principal
quantum number increases) which outweighs the effect of increased nuclear charge. Remember that He and Fr are
smallest and largest atom respectively.

(2) Ionic radius : It is the effective distance from the nucleus of an ion upto which it has its
influence on its electron cloud.

A cation (formed by the removal of one or more electrons from an atom) is always much smaller than the
corresponding atom. Further, more the number of electrons removed smaller will be the size of the resulting positive
ion. For example, rFe > rFe2+ > rFe3+ . This is due to following two factors

1.26 0.76 0.64

(i) A cation formed by the loss of electrons may result in the complete disappearance of the outer shell and
since the remaining inner shells do not extend so far in space, the cation is much smaller than the metal atom.
For example, Sodium atom (Na) → Sodium ion (Na+)

(2, 8, 1) (2, 8)

(ii) Whenever a cation is formed, the ratio of nuclear charge to the number of electrons (Z/e ratio) is increased
with the result the effective nuclear charge is increased and the electrons are pulled towards the nucleus. Consequently,
the cation becomes smaller. An anion (formed by gain of one or more electrons to the atom) is always larger than
the corresponding atom. For example, Atomic radius of I = 1.23 ; Ionic radii of I– = 2.16

This is again due to following two factors : (a) Since in the formation of an anion, one or more electrons are
added, the electron cloud expands and the ionic size increases. (b) In the formation of anion, the effective nuclear
charge decreases with the result the electrons get away from the nucleus and thus the anion becomes larger than the
corresponding atom.

In any particular group, the ions of elements increase in size on moving from top to bottom.

In case of isoelectronic ions (ions having same number of electrons but different nuclear charge); the greater
the nuclear charge, the greater is the attraction for electrons and smaller is ionic radius. Hence size of such ions
decreases from top to bottom as shown in the given table,

Variation of Radii of Iso–electronic Ions (Atom)

Ion / Atom At. number (Nuclear charge) No. of electrons in the ion Z/e ratio Ionic (atomic) radii

C4– 6 10 0.6 2.60
N3– 7 10 0.7 1.71
O2– 8 10 0.8 1.40
F– 9 10 0.9 1.36
Ne 10 10 1.0 1.12
Na+ 11 10 1.1 0.95
Mg2+ 12 10 1.2 0.65
Al3+ 13 10 1.3 0.50
Si4+ 14 10 1.4 0.41
P5+ 15 10 1.5 0.34
S6+ 16 10 1.6 0.29
Cl7+ 17 10 1.7 0.26


(3) Van der Waal's radius : It is one–half of the distance between the nuclei of two adjacent
atoms belonging to two neighbouring molecules of an element in the solid state.

The covalent radius is always smaller than the corresponding van der Waal's radius. This is because of the fact
that in the formation of a chemical bond, the two atoms have to come closer to each other. This also explains why
covalent bonds are much stronger than the van der Waal's forces. It is important to note that since the noble gases
ordinarily do not form any covalent bond, in crystals of noble gases, no chemical forces are operating between the
atoms. Hence the van der Waal's forces are the only attractive forces in these cases. In other words, the van der
Waal's radii constitute the atomic radii of noble gases and since van der Waal's radii are larger than covalent
radii, atomic radii of noble gases are largest in their respective periods (anomaly).

Ionisation Potential or Ionisation Energy .

The electrons in an atom are attracted by the nucleus. When an electron is to be removed then work is done
against this nuclear attraction. In other words energy is required to remove an electron from an atom. To
understand the details of chemical behaviour of an element we must have an indication of the energy with which an
atom binds its electrons. This is obtained by the measurement of ionisation potential or ionisation energy. It may be
defined as the energy required to remove an electron from the outermost orbit of an isolated gaseous
atom in its ground state. It is expressed in electron volts (eV) or kilo calories per gram atom. In an atom, the
energy required to remove first electron from a gaseous atom is called first ionisation energy. The energy required to
remove one electron from a unipositive ion to form a bipositive ion is called second ionisation energy. Second
ionisation energy is higher than the first. The reason is that in unipositive ion left after the removal of one electron
from the atom, the electrons are more firmly bound to the nucleus than in the atom. Hence more energy is needed
to remove the second electron.

A → A+ + e −1 (First I.E.)

A + → A +2 + e −1 (Second I.E.)

A+2 → A+3 + e −1 (Third I.E.)

Similarly, third ionisation energy is even more than second ionisation energy.

(1) Variation of ionisation energy in periodic table

(i) Ionisation energy decreases in a group as the atomic number increases. It is based on the fact
that as we move down a group, the size of atom increases, and the outer electrons become farther away from the
nucleus thus reducing the force of attraction and hence ionisation energy decreases.

Li Na K Rb Cs

5.4 eV 1. eV 4.3 eV 4.2 eV 3.9 eV

(ii) Ionisation energy increases along a period with increase in atomic number. This is due to the
size of atom since it decreases along a period and outer electrons are most strongly attracted by the nucleus and
hence more energy is required to remove the electron.

Li Be B CN O F Ne

5.4 eV 9.3 eV 8.3 eV 11.3 eV 14.6 eV 13.6 eV 17.0 eV 21.6 eV


(iii) The ionisation energies of inert gases are greater than that of their immediate neighbour. It is due to their
complete octet ns2p6 configuration which is highly stable. Therefore, it is very difficult to remove an electron from

the outermost orbit of an inert gas.

(2) Factors affecting the value of ionisation energy

(i) Size of atom : With an increase in atomic size, the ionisation potential is reduced, since the distance of
the outermost electron from the nucleus increases and hence the force of attraction decreases.

(ii) The charge on the nucleus : With an increase in the nuclear charge, there is an increase in force of
attraction of nucleus for electrons making the removal of the electrons more difficult. Thus an increase in nuclear
charge increases the ionisation potential.

(iii) The shielding or screening effect of inner shells : The valence electrons in a multi–electron atom
are pulled by the nucleus but are repelled by the electrons of the inner shells. The valence electrons, therefore, do
not experience the total pull of the nucleus. Instead the total pull of the nucleus is reduced by the electrons in inner
shells. This effect of reducing the force of attraction of nucleus by the inner shells is called screening effect. This
effect is exhibited maximum by s2p6 (the most stable) shell. Therefore, the ionisation energy of K is much less than
Cu, however, both have one electron in their fourth shell.

K 19 = 1s 2 , 2s 2 p6 , 3s 2 p6 , 4s1 ; Cu 29 = 1s 2 , 2s 2 p6 , 3s 2 p6d10 , 4s1

The ionisation energy of K is 4.33 eV while that of Cu is 7.72 eV. This is due to a large screening effect of
s 2 p6, penultimate orbit in K while s 2 p6d10 , penultimate orbit in Cu which exhibit little screening effect.

(iv) Type of electrons involved : Ionisation energy also depends upon the type, i.e., s, p, d or f, electrons
which are to be removed, s –electrons are closer to the nucleus and are more tightly held as compared to p, d or f
electrons. Hence, ionisation energy decreases in the order of s > p > d > f orbitals.

(v) Completely filled or half filled sub-shells : According to Hund's rule, completely filled or half filled
orbitals are more stable. Therefore, it is comparatively difficult to remove the electrons from these shells. The
ionisation energy of Be (9.3 eV) is more than B (8.3 eV) because Be has 2s2 configuration of the outermost orbit
which is fully filled. Similarly, nitrogen (14.6 eV) has more ionisation energy than oxygen (13.6 eV) because
nitrogen has outermost shell configuration as 2s2p3 in which p shell is half filled and is more stable. Similarly,
ionisation energy of Mg is more than Al and that of P is more than S.

(3) Relative ionisation energies IE1 and IE2 of the 3rd period elements
IE1 and IE2 of the 2nd period elements IE1 : Na < Al < Mg < Si < S < P < Cl < Ar
IE1 : Li < B < Be < C < O < N < F < Ne IE2 : Mg < Si < Al < P < S < Cl < Ar < Na
IE2 : Be < C < B < N < F < O < Ne < Li
IE1 of elements with very high values
Cl < H < O < Kr < N < Ar < F < Ne < He

(4) Importance of ionisation energy
(i) Lower is the ionisation potential of an element, more would be its reducing power and also reactivity.

(ii) It gives rough estimate about the basic character of the elements.

(iii) The relative values of ionisation potential and electron affinity of two elements are related to the nature of
bond formed during their combination.


(iv) The ionisation potentials provide an indication about the number of valence electron(s) in an atom; the
abnormally high value indicates that the electron removed is other than the valence electron.

For example, IE1, IE2 and IE3 values are 5.39, 75.62 and 122.42 eV. Since the values shows sudden jump, it
indicates that the number of valence electron in its atom is one. Similarly, values of IE1, IE2 and IE3 as 9.32, 18.21
and 153.85 eV indicate that the number of valence electrons in its atom is two.

Electron Affinity .

Those atoms whose nuclear forces are not completely screened by electronic shells, offer attraction for
electrons. Such atoms capture electrons if these are available with in their effective fields to neutralise the
electrostatic forces of the nucleus. Energy is always liberated whenever there is a force of attraction offered by an
atom or ion, and this energy is called electron affinity. This may be defined as, ''the energy released when an
extra electron is added to a neutral gaseous atom''.

This is first electron affinity. Similarly, second electron affinity will be the energy released (absorbed) when a
second electron is added. X(g) + e −1 → X − (g) + E1 (Exoenergic) ; X − (g) + e −1 → X −2 (g) + E2

But the third electron affinity corresponds to energy absorbed instead of energy evolved in the process.

X −2 (g) + e −1 → X −3 (Endoenergic)

For example, chlorine atom readily accepts one electron and 86.5 Kcal/gm energy is released.

Cl + e–1 → Cl– + 86.5 Kcal/gm

Thus, higher the energy released in the process of taking up an extra electron, the higher will be the electron
affinity. Higher the value of electron affinity of an atom, the more is its tendency to change into anion. It is very
difficult to determine the electron affinity experimentally. The values have been calculated on the basis of
thermodynamic concepts. It is expressed in electron volts. The values of inert gases are assumed to be zero because
they have stable ns2p6 configuration and unable to accept any electron. The values for alkali metals are between
zero and one.

Electron Affinity of Elements (For one electron only at 25o C)

Li Be B C N O F Ne

0.54 0 0.3 1.13 0.20 1.48 3.62 0

Na Mg Al Si P S Cl Ar

0.74 0 0.4 1.9 0.80 2.07 3.79 0

The electron affinities of Be, Mg and zero since they have complete ns2 configuration which can not

accommodate extra electron. Similarly, the values for N and P are very low because they also have completely half–

filled p orbitals (ns2p3) and are more stable.

(1) Factors affecting the value of electron affinity
(i) Atomic size : The value of electron affinity decreases with the increase in the size of atom since the
nuclear attraction decreases down a group as the atomic number increases. Its value increases as we move along a
period since the size of atoms decreases along a period. The lower value of F than Cl is due to the very small size of
F in which negative charge is highly concentrated and repels the incoming electron thereby reducing the force of
attraction of the nucleus towards the adding electron and hence decreasing the electron affinity. Thus, chlorine has
a highest value of electron affinity.

(ii) Nuclear charge : The value of electron affinity increases with increasing nuclear charge. Thus, its value
increases with increase in nuclear charge along a period.


(iii) Screening or shielding effect : The value of electron affinity increases with the decrease in shielding
effect of inner electrons. Besides, the value of electron affinity also depends to some extent upon the type of orbital
in which electron is added. The value is greater when electron enters 's' orbital and decreases successively for p, d
and f orbitals.

(2) Importance of electron affinity : Certain properties of the elements are predicted on the basis of values
of electron affinity.

(i) The elements having high value high values of electron affinity are capable of accepting electron easily.
They form anions and electrovalent compounds. These elements are electronegative in nature.

(ii) The elements having high values of electron affinity act as strong oxidising agents, for example,
F, Cl, Br,O, S, etc.

On the basis of the general trend of ionisation potential and electron affinity, the following properties can be
predicted,

(a) Metallic nature decreases in a period while nonmetallic nature increases. Metallic nature increases in a
group while non-metallic nature decreases. The arrow (↓) represents a group and (→) represents a period.

Metallic Non-Metallic

Metallic Decreases Non-Metallic Increases
(Electro +Ve) (Electro –Ve)

Increase Decreases

(b) Reducing nature decreases in a period while oxidising nature increases. The reducing nature increases in a
group while oxidising nature decreases.

Reducing nature Oxidising nature

Reducing Decreases Oxidising Increases
nature nature

Increase Decreases

(c) Stability of metal increases while activity of the metal decreases in a period and in a group stability

decreases while activity increases.

Stability of the metal Activity of metal

Stability of increases Activity of Decreases
the metal the metal

Decrease Increases

This trend is observed especially in IA, IIA and IIIA elements.


(d) The basic nature of the oxides decreases in a period while acidic nature increases. In a group, basic nature
increases while acidic nature decrases.

Basic nature of oxides Acidic nature of oxide

Basic nature Decreases Acidic nature Increases
of oxides of oxides

Increase Decreases

Electronegativity .

The tendency of an atom in a compound to attract a pair of bonded electrons towards itself is
known as electronegativity of the atom. It is important to note that electron affinity and electronegativity both
measure the electrons attracting power but the former refers to an isolated gaseous atom while the latter to an atom
in a compound. Thus electron affinity is attraction for a single electron while electronegativity is for a pair of bonded
electrons. Further electron affinity is energy while electronegativity is a tendency.

(1) Factors affecting the value of electronegativity
(i) The size of the atom.

(ii) Electronic configuration. Small atoms attract electrons more than the larger one and are therefore more
electronegative. Secondly, atoms with nearly filled shell of electrons, will tend to have higher electronegativity than
those sparsely occupied ones.

(2) Variation of electronegativity in the periodic table
(i) In a period, electronegativity increases from left to right. This is due to decrease in size and increase in
nuclear charge. Thus the alkali metals possess the lowest value, while the halogens have the highest. Inert gases
have zero electronegativity.

(ii) In a group, electronegativity decreases from top to bottom. This is due to increase in atomic size.

If an element exhibits various oxidation state, the atom in the higher oxidation state will be more negative due
to greater attraction for the electron, e.g., Sn II (1.30) and Sn IV (1.90).

(3) Electronegativity may be expressed on the following three scales

(i) Mulliken's scale : Mulliken regarded electronegativity as the average value of ionization potential and
electron affinity of an atom.

Electronegativity = Ionization potential + Electron affinity
2

(ii) Allred-Rochow scale : Allred and Rochow defined electronegativity as the electrostatic force exerted by

the nucleus on the valence electrons. Thus χ = 0.359 Z + 0.744 where Z is the effective nuclear charge and r is the
r2

covalent radius of the atom in Å.

(iii) Pauling scale : Pauling scale of electronegativity is the most widely used. It is based on excess bond
energies. He determined electronegativity difference between the two atoms and then by assigning arbitrary values
to few elements (e.g. 4.00 to fluorine, 2.5 to carbon and 2.1 to hydrogen), he calculated the electronegativity of the

other elements. χ A − χ B = 0.208 ∆E where χ A and χ B are electronegativities of the atoms A and B respectively,


the factor 0.208 arises from the conversion of kcal to electron volt (1 eV = 23.0 kcal/mole), while ∆E = Actual bond
energy − (E A−A × EB−B )

Pauling and Mulliken values of electronegativities are related as below χ (Pauling) = 0.34χ (Mulliken) – 0.2

(4) Importance of electronegativity : The following predictions can be made from value of
electronegativity,

(i) Nature of the bond between two atoms can be predicted from the eelctronegativity difference of the two atoms.

(a) The difference X A − XB = 0, i.e., X A = XB the bond is purely covalent.
(b) The difference X A − XB is small, i.e., XA > XB , the bond is polar covalent.
(c) The difference XA − XB is 1.7, the bond is 50% covalent and 50% ionic.
(d) The difference XA − XB is very high, the bond is more ionic and less covalent. The molecule will be
represented in such case as BA (B+ A−) .

Percentage ionic character may be calculated as,

Percentage of ionic character = 16| X A − XB |+ 3.5(X A − XB)2
where XA and XB represents electronegativity of bonded atoms A and B.
This relation was given by A.L. Allerd (1961).

(ii) Greater the value of difference (XA − XB) more stable will be the bond.

H−F H − Cl H − Br H−I

(XA − XB) 1.9 0.9 0.7 0.4

Stability decreases

Stability of compounds in which X A − X B is very small are unstable in nature, SiH4 (o.3), NCl3 (0.0) ,
PH 3 (0), AsH 3 (0.1) are unstable.

(iii) (XO − X A ) difference predicts the nature of the oxides formed by the element A. XO is the
electronegativity of oxygen.

XO − X A is large, the oxide shows basic nature, (e.g., Na2O ).

XO − X A is small, the oxide shows acidic nature, (e.g., SO2 ).

(iv) Ionic compounds having percentage ionic character less than 20% were found coloured, e.g.,

AgCl AgBr AgI Ag2S

22% 18% 11% 8%

White Light yellow Dark yellow Black

Lesser the percentage ionic character, darker will be the colour.

Some Other Periodic Properties .

(1) Atomic volume : It is defined as the volume occupied by one gram atom of an element. Mathematically,

Atomic volume = Gram atomic weight
Density in solid state


Units of atomic volume are c.c./mole. Atomic volume signifies the volume occupied by one mole (Avogadro
number) of atoms of the given element in solid state. Lower atomic volume generally leads to higher density,
increased hardness and brittleness, higher melting and boiling points, less malleability and ductility.

(i) while descending a group, the atomic volume generally increases which is due to increase in the number
of shells though the valence electrons in a given group remains constant.

(ii) While going left to right across a period the atomic volume first decreases to a minimum and then
increases. Francium has the highest atomic volume and boron has lowest atomic volume.

(2) Density : The density of the elements in solid state varies periodically with their atomic numbers. At first,
the density increases gradually in a period and becomes maximum somewhere for the central members and then

starts decreasing afterwards gradually.

(3) Melting and boiling points : The melting points of the elements exhibit some periodicity with rise of
atomic number. It is observed that elements with low values of atomic volumes have high melting points while

elements with high values of atomic volumes have low melting points. In general, melting points of elements in any

periodic at first increase and become maximum somewhere in the centre and thereafter begins to decreases.

Tungsten has the maximum melting point (3410°C) amongst metals and carbon has the maximum melting
point (3727°C) amongst non-metals. Helium has the minimum melting point (–270°C). The metals, Cs, Ga and Hg
are known in liquid state at 30°C.

The boiling points of the elements also show similar trends, however, the regularities are not so striking as
noted in the case of melting points.

(4) Oxidation state (Oxidation number, O.N.) : Oxidation number of an element in a compound is
the total number of electrons it appears to have gained or lost (negative and positive oxidation states

respectively) during the formation of that particular compound.

Trend of oxidation states in various groups

(1) Group 1 (ns1) and group II (ns2) elements attain the inert gas configuration by the loss of outer s electrons.
The group 1 and 2 elements have oxidation states of I and II respectively.

(2) Groups 13 elements (ns2p1) can exhibit oxidation states of I (by the loss of the np1 electron) and III (by the
loss of np1 + ns2 electrons). The stability of the lowest state (I) increases while that of higher state (III) decreases with
increase in atomic number. Thus B, Al and Ga show only III oxidation state. In shows both mono and trivalent
states while in Tl monovalent state is more stable.

(3) Group 14 elements (ns2p2) show oxidation states of II (by the loss of np2 electrons) and IV (by the loss of
ns2p2 electrons). The stability of II state increases while that of IV decreases with the increase in atomic number.
Certain elements of this group (C, Si and Ge) also show – IV state in which elements attain inert gas configuration
(ns2p6) by acquiring four elements. This state can exist only with the highly electropositive elements in carbides,
silicides and germanides which are covalent.

(4) Group 15 elements (ns2p3) show oxidation state of –III (by the gain of three electrons), III (by the loss of

np3 electrons) and V (by the loss of np3 + ns2 electrons). The stability of –III and V states decreases while that of III

state increases with increasing atomic number. Nitrogen is unique in having a large number of oxidation states,

including a fractional one (e.g. – 1 in NH 3 ).
3


(5) Group 16 elements (ns2p4) show mainly – II oxidation state which can be easily achieved by the gain of
two electrons short of the next inert gas configuration. Hence the elements of 16 group are electronegative. The
stability of – II state decreases down the group. Other important oxidation states of these elements are II, IV and VI
(II oxidation state is not very stable). The different oxidation states may be explained as below,

M ns 2 p 2 p 1 p1z
x y

M [–II] ns 2 p 2 p 2 p 2
x y z

M [II] ns 2 p 2
M [IV] x

ns2

M [VI] No electron in the outermost shell of the atom.

(6) Group 17 elements (ns2p5) show mainly –I oxidation state and hence the elements are extremely high
electronegative. Fluorine is the most electronegative and cannot give a positive oxidation state. Other members can
show the oxidation states of I, III, V and VII; the stability of these states decreases in the order : V > VII > III > I.
Moreover, chlorine shows oxidation states of IV (in ClO2) and VI (in ClO3).

(7) Group zero elements (ns2p6) are stable. However, after 1962 compounds of inert gases with F and O (most
reactive elements) have been characterised. In these compounds inert gases show oxidation states of IV, VI, II and
VIII (in order of decreasing stability).

(8) Oxidation states in transition elements : The general electronic configuration (n – 1) dx ns2 of the
transition elements suggests,

(i) II (achieved by the loss of ns2 electrons) as the lowest common oxidation state, and

(ii) 2 + x [achieved by the loss of ns2 + (n – 1) dx electrons] as the maximum oxidation state of these
elements.

(a) In the 3d series, the Ti (II) is unstable and strongly reducing. The stability of II state increases (with a
decrease in reducing character) with the addition of electrons in (n–1) d orbital. In 4d. and 5d series, II state is
unstable except for Pt (II) and Pd (II).

(b) In the 3d series, the highest oxidation sate increases from III in Sc to VII in Mn. Beyond Mn (d5 in the
ground state), electron pairing commences in the d orbitals and the highest oxidation state for the subsequent
members does not exceed VI. In the 4d and 5d series, the higher states are more stable and less oxidising than
those in the d series. Thus Nb (V), Ta (V), Mo (VI), W (VI), Tc (VII) and Re (VII) are almost non–oxidising. In
general, the stability of the higher states is in the order, 3 d << 4d < 5d

(5) Oxidising and Reducing Power : Tendency to lose electrons and reducing power are directly related to
each other; while tendency to gain electrons and oxidising power are directly related to each other.

(i) In a group, the reducing character generally increases. Thus among alkali metals (except francium), cesium
is the most powerful reducing element (in absence of water).

(ii) The reducing character decreases and oxidising character increases along the period from left to right.
Thus alkali metals are strong reducing agents while halogens are oxidising agents among halogens, fluorine is the
most powerful oxidising agent; the oxidising property decreases from F to I and actually iodine shows reducing
properties.


(iii) Inert gases are neither oxidising nor reducing agents. (iv) Generally, metals are reducing agents and non-
metals are oxidising agents.

(6) Paramagnetism, Diamagnetism, and Ferromagnetism : Magnetic properties of matter depend on the
properties of the individual atoms. A substance (atom, ion or compound) capable of being attracted into a magnetic
field is known as paramagnetic. The paramagnetic substances have a net magnetic moment which in turn is due to
the presence of unpaired electron(s) in atoms, ions or molecules. Since most of the transition metal ions have
unpaired d–electrons, they show paramagnetic behaviour. The exceptions are Sc 3+ , Ti 4+ , Zn2+ , Cu+ , etc. which do
not contain any unpaired electron and hence are diamagnetic.

On the other hand, a substance which is repelled by a magnetic field is known as diamagnetic.
Such substances do not have any net magnetic moment because they do not have any unpaired electron. Electrons
determine the magnetic properties of matter in two ways,

(i) Each electron can be treated as a small sphere of negative charge spinning on its axis. The spinning of
charge produces magnetic moment.

(ii) An electron travelling in closed path around a nucleus will also produce magnetic moment just as does
electric current travelling in a loop of wire.

The observed magnetic moment is therefore the sum of the two moments: the spin moment and the orbital
moment. It is expressed in units called Bohr Magnetons (BM). In terms of n (number of unpaired electron), magnetic
moment is given by the formula, µ = n(n + 2)

Thus when n = 1 µ = 1 × 3 = 1.73 BM

n = 2 µ = 2 × 4 = 2.83 BM

n = 3 µ = 3.87 BM

n = 4 µ = 4.90 BM

n = 5 µ = 5.92 BM

Thus greater the number of unpaired electrons in a substance, the greater is the magnetic moment of the
substance. The value of magnetic moment has been used to calculate the number of unpaired electrons in an ion.
In some cases, even the structure of the molecule or complex is indicated by its magnetic moment.

Paramagnetism is generally measured by a simple device known as Guoy's balance which involves
weighing the species in presence of a magnetic field. Diamagnetic substances show a decrease in weight whereas
paramagnetic substances show increase in weight. The larger the number of unpaired electrons in a substance, the
greater is the increase in its weight in a magnetic field.

Ferromagnetism is a special property observed in some substances in the solid state (not in solution). Such
substances are strongly attracted to magnetic field and may retain the magnetic properties for some time even after
the removal of the field. The most common example is of Fe followed by Co and Ni.

(7) Hydration and hydration energy
(i) Hydration energy is the enthalpy change that accompanies the dissolving of 1 mol of gaseous ions in water.
(ii) Size of ions and its charge determines extent of hydration. Greater the charge smaller the size of the ion,
greater the attraction for the lone pair of O of H2O , hence greater the extent of hydration energy.


(a) Size of the hydration ion increases.
(b) Ionic mobility decreases i.e. heavier (hydrated) ions moves slower.

(8) Acid-base-character of oxides
(i) On moving across a period, the basic character of the oxides gradually changes first into amphoteric and
finally into acidic character.
(ii) On moving down a group, reverse behaviour is observed i.e., from more acidic to more basic.
(iii) Stability of oxides decreases across a period.

(9) Hydrides
(i) Hydrogen combines with a number of other elements including metals and non-metals to form compounds
called hydrides (with H − ).
(ii) Covalent nature of hydrides increases across a period and decreases down the group.
(iii) Ionic hydride are better reducing agents than covalent hydride and reducing nature of hydride decreases
across a period and increases down the group.
(iv) Covalent and ionic hydrides are classified as follows,

(A) (B) (C) LiH NaH MgH2 AlH3 SiH3
NaH (A)
KH covalent (B)
RbH (C)
ionic
Maximum
reducing

CsH (Minimum

Diagonal Relationship.

Certain elements of 2nd period show similarity with their diagonal elements in the 3rd period as shown below :

Group 1 Group 2 Group 13 Group 14

2nd period Li Be B C

3rd period Na Mg Al Si

Thus, Li resembles Mg, Be resembles Al and B resembles Si. This is called diagonal relationship and is due to

the reason that these pairs of element have almost identical ionic radii and polarizing power (i.e. charge/size ratio).

Element of second period are known as bridge elements.

Anomalous behaviour of the first elements of a group : The first element of a group differs
considerably from its congeners (i.e. the rest of the element of its group). This is due to (i) small size (ii) high
electronegativity and (iii) non availability of d-orbitals for bonding. Anomalous behaviour is observed among the
second row elements (i.e. Li to F).


Chemical bonding

Formation of bonding and antibonding molecular orbitals

(Linear combination of atomic orbitals – LCAO)

When two atomic orbitals overlap they can be in phase (added) or out of phase (subtracted). If they overlap in

phase, constructive interaction occurs in the region

between two nuclei and a bonding orbital is Antibonding M.O. Subtraction
produced. The energy of the bonding orbitals is Repulsive
always lower (more stable) then the energies of

the combining atomic orbitals. When they overlap Energy Atomic Atomic
out of phase, destructive interference reduces the orbital orbital

probability of finding an electron in the region Bonding M.O. Addition
between the nuclei and antibonding orbital is
produced. The energy of an antibonding orbital is Attractive
higher (less stable) than the energies of the Formation of bonding and antibonding molecular orbitals

combining atomic orbitals. Thus, the number of molecular orbitals formed from atomic orbitals is equal to the

number of atomic orbitals responsible for their formation.

According to LCAO method, the following combination is not allowed. [consider the z-axis as the molecular
axis). npx + npy, npx + npz or npy + npz .

Table : Difference in Bonding molecular orbitals and Antibonding molecular orbitals.

Bonding molecular orbitals Antibonding molecular orbitals
It is formed by linear combination of two atomic orbitals
It is formed by linear combination of two atomic when their wave functions are subtracted.
orbitals when their wave functions are added. i.e., Ψa = ΨA − ΨB
i.e., Ψb = ΨA + ΨB Its energy is more than the combining atomic orbitals.

Its energy is less than the combining atomic It decreases the electron density between the nuclei. It
orbitals. therefore destabilises the molecule.
It increases the electron density between the nuclei. It has nodal plane.
It therefore stabilises the molecule.
It has no nodal plane (plane where electron density It is symmetrical about internuclear axis and about a line
is zero). perpendicular to it.
It is symmetrical about internuclear axis.

σ and π – Molecular orbitals

A sigma (σ) M.O. is one that has cylinderical symmetry around the internuclear axis. It does not show any

change of sign or rotation through 180o about the axis. Sigma M.O. has no nodal plane (in which electron density

is zero) along the internuclear axis. The bonding M.O. is designated by σ and antibonding by σ * .

++ + ++ + +
•• • •• ••

1s 1s σ(1s)
Bonding M.O.

+– +– +• • –
•• ••

1s 1s σ*(1s)
Antibonding M.O.
Formation of σ (1s) and σ * (1s) M.O. from 1s atomic orbital


Sigma (σ) M.O. is also formed when two p-atomic orbitals overlap in head on (along their axes) position.

– • ++ • – – • +– •+
2pz
2pz 2pz

– • ++ • – – • +– • +

–• + •– – • +– • +
σ(2pz) σ*(2pz)

Formation of σ (2pz and σ * (2pz ) molecular orbitals.

π-molecular orbital is formed by the sideways overlapping of the p-atomic orbitals. It consists of two electrons

clouds, one lying above and the other lying below a plane passing through the nuclei. It has nodal planes. Since the

energy of a M.O. is directly related to the nuclei of nodal planes. The π-MO is more energetic than the σ-MO. This

explains why a π-bond is a weaker bond than a σ-bond.

++ ++ +

• • •• • •

– – –
2px 2px
(or 2py) (or 2py) – – Nodal Plane

π (2px)
or π (2py)

+– +– +–

•• •• ••

–+ –+ –+

2px 2px Nodal Plane π* (2px)
(or 2py) (or 2py) or π* (2py)

Formation of π (2px ) or π (2py ) and π *(2px ) or π * (2py ) molecular orbitals.

Relative energies of Molecular orbitals : Initial energy of the atomic orbitals and the extent of their
overlap is the criteria which determines the energy of the M.O.

(1) It is obvious that molecular orbitals formed from lower energy atomic orbitals have lower energy than the
molecular orbitals formed from higher energy atomic orbitals.

(2) As theσ overlap is much more effective than π-overlap, σ p-molecular orbital is of lowest energy, even
though originally all the three p-orbitals are of equal energy.

(3) The relative energies of the M.O. are obtained experimentally from spectroscopic data.

(4) The sequence in the order of increasing energy for O2 , F2 and Ne2

σ 1s < σ * 1s < σ 2s < σ * 2s < σ 2pz < π 2px = π 2py < π * 2px = π * 2py < σ * 2pz

(5) It may be noted that π 2py an π 2px bonding molecular orbitals are degenerate (i.e. have same energy).

Similarly π * 2py and π * 2px antibonding molecular orbitals are also degenerate (have the same energy).

(6) The main difference between the two types of sequences in energy level is that for molecules O2 , F2 and
Ne2 (Hypothetical) the σ 2pz M.O. is lower in energy than π 2px and π 2py .


(7) It has been found experimentally that in some of the diatomic molecules such as Li2 , B2 , C2 and N 2
σ 2pz M.O. is higher in energy than π 2py and π 2px M.O.’s. Therefore, the order of increasing energy of these

M.O.s changes to σ 1s < σ * 1s < σ 2s < σ * 2s < π 2px = π 2py < σ 2pz < π * 2px = π * 2py < σ * 2pz .

Cause of exceptional behaviour of MO’s in B2 , C2 and N 2

In atoms with Z upto 7, energy of 2s and 2p atomic orbitals lie fairly close. As a result the interaction between
2s and 2p orbitals is quite large. Thus σ 2s and σ * 2s MO’s become more stable with less energy at the cost of
σ 2px and σ * 2px which gets unstabilised (higher energy).

Stability of the molecules : Stabitity of the molecules can be explain in following ways.

(1) Stability of molecules in terms of bonding and antibonding electrons

Since electrons in bonding orbitals (Nb ) increase the stability of the molecule, on the whole, depends on their
relative numbers. Thus

(i) If Nb > Na , the molecule is stable. (ii) If Nb < Na , the molecule is unstable.

(iii) Even if Nb = Na , the molecule is unstable. This is due to the fact the anti bonding effect is some what
stronger than the bonding effect.

(2) Stability of molecules in terms of bond order

(i) The relative stability of a molecule is further evaluated by a parameter known as bond order.

(ii) It can be defined as number of covalent bonds present between two atoms in a molecule.

(iii) It is given by one half of the differences between the number of electrons in bonding molecular
orbitals and those in antibonding molecular orbitals.

(iv) Bond order = 1 [No. of electrons in bonding molecular orbitals – No. of electrons in antibonding
2

molecular orbitals].

(v) The bond order of 1, 2 and 3 corresponds to single, double and triple bonds respectively. It may be
mentioned that according to M.O theory, even a fractional bond order is possible, but cannot be
negative.

(vi) bond order ∝ Stability of molecule ∝ Dissociation energy ∝ 1
Bond length

(vii) If all the electrons in a molecule are paired then the substance is a diamagnetic on the other hand if
there are unpaired electrons in the molecule, then the substance is paramagnetic. More the number of
unpaired electron in the molecule greater is the paramagnetism of the substance.

MO energy level diagrams of some molecules : Energy level diagrams of some molecules are given below.

H2 Molecule : H 2 molecule is formed from 1s1 atomic orbitals of two H- atoms. The atomic orbitals (1s1 ) will
combine to form two molecular orbitals σ (1s) and σ * (1s) . Two electrons are accommodated in σ (1s) and σ * (1s)

remains vacant. Thus bond order for H2 = 1 (2 − 0) = 1 . It is stable and diamagnetic in nature.
2

Increasing energy σ*1s

1s 1s
HH

σ1s
H2


He2 molecule : If two atoms of He (1s 2) combine to form He2 , the probability of the formation of molecular

orbitals is as shown in the figure. Increasing energy σ*(1s2)
Filling of electrons is as follows

He 2 = σ (1s)2 , σ * (1s)2 1s2 1s2

Thus bond order 1 (2 2) 0 He He
2
= − =

Hence there is no possibility for the existence of He2 molecule. σ(1s)2

N2 molecule : Total number of electrons in N2 are 14, of which 4 are in K shell and the 10 electrons are

arranged as, KK(σ 2s)2(σ * 2s)2(π 2px )2(π 2py)2(σ 2pz )2 σ * (2pz )

Bond order = 1 (10 − 4) = 3 2p 2p
2
π*(2px) π*(2py)

Increasing energy σ (2pz)

π (2px)=π (2py)

σ* (2s)

2s 2s

N (AO) N (AO)

σ (2s)

M.O. energy level diagraNm2 (MfoOr )N2 molecule.

O2 molecule : Total number of electrons in O2 = 16

Electronic arrangement in M.O.’s , [KK(σ 2s)2 (σ * 2s)2 (σ 2pz )2 (π 2px )2 (π z 2py )2 (π * 2px )1(π * 2py )1 ]

Bond order = 1 (8 − 4) = 2 σ * (2pz )
2

π* 2px π* 2py

Increasing energy π 2px=π 2py

σ 2pz
σ* 2s

O (AO) O (AO)

σ 2s

M.O. Energy level diagrOa2m(MfoOr) O2 molecule


Table : Bond order and magnetic nature of some molecules & ions

Molecule Molecular Orbital Configuration Valence Nb Na Magnetic B.O.
electrons Nature

H2 (σ1s)2 2 2 0 Diamagnetic 1
(σ1s)1
H + 1 1 0 Paramagnetic 0.5
2

H − (σ1s)2(σ* 1s)1 3 2 1 Paramagnetic 0.5
2

He 2 (σ1s)2  σ∗1s2 4 2 2 Molecule does 0
(σ1s)2 (σ* 1s)1 not exist

He + 3 2 1 Paramagnetic 0.5
2

He − KK ' (σ2s)1 1 1 0 Paramagnetic 0.5
2 KK' (σ2s)2 (σ* 2s)2

Be2 4 2 2 Diamagnetic 0

B2 KK ' (σ 2s)2 (σ* 2s)2 (π 2px )1 (π 2py)1 6 4 2 Paramagnetic 1

C2 ( )KK ' (σ 2s)2(σ*2s)2(π2px )2 π2py 2 8 6 2 Diamagnetic 2

N2 ( )KK ' (σ 2s)2(σ*2s)2(π2px )2 π2py 2(σ 2pz )2 10 8 2 Diamagnetic 3

N + ( )KK ' (σ 2s)2(σ*2s)2(π2px )2 π2py 2(σ 2pz )1 9 7 2 Paramagnetic 2.5
2 ( )KK ' (σ 2s)2(σ*2s)2(π2px )2 π2py 2(σ 2pz )2(π * 2px )1 11
( ) ( )KK ' (σ 2s)2(σ*2s)2(σ 2pz )2(π2px )2 π2py 2(π * 2px )1 π * 2py 1 12
N − ( ) ( )KK′ (σ 2s)2(σ*2s)2(σ 2pz )2(π2px )2 π2py 2(π*2px )1 π*2py 0 11 8 3 Paramagnetic 2.5
2

O2 8 4 Paramagnetic 2

O2+ 8 3 Paramagnetic 2.5

O 2+ KK ' (σ 2s)2(σ * 2s)2(σ 2pz )2(π 2px )2(π 2py)2 10 8 2 Diamagnetic 3
2

O2− ( ) ( )KK′ (σ2s)2(σ*2s)2(σ 2pz )2(π2px )2 π2py 2(π * 2px )2 π * 2py 1 13 8 5 Paramagnetic 1.5

O 2− ( ) ( )KK′ (σ 2s)2(σ*2s)2(σ 2pz )2(π2px )2 π2py 2(π * 2px )2 π * 2py 2 14 8 6 Diamagnetic 1
2 ( ) ( )KK′ (σ 2s)2(σ*2s)2(σ 2pz )2(π2px )2 π2py 2(π*2px )2 π*2py 2 14 8 6 Diamagnetic 1

F2

Ne 2 ( ) ( )KK ' 16 8 8 Molecule does 0
(σ 2s)2(σ*2s)2(σ 2pz )2(π2px )2 π 2 py 2(π*2px )2 π*2py 2 10 not exist
(π*2pz )2

CO ( )KK ' (σ 2s)2(σ*2s)2(σ 2pz )2(π2px )2 π2py 2 8 2 Diamagnetic 3
NO
NO + ( ) ( )KK ' (σ 2s)2(σ*2s)2(σ 2pz )2(π2px )2 π2py 2 π∗ 2px 1 11 8 3 Paramagnetic 2.5

( )KK ' (σ 2s)2(σ*2s)2(σ 2pz )2(π2px )2 π2py 2 10 8 2 Diamagnetic 3


Hydrogen bonding.

In 1920, Latimer and Rodebush introduced the idea of “hydrogen bond” to explain the nature of
association of substances in liquid state like water, hydrogen fluoride, ammonia, formic acid, etc.

(1) Conditions for the formation of hydrogen bonding

(i) High electronegativity of atom bonded to hydrogen : The molecule should contain an atom of high
electronegativity such as F, O or N bonded to hydrogen atom. The common examples are H 2O ,
NH 3 and HF .

δ+ δ− δ+ δ− δ+ δ−
H— X− − − H— X− − − H— X

(ii) Small size of the electronegative atom : The size of the electronegative atom should be quite small.
This is due to the fact that the smaller the size of electronegative atom, the greater will be its attraction
for the bonded electron pair. This will cause greater polarity in the bond between H and
electronegative atom, which results in stronger hydrogen bond.

(2) Types of hydrogen bonding

(i) Intermolecular hydrogen bond : Intermolecular hydrogen bond is formed between two different
molecules of the same or different substances. For example:

(a) Hydrogen bond between the molecules of hydrogen fluoride.

(b) Hydrogen bond in alcohol and water molecules

(ii) Intramolecular hydrogen bond (Chelation) : Intramolecular hydrogen bond is formed between the
hydrogen atom and the highly electronegative atom (F, O or N) present in the same molecule.
Intramolecular hydrogen bond results in the cyclisation of the molecules and prevents their

association. Consequently, the effect of intramolecular hydrogen bond on the physical properties is

OH
|| |
O CO CO
H
H

NO NO OH
|| ||
O O Salicyldehyde
(o-hydroxy benzaldehyde)
Ortho nitrophenol Ortho nitrobenzoic acid

negligible. For example : Intramolecular hydrogen bonds are present in molecules such as o-nitrophenol,
o-nitrobenzoic acid, etc.

The extent of both intramolecular and intermolecular hydrogen bonding depends on temperature.

(3) Effects of hydrogen bonding
Hydrogen bond helps in explaining the abnormal physical properties in several cases. Some of the properties
affected by H-bond are given below:

(i) Dissociation : In aqueous solution, hydrogen fluoride dissociates and gives the difluoride ion (HF2− )
instead of fluoride ion (F − ). This is due to H-bonding in HF. This explains the existence of KHF2 . On
the other hand, the molecules of HCl , HBr , HI do not have H-bonding (because Cl, Br, I are not
so highly electronegative). This explains the non-existence of compounds like KHCl2 , KHBr2 or


KHI 2 . H-bond formed is usually longer than the covalent bond present in the molecule (e.g. in H2O,

O–H bond = 0.99 Å but H-bond = 1.77 Å).

(ii) Association : The molecules of carboxylic acids exist as dimers because of the hydrogen bonding.
The molecular masses of such compounds are found to be double than those calculated from their
simple formulae. For example, molecular mass of acetic acid is found to be 120.

(iii) High melting and boiling point : The compounds having hydrogen bonding show abnormally
high melting and boiling points.

The high melting points and boiling points of the compounds (H 2O, HF and NH3 ) containing
hydrogen bonds is due to the fact that some extra energy is needed to break these bonds. A few
examples are given below:
(a) H 2O is a liquid whereas H 2S, H 2Se and H 2Te are all gases at ordinary temperature. The reason
for this is that in case of water, hydrogen bonding causes association of the H 2O molecules with the
result that the boiling point of water is more than that of the other compounds. On the other hand,
there is no such hydrogen bonding in H 2S, H 2Se and H2Te.

(b) NH 3 has higher boiling point than PH3. This is again because there is hydrogen bonding in NH 3 but
not in PH3.

(c) Ethanol has higher boiling point than diethyl ether because there is hydrogen bonding in the former
but there is no hydrogen bonding in the later.
Intramolecular hydrogen bonding is not possible in case of m - and p -isomers because of the size of
the ring which would be formed. Thus, here the intermolecular hydrogen bonding takes place which
causes some degree of association with the result the m -and p -isomers melt and boil at higher
temperatures.

(iv) Solubility : The compound which can form hydrogen bonds with the covalent molecules are soluble
in such solvents. For example, lower alcohols are soluble in water because of the hydrogen bonding
which can take place between water and alcohol molecules as shown below :

δH+− δO− ................ δH+− δO− ............... δH+− δO−

C2H5 H C2H5

Similarly, ammonia (NH 3 ) is soluble in water because of hydrogen bonding as represented below:

Hδ + Hδ +
Hδ + δN−−− ....... Hδ + — Oδ −− ............... Hδ + δN−−− ....... Hδ + — Oδ −−
Hδ + H Hδ +
Hδ +

The intermolecular hydrogen bonding increases solubility of the compound in water while, the

intramolecular hydrogen bonding decreases. This is due to the fact that the formation of internal

hydrogen bond prevents hydrogen bonding between the compound and water which thus reduced

solubility of the compound in water. H−O−−− H−O−H
OH

NO O← N =O
||
o- NitropOhenol p − Nitrophenol

Due to chelation, – OH group is not available to form hydrogen – OH group available to form hydrogen bond with
bond with water hence it is sparingly soluble in water. water, hence it is completely soluble in water.


(v) Volatility : As the compounds involving hydrogen bonding between different molecules
(intermolecular hydrogen bonding) have higher boiling points, so they are less volatile.

(vi) Viscosity and surface tension : The substances which contain hydrogen bonding exist as
associated molecules. So their flow becomes comparatively difficult. In other words, they have higher
viscosity and high surface tension.

(vii) Explanation of lower density of ice than water and maximum density of water at 277K :
In case of solid ice, the hydrogen bonding gives rise to a cage like structure of water molecules as
shown in following figure. As a matter of fact, each water molecule is linked tetrahedrally to four other
water molecules. Due to this structure ice has lower density than water at 273K. That is why ice floats
on water. On heating, the hydrogen bonds start collapsing, obviously the molecules are not so closely
packed as they are in the liquid state and thus the molecules start coming together resulting in the
decrease of volume and hence increase of density. This goes on upto 277K. After 277 K, the increase
in volume due to expansion of the liquid water becomes much more than the decrease in volume due
to breaking of H-bonds. Thus, after 277K , there is net increase of volume on heating which means
decrease in density. Hence density of water is maximum 277K .

0.90 Å H 1.77 Å
(99 pm) O (177 pm)

HH H Vacant
O O Spaces

HH HH

OO

HH HH
O O

HH HH

O O
H H

Cage like structure of H2O in the ice

Important Tips

 Hydrogen bonding is strongest when the bonded structure is stabilised by resonance.
 The bond length of hydrogen bond is the order of 250 to 275 pm.
 The bond that determines the secondary structure of protein is hydrogen bond.
 Pairs of DNA are held together by hydrogen bonds.

 Chlorine has the same electronegativity as nitrogen but does not form strong hydrogen bonds. This is because of the larger size than

that of nitrogen.

\


Types of bonding and forces in solids.

(1) Ionic bonding : Solid containing ionic bonds consists of any array or a net work of positive and negative
ions arranged systematically in a characteristic pattern. The binding forces are strong electrostatic bonds between
positive and negative ions. e.g., Compounds of elements of group 1 and 2 with elements of group 16 and 17 e.g.,
NaCl, CaS etc.

(2) Covalent bonding : The solid containing covalent bonding consists of an array of atoms that share
electrons with their neighbouring atoms. The atoms are linked together by strong covalent bonds extending into
three dimensional structure. e.g., Diamond, Silicon carbide, Silicon dioxide etc.

(3) Molecular bonding : The solid containing molecular bonding consists of symmetrical aggregates of
discrete molecules. However, these molecules are further bound to other molecules by relatively weak force such as
dipole-dipole forces (Vander Waal forces), dispersion forces or H-bonds depending upon the nature of molecules.
The existence of weak attractive forces among the non polar molecules was first proposed by S.D. Vander waal.
Vander waal's forces are non-directional, non valence force of attraction. Vander Waal force ∝ molecular mass
∝ Boiling point ∝ Size of atom or molecule. The forces present in the crystals of naphthalene, Iodine and dry ices
solid CH 4 , solid hydrogen are Vander Waals forces. SiO2 Possesses giant covalent molecular structure due to
tetravalency and catenation nature of Si .

(4) Metallic Bond

The constituent particles in metallic solids are metal atoms which are held together by metallic bond. Lorentz
proposed a simple theory of metallic bond. This theory is known as electron gas model or electron sea model.

A metal atom consists of two parts, valence electrons and the remaining part (the nucleus and the inner shell
electrons) called kernel. The kernels of metal atoms occupy the lattice sites while the space in-between is occupied
by valence electrons. These electrons are not localized but are mobile. The attraction between the kernels and the
mobile electrons, which hold the kernel together, is known as metallic bond. Low ionisation energy and sufficient
number of vacant orbital in the valency shell are essential conditions for metallic bonding. Metallic bond is
electrostatic in nature. The strength of the metallic bond depends on the number of valency electron and the charge
on the nucleus. As the number of valency electron and the charge increase, the metallic bond becomes strong. Due
to this fact alkali metals are soft and have low melting and boiling points while transition metals and hard and have
high melting and boiling points. Strong metallic bonding is also favoured by smaller size of kernel. Ge, Cu, Zn has
metallic bonding while brass etc does not have metallic bonding.

Metals have properties like metallic lustre, thermal and electrical conductivity due to delocalized mobile
electrons. Thermal conductivity of metal decreases with increase in temperature because the kernels start vibrating.

Since the metallic bond is non-directional; metals can be twisted, drawn into wires or beaten into sheets. This
is because the kernels can slip over each other when a deforming force is applied.

The relative strength of various bonds is Ionic >Covalent>Metallic>H-bond>vander waal forces.


Gaseous state

The state of matter in which the molecular forces of attraction between the particles of matter are minimum, is
known as gaseous state. It is the simplest state and shows great uniformity in behaviour.

Characteristics of gases.

(1) Gases or their mixtures are homogeneous in composition.
(2) Gases have very low density due to negligible intermolecular forces.
(3) Gases have infinite expansibility and high compressibility.
(4) Gases exert pressure.
(5) Gases possess high diffusibility.
(6) Gases do not have definite shape and volume like liquids.
(7) Gaseous molecules move very rapidly in all directions in a random manner i.e., gases have highest kinetic energy.
(8) Gaseous molecules are loosely packed having large empty spaces between them.
(9) Gaseous molecules collide with one another and also with the walls of container with perfectly elastic collisions.
(10) Gases can be liquified, if subjected to low temperatures (below critical) or high pressures.
(11) Thermal energy of gases >> molecular attraction.
(12) Gases undergo similar change with the change of temperature and pressure. In other words, gases obey
certain laws known as gas laws.

Measurable properties of gases.

(1) The characteristics of gases are described fully in terms of four parameters or measurable properties :
(i) The volume, V, of the gas.
(ii) Its pressure, P
(iii) Its temperature, T
(iv) The amount of the gas (i.e., mass or number of moles).

(2) Volume : (i) Since gases occupy the entire space available to them, the measurement of volume of a gas
only requires a measurement of the container confining the gas.

(ii) Volume is expressed in litres (L), millilitres (mL) or cubic centimetres (cm3 ) or cubic metres (m3 ) .

(iii) 1L = 1000 mL ; 1mL = 10 −3 L

1 L = 1dm3 = 103 cm3

1m3 = 103 dm3 = 106 cm3 = 106 mL = 103 L

(3) Mass : (i) The mass of a gas can be determined by weighing the container in which the gas is enclosed

and again weighing the container after removing the gas. The difference between the two weights gives the mass of
the gas.

(ii) The mass of the gas is related to the number of moles of the gas i.e.

moles of gas (n) = Mass in grams = m
Molar mass M

(iii) Mass is expressed in grams or kilograms, 1 Kg = 103 g


(4) Temperature : (i) Gases expand on increasing the temperature. If temperature is increased twice, the
square of the velocity (v2) also increases two times.

(ii) Temperature is measured in centigrade degree ( oC) or celsius degree with the help of thermometers.
Temperature is also measured in Fahrenheit (Fo).

(iii) S.I. unit of temperature is kelvin (K) or absolute degree.

K = oC + 273

(iv) Relation between F and oC is oC = Fo − 32
5 9

(5) Pressure : (i) Pressure of the gas is the force exerted by the gas per unit area of the walls of the container

in all directions. Thus, Pressure (P) = Force(F) = Mass(m) × Acceleration(a)
Area(A) Area(a)

(ii) Pressure exerted by a gas is due to kinetic energy (KE = 1 mv 2 ) of the molecules. Kinetic energy of
2

the gas molecules increases, as the temperature is increased. Thus, Pressure of a gas ∝ Temperature (T).

(iii) Pressure of a pure gas is measured by manometer while that of a mixture of gases by barometer.

(iv) Commonly two types of manometers are used,

(a) Open end manometer; (b) Closed end manometer

(v) The S.I. unit of pressure, the pascal (Pa), is defined as 1 newton per metre square. It is very small unit.

1Pa = 1Nm−2 = 1kg m−1s −2

(vi) C.G.S. unit of pressure is dynes cm−2 .
(vii) M.K.S. unit of pressure is kgf / m2 . The unit kgf / cm2 sometime called ata (atmosphere technical
absolute).
(viii) Higher unit of pressure is bar, KPa or MPa.

1bar = 105 Pa = 105 Nm−2 = 100KNm−2 = 100KPa

(ix) Several other units used for pressure are,

Name Symbol Value

bar bar 1bar = 105 Pa
atmosphere atm 1atm = 1.01325 × 105 Pa

Torr Torr 1 Torr = 101325 Pa = 133.322 Pa
760

millimetre of mercury mm Hg 1mm Hg = 133.322 Pa

(x) The pressure relative to the atmosphere is called gauge pressure. The pressure relative to the perfect
vacuum is called absolute pressure.

Absolute pressure = Gauge pressure + Atmosphere pressure.

(xi) When the pressure in a system is less than atmospheric pressure, the gauge pressure becomes
negative, but is frequently designated and called vacuum. For example, 16 cm vacuum will be

76 − 16 × 1.013 = 0.80 bar .
76


(xii) If ‘h’ is the height of the fluid in a column or the difference in the heights of the fluid columns in the
two limbs of the manometer d the density of the fluid (Hg = 13.6 × 103 Kg / m3 = 13.6 g / cm3 ) and g is the gravity,
then pressure is given by, Pgas = Patm + h dg

Gas ρgas Vacuum Height (h) of
Hg h mercury column
ρgas Atmospheric
ρgas=ρgashdg pressure

An open arm manometer Mercury Barometer

(xiii) Two sets of conditions are widely used as 'standard' values for reporting data.

Condition T P Vm (Molar volume)
S.T.P./N.T.P. 273.15 K 1 atm 22.414 L
298.15 K 1 bar 24.800 L
S.A.T.P*.

* Standard Ambient temperature and pressure.

Boyle's law.

(1) In 1662, Robert Boyle discovered the first of several relationships among gas variables (P, T, V).

(2) It states that, “For a fixed amount of a gas at constant temperature, the gas volume is inversely
proportional to the gas pressure.”

Thus, P ∝ 1 at constant temperature and mass
V

or P= K (where K is constant)
V

or PV = K

For two or more gases at constant temperature and mass.

P1V1 = P2V2 = ....... = K

Boyle's law can also be given as,  dP  K
 dV  = − v2
T

(3) Graphical representation of Boyle's law : Graph between P and V at constant temperature is called

isotherm and is an equilateral (or rectangular) hyperbola. By plotting P versus 1 , this hyperbola can be converted
V

to a straight line. Other types of isotherms are also shown below,


T1< T2< T3 T3 T1< T2< T3
P T2 PV T3

P T1 T2 log P
T1

O V or 1/d T3 T1< T2< T3 O P O log 1/V
T2 1/V or d
T1 O

Note :  The isotherms of CO2 were first studied by Andrews.

(4) At constant mass and temperature density of a gas is directly proportional to its pressure and inversely
proportional to its volume.

Thus, d∝ P ∝ 1  V = mass 
V d 

or d1 = P1 = V2 = ....... = K
d2 P2 V1

(5) At altitudes, as P is low d of air is less. That is why mountaineers carry oxygen cylinders.

(6) Air at the sea level is dense because it is compressed by the mass of air above it. However the density and
pressure decreases with increase in altitude. The atmospheric pressure at Mount Everest is only 0.5 atm.

Example : 1 A sample of a given mass of a gas at a constant temperature occupies 95cm2 under a pressure of
Solution: (b) 9.962 × 10 4 Nm−2 . At the same temperature, it volume at a pressure of 10.13 × 10 4 Nm−2 is [Bihar CEE 1992]
Example : 2
Solution: (b) (a) 190 cm3 (b) 93 cm3 (c) 46.5 cm3 (d) 47.5 cm3

Example : 3 P1V1 = P2V2
Solution: (b)
9.962 × 10 4 × 95 = 10.13 × 10 4 × V2

V2 = 93 cm3

A gas occupied a volume of 250 ml at 700 mm Hg pressure and 25o C . What additional pressure is required
to reduce the gas volume to its 4/5th value at the same temperature

(a) 225 mm Hg (b) 175 mm Hg (c) 150 mm Hg (d) 265 mm Hg

P1V1 = P2V2

700 × 250 = P2 ×  4 × 250 ; P2 = 875 mm Hg
 5 

Additional pressure required = 875 – 700 = 175 mm Hg

At constant temperature, if pressure increases by 1%, the percentage decrease of volume is

(a) 1% (b) 100/101% (c) 1/101% (d) 1/100%

P1V1 = P2V2

If P1 = 100 mm , P2 will be 101 mm

Hence 100 × V = 101 × V2 ,


V2 = 100 × V ,
101

Decrease in volume = V − 100V = 1 of V i.e. 100 %
101 101 101

Charle's law.

(1) French chemist, Jacques Charles first studied variation of volume with temperature, in 1787.

(2) It states that, “The volume of a given mass of a gas is directly proportional to the absolute temperature
(= o C + 273) at constant pressure”.

Thus, V ∝ T at constant pressure and mass

or V = KT = K(t( oC) + 273.15) , (where k is constant),

V =K
T

For two or more gases at constant pressure and mass

V1 = V2 = ......K ,
T1 T2

Charle's law can also be given as,  dV  = K .
 dT 
P

(3) If t = 0o C , then V = V0

hence, V0 = K × 273.15

∴ K = V0
273.15

V = V0 [t + 273.15] = V0 1 + t = V0 [1 + α v t]
273.15 273.15 

where αv is the volume coefficient, αv = V − V0 = 1 = 3.661× 10−3 Co −1
tV0 273.15

Thus, for every 1o change in temperature, the volume of a gas changes by 1  ≈ 1  of the
273.15  273 

volume at 0o C .

(4) Graphical representation of Charle's law : Graph between V and T at constant pressure is called
isobar or isoplestics and is always a straight line. A plot of V versus t( oC) at constant pressure is a straight line

cutting the temperature axis at − 273.15o C . It is the lowest possible temperature.


1/d or V

1/d or V T(k) –273.15oC 22.4 L mol–1 = V0
O 0oC t(oC)

(5) To lower the temperature of a substance, we reduce the thermal energy. Absolute zero (0K) is the
temperature reached when all possible thermal energy has been removed from a substance. Obviously, a substance
cannot be cooled any further after all thermal energy has been removed.

(6) At constant mass and pressure density of a gas is inversely proportional to it absolute temperature.

Thus, d ∝ 1 ∝ 1  V = mass  or d1 = T2 = V2 = ...... = K
T V d  d2 T1 V1

(7) Use of hot air balloons in sports and meteorological observations is an application of Charle's law.

Example : 4 When the temperature of 23 ml of dry CO2 gas is changed from 10o to 30o C at constant pressure of 760

Solution: (c) mm, the volume of gas becomes closest to which one of the following [CPMT 1992]
Example : 5
(a) 7.7 ml (b) 25.5 ml (c) 24.6 ml (d) 69 ml
Solution: (b)
Example : 6 V1 = V2 i.e. 23 = V2 ; V2 = 24.6 ml
Solution: (d) T1 T2 283 K 303 K
Example : 7
The volume of a gas is 100 ml at 100o C . If pressure remains constant then at what temperature it will be

about 200 ml [Roorkee 1993]

(a) 200o C (b) 473o C (c) 746o C (d) 50o C

V1 = V2 i.e. 100 200
T1 T2 373 K = T2

T2 = 746 k = 473o C

If 300 ml of a gas at 27o C is cooled to 7o C at constant pressure, its final volume will be [AIIMS 2000]

(a) 135 ml (b) 540 ml (c) 350 ml (d) 280 ml

V1 = V2 i.e. 300 = V2
T1 T2 300 K 280K

V2 = 280 ml

A flask containing air (open to atmosphere) is heated from 300 K to 500 K. the percentage of air escaped to

the atmosphere is nearly [CBSE PMT 1991]

(a) 16.6 (b) 40 (c) 66 (d) 20

Solution: (c) V1 = V2 i.e. V1 = V2 ; V2 = 5 V = 1.66 V
T1 T2 300 500 3


Volume escaped = 1.66 V − V = 0.66 V = 66% of V

Example : 8 According to Charle’s law, at constant pressure, 100 ml of a given mass of a gas with 10o C rise in

temperature will become  1 = 0.00366 
 273 

(a) 100.0366 (b) 99.9634 (c) 103.66 (d) 100.366

Solution: (a) Vt = V0 + V0 × t = 100 + 1 × 10 = 100 + 0.0366 = 100.0366 ml
273 273

Gay-Lussac's law (Amonton's law).

(1) In 1802, French chemist Joseph Gay-Lussac studied the variation of pressure with temperature and
extende the Charle’s law so, this law is also called Charle’s-Gay Lussac’s law.

(2) It states that, “The pressure of a given mass of a gas is directly proportional to the absolute temperature
(= o C + 273) at constant volume.”

Thus, P ∝ T at constant volume and mass

or P = KT = K(t(o C) + 273.15) (where K is constant)

P = K
T

For two or more gases at constant volume and mass

P1 = P2 = ....... = K
T1 T2

(3) If t = 0o C , then P = P0

Hence, P0 = K × 273.15

∴ K = P0
273.15

P = P0 [t + 273.15] = P0 1 + t = P0 [1 + αt]
273.15 273.15 

where αP is the pressure coefficient, αP = P − P0 = 1 = 3.661 × 10−3 o C −1
tP0 273.15

Thus, for every 1o change in temperature, the pressure of a gas changes by 1  ≈ 1  of the
273.15  273 

pressure at 0o C .

(4) Graphical representation of Gay-Lussac's law : A graph between P

and T at constant V is called isochore. P

O
T(k)


Note :  This law fails at low temperatures, because the volume of the gas molecules become significant.

Example : 9 A sealed tube which can withstand a pressure of 3 atmosphere is filled with air at 27 o C and 760 mm
pressure. The temperature above which the tube will burst will be

(a) 900o C (b) 627o C (c) 627o C (d) 1173o C

Solution: (b) The tube will burst when the final pressure exceeds 3 atm. at constant volume,

P1 = P2 i.e. 760 = 3 × 760
T1 T2 300 K T2

T2 = 900 K = 627o C

Avogadro's law.

(1) According to this law, “Equal volumes of any two gases at the same temperature and pressure contain the
same number of molecules.”

Thus, V ∝ n (at constant T and P)

or V = Kn (where K is constant)

or V1 = V2 = ....... = K
n1 n2

Example, 2H2(g)+ O2(g) → 2H2O(g)
2 moles 1mole 2 moles

2volumes 1volume 2volumes

2 litres 1litre 2 litres

1litre 1 / 2litre 1litre

1n litre 1 / 2nlitre 1n litre

(2) One mole of any gas contains the same number of molecules (Avogadro's number = 6.02 × 1023 ) and by
this law must occupy the same volume at a given temperature and pressure. The volume of one mole of a gas is

called molar volume, Vm which is 22.4 L mol −1 at S.T.P. or N.T.P.
(3) This law can also express as, “The molar gas volume at a given temperature and pressure is a specific

constant independent of the nature of the gas”.

Thus, Vm = specific constant = 22.4 L mol −1 at S.T.P. or N.T.P.
(4) This law is widely applicable to solve the problems of reactive gaseous system.

Note :  Loschmidt number : It is the number of molecules present in 1 c.c. of a gas or vapour at S.T.P.

Its value is 2.687 × 1019 per c.c.

Ideal gas equation.

(1) The simple gas laws relating gas volume to pressure, temperature and amount of gas, respectively, are

stated below :

Boyle's law : P ∝ 1 or V ∝ 1 (n and T constant)
V P

Charle's law : V∝T (n and P constant)

Avogadro's law : V ∝ n (T and P constant)

If all the above law's combines, then


V ∝ nT
P

or V = nRT
P

or PV = nRT

This is called ideal gas equation. R is called ideal gas constant. This equation is obeyed by isothermal
and adiabatic processes.

(2) Nature and values of R : From the ideal gas equation, R = PV = Pressure × Volume
nT mole × Temperature

Force × Volume Force × Length Work or energy .
Area mole × Temperature mole × Temperature
= = =
mole × Temperature

So, R is expressed in the unit of work or energy mol −1 K −1 .

Different values of R are summarised below :

R = 0.0821 L atm mol −1 K −1

= 8.3143 × 107 erg mol −1 K −1

= 8.3143 joule mol −1 K −1 (S.I. unit)
= 8.3143 Nm mol −1 K −1
= 8.3143 KPa dm3 mol −1 K −1
= 8.3143 MPa cm3 mol −1 K −1
= 8.3143 × 10 −3 kJ mol −1 K −1
= 5.189 × 1019 eV mol −1 K −1
= 1.99 cal mol −1 K −1
= 1.987 × 10 −3 K cal mol −1 K −1

Note :  Although R can be expressed in different units, but for pressure-volume calculations, R must be

taken in the same units of pressure and volume.

(3) Gas constant, R for a single molecule is called Boltzmann constant (k)

k = R = 8.314 × 107 ergs mole −1 degree −1
N 6.023 × 10 23

= 1.38 × 10−16 ergs mol −1 degree −1 or 1.38 × 10−23 joule mol −1 degree −1

(4) Calculation of mass, molecular weight and density of the gas by gas equation

PV = nRT = m RT  n = mass of the gas (m) (M) 
M Molecular weight of the gas


∴ M = mRT
PV

d = PM  d = m 
RT  V 

or dT = M
P R

Since M and R are constant for a particular gas,

Thus, dT = constant
P

Thus, at two different temperature and pressure

d1T1 = d 2 T2
P1 P2

(5) Gas densities differ from those of solids and liquids as,

(i) Gas densities are generally stated in g/L instead of g / cm3 .

(ii) Gas densities are strongly dependent on pressure and temperature as,

d∝P

d ∝ 1
T

Densities of liquids and solids, do depend somewhat on temperature, but they are far less dependent on
pressure.

(iii) The density of a gas is directly proportional to its molar mass. No simple relationship exists between the
density and molar mass for liquid and solids.

(iv) Density of a gas at STP = molar mass
22.4

d(N 2 ) at STP = 28 = 1.25 g L−1
22.4

d(O2 ) at STP = 32 = 1.43 g L−1
22.4

Example : 10 The pressure of 2 moles of an ideal gas at 546 K having volume 44.8 L is [CPMT 1995]

(a) 2 atm (b) 3 atm (c) 4 atm (d) 1 atm

Solution: (a) PV = nRT, P × 44.8 = 2 × 0.082 × 546

P = 2 atm

Example : 11 The number of moles of H 2 in 0.224 litre of hydrogen gas at STP (273 K, 1 atm.) is [MLNR 1994]

(a) 1 (b) 0.1 (c) 0.01 (d) 0.001

Solution: (c) PV = nRT , 1 × 0.224 = n × 0.082 × 273 n = 0.01mol

Example : 12 120 g of an ideal gas of molecular weight 40 mole −1 are confined to a volume of 20 L at 400 K. Using

R = 0.0821 L atm K −1 mole −1 , the pressure of the gas is [Pb. CET 1996]

(a) 4.90 atm (b) 4.92 atm (c) 5.02 atm (d) 4.96 atm


Solution: (b) 120 g = 120 = 3 moles
40

P = nRT = 3 × 0.0821 × 400 = 4.92 atm.
V 20

Example : 13 The volume of 2.8 g of carbon monoxide at 27o C and 0.821 atm pressure is (R = 0.0821lit atm K −1 mol −1)

(a) 0.3 litre (b) 1.5 litre (c) 3 litre [Manipal PMT 2001]

(d) 30 litre

Solution: (c) 2.8 g CO = 2.8 = mol = 0.1 mol
28
Example: 14
Solution: (b) PV = nRT

or V = nRT = 0.1 × 0.0821 × 300 = 3 litre
P 0.821

3.2 g of oxygen (At. wt. = 16) and 0.2 g of hydrogen (At. wt. = 1) are placed in a 1.12 litre flask at 0o C . The

total pressure of the gas mixture will be [CBSE PMT 1992]

(a) 1 atm (b) 4 atm (c) 3 atm (d) 2 atm

3.2 g O2 = 0.1mol , 0.2 g H 2 = 0.1mol,

Total n = 0.2 mol ,

P = nRT = 0.2 × 0.082 × 273 = 4atm
V 1.12

Example : 15 The density of methane at 2.0 atmosphere pressure and 27o C is [BHU 1994]

(a) 0.13 g L−1 (b) 0.26 g L−1 (c) 1.30 g L−1 (d) 2.60 g L−1

Solution: (c) d = PM = 2 × 16 = 1.30 g L−1
RT 0.082 × 300

Example : 16 The volume of 0.0168 mol of O2 obtained by decomposition of KClO3 and collected by displacement of

water is 428 ml at a pressure of 754 mm Hg at 25o C . The pressure of water vapour at 25o C is[UPSEAT 1996]

(a) 18 mm Hg (b) 20 mm Hg (c) 22 mm Hg (d) 24 mm Hg

Solution: (d) Volume of 0.0168 mol of O2 at STP = 0.0168 × 22400 cc = 376.3 cc

V1 = 376.3 cc , P1 = 760 mm , T1 = 273 K

V2 = 428 cc , P2 = ? , T2 = 298 K

P1 V1 = P2 V2 gives P2 = 730 mm (approx.)
T1 T2

∴ Pressure of water vapour = 754 − 730 = 24 mm Hg

Dalton's law of partial pressures.

(1) According to this law, “When two or more gases, which do not react chemically are kept in a closed vessel,
the total pressure exerted by the mixture is equal to the sum of the partial pressures of individual gases.”

Thus, Ptotal = P1 + P2 + P3 + .........

Where P1, P2, P3 ,...... are partial pressures of gas number 1, 2, 3 .........

(2) Partial pressure is the pressure exerted by a gas when it is present alone in the same container and at
the same temperature.

Partial pressure of a gas (P1 ) = Number of moles of the gas (n1) × PTotal = Mole fraction (X1)× PTotal
Total number of moles (n) in the mixture


(3) If a number of gases having volume V1, V2, V3 ...... at pressure P1, P2, P3 ........ are mixed together in
container of volume V, then,

PTotal = P1 V1 + P2V2 + P3 V3 .....
V

or = (n1 + n2 + n3 .....) RT ( PV = nRT) or = n RT ( n = n1 + n2 + n3 .....)
V V

(4) Applications : This law is used in the calculation of following relationships,

(i) Mole fraction of a gas (X1) in a mixture of gas = Partial pressure of a gas (P1)
PTotal

(ii) % of a gas in mixture = Partial pressure of a gas (P1 ) × 100
PTotal

(iii) Pressure of dry gas collected over water : When a gas is collected over water, it becomes moist due to
water vapour which exerts its own partial pressure at the same temperature of the gas. This partial perssure of water
vapours is called aqueous tension. Thus,

Pdry gas = Pmoist gas or PTotal − Pwater vapour

or Pdry gas = Pmoist gas − Aqueous tension (Aqueous tension is directly proportional to absolute temperature)

(iv) Relative humidity (RH) at a given temperature is given by :

RH = Partial pressure of water in air .
Vapour pressure of water

(5) Limitations : This law is applicable only when the component gases in the mixture do not react with
each other. For example, N 2 and O2 , CO and CO2 , N 2 and Cl2 , CO and N 2 etc. But this law is not applicable
to gases which combine chemically. For example, H 2 and Cl2 , CO and Cl2 , NH3 , HBr and HCl, NO and O2
etc.

Note :  N 2 (80%) has the highest partial pressure in atmosphere.

(6) Another law, which is really equivalent to the law of partial pressures and related to the partial volumes of
gases is known as Law of partial volumes given by Amagat. According to this law, “When two or more gases, which
do not react chemically are kept in a closed vessel, the total volume exerted by the mixture is equal to the sum of
the partial volumes of individual gases.”

Thus, VTotal = V1 + V2 + V3 + ......

Where V1, V2 , V3 ,...... are partial volumes of gas number 1, 2, 3.....
Example: 17 What will be the partial pressure of H 2 in a flask containing 2 g of H 2 , 14 g of N 2 and 16 g of O2

[Assam JET 1992]

(a) 1/2 the total pressure (b) 1/3 the total pressure(c) 1/4 the total pressure (d) 1/16 the total pressure

Solution: (a) n(H 2 ) = 2 = 1, n(N 2 ) = 14 = 0.5 , n(O2 ) = 16 = 0.5, p(H 2 ) = 1+ 1 0.5 p = 1 p
2 28 32 0.5 + 2


Example : 18 Equal weights of methane and oxygen are mixed in an empty container at 25o C . The fraction of the total

pressure exerted by oxygen is [IIT 1981]

(a) 1/3 (b) 1/2 (c) 2/3 (d) 1 / 3 × 273 / 298

Solution: (a) n(CH 4 ) = w = 1, n(O2 ) = w
16 32

p(O2 ) = w / 32 = 1 = 1
w / 16 + w / 32 2+1 3

Example : 19 In a flask of volume V litres, 0.2 mol of oxygen, 0.4 mol of nitrogen, 0.1 mol of ammonia and 0.3 mol of

helium are enclosed at 27o C . If the total pressure exerted by these non-reating gases is one atmosphere, the

partial pressure exerted by nitrogen is [Kerala MEE 2001]

(a) 1 atm (b) 0.1 atm (c) 0.2 atm (d) 0.4 atm

Solution: (d) PN2 = Mol. fraction of N 2 × Total perssure = 0.2 + 0.4 + 0.3 × 1 atm = 0.4 atm .
0.4 + 0.1

Example : 20 Equal weights of ethane and hydrogen are mixed in an empty container at 25o C . The fraction of the total

pressure exerted by hydrogen is [IIT 1993]

(a) 1 : 2 (b) 1 : 1 (c) 1 : 16 (d) 15 : 16

Solution: (d) n(C2 H 6 ) = w , n (H 2 ) = w
30 2

p(H2) = w/2 = 1 = 15
w / 2 + w / 30 16
1 + 1
15

Example : 21 A gaseous mixture contains 56 g of N 2 , 44 g of CO2 and 16 g of CH 4 . The total pressure of the mixture is

720 mm Hg. The partial pressure of CH 4 is [IIT 1993]

(a) 180 mm (b) 360 mm (c) 540 mm (d) 720 mm

Solution: (a) p(CH 4 ) = 5 / 28 + 16 / 16 × 720 = 2 + 1 + 1 × 720 = 1 × 720 = 180 mm .
44 / 44 + 16 / 16 1 4

Graham's law of diffusion and effusion.

(1) Diffusion is the process of spontaneous spreading and intermixing of gases to form homogenous mixture

irrespective of force of gravity. While Effusion is the escape of gas molecules through a tiny hole such as pinhole in

a balloon.

• All gases spontaneously diffuse into one another when they are brought into contact.

• Diffusion into a vacuum will take place much more rapidly than diffusion into another place.

• Both the rate of diffusion of a gas and its rate of effusion depend on its molar mass. Lighter gases
diffuses faster than heavier gases. The gas with highest rate of diffusion is hydrogen.

(2) According to this law, “At constant pressure and temperature, the rate of diffusion or effusion of a gas is

inversely proportional to the square root of its vapour density.”

Thus, rate of diffusion (r) ∝ 1 (T and P constant)
d

For two or more gases at constant pressure and temperature,


r1 = d2
r2 d1

Note :  Always remember that vapour density is different from absolute density. The farmer is

independent of temperature and has no unit while the latter depends upon temperature and expressed in gm−1 .

(3) Graham's law can be modified in a number of ways as,

(i) Since, 2 × vapour density (V.D.) = Molecular weight

then, r1 = d2 = d2 ×2 = M2
r2 d1 d1 ×2 M1

where, M1 and M 2 are the molecular weights of the two gases.

(ii) Since, rate of diffusion (r) = Volume of a gas diffused
Time taken for diffusion

then, r1 = V1 / t1 = d2
r2 V2 / t2 d1

(a) When equal volume of the two gases diffuse, i.e. V1 = V2

then, r1 = t2 = d2
r2 t1 d1

(b) When volumes of the two gases diffuse in the same time , i.e. t1 = t2

then, r1 = V1 = d2
r2 V2 d1

(iii) Since, r ∝ p (when p is not constant)

then, r1 = P1 = M2  r ∝ 1 
r2 P2 M1 M

Note :  It should be noted that this law is true only for gases diffusing under low pressure gradient.

  CO2 > SO2 > SO3 > PCl3 is order of rates of diffusion.

(4) Rate of diffusion and effusion can be determined as,

(i) Rate of diffusion is equal to distance travelled by gas per unit time through a tube of uniform cross-
section.

(ii) Number of moles effusing per unit time is also called rate of diffusion.

(iii) Decrease in pressure of a cylinder per unit time is called rate of effusion of gas.

(iv) The volume of gas effused through a given surface per unit time is also called rate of effusion.

(5) Applications : Graham's law has been used as follows :

(i) To determine vapour densities and molecular weights of gases.

(ii) To prepare Ausell’s marsh gas indicator, used in mines.

(iii) Atmolysis : The process of separation of two gases on the basis of their different rates of diffusion due to
difference in their densities is called atmolysis. It has been applied with success for the separation of


isotopes and other gaseous mixtures. Example, this process was used for the large-scale separation of
gaseous 235UF6 and 238UF6 during the second world war.

Example : 22 The time taken for a certain volume of a gas ‘X’ to diffuse through a small hole is 2 minutes. It takes 5.65

minutes for oxygen to diffuse under the similar conditions. The molecular weight of ‘X’ is [NCERT 1990]

(a) 8 (b) 4 (c) 16 (d) 32

Solution: (b) rX = M O2
rO2 MX

v/2 = 32 , 5.65 = 32 ,MX =4
v / 5.65 MX 2 MX

Example : 23 The rate of diffusion of methane at a given temperature is twice that of gas X. The molecular weight of X is

[IIT 1990]

(a) 64.0 (b) 32.0 (c) 4.0 (d) 8.0

Solution: (a) rCH 4 = MX ⇒ 2= MX ⇒ MX = 64 .
rX M CH 4 16

Example : 24 Density ratio of O2 and H 2 is 16 : 1. The ratio of their r.m.s. velocities will be [AIIMS 2000]

(a) 4 : 1 (b) 1 : 16 (c) 1 : 4 (d) 16 : 1

Solution: (c) r1 = v1 = d2 = 1 = 1: 4.
r2 v2 d1 16

Example : 25 The rate of diffusion of a gas having molecular weight just double of nitrogen gas is 56 ml s −1 . The rate of

diffusion of nitrogen will be [CPMT 2000]

(a) 79.19 ml s −1 (b) 112.0 ml s −1 (c) 56 ml s −1 (d) 90.0 ml s −1

Solution: (a) rX = M N2 = 28 = 1 ; 56 1 or rN2 = 56 2 = 79.19 ms −1
rN2 MX 56 2 rN2 = 2

Example : 26 50 ml of gas A diffuse through a membrane in the same time as 40 ml of a gas B under identical pressure-

temperature conditions. If the molecular weight of A is 64, that of B would be [CBSE PMT 1992]

(a) 100 (b) 250 (c) 200 (d) 80

Solution: (a) r1 = M2 ⇒ 50 / t = M2
r2 M1 40 / t 64

50 = M2 or 5 = M2 or 25 = M2 or M 2 = 100
40 64 4 64 16 64

Barometric distribution law.

(1) For gaseous systems, gravitational force is negligible but this is not true for the gases of high molecular
masses such as polymer. In this case, the pressure will be different in vertical positions in a container. The variation
of pressure with altitude is given by the so-called Barometric formula.

P = P o e −Mgh / RT


where, P o and P are the pressure of the gas at the ground level and at a height 'h' from the ground respectively.

M is molecular mass of the gas, g is acceleration due to gravity, R is gas constant and T is temperature in kelvin.

(2) Since number of moles of gas 'n' and density of the gas 'd' are directly proportional to pressure hence the
above equation may be expressed as, d = d o e −Mgh / RT and n = no e −Mgh / RT .

(3) The above equations may be expressed as,

log P = log d = log n = − 1 × Mgh
Po do no 2.303 RT

Kinetic theory of gases.

(1) Kinetic theory was developed by Bernoulli, Joule, Clausius, Maxwell and Boltzmann etc. and represents
dynamic particle or microscopic model for different gases since it throws light on the behaviour of the
particles (atoms and molecules) which constitute the gases and cannot be seen. Properties of gases which we
studied earlier are part of macroscopic model.

(2) Postulates

(i) Every gas consists of a large number of small particles called molecules moving with very high
velocities in all possible directions.

(ii) The volume of the individual molecule is negligible as compared to the total volume of the gas.

(iii) Gaseous molecules are perfectly elastic so that there is no net loss of kinetic energy due to their collisions.

(iv) The effect of gravity on the motion of the molecules is negligible.

(v) Gaseous molecules are considered as point masses because they do not posses potential energy. So
the attractive and repulsive forces between the gas molecules are negligible.

(vi) The pressure of a gas is due to the continuous bombardment on the walls of the containing vessel.

(vii) At constant temperature the average K.E. of all gases is same.

(viii) The average K.E. of the gas molecules is directly proportional to the absolute temperature.

(3) Kinetic gas equation : On the basis of above postulates, the following gas equation was derived,

PV = 1 mnur2ms
3

where, P = pressure exerted by the gas, V = volume of the gas, m = average mass of each molecule,

n = number of molecules, u = root mean square (RMS) velocity of the gas.
(4) Calculation of kinetic energy

We know that,

K.E. of one molecule = 1 mu 2
2

K.E. of n molecules = 1 mnu 2 = 3 PV ( PV = 1 mnu 2 )
2 2 3


n = 1, Then K.E. of 1 mole gas = 3 RT ( PV = RT)
2

= 3 × 8.314 ×T = 12.47 T Joules .
2

= Average K.E. per mole = 3 RT = 3 KT  K = R = Boltzmann constant 
N(Avogadro number) 2 N 2  N 

This equation shows that K.E. of translation of a gas depends only on the absolute temperature. This is
known as Maxwell generalisation. Thus average K.E. ∝ T.

If T = 0K (i.e., − 273.15o C) then, average K.E. = 0. Thus, absolute zero (0K) is the temperature at which
molecular motion ceases.

(5) Kinetic gas equation can be used to establish gas laws.

Example : 27 The kinetic energy for 14 grams of nitrogen gas at 127o C is nearly (mol. mass of nitrogen = 28 and gas

constant = 8.31 JK −1 mol −1 ) [CBSE PMT 1999]

(a) 1.0 J (b) 4.15 J (c) 2493 J (d) 3.3 J

Solution: (c) K.E. = 3 RT mol −1
2

or K.E. = 3 nRT = 3 × 14 × 8.31 × 400 J = 2493 J
2 2 28

Molecular collisions.

(1) The closest distance between the centres of two molecules taking part in a collision is called molecular or
collision diameter (σ). The molecular diameter of all the gases is nearly same lying in the
order of 10−8 cm .

(2) The number of collisions taking place in unit time per unit volume, called collision σ
frequency (z). Molecular diameter

(i) The number of collision made by a single molecule with other molecules per unit time are given by,

Z A = 2πσ 2uav.n
where n is the number of molecules per unit molar volume,

n = Avogadro number(N0 ) = 6.02 × 10 23 m−3
Vm 0.0224

(ii) The total number of bimolecular collision per unit time are given by,

Z AA = 1 πσ 2uav. n 2
2

(iii) If the collisions involve two unlike molecules, the number of bimolecular collision are given by,


8πRT (M A +M )  1 / 2
 M AMB 
Z AB = σ 2 B 
AB

where, σ AB = σA +σB and M A , M B are molecular weights (M = mN0 )
2

(iv) (a) At particular temperature; Z ∝ p 2

(b) At particular pressure; Z ∝ T −3 / 2

(c) At particular volume; Z ∝ T1/ 2

(3) During molecular collisions a molecule covers a small distance before it gets deflected. The average
distance travelled by the gas molecules between two successive collision is called mean free path (λ).

λ = No. of Average distance travelled per unit time(uav ) = uav = 1.
collisions made by single molecule per unit time (Z A ) 2πσ 2uavr.n 2πnσ 2

(4) Based on kinetic theory of gases mean free path, λ ∝ T . Thus,
P

(i) Larger the size of the molecules, smaller the mean free path, i.e., λ ∝ 1
(radius)2

(ii) Greater the number of molecules per unit volume, smaller the mean free path.

(iii) Larger the temperature, larger the mean free path.

(iv) Larger the pressure, smaller the mean free path.

(5) Relation between collision frequency (Z) and mean free path (λ) is given by,
Z = urms
λ

Molecular speeds or velocities.

(1) At any particular time, in the given sample of gas all the molecules do not possess same speed, due to the
frequent molecular collisions with the walls of the container and also with one another, the molecules move with
ever changing speeds and also with ever changing direction of motion.

(2) According to Maxwell, at a particular temperature the distribution of speeds remains constant and this
distribution is referred to as the Maxwell-Boltzmann distribution and given by the following expression,

dn0 = 4π  M  3 / 2 / 2 RT .u 2 dc
n  2πRT 
.e −Mu2

where, dn0 = Number of molecules out of total number of Fraction of molecules Ump
molecules n, having velocities between c and c + dc , dn0 / n = Fraction of
the total number of molecules, M = molecular weight, T = absolute 300 K (T1) T1<T2<T3

Ump
Uav Urms Ump

temperature. The exponential factor e −Mu2 / 2RT is called Boltzmann factor. 1500 K (T2) 1800 K (T3)

(3) Maxwell gave distribution curves of molecular speeds for CO2 at Molecular speed


different temperatures. Special features of the curve are :

(i) Fraction of molecules with two high or two low speeds is very small.
(ii) No molecules has zero velocity.

(iii) Initially the fraction of molecules increases in velocity till the peak of the curve which pertains to most
probable velocity and thereafter it falls with increase in velocity.

(4) Types of molecular speeds or Velocities :

(i) Root mean square velocity (urms) : It is the square root of the mean of the squares of the velocity of
a large number of molecules of the same gas.

urms = u12 + u22 + u32 + .....un2
n

urms = 3PV 3RT 3RT = 3kT = 3P
(mN 0 ) = M = (mN 0 ) = M = M m d

where k = Boltzmann constant = R
N0

(a) For the same gas at two different temperatures, the ratio of RMS velocities will be, u1 = T1
u2 T2

(b) For two different gases at the same temperature, the ratio of RMS velocities will be, u1 = M2
u2 M1

(c) RMS velocity at any temperature t oC may be related to its value at S.T.P. as, ut = 3P(273 + t) .
273d

Note :  RMS velocity explained the non-existence of gases in the atmosphere of moon.

When temperature alone is given then, urms = 1.58 × T × 104 cm / sec .
M

If P and T both are given, use equation in terms of temperature, i.e., use urms = 3RT and not 3PV
M M

(ii) Average velocity (vav ) : It is the average of the various velocities possessed by the molecules.

vav = v1 + v2 + v3 + ......vn
n

vav = 8 RT 8kT
πM = πm

(iii) Most probable velocity (αmp ) : It is the velocity possessed by maximum number of molecules of a
gas at a given temperature.

α mp = 2RT = 2PV = 2P
M M d


(5) Relation between molecular speeds or velocities,

(i) Relation between urms and vav : vav = 0.9213 × urms or urms = 1.085 × vav
(ii) Relation between α mp and urms : α mp = 0.816 × urms or urms = 1.224 × α mp

(iii) Relation between α mp and vav : vav = 1.128 × α mp

(iv) Relation between α mp , vav and urms :

α mp : vav : urms

2RT : 8 RT : 3RT
M πM M

2 :8 :3
π
1.414
1 : 1.595 : 1.732

: 1.128 : 1.224 i.e., α mp < vav < urms

Example : 28 The rms velocity of CO2 at a temperature T (in kelvin) is x cm s −1 . At what temperature (in kelvin) the rms

velocity of nitrous oxide would be 4x cm s −1 [EAMCET 2001]

(a) 16 T (b) 2 T (c) 4 T (d) 32 T

Solution: (a) u= 3RT
M

∴ uCO2 = TCO2 × M N2O
u N 2O M CO2 TN2O

i.e., x = T 44 or 1 = T or TN2O = 16 T
4x 44 × TN2O 4 TN2O

Example : 29 The rms velocity of an ideal gas at 27o C is 0.3 ms−1 . Its rms velocity at 927o C (in ms−1 ) is

[IIT 1996; EAMCET 1991]

(a) 3.0 (b) 2.4 (c) 0.9 (d) 0.6

Solution: (d) u= 3RT
M

For the same gas at two different temperatures,

u1 = T1 ; 0.3 300 = 1 , u2 = 0.6 ms−1
u2 T2 u2 = 1200 2

Example : 30 The rms velocity of hydrogen is 7 times the rms velocity of nitrogen. If T is the temperature of the gas

[IIT 2000]

(a) T(H2) = T(N2) (b) T(H2) > T(N2) (c) T(H2) < T(N2) (d) T(H2) = 7T(N2)


Solution: (c) u= 3RT ; ∴ u(H 2 ) = T(H 2 ) × M(N 2 ) or 7= T(H 2 ) × 28 or 7= T(H 2 ) × 14 or T(H 2 ) = 1
M u (N 2 ) M(H 2 ) T(N 2 ) T(N 2 ) 2 T(N 2 ) T(N 2 ) 2

or T(N 2 ) = 2 × T(H 2 ) i.e., T(N 2 ) > T(H 2 )

Example : 31 If the average velocity of N 2 molecules is 0.3 m/s at 27o C , then the velocity of 0.6 m/s will take place at

[Manipal PMT 2001]

(a) 273 K (b) 927 K (c) 1000 K (d) 1200 K

Solution: (d) v = 0.921u

∴ v1 = u1 = T1
v2 u2 T2

∴ 0.3 300 or 1 300 or T2 = 300 × 4 = 1200 K
0.6 = T2 2= T2

Example : 32 The temperature of an ideal gas is reduced from 927o C to 27o C . The rms velocity of the molecules

becomes [Kerala CEE 2001]

(a) Double the initial value (b) Half of the initial value

(c) Four times the initial value (d) Ten times the initial value

Solution: (b) u= 3RT
M

∴ u1 = T1 = 927 + 273 = 1200 4 =2
u2 T2 27 + 273 300 =

∴ u2 = 1 u1
2

Real and ideal gases.

(1) Gases which obey gas laws or ideal gas equation (PV = nRT) at all temperatures and pressures are called
ideal or perfect gases. Almost all gases deviate from the ideal behaviour i.e., no gas is perfect and the concept of
perfect gas is only theoretical.

(2) Gases tend to show ideal behaviour more and more as the temperature rises above the boiling point of
their liquefied forms and the pressure is lowered. Such gases are known as real or non ideal gases. Thus, a “real gas
is that which obeys the gas laws under low pressure or high temperature”.

(3) The deviations can be displayed, by plotting the P-V isotherms of real gas and ideal gas.

real gas

p ideal gas

V
Plot of pressure against volume for ideal

and real gases

(4) It is difficult to determine quantitatively the deviation of a real gas from ideal gas behaviour from the P-V
isotherm curve as shown above. Compressibility factor Z defined by the equation,


PV = ZnRT or Z = PV / nRT = PVm / RT

is more suitable for a quantitative description of the deviation from ideal gas behaviour.

(5) Greater is the departure of Z from unity, more is the deviation from ideal behaviour. Thus, when

(i) Z = 1, the gas is ideal at all temperatures and pressures. In case of N 2 , the value of Z is close to 1 at
50o C . This temperature at which a real gas exhibits ideal behaviour, for considerable range of pressure, is known
as Boyle's temperature or Boyle's point (TB ) .

(ii) Z > 1 , the gas is less compressible than expected from ideal behaviour and shows positive deviation, usual
at high P i.e. PV > RT .

(iii) Z < 1 , the gas is more compressible than expected from ideal behaviour and shows negative deviation,
usually at low P i.e. PV < RT .

(iv) Z > 1 for H 2 and He at all pressure i.e., always shows positive deviation.

(v) The most easily liquefiable and highly soluble gases (NH3 , SO2 ) show larger deviations from ideal
behaviour i.e. Z << 1 .

(vi) Some gases like CO2 show both negative and positive deviation.

(6) Causes of deviations of real gases from ideal behaviour : The ideal gas laws can be derived from
the kinetic theory of gases which is based on the following two important assumptions,

(i) The volume occupied by the molecules is negligible in comparison to the total volume of gas.

(ii) The molecules exert no forces of attraction upon one another. It is because neither of these assumptions
can be regarded as applicable to real gases that the latter show departure from the ideal behaviour.

Vander Waal's equation.

(1) To rectify the errors caused by ignoring the intermolecular forces of attraction and the volume occupied by
molecules, Vander Waal (in 1873) modified the ideal gas equation by introducing two corrections,

(i) Volume correction
(ii) Pressure correction

(2) Vander Waal's equation is obeyed by the real gases over wide range of temperatures and pressures, hence
it is called equation of state for the real gases.

(3) The Vander Waal's equation for n moles of the gas is,

 P + n2a  [V − nb] = nRT
V2
Volume correction for
Pressure correction finite size of molecules
for molecular attraction

a and b are Vander Waal's constants whose values depend on the nature of the gas. Normally for a gas a >> b .


(i) Constant a : It is a indirect measure of magnitude of attractive forces between the molecules. Greater
is the value of a, more easily the gas can be liquefied. Thus the easily liquefiable gases (like
SO2 > NH3 > H 2S > CO2 ) have high values than the permanent gases (like N 2 > O2 > H 2 > He) .

Units of 'a' are : atm. L2 mol −2 or atm. m6mol−2 or N m4 mol −2 (S.I. unit).

(ii) Constant b : Also called co-volume or excluded volume,

b = 4 N 0 v = 4 πr 3 
 3 

It's value gives an idea about the effective size of gas molecules. Greater is the value of b, larger is the
size and smaller is the compressible volume. As b is the effective volume of the gas molecules, the constant
value of b for any gas over a wide range of temperature and pressure indicates that the gas molecules are
incompressible.

Units of 'b' are : L mol −1 or m3 mol −1 (S.I. unit)

(iii) Vander Waal's constant for some gases are,

Name of gas a b

Hydrogen atm litre2mol−2 Nm4mol −2 litre mol−1 m3mol −1
Oxygen 0.245 0.0266 0.0266 0.0266
Nitrogen 1.360 0.1378 0.0318 0.0318
Chlorine 1.390 0.1408 0.039 0.0391
Carbon dioxide 6.493 0.6577 0.0562 0.0562
Ammonia 3.590 0.3637 0.0428 0.0428
Sulphur dioxide 4.170 0.4210 0.0371 0.0371
Methane 6.170 0.678 0.0564 0.0564
2.253 0.0428

(iv) The two Vander Waal's constants and Boyle's temperature (TB ) are related as, TB = a
bR

(4) Vander Waal's equation at different temperature and pressures :
(i) When pressure is extremely low : For one mole of gas,

 P + a  (V − b) = RT or PV = RT − a + Pb + ab
 V2  V V2

(ii) When pressure is extremely high : For one mole of gas,

PV = RT + Pb ; PV = 1 + Pb or Z =1+ Pb
RT RT RT

where Z is compressibility factor.

(iii) When temperature is extremely high : For one mole of gas,


PV = RT .
(iv) When pressure is low : For one mole of gas,

 P + a  (V − b) = RT or PV = RT + Pb − a + ab or PV =1− a or Z =1− a
 V2  V V2 RT VRT VRT

(v) For hydrogen : Molecular mass of hydrogen is small hence value of 'a' will be small owing to smaller

intermolecular force. Thus the terms a and ab may be ignored. Then Vander Waal's equation becomes,
V V2

PV = RT + Pb or PV =1+ Pb or Z =1+ Pb
RT RT RT

In case of hydrogen, compressibility factor is always greater than one.

(5) Merits of Vander Waal's equation :

(i) The Vander Waal's equation holds good for real gases upto moderately high pressures.

(ii) The equation represents the trend of the isotherms representing the variation of PV with P for various

gases.

(iii) From the Vander Waal's equation it is possible to obtain expressions of Boyle's temperature, critical
constants and inversion temperature in terms of the Vander Waal's constants 'a' and 'b'.

(iv) Vander Waal's equation is useful in obtaining a 'reduced equation of state' which being a general
equation of state has the advantage that a single curve can be obtained for all gases when the equation if

graphically represented by plotting the variables.

(6) Limitations of Vander Waal's equation :

(i) This equation shows appreciable deviations at too low temperatures or too high pressures.

(ii) The values of Vander Waal's constants a and b do not remain constant over the entire ranges of T and
P, hence this equation is valid only over specific range of T and P.

(7) Other equations of state : In addition to Vander Waal's equation, there are also equations of state
which have been used to explain real behaviour of gases are,

(i) Clausius equation : P + T(V a c) 2  (V − b) = RT . Here 'c' is another constant besides a, b and R.
 + 


(ii) Berthelot equation :  P + a  (V − b) = RT .
 TV 2 

(iii) Wohl equation : P = RT − a b) + c
(V − b) V(V − V2

(iv) Dieterici equation : P = RT .e −a / RTV . The expression is derived on the basis of the concept that
V −b

molecules near the wall will have higher potential energy than those in the bulk.

(v) Kammerlingh Onnes equation : It is the most general or satisfactory expression as equation

expresses PV as a power series of P at a given temperature.

PV = A + BP + CP 2 + DP 3 + ......
Here coefficients A, B, C etc. are known as first, second and third etc. virial coefficients.


(a) Virial coefficients are different for different gases.
(b) At very low pressure, first virial coefficient, A = RT.
(c) At high pressure, other virial coefficients also become important and must be considered.

The critical state.

(1) A state for every substance at which the vapour and liquid states are indistinguishable is known as critical
state. It is defined by critical temperature and critical pressure.

(2) Critical temperature (Tc) of a gas is that temperature above which the gas cannot be liquified however

large pressure is applied. It is given by, Tc = 8a
27Rb

(3) Critical pressure (Pc) is the minimum pressure which must be applied to a gas to liquify it at its critical

temperature. It is given by, Pc = a
27b 2

(4) Critical volume (Vc) is the volume occupied by one mole of the substance at its critical temperature and
critical pressure. It is given by, Vc = 3b

(5) Critical compressibility factor (Zc) is given by, Zc = Pc Vc = 3 = 0.375
RTc 8

A gas behaves as a Vander Waal’s gas if its critical compressibility factor (Zc ) is equal to 0.375.

Note :  A substance in the gaseous state below Tc is called vapour and above Tc is called gas.

The principle of corresponding states.

(1) In 1881, Vander Waal’s demonstrated that if the pressure, volume and temperature of a gas are expressed
in terms of its Pc , Vc and Tc , then an important generalization called the principle of corresponding states would
be obtained.

(2) According to this principle, “If two substances are at the same reduced temperature (θ) and pressure (π)

they must have the same reduced volume (φ),” i.e. π + 3  (3φ − 1) = 8θ
φ2

where, φ = V / Vc or V = φVc ; π = P / Pc or P = πPc ; θ = T / Tc or T = θTc

This equation is also called Vander Waal's reduced equation of state. This equation is applicable to all
substances (liquid or gaseous) irrespective of their nature, because it is not involving neither of the characteristic
constants.

(3) This principle has a great significance in the study of the relationship between physical properties and
chemical constitution of various liquids.

Degrees of freedom of a gaseous molecule.

(1) The motion of atoms and molecules is generally described in terms of the degree of freedom which they
possess.


(2) The degrees of freedom of a molecule are defined as the independent number of parameters required to
describe the state of the molecule completely.

(3) When a gaseous molecule is heated, the energy supplied to it may bring about three kinds of motion in it,
these are,

(i) The translational motion (ii) The rotational motion (iii) The vibrational motion.

This is expressed by saying that the molecule possesses translational, rotational and vibrational degrees of freedom.

(4) For a molecule made up of N atoms, total degrees of freedom = 3N. Further split up of these is as follows :

Translational Rotational Vibrational

For linear molecule : 3 2 3N – 5

For non-linear molecule : 3 3 3N – 6

Specific and Molar heat capacity of Gases.

(1) Specific heat (or specific heat capacity) of a substance is the quantity of heat (in calories, joules, kcal,
or kilo joules) required to raise the temperature of 1g of that substance through 1o C . It can be measured at
constant pressure (c p ) and at constant volume (cv ).

(2) Molar heat capacity of a substance is the quantity of heat required to raise the temperature of 1 mole of
the substance by 1o C .

∴ Molar heat capacity = Specific heat capacity × Molecular weight, i.e.,

Cv = cv × M and Cp = c p × M .

(3) Since gases upon heating show considerable tendency towards expansion if heated under constant
pressure conditions, an additional energy has to be supplied for raising its temperature by 1o C relative to that
required under constant volume conditions, i.e.,

Cp > Cv or Cp = Cv + Work done on expanson, P∆V(= R)

where, Cp = molar heat capacity at constant pressure; Cv = molar heat capacity at constant volume.

Note :  Cp and Cv for solids and liquids are practically equal. However, they differ considerable in

case of gas because appreciable change in volume takes place with temperature.

(4) Some useful relations of Cp and Cv
(i) Cp − Cv = R = 2 calories = 8.314 J

(ii) Cv = 3 R (for monoatomic gas) and Cv = 3 + x (for di and polyatomic gas), where x varies from gas to gas.
2 2

(iii) Cp =γ (Ratio of molar capacities)
Cv

(iv) For monoatomic gas Cv = 3 calories whereas, Cp = Cv + R = 5calories


Cp 5 R
Cv 2
(v) For monoatomic gas, (γ ) = = 3 = 1.66 .
2
R

Cp 7 R
Cv 2
(vi) For diatomic gas (γ ) = = 5 = 1.40
2
R

(vii) For triatomic gas (γ ) = Cp = 8R = 1.33
Cv 6R

Values of Molar heat capacities of some gases,

Gas Cp Cv Cp– Cv Cp/Cv= γ Atomicity
3.01 1.99 1.661 1
He 5

N2 6.95 4.96 1.99 1.4 2
O2 6.82 4.83 1.99 1.4 2
CO2 8.75 6.71 2.04 1.30 3
H2S 8.62 6.53 2.09 1.32 3

Liquefaction of gases.

(1) A gas may be liquefied by cooling or by the application of high pressure or by the combined effect of both.
The first successful attempt for liquefying gases was made by Faraday (1823).

(2) Gases for which the intermolecular forces of attraction are small such as H 2 , N 2 , Ar and O2 , have low
values of Tc and cannot be liquefied by the application of pressure are known as “permanent gases” while the
gases for which the intermolecular forces of attraction are large, such as polar molecules NH3 , SO2 and H2O
have high values of Tc and can be liquefied easily.

(3) Methods of liquefaction of gases : The modern methods of cooling the gas to or below their Tc and
hence of liquefaction of gases are done by Linde's method and Claude's method.

(i) Linde's method : This process is based upon Joule-Thomson effect which states that “When a gas
is allowed to expend adiabatically from a region of high pressure to a region of extremely low pressure, it is
accompained by cooling.”

(ii) Claude's method : This process is based upon the principle that when a gas expands adiabatically
against an external pressure (as a piston in an engine), it does some external work. Since work is done by the
molecules at the cost of their kinetic energy, the temperature of the gas falls causing cooling.

(iii) By adiabatic demagnetisation.

(4) Uses of liquefied gases : Liquefied and gases compressed under a high pressure are of great
importance in industries.


(i) Liquid ammonia and liquid sulphur dioxide are used as refrigerants.

(ii) Liquid carbon dioxide finds use in soda fountains.

(iii) Liquid chlorine is used for bleaching and disinfectant purposes.

(iv) Liquid air is an important source of oxygen in rockets and jet-propelled planes and bombs.

(v) Compressed oxygen is used for welding purposes.

(vi) Compressed helium is used in airships.

(5) Joule-Thomson effect : When a real gas is allowed to expand adiabatically through a porous plug or a
fine hole into a region of low pressure, it is accompanied by cooling (except for hydrogen and helium which get
warmed up).

Cooling takes place because some work is done to overcome the intermolecular forces of attraction. As a
result, the internal energy decreases and so does the temperature.

Ideal gases do not show any cooling or heating because there are no intermolecular forces of attraction
i.e., they do not show Joule-Thomson effect.

During Joule-Thomson effect, enthalpy of the system remains constant.

Joule-Thomson coefficient. µ = (∂T / ∂P)H . For cooling, µ = +ve (because dT and dP will be − ve ) for
heating µ = −ve (because dT = +ve, dP = −ve) . For no heating or cooling µ = 0 (because dT = 0).

(6) Inversion temperature : It is the temperature at which gas shows neither cooling effect nor heating effect
i.e., Joule-Thomson coefficient µ = 0 . Below this temperature, it shows cooling effect and above this temperature, it
shows heating effect.

Any gas like H2, He etc, whose inversion temperature is low would show heating effect at room
temperature. However, if these gases are just cooled below inversion temperature and then subjected to Joule-
Thomson effect, they will also undergo cooling.


Thermodynamics and thermochemistry

Thermodynamics

Thermodynamics (thermo means heat and dynamics means motion) is the branch of science which deals
with the study of different forms of energy and the quantitative relationships between them.

The complete study of thermodynamics is based upon three generalizations celled first, second and third
laws of thermodynamics. These laws have been arrived purely on the basis of human experience and there is no
theoretical proof for any of these laws.
Basic Terms of Thermodynamics.

(1) System, surroundings and Boundary : A specified part of the universe which is under observation is
called the system and the remaining portion of the universe which is not a part of the
system is called the surroundings. Surroundings
Boundary
The system and the surroundings are separated by real or imaginary

boundaries. The boundary also defines the limits of the system. The system and the

surroundings can interact across the boundary. System

(2) Types of systems

(i) Isolated system : This type of system has no interaction with its surroundings.
The boundary is sealed and insulated. Neither matter nor energy can be exchanged with surrounding. A substance
contained in an ideal thermos flask is an example of an isolated system.

(ii) Closed system : This type of system can exchange energy in the form of heat, work or radiations but not
matter with its surroundings. The boundary between system and surroundings is sealed but not insulated. For
example, liquid in contact with vapour in a sealed tube forms a closed system. Another example of closed system is
pressure cooker.

(iii) Open system : This type of system can exchange matter as well as energy with its surroundings. The
boundary is not sealed and not insulated. Sodium reacting with water in an open beaker is an example of open system.

Surroundings Surroundings Surroundings
Matter
Matter Energy Matter
Energy Energy
(Work/Heat)

Open system Closed system Isolated system

Illustration of characteristics of different types of systems

(iv) Homogeneous system : A system is said to be homogeneous when it is completely uniform
throughout. A homogeneous system is made of one phase only. Examples: a pure single solid, liquid or gas,
mixture of gases and a true solution.

(v) Heterogeneous system : A system is said to be heterogeneous when it is not uniform throughout, i.e.,
it consist of two or more phases. Examples : ice in contact with water, two or more immiscible liquids, insoluble
solid in contact with a liquid, a liquid in contact with vapour, etc.

(vi) Macroscopic system : A macroscopic system is one in which there are a large number of particles
(may be molecules, atoms, ions etc. )

Note : All physical and chemical processes taking place in open in our daily life are open systems

because these are continuously exchanging matter and energy with the surroundings.


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