Chemical Bonding

It is observed that the atoms of all the elements, except those for noble gases, tend to remain in combined state with the atoms of same or other element. They do not exist as single atoms under ordinary conditions. Such atomic aggregates occur as molecules or as giant network. A molecule may be defined as a small electrically neutral cluster or group of mutually bonded atoms. A chemical bond may be defined as “binding force between two atoms in a molecule”.

Cause of Chemical Bonding

Inert gases are very stable and show very few chemical reactions. On observing the electronic configuration of these gases, it was found that in all noble gases (except helium), the outermost shall contains eight electrons (an octet). Helium contains two electrons in the outermost shell.

It was concluded that the chemical stability of noble gases is due to the presence of eight electrons (two in case of helium) in the outermost shell. This electronic arrangement was, therefore, regarded as a stable configuration. In case of all other elements which are chemically reactive, the number of electrons present in outermost shell is less than eight. These elements combine with each other in order to complete their octet and hence, attain a stable configuration.

Thus, the cause of chemical combination is to attain a stable configuration of eight electrons in their outermost shell (octet rule). However, for hydrogen and helium, the stable configuration is of two electrons and is known as duplet. The atoms complete their octets by losing, gaining or sharing electrons.

Limitations of octet rule

Formation of electron deficient compounds like $\mathrm{BeCl} _{3}, \mathrm{BF} _{3}, \mathrm{AlCl} _{3}$. Central atom has less than $8 \mathrm{e}$.

Formation of hypervalent compounds like $\mathrm{PCl} _{5}, \mathrm{SF} _{6}, \mathrm{IF} _{7}, \mathrm{H} _{2} \mathrm{SO} _{4}$ in which central atom has more than 8 electrons.

Formation of compounds of noble gases like $\mathrm{XeF} _{2}, \mathrm{XeF} _{4}, \mathrm{XeF} _{6}$.

Odd electron molecules like $\mathrm{NO}, \mathrm{NO} _{2}$.

Types of chemical Bonds

Broadly speaking, the chemical bonds are classified into following types:

(i) lonic bond

(ii) Covalent bond

(iii) Dative bond or Co-ordinate bond

(iv) Hydrogen bond

(v) Metallic bond

Ionic Bond

When a complete transfer of electrons takes place from one atom to the other so as to complete their octet, the two atoms become oppositely charged ions. These ions are held together by electrostatic force of attraction. The bond thus formed between the two atoms is called an ionic bond or electrovalent linkage. The number of electrons lost or gained is known as the electro-valency.

Factors favouring formation of ionic bond

Low ionization enthalpy (IE)

More negative electron gain enthalpy (EGE)

High lattice enthalpy (LE)

If $L E+E G E>I E$, ionic bond is formed.

Properties of ionic compounds

(a) Non-directional nature of ionic bond : These compounds do not exist as individual molecules. Instead they are often hard solids, always made up of ions held together by strong electrostatic forces of attraction between ions with opposite charge.

(b) Electrical conductivity : When put into water, the ions fall apart. The solution can thus conduct electricity. Similarly, these compounds conduct electricity in the fused state also.

(c) Melting and boiling points : These compounds generally have high melting points. This is because a lot of thermal energy is required to break down the inter-ionic forces and form a liquid. Similarly, they possess high boiling points.

(d) Solubility: These compounds are usually soluble in water but insoluble in benzene or other organic solvents.

Covalent Bond

When a bond is formed by sharing of electrons by the two atoms the bond formed between them is called covalent bond. These mutually shared electrons become the common property of both atoms and also counted towards their stable configuration. Each pair of shared electrons is indicated by a line (一).

For example during formation of $\mathrm{H} _{2}$ molecule, the two hydrogen atoms mutually share their electron. It is represented as.

$$ \dot{\mathrm{H}}+\dot{\mathrm{H}} \longrightarrow \mathrm{H}: \mathrm{H} \text { or } \mathrm{H}-\mathrm{H} $$

Once the covalent bond has been formed, the two bonding electrons are attracted by the two nuclei instead of one, so the bonded state becomes more stable than the non-bonded state. The number of covalent bonds formed by an atom is termed as its covalency. The covalency of an atom will be equal to the number of electrons, the atom needs to become iso-electronic with a noble gas. For example, in the formation of water molecule $\left(\mathrm{H} _{2} \mathrm{O}\right)$, covalency of oxygen is two and that of hydrogen is one as shown below.

$$ 2 \dot{\mathrm{H}}+\dot{\mathrm{O}} \longrightarrow \mathrm{H}: \mathrm{O}: \mathrm{H} \text { or } \mathrm{H}-\mathrm{O}-\mathrm{H} $$

The covalent bond formed by the mutual sharing of one pair of electrons between two atoms is called a single bond. It the electron pairs shared between two atoms are two or three then the bond is said to be double or triple, respectively.

Covalent Bond Parameters

Bond Order: The bond order is the number of bonds present between two atoms in a molecule or ion. The bond order is said to be I when there is only one covalent bond formed between two atoms. For example

The bond order is said to be 2 , when there is a double bond between two atoms. For example

Similarly bond order is 3 when there is a triple bond present between two atoms, as in $\mathrm{C} \equiv \mathrm{N}^{-}$ion. Bond order can be a fraction number also, which is generally in molecules which have resonating structures. As in Ozone $\left(\mathrm{O} _{3}\right)$ molecule

There are total three bonds for two oxygen-oxygen links. Therefore Bender order $=3 / 2$

Bond Length :

It is defined as the distance between the nuclei of two bonded atoms. Bond lengths depend upon the relative sizes of the atoms and the bond order. For example, the bond lengths of hydrogen halides follow the sequence $\mathrm{H}-\mathrm{F}<\mathrm{H}-\mathrm{Cl}<\mathrm{H}-\mathrm{Br}<\mathrm{H}-\mathrm{I}$ because the size of the halogens increases down the periodic group as $\mathrm{F}<\mathrm{Cl}<\mathrm{Br}<\mathrm{I}$. In case of atoms joined by multiple bonds, greater is the bond order, smaller will be the bond length.

Bond Enthalpy (Bond Energy)

Energy is released during the process of covalent bond formation between atoms which leads to the lowering of energy and energy is required to break a bond. Bond enthalpy (or bond energy) is defined as the amount of heat energy that must be supplied to one mole of diatoms molecules in gaseous state to separate its atoms in gaseous state. For poly atomic molecules, average bond enthalpy is normally considered, which is average enthalpy per bond required to dissociate 1 mole of gaseous molecules into their constituent atoms.

Properties of Covalent Compounds

The covalent compounds are found to possess the following general properties, which are opposite to those of ionic compounds :-

(a) These substances are made up of individual covalent molecules with weak inter-molecular forces.

(b) Because of the weak inter-molecular forces, the substances are gases, liquids or soft solids at room temperature.

(c) These substances when put into water do not undergo ionisation. Hence these substances in the fused state or in solution do not conduct electricity. However, in presence of water, some polar covalent compounds can ionise and conduct electricity. (d) These substance have low melting and boiling points because less energy is needed to break down the weak inter-molecular forces.

(e) These substances are commonly soluble in benzene and other organic solvents but insoluble in water.

Polarity of Bonds and Molecules

Non-polar covalent bond

If two similar atoms come close to each other and form a bond by sharing their electrons the shared electrons are equally attracted by the atoms as their electronegativity is same. Hence, no poles are developed. eg. $\mathrm{Cl} _{2}, \mathrm{~N} _{2}, \mathrm{O} _{2}, \mathrm{H} _{2}$.

Polar covalent bond

When two dissimilar atoms combine together to form a covalent bond, shared pair of electrons shifts towards atom having high electronegativity. As a result, more electronegative atom acquires - ve charge and less electronegative atom acquires + ve charge forming dipoles eg. $\mathrm{HCl}$.

The polar character of covalent bonds is measured with the help of a property called dipole moment $(\mu)$. It is defined as the product of the magnitude of charge and the distance between the two atoms.

$\mu=q \times d$ Debye

Dipole moment is a vector quantity and is represented by $\longrightarrow$ with the arrow pointing towards the negative end.

Dipole moment of a molecule is the vectorial sum of the dipole moments of individual bonds.

Ionic bond as an extreme case of polar covalent bond

If electronegativity difference between two atoms bonded by a covalent bond is very high, electron can be completely transferred from less electronegative atom to more electronegative atom. This results in formation of oppositely charged ions bonded by electrostatic forces of attraction. This type of linkage is ionic bond.

Fajan’s Rule

Fajan Discussed the covalent character in ionic compounds and the rule can be summarised in following points

(i) Smaller the cation, greater is its polarising power due to more effective nuclear charge, higher will be the covalent character. For example the polarising power of alkali metal cations follows the order $\mathrm{Li}^{+}>\mathrm{Na}^{+}>\mathrm{K}^{+}>\mathrm{Rb}^{+}>\mathrm{Cs}^{+}$and the covalent character in their chlorides also follows the order $\mathrm{LiCl}>\mathrm{NaCl}>\mathrm{KCl}>\mathrm{RbCl}>\mathrm{CsCl}$.

(ii) Larger the anion, more is the polarisability, hence more is the covalent character. For example covalent character in lithium halides follows the order $\mathrm{LiF}<\mathrm{LiCl}<\mathrm{LiBr}<\mathrm{Lil}$.

(iii) Higher is the oxidation state of the cation, again power of polarisation increases hence the covalent character also increases. For example $\mathrm{FeCl} _{3}$ has more covalent character than $\mathrm{FeCl} _{2}$.

$\stackrel{+3}{\mathrm{AlCl} _{3}}>\mathrm{M}^{+2} \mathrm{MCl} _{2}>\mathrm{NaCl}$ (decreasing covalent character)

The valence shell electron pair repulsion (VSEPR) Theory

Shapes of a variety of polyatomic molecules are easily predicted with the help of this theory. In any molecule, valence electrons can be classified as bonding pairs and non-bonding pairs. Non-bonding electron pairs are also known as lone pairs of electrons or unshared pair of electrons.

The shapes of the molecules or ions are primarily due to repulsions between electron pairs (bonding and non-bonding). These pairs try to be as far apart as possible from each other. The shapes are not due to repulsions between atoms constituting the chemical species. Repulsion between various electron pairs follows the sequence lone pair-lone pair > lone pair-bond pair > bond pair-bond pair.

A molecule with two electron pairs around the central atom eg. $\mathrm{BeCl} _{2}$, will have linear arrangement since this places the electron pairs at maximum possible distance.

$: C \mathrm{Ci}: \mathrm{Be}: \mathrm{Ci}$ : or $\mathrm{Cl}-\mathrm{Be}-\mathrm{Cl}$

The shapes of various $A B _{n}$ type of molecules without lone pairs of electrons are summarised in table 4.1.

Table 4.1 Shape of the Molecules and lons Without Lone Pairs of electrons on the Central Atom

The geometry of molecules or ions is determined by considering both the types of electron pairs but for actual shape, lone pairs are not to be considered. For example in ammonia $\left(\mathrm{NH} _{3}\right)$ molecule, the nitrogen atom contains three bond pairs and one lone pair around it. The molecule is of $\mathrm{AB} _{3} \mathrm{E}$ type. Therefore, the actual shape of the $\mathrm{NH} _{3}$ molecule is trigonal pyramidal. However, the $\mathrm{H}-\mathrm{N}-\mathrm{H}$ angles are less than the normal tetrahedral angle $109.5^{\circ}$. This decrease in angle in due to the presence of lone pair of electrons on nitrogen which repels the bond pairs more strongly than they repel each other.

Shapes of various $A B _{n} E _{m}$ type of molecules are summarised in table 4.2. In the above formula $E$ represents the lone pair of electrons.

Table 4.2 Shapes of the Molecules or lons with Lone Pairs of electrons on the Central Atom

Valence Bond Theory :

This theory was developed by Linus Pauling and it explains the covalent bond formation between two atoms as the interaction or overlap of atomic orbitals. This theory can be described in following points.

(i) A covalent bond is formed when the orbital of one atom is situated in such a way that it overlaps with the orbital of another atom, each of them containing one unpaired electron.

(ii) The atomic orbitals overlap and the overlapped region is occupied by both the electrons. These two electron must have opposite spin.

(iii) As a result of this overlapping, there is maximum electron density between the two atoms. A large part of the binding force of covalent bond results from the attraction of these electrons by the nuclei of both the atoms.

(iv) Atoms maintain their individuality. When the bond is formed, only valence electrons from each bonded atom are involved and the inner atomic orbitals of each atom remain undisturbed.

The simplest example of bond formation by atomic orbital overlap is the formation of $\mathrm{H} _{2}$ molecule. For two hydrogen atoms (say $\mathrm{H} _{\mathrm{a}}$ and $\mathrm{H} _{b}$ ) which are infinitely apart i.e. when there is no interaction between them, their potential energy is taken as zero. As the two atoms approach each other, the electron of $\mathrm{H} _{a}$ is attracted by the nucleus of $\mathrm{H} _{b}$ and electron of $\mathrm{H} _{b}$ is similarly attracted by the nucleus of $\mathrm{H} _{\mathrm{a}}$. In an isolated hydrogen atom the electron is attracted by only a single nucleus, but when two hydrogen atoms are close together, each of the two electrons experiences the attractive pull of two nuclei and energy is lowered. There will also be some repulsion between the two electrons and the two nuclei. At a certain distance between two nuclei these attractive and repulsive forces balance each other. At this distance the two hydrogen atoms form a stable grouping called a hydrogen molecule. This critical distance corresponding to the minimum energy and maximum stability is called the bond length. The energy decreased in this process is called bond energy.

This idea of electron sharing can also be expressed in terms of orbital overlap. When the two atoms are for apart, each electron occupies 1s orbital. As the two atoms approach near to each other, their orbitals overlap partly. The two electrons can now be shared through this overlapped region.

Resonance

The concept of resonance is the outcome of valence bond theory. Resonance is defined as a phenomenon in which a molecule or ion can be represented by more than one electronic arrangements but none of these structures can explain all the properties of the molecule or ion. Such structures are called contributing structures and the molecule is said to have resonance. The actual structure, which

can explain all the properties of the molecule or ion, can not be drawn and is called the resonance hybrid. The resonance hybrid is more stable than any one of the contributing structures. This extra stability is known as resonance energy. The contributing structures are written by keeping the following points is mind.

(i) In each contributing structure, the position of atoms should be same, only the arrangement of electrons may change.

(ii) Each structure must have the same number of unpaired electrons.

(iii) The contributing structures should have energies close to each other. The contribution of a less stable contributing structure is also less.

(iv) A more stable contributing structure is one in which the negative charge resides on most electronegative atom and positive charge on most electropositive atom.

(v) Various contributing structures are written with a double headed arrow between them.

Hybridization

It is a hypothetical phenomenon which is used to explain the shape of molecules, bond length and bond energy. It is considered to be a process in which the atomic orbitals having a little difference in their energy mix up and form equal number of orbitals identical in all respects (eg., shape, size and energy). These hybridized orbitals of one atom overlap with the orbitals of other atoms to give a final shape to the molecule. Hybridization is named on the basis of orbitals involved in the process as summarized in the following table 4.3

Dative Bond or Coordinate Bond

When in the formation of a bond between two atoms, only one atom contributes the pair of electrons and the other atom simply participates in sharing, the bond formed is called dative bond or Co-ordinate bond. This may be explained as under :

In this case, one atom has its octet complete and has at least one lone pair (unshared) of electrons. The Ione pair of electrons belonging to the atom with the complete octet is shared with the other atom containing an empty orbital and thus a link is established. The atom contributing the shared pair of electrons is called the donor while the atom which accepts these shared electrons is called the acceptor. This linkage set up between two atoms is represented by an arrow starting from the donor to acceptor.

Effect of Electronegativity on the shapes of molecules

1. Effect of change of electronegativity of the central atom

Consider the hydrides of Group VI elements and the bond angles in them.

$ \begin{array}{llll} \mathrm{H} _{2}\mathrm{O} & \mathrm{H} _{2}\mathrm{S} & \mathrm{H} _{2}\mathrm{Se} & \mathrm{H} _{2}\mathrm{Te} \\ 104.5^{\circ} & 92.6^{\circ} & 91^{\circ} & 90^{\circ} \\ \end{array} $

The central atom involves $s p^{3}$ hybridisation and is surrounded by four electron pairs, two lone pairs and two bond pairs. Due to the large lone pair-lone pair repulsions, the bonded electron pairs are forced closer to each other and therefore, the bond angle decreases from normal tetrahedral angle of $109.5^{\circ}$.

As we move down the group from oxygen to tellurium, the size of the central atom goes on increasing and its electronegativity goes on decreasing. As a result, the force of repulsion between the bonded pairs of electrons in $\mathrm{H} _{2} \mathrm{O}$ is more than the in $\mathrm{H} _{2} \mathrm{~S}$. The force of repulsion between the bonded pairs of electrons decreases as we move from $\mathrm{H} _{2} \mathrm{O}$ to $\mathrm{H} _{2}$ Te and therefore, the bond angle also decreases in the same order.

Similar behaviour is observed in the case of hydrides of group $V$ elements as given below:

$ \begin{array}{lllll} \mathrm{NH} _{3} & \mathrm{PH} _{3} & \mathrm{AsH} _{3} & \mathrm{SbH} _{3} \\ 107.3^{\circ} & 93.3^{\circ} & 91.8^{\circ} & 91.3^{\circ} \\ \end{array} $

2. Effect of change of electronegativity of the bonded atom

$ \begin{array}{lllll} \mathrm{PF} _{3} & \mathrm{PCl} _{3} & \mathrm{PBr} _{3} & \mathrm{Pl} _{3} \\ 97^{\circ} & 100^{\circ} & 101.5^{\circ} & 102^{\circ} \\ \end{array} $

In these molecules, $\mathrm{P}$ involves $\mathrm{sp}^{3}$ hybridisation forming three sigma bonds and one position is occupied by a lone pair. As we move from $\mathrm{PF} _{3}$ to $\mathrm{PI} _{3}$, the electronegativity of the atom attached to the central phosphorus atom goes on decreasing. As a result, the bonding pair moves closer to the central atom as shown. Therefore, the repulsive effect between the bonding pairs increases and hence bond angle increases.

Effect of Lone pairs on bond angle

The bond angle in $\mathrm{NO} _{2}, \mathrm{NO} _{2}^{+}$and $\mathrm{NO} _{2}^{-}$decreases as :

$\mathrm{NO} _{2}^{+}>\mathrm{NO} _{2}>\mathrm{NO} _{2}^{-}$

In $\mathrm{NO} _{2}^{+}$, there is no unshared electron on $\mathrm{N}$ while in $\mathrm{NO} _{2}^{-}$, there is an unshared electron pair on $\mathrm{N}-$ atom. The unshared electron pair causes more repulsion than an unshared electron. Therefore, the bond pairs in $\mathrm{NO} _{2}^{-}$are forced more closer than in $\mathrm{NO} _{2}^{+}$.

Molecular Orbital Theory

Features

1. When two atomic orbitals combine or overlap, they lose their identity and form new orbitals. The new orbitals formed are called molecular orbitals.

2. Only those atomic orbitals can combine to form molecular orbitals which have comparable energies and proper orientations.

3. The number of molecular orbitals formed is equal to the number of combining atomic orbitals.

4. When two atomic orbitals combine, they form two new orbitals called ‘bonding molecular orbital’ and ‘antibonding molecular orbital’.

5. The bonding molecular orbital has lower energy and hence greater stability than the corresponding antibonding molecular orbital.

6. The bonding molecular orbitals are represented by $\sigma, \pi, \delta$ and antibonding are represented by $\sigma^{\star}, \pi^{\star}, \delta^{\star}$.

7. The shapes of the molecular orbitals formed depend upon the type of the combining atomic orbitals.

8. Filling of $\mathrm{MO}$ takes place according to following rules:

Aufbau principle, Pauli’s exclusion principle, Hund’s rule of maximum multiplicity.

Conditions for combination to atomic orbitals to form molecular orbitals.

1. The combining atomic orbitals should have comparable energy

2. The combining atomic orbitals must have proper orientation

3. The extent of over lapping should be large

The energy level diagram for molecules of hydrogen to nitrogen is as shown in fig. 1. But it becomes different for oxygen, fluorine and neon as show in fig. 2.

Bond order $=\frac{1}{2}\left(N _{b}-N _{a}\right)$

$\mathrm{N} _{\mathrm{a}}=$ No. electrons in antibonding orbitals.

$\mathrm{N} _{\mathrm{b}}=$ No. of electrons in bonding orbitals.

Bond order of $\mathrm{N} _{2}^{+}$

$\sigma 1 s^{2} \sigma^{*} 1 s^{2} \sigma-2 s^{2} \sigma^{\star} 2 s^{2} \pi 2 p _{x}{ }^{2} \pi 2 p _{y}{ }^{2} \sigma 2 p _{z}{ }^{1}$

$N _{b}=9, N _{a}=4$

(b)0. $=\frac{1}{2}(9-4)=\frac{5}{2}=2.5$

Bond order of $\mathrm{N} _{2}^{-}$

$ \begin{aligned} & \sigma 1 s^{2} \sigma^{\star} 1 s^{2} \sigma 2 s^{2} \sigma^{\star} 2 s^{2} 2 p _{x}^{2} \pi 2 p _{y}^{2} \sigma 2 p _{z}^{2} \pi^{\star} 2 p _{x}^{1} \\ & N _{b}=10, N _{a}=5 \\ & \text { B.0. }=\frac{1}{2}(10-5)=\frac{5}{2}=2.5 \end{aligned} $

Bond order of $\mathrm{O} _{2}^{+}$

$ \begin{aligned} & \sigma 1 s^{2} \sigma^{\star} 1 s^{2} \sigma-2 s^{2} \sigma^{\star} 2 s^{2} \sigma 2 p _{z}^{2} \pi 2 p _{x}^{2} \pi 2 p _{y}^{2} \pi^{\star} 2 p _{x}^{1} \\ & N _{b}=10, N _{a}=5 \\ & \text { B.0. }=\frac{1}{2}(10-5)=\frac{5}{2}=2.5 \end{aligned} $

Bond order of $\mathbf{O} _{2}^{2-}$

$ \begin{aligned} & \sigma 1 s^{2} \sigma^{\star} 1 s^{2} \sigma 2 s^{2} \sigma^{\star} 2 s^{2} \sigma 2 p _{z}^{2} \pi 2 p _{x}^{2} \pi 2 p _{y}^{2} \pi^{\star} 2 p _{x}^{2} \pi^{\star} 2 p _{y}^{2} \\ & \text { B.0. }=\frac{1}{2}(10-8)=1 \end{aligned} $

Stability

(b) $0 \propto$ Stability

Bond dissociation energy

(b) $0 \propto$ Bond dissociation energy

Magnetic properties

Paramagnetic - if unpaired electrons are present.

Diamagnetic - If all electrons are paired

Hydrogen Bond

Whenever a substance contains a hydrogen atom linked to a highly electronegative atom $\mathrm{N}, 0$ or $\mathrm{F}$, the latter attracts the pair of electrons more and becomes slightly negative whereas the hydrogen atom becomes a slightly positive end. The negative end of the molecule is attracted by the positive end (i.e., Hydrogen atom) of the other molecule and in this way a bond is formed. Such a bond is called a hydrogen bond and is represented by a dotted line. Thus the hydrogen bond between the hydrogen atom bonded to the electronegative atom $\mathrm{X}$ may be represented as follows:

$$ —X-H—X-H—X-H— $$

In this situation the hydrogen atom is attached simultaneously to two electronegative atoms. Hence it acts as a bridge between the two and is, therefore, called the hydrogen bridge.

Inter-molecular hydrogen bonding

When hydrogen bonding occurs between two or more molecules, it is called inter molecular hydrogen bonding. Some examples are-

(a) Hydrides of Fluorine, Oxygen and Nitrogen :

In case of HF, fluorine is highly electronegative atom. Thus the fluorine atom attracts the pair of electrons so that $\mathrm{F}$ end becomes negative and the $\mathrm{H}$ end becomes positive. The hydrogen atom of one molecule will attract the fluorine atom of the second molecule resulting in the formation of the associated molecule $(\mathrm{HF}) _{n}$ as shown below:-

$\mathrm{H}-\mathrm{F} \ldots \mathrm{H}-\mathrm{F} \ldots \mathrm{H}-\mathrm{F} . . \mathrm{H}-\mathrm{F}$

Similar association is found to occur in water $\left(\mathrm{H} _{2} \mathrm{O}\right)$ and ammonia $\left(\mathrm{Nh} _{3}\right)$. The existence of hydrogen bonds in these compounds is proved by the fact that the compounds show abnormally high melting and boiling points as compared with other hydrides in the same group of the periodic table.

It may be noted that since chlorine, bromine and iodine are not as highly electronegative as fluorine, therefore $\mathrm{HCl}, \mathrm{HBr}$ and $\mathrm{HI}$ molecules do not show hydrogen bonding.

(b) The water molecules in ice form hydrogen bonds causing association due to which it has very open structure. This explains the low density of ice than that of water. Similarly, intermolecular hydrogen bonding exists in alcohols and carboxylic acids also.

Intra-molecular hydrogen bonding

When hydrogen bonding occurs within a molecule, it is called intramolecular hydrogen bonding. It usually involves linking of two groups leading to the formation of a ring structure. An important example of intra-molecular hydrogen bonding is that of o-nitrophenol. Boiling point of 0 -nitrophenol is $214^{\circ} \mathrm{C}$ as compared to $279^{\circ} \mathrm{C}$ for p-nitrophenol. Further 0 -nitrophenol is more volatile in steam and less soluble in water than the other two isomers.

All the above facts can be explained on the assumption that 0 -nitrophenol contains hydrogen bond represented as

Due to the intra-molecular hydrogen bonding, the solubility of 0 -nitrophenol in water is also reduced. In $\mathrm{m}$ - and $\mathrm{p}$ - isomers, intra-molecular hydrogen bonding is not possible so inter-molecular hydrogen bonding takes place. This explains higher boiling points of $\mathrm{m}$-and $\mathrm{p}$-isomers.

Conditions for Hydrogen Bonding

In case of inter molecular hydrogen bonding, the most important condition is that the molecules must contain one hydrogen atom linked to one highly electronegative atom.

In case of intra-molecular hydrogen bonding, the following conditions are favourable for hydrogen bonding:

(i) The molecule should contain two groups such that one group contains $\mathrm{H}$-atom linked to a highly electro-negative atom and the other group should also contain a highly electronegative atom linked to a lesser electronegative atom.

(ii) The molecule should be planar.

Strength of Hydrogen Bond

The hydrogen bond is a weak bond. The strength of a hydrogen bond is $20-40 \mathrm{~kJ} \mathrm{~mol}^{-1}$ as compared with strength of $200-400 \mathrm{~kJ} \mathrm{~mol}^{-1}$ for normal covalent bonds.

Importance of Hydrogen Bond

(i) Hydrogen Bonds are important in fixing properties such as solubilities, melting points and boiling points and in determining the form and stability of crystalline structures.

(ii) Hydrogen bonds are important in biological systems. Proteins, for example, contain both

$\searrow \mathrm{CO}$ and $\mathrm{NH}$ group and hydrogen bonds can be formed to bridge the space between $\mathrm{H}$ and 0 . The structure and hence the properties of proteins depend upon the existence of hydrogen bonds.

(iii) Hydrogen bonding within a single molecule is one of the chief factors in determining the structure of important biological substances.

Metallic Bonding, Electron-sea model

In case of metals, the electrons in the outermost shell (valence electrons) are loosely bound, hence they are also called as free electrons. The remainder portion of the atom is known as Kernel (which is positively charged sphere). These positively charged spheres are packed in a regular fashion. The free electrons are mobile in nature and move from one kernel to another throughout the metal lattice. Thus the metal crystal may be pictured as an arrangement of positive ions immersed in a ‘Sea of mobile electrons’. The electrons move in such a way that they are simultaneously near to two or more kernels and bind them together. The bond thus formed between the metal atoms is known as Metallic bond. Hence a metallic bond may be defined as the bond formed as a result of simultaneous attraction of an electron by two or more than two positive ions of the metal.

SOLVED PROBLEMS

Question 1. Which of the following contains both ionic and covalent bonds

(a) $\mathrm{H} _{2} \mathrm{O}$

(b) $ \mathrm{NaOH}$

(c) $ \mathrm{C} _{6} \mathrm{H} _{5} \mathrm{Cl}$

(d) $ \mathrm{CO} _{2}$

Question 2. Compound having $\mathrm{sp}^{3} \mathrm{~d}$ hybridisation is

(a) $\mathrm{BF} _{3}$

(b) $ \mathrm{PF} _{5}$

(c) $\mathrm{SF} _{6}$

(d) $ \mathrm{IF} _{7}$

Question 3. Which one of the following statements is not true for salicylaldehyde?

(a) It is intra molecularly hydrogen bonded

(b) It is steam volatile

(c) has higher solubility in organic solvents

(d) has high malting and boiling points

Show Answer

Solution: $d$

It will have low melting and boiling points because it has intramolecular hydrogen bonding.

Q.4. The molecule which has pyramidal shape is

(a) $ \mathrm{PCl} _{3}$

(b) $ \mathrm{SO} _{3}$

(c) $\mathrm{CO} _{3}{ }^{2-}$

(d) $ \mathrm{NO} _{3}^{-}$

Question 5. An example of non-polar molecule is

(a) $\mathrm{BF} _{3}$

(b) $\mathrm{ClF} _{3}$

(c) $\mathrm{PCl} _{3}$

(d) $\mathrm{SO} _{2}$

Question 6. In solid argon, the atoms are hold together by

(a) Ionic bond

(b) Hydrogen bond

(c) Vander waals forces

(d) Hydrophobic bonds

Question 7. $\mathrm{NH} _{3}$ and $\mathrm{BF} _{3}$ form an adduct which insolves

(a) Ionic bond

(b) Hydrogen bond

(c) Covalent bond

(d) Coordinate bond

Question 8. The internuclear distance in $\mathrm{H} _{2}$ and $\mathrm{Cl} _{2}$ molecules are 74 and $198 \mathrm{pm}$ respectively. The bond length of $\mathrm{H}-\mathrm{Cl}$ is

(a) $ 272 \mathrm{pm}$

(b) $ 124 \mathrm{pm}$

(c) $ 136 \mathrm{pm}$

(d) $ 248 \mathrm{pm}$

Question 9. Outermost shell of two elements $X$ and $Y$ have two and six electrons respectively. It they combine, the expected formula of the compound will be

(a) $X Y$

(b) $X _{2} Y$

(c) $ X _{2} Y _{3}$

(d) $ \mathrm{Xy} _{2}$

Question 10. What is the bond order in $\mathrm{H} _{2}^{-}$

(a) $-1 / 2$

(b) $1 / 2$

(c) 0

(d) 1

PRACTICE QUESTIONS

1. Which one of the following molecules is planar?

(a) $ \mathrm{NF} _{3}$

(b) $ \mathrm{NCl}$

(c) $\mathrm{PH} _{3}$

(d) $\mathrm{BF} _{3}$

2. Among the following species, identify the isostructural pairs.

$\mathrm{NF} _{3}, \mathrm{NO} _{3}^{-}, \mathrm{BF} _{3}, \mathrm{H} _{3} \mathrm{O}^{+}, \mathrm{N} _{3} \mathrm{H}$

(a) $\left[\mathrm{NF} _{3}, \mathrm{NO} _{3}^{-}\right]$and $\left[\mathrm{BF} _{3}, \mathrm{H} _{3} \mathrm{O}^{+}\right]$

(b) $\left[\mathrm{NF} _{3}, \mathrm{~N} _{3} \mathrm{H}\right]$ and $\left[\mathrm{NO} _{3}^{-}, \mathrm{BF} _{3}\right]$

(c) $\left[\mathrm{NF} _{3}, \mathrm{H} _{3} \mathrm{O}^{+}\right]$and $\left[\mathrm{NO} _{3}^{-}, \mathrm{BF} _{3}\right]$

(d) $\left[\mathrm{NF} _{3}, \mathrm{H} _{3} \mathrm{O}^{+}\right]$and $\left[\mathrm{N} _{3} \mathrm{H}, \mathrm{BF} _{3}\right]$

3. The number and type of bonds in allyl cyanide are

(a) $ 9(\sigma)$ and $4(\pi)$ bonds

(b) $ 9(\sigma)$ and $3(\pi)$ bonds

(c) $8(\sigma)$ and $5(\pi)$ bonds

(d) $ 6(\sigma)$ and $2(\pi)$ bonds

4. Which contains both polar and non-polar bonds?

(a) $ \mathrm{NH} _{4} \mathrm{Cl}$

(b) $\mathrm{HCN}$

(c) $\mathrm{H} _{2} \mathrm{O} _{2}$

(d) $\mathrm{CH} _{4}$

5. The molecule that has linear structure is

(a) $\mathrm{CO} _{2}$

(b) $ \mathrm{NO} _{2}$

(c) $\mathrm{SO} _{2}$

(d) $\mathrm{SiO} _{2}$

6. The $\mathrm{Cl}-\mathrm{C}-\mathrm{Cl}$ angle in 1, 1, 2, 2-tetrachloroethene and tetrachloromethane respectively will be about

(a) $ 120^{\circ}$ and $109.5^{\circ}$

(b) $ 90^{\circ}$ and $109.5^{\circ}$

(c) $109^{\circ}$ and $90^{\circ}$

(d) $ 109.5^{\circ}$ and $120^{\circ}$

7. The molecule which has pyramidal shape is

(a) $ \mathrm{PCl} _{3}$

(b) $ \mathrm{SO} _{3}$

(c) $\mathrm{CO} _{3}{ }^{2-}$

(d) $\mathrm{NO} _{3}^{-}$

8. Which of the following are isoelectronic and isostructural ? $\mathrm{NO} _{3}, \mathrm{CO} _{3}^{2-}, \mathrm{ClO} _{3}^{-}, \mathrm{SO} _{3}$

(a) $ \mathrm{NO} _{3}^{-}, \mathrm{CO} _{3}{ }^{2-}$

(b) $ \mathrm{SO} _{3}, \mathrm{NO} _{3}^{-}$

(c) $\mathrm{ClO} _{3}^{-}, \mathrm{CO} _{3}{ }^{2-}$

(d) $\mathrm{ClO} _{3}^{-}, \mathrm{SO} _{3}$

9. Number of lone pair (s) of electrons in the valence shell of $\mathrm{Xe}$ in $\mathrm{XeOF} _{4}$ is/are

(a) 0

(b) 1

(c) 2

(d) 3

10. Which of the following contains maximum number of lone pairs on the central atom?

(a) $\mathrm{ClO} _{3}^{-}$

(b) $ \mathrm{XeF} _{4}$

(c) $\mathrm{SF} _{4}$

(d) $\mathrm{I} _{3}^{-}$

11. The species having bent-T-shape is

(a) $\mathrm{SO} _{3}$

(b) $ \mathrm{BrF} _{3}$

(c) $\mathrm{SiO} _{3}{ }^{2-}$

(d) $ \mathrm{OF} _{2}$

12. Molecular Shape of $\mathrm{SF} _{4}, \mathrm{CF} _{4}$ and $\mathrm{XeF} _{4}$ are

(a) the same with 2, 0 and 1 lone pair of electrons respectively

(b) the same with 1,1 and 1 lone pair of electros respectively

(c) different with 0,1 and 2 lone pair of electrons respectively

(d) different with 1, 0 and 2 lone pair of electrons respectively

13. Which of the following molecular species has unpaired electrons?

(a) $\mathrm{N} _{2}$

(b) $ F _{2}$

(c) $\mathrm{O} _{2}^{-}$

(d) $ \mathrm{O} _{2}^{2-}$

14. Among the following, the molecule with the highest dipole moment is

(a) $\mathrm{CH} _{3} \mathrm{Cl}$

(b) $\mathrm{CH} _{2} \mathrm{Cl} _{2}$

(c) $\mathrm{CHCl} _{3}$

(d) $\mathrm{CCl} _{4}$

15. The type of hybrid orbitals used by the chlorine atom in $\mathrm{ClO} _{2}^{-}$is

(a) $s p^{3}$

(b) $ \mathrm{sp}^{2}$

(c) $s p$

(d) None of these

16. The species having bond order different from that in $\mathrm{CO}$ is

(a) $\mathrm{NO}^{-}$

(b) $\mathrm{NO}^{+}$

(c) $\mathrm{CN}^{-}$

(d) $\mathrm{N} _{2}$

17. The correct order of hybridisation of the central atom in the following species $\mathrm{NH} _{3},\left[\mathrm{PtCl} _{4}\right]^{2}, \mathrm{PCl} _{5}$ and $\mathrm{BCl} _{3}$ is

(a) $ d p^{2}, d s p^{3}, s p^{2}$ and $s p^{3}$

(b) $ s p^{3}, d s^{2}, s^{3}$ dand $s p^{2}$

(c) $ d p^{2}, s p^{2}, s p^{3}, d s p^{3}$

(d) $ d p^{2}, s p^{3}, s p^{2}, d s p^{3}$

18. Specify the coordination geometry around and hybridisation of $\mathrm{N}$ and $\mathrm{B}$ atoms in a $1: 1$ complex of $\mathrm{BF} _{3}$ and $\mathrm{Nh} _{3}$

(a) $ \mathrm{N}$ : tetrahedral, $s p^{3} ; \mathrm{B}:$ tetrahedral, $s p^{3}$

(b) $ \mathrm{N}$ : pyramidal, $s p^{3} ; \mathrm{B}$ : pyramidal, $\mathrm{sp}^{3}$

(c) $ \mathrm{N}$ : pyramidal, $s p^{3} ; \mathrm{B}:$ planar, $^{2} \mathrm{sp}^{2}$

(d) $ \mathrm{N}$ : pyramidal, $\mathrm{sp}^{3} ; \mathrm{B}$ : tetrahedral, $\mathrm{sp}^{3}$

19. The hybridisation of sulphur in sulphur dioxide is

(a) $\mathrm{sp}$

(b) $ s p^{3}$

(c) $ s p^{2}$

(d) $ d p^{2}$

20. Of the following compounds, which will have a zero dipole moment?

(a) 1,1-dichloroethylene

(b) cis-1,2-dichloroethylene

(c) trans-1,2-dichloroethylene

(d) None of the above

21. The species in which the central atom uses $\mathrm{sp}^{2}$-hybrid orbitals in its bonding is

(a) $\mathrm{PH} _{3}$

(b) $ \mathrm{NH} _{3}$

(c) $\mathrm{CH} _{3}^{+}$

(d) $\mathrm{SbH} _{3}$

22. Which one of the following compounds has $\mathrm{sp}^{2}$-hybridisation?

(a) $\mathrm{CO} _{2}$

(b) $ \mathrm{SO} _{2}$

(c) $\mathrm{N} _{2} \mathrm{O}$

(d) $ \mathrm{CO}$

23. The geometry and the type of hybrid orbital present about the central atom in $\mathrm{BF} _{3}$ is

(a) linear, $\mathrm{sp}$

(b) trigonal planar, $s p^{2}$

(c) tetrahedral, $\mathrm{sp}^{3}$

(d) pyramidal, $s p^{3}$

24. The correct order of increasing $\mathrm{C}-0$ bond length of $\mathrm{CO}, \mathrm{CO} _{3}{ }^{2-}, \mathrm{CO} _{2}$ is

(a) $ \mathrm{CO} _{3}^{2-}<\mathrm{CO} _{2}<\mathrm{CO}$

(b) $ \mathrm{CO} _{2}<\mathrm{CO} _{3}{ }^{2-}<\mathrm{CO}$

(c) $\mathrm{CO}<\mathrm{CO} _{3}^{2-}<\mathrm{CO} _{2}$

(d) $\mathrm{CO}<\mathrm{CO} _{2}<\mathrm{CO} _{3}{ }^{2-}$

25. The geometry of $\mathrm{H} _{2} \mathrm{~S}$ and its dipole moment are

(a) angular and non-zero

(b) linear and non-zero

(c) linear and non-zero

(d) linear and zero

26. The hybridisation of atomic orbitals of nitrogen in $\mathrm{No} _{2}{ } _{2}^{+}, \mathrm{NO} _{3}{ }^{-}$, and $\mathrm{NH} _{4}{ }^{+}$are

(a) $ s p, s p^{3}$ and $s p^{2}$ respectively

(b) $ s p, s p^{2}$ and $s p^{3}$ respectively

(c) $ s p^{2}, s p$ and $s p^{3}$ respectively

(d) $ s p^{2}, s p^{3}$ and $s p$ respectively

27. Amongst $\mathrm{H} _{2} \mathrm{O}, \mathrm{H} _{2} \mathrm{~S}, \mathrm{H} _{2} \mathrm{Se}$ and $\mathrm{H} _{2}$ Te. the one with the highest boiling point is

(a) $\mathrm{H} _{2} \mathrm{O}$ because of hydrogen bonding

(b) $\mathrm{H} _{2}$ Te because of higher molecular mass

(c) $\mathrm{H} _{2} \mathrm{~S}$ because of hydrogen bonding

(d) $\mathrm{H} _{2} \mathrm{Se}$ because of lower molecular mass

28. The molecule which has zero dipole moment is

(a) $\mathrm{CH} _{2} \mathrm{Cl} _{2}$

(b) $\mathrm{BF} _{3}$

(c) $\mathrm{NF} _{3}$

(d) $\mathrm{ClO} _{2}$

29. Which of the following is paramagnetic?

(a) $\mathrm{O} _{2}^{-}$

(b) $\mathrm{CN}^{-}$

(c) $\mathrm{CO}$

(d) $ \mathrm{NO}^{+}$

30. According to MO theory.

(a) $ \mathrm{O} _{2}{ }^{+}$is paramagnetic and bond order greater than $\mathrm{O} _{2}$

(b) $ \mathrm{O} _{2}^{+}$is paramagnetic and bond order less than $\mathrm{O} _{2}$

(c) $ \mathrm{O} _{2}{ }^{+}$is diamagnetic and bond order is less than $\mathrm{O} _{2}$

(d) $ \mathrm{O} _{2}^{+}$is diamagnetic and bond order is more than $\mathrm{O} _{2}$

31. In the compound $\mathrm{CH} _{2}=\mathrm{CH}-\mathrm{CH} _{2}-\mathrm{CH} _{2}-\mathrm{C} \equiv \mathrm{CH}$, the hybridisation of carbon atoms involved in $\mathrm{C} _{2}-\mathrm{C} _{3}$ bond is

(a) $s p-s p^{2}$

(b) $ s p^{3}-s p^{2}$

(c) $ s p-s p^{3}$

(d) $ s p^{2}-s p^{3}$

32. Which one of the following molecules is expected to exhibit diamagnetic behaviour?

(a)$\mathrm{C} _{2}$

(b) $ \mathrm{N} _{2}^{-}$

(c) $\mathrm{O} _{2}$

(d) $ \mathrm{O} _{2}^{-}$

33. Number of $\pi$ bonds and $\sigma$ bonds in the following structure is

(a)6,19

(b)4,20

(c)5,19

(d)5,20

34. In $\mathrm{PO} _{4}{ }^{3-}$ ion, formal charge on the oxygen atom of $\mathrm{P}-\mathrm{O}$ bond is

(a) +1

(b) -1

(c) -0.75

(d) $+0.75$

35. Hydrogen bond is strongest in

(a) $ \mathrm{S}-\mathrm{H} \ldots \mathrm{O}$

(b) $ \mathrm{O}-\mathrm{H} \ldots \mathrm{S}$

(c) $\mathrm{F}-\mathrm{H} \ldots \mathrm{F}$

(d) $\mathrm{F}-\mathrm{H} \ldots \mathrm{O}$

36. Formal charge on two 0 atoms in

(a) $-1,+1$

(b) $-1,0$

(c) $0,+1$

(d) $-1,-1$

37. The pair likely to form strongest hydrogen bonding is

(a) $\mathrm{H} _{2} \mathrm{O} _{2}$ and $\mathrm{H} _{2} \mathrm{O}$

(b) $\mathrm{HCOOH}$ and $\mathrm{CH} _{3} \mathrm{COOH}$

(c) $\mathrm{CH} _{3} \mathrm{COOH}$ and $\mathrm{CH} _{3} \mathrm{COOCH} _{3}$

(d) $\mathrm{SiH} _{4}$ and $\mathrm{SiCl} _{4}$

38. Which of the following has highest boiling point?

(a) $ \mathrm{NH} _{3}$

(b) $\mathrm{PH} _{3}$

(c) $\mathrm{SbH} _{3}$

(d) $\mathrm{AsH} _{3}$

39. Ratio of $\pi$ to $\sigma$ bonds in benzene is

(a) $1: 2$

(b) $1: 6$

(c) $1: 4$

(d) $1: 1$

40. The correct decreasing order of the boiling points of $\mathrm{H} _{2} \mathrm{O}, \mathrm{HF}$ and $\mathrm{NH} _{3}$ is

(a) $\mathrm{HF}>\mathrm{H} _{2} \mathrm{O}>\mathrm{NH} _{3}$

(b) $\mathrm{H} _{2} \mathrm{O}>\mathrm{HF}>\mathrm{NH} _{3}$

(c) $ \mathrm{NH} _{3}>\mathrm{HF}>\mathrm{H} _{2} \mathrm{O}$

(d) $ \mathrm{NH} _{3}>\mathrm{H} _{2} \mathrm{O}>\mathrm{HF}$

41. In which of the following substances will hydrogen bond be strongest?

(a) $\mathrm{HCl}$

(b) $\mathrm{H} _{2} \mathrm{O}$

(c) $\mathrm{HI}$

(d) $\mathrm{H} _{2} \mathrm{~S}$

42. The number of types of bonds between two carbon atoms in calcium carbide is

(a) one sigma, one pi

(b) two sigma, one pi

(c) two sigma, two pi

(d) one sigma, two pi

43. Which of the following has highest bond angle?

(a) $ \mathrm{NO} _{2}^{+}$

(b) $ \mathrm{NO} _{2}^{-}$

(c) $\mathrm{NO} _{2}$

(d) $\mathrm{NO} _{3}^{-}$

44. The correct order of bond angles (smallest first) in

(a) $\mathrm{H} _{2} \mathrm{~S}<\mathrm{NH} _{3}<\mathrm{SiH} _{4}<\mathrm{BF} _{3}$

(b) $ \mathrm{NH} _{3}<\mathrm{H} _{2} \mathrm{~S}<\mathrm{SiH} _{4}<\mathrm{BF} _{3}$

(c) $\mathrm{H} _{2} \mathrm{~S}<\mathrm{SiH} _{4}<\mathrm{NH} _{3}<\mathrm{BF} _{3}$

(d) $\mathrm{H} _{2} \mathrm{~S}<\mathrm{NH} _{3}<\mathrm{BF} _{3}<\mathrm{SiH} _{4}$

45. The correct order in which the $0-0$ bond length increases in the following is

(a) $\mathrm{O} _{2}<\mathrm{H} _{2} \mathrm{O} _{2}<\mathrm{O} _{3}$

(b) $\mathrm{O} _{3}<\mathrm{H} _{2} \mathrm{O} _{2}<\mathrm{O} _{2}$

(c) $\mathrm{H} _{2} \mathrm{O} _{2}<\mathrm{O} _{2}<\mathrm{O} _{3}$

(d) $\mathrm{O} _{2}<\mathrm{O} _{3}<\mathrm{H} _{2} \mathrm{O} _{2}$

46. Which one of the following compounds has the smallest bond angle in its molecule?

(a) $\mathrm{OH} _{2}$

(b) $\mathrm{SH} _{2}$

(c) $\mathrm{NH} _{3}$

(d) $\mathrm{SO} _{2}$

47. The correct order of $\mathrm{C}-0$ bond lengths among $\mathrm{CO} _{2} \mathrm{CO} _{3}{ }^{2-}$ and $\mathrm{CO} _{2}$ is

(a) $\mathrm{CO}<\mathrm{CO} _{2}<\mathrm{CO} _{3}^{2-}$

(b) $\mathrm{CO} _{2}<\mathrm{CO} _{3}^{2-}<\mathrm{CO}$

(c) $\mathrm{CO}<\mathrm{CO} _{3}{ }^{2}<\mathrm{CO} _{2}$

(d) $\mathrm{CO} _{3}{ }^{2-}<\mathrm{CO} _{2}<\mathrm{CO}$

48. Dipole moment of

is $1.5 \mathrm{D}$. The dipole moment of

(a) $1.5 \mathrm{D}$ (b) $2.25 \mathrm{D}$ (c) $1 \mathrm{D}$ (d) $3 \mathrm{D}$

49. Likely bond angles of $\mathrm{SF} _{4}$ molecule are

(a) $ 89^{\circ}, 117^{\circ}$

(b) $ 120^{\circ}, 180^{\circ}$

(c) $45^{\circ}, 118^{\circ}$

(d) $117^{\circ}, 92^{\circ}$

50. Which of the two ions from the list given below have the geometry that is explained by the same hybridization of orbitals: $\mathrm{NO} _{2}^{-}, \mathrm{NO} _{3}^{-}, \mathrm{NH} _{2}^{-}, \mathrm{Nh} _{4}^{+}, \mathrm{SCN}^{-}$?

(a) $ \mathrm{NO} _{2}^{-}$and $\mathrm{NO} _{3}^{-}$

(b) $ \mathrm{NH} _{4}^{+}$and $\mathrm{NO} _{3}^{-}$

(c) $\mathrm{SCN}^{-}$and $\mathrm{NH} _{2}^{-}$

(d) $ \mathrm{NO} _{2}^{-}$and $\mathrm{NH} _{2}^{-}$