In order to explain the characteristic geometrical shapes of polyatomic molecules like CH4, NH3 and H2O etc. , Pauling introduced the concept of hybridization. According to him, the atomic orbitals combine to form new set of equivalent orbitals known as hybrid orbitals. Unlike pure orbitals, the hybrid orbitals are used in bond formation. The phenomenon is known as hybridization which can be defined as the process of intermixing of the orbitals of slightly different energies so as to redistribute their energies , resulting in the formation of new set of orbitals of equivalent energies and shape. For example, when one 2s and three 2p-orbitals of carbon hybridise, there is the formation of four new sp3 hybrid orbitals.
Salient features of Hybridisation
Hybridisation is a theoretical concept which has been introduced to explain some structural properties such as shapes of molecules or equivalence of bonds, etc. which cannot be explained by simple theories of valency. Some salient features of hybridisation are :
(i) The orbitals taking part in hybridisation must have only small difference of energies.
(ii) Both half-filled as well as completely filled orbitals can take part in hybridisation. It implies that promotion of the electrons from lower sub-shell to higher sub-shell is not essential during hybridisation.
(iii) The hybrid orbital has electron density concentrated on one side of the nucleus i.e., it has one lobe relatively larger than the other as shown in Fig.
(iv) The number of hybrid orbitals is equal to the number of orbitals taking part in hybridisation.
(v) The hybridised orbitals have equivalent energies and identical shapes .
(vi) The hybrid orbitals can form more stronger bonds as compared to pure atomic orbitals because they can overlap to a greater extent.
(vii) The hybrid orbitals are directed in space in some preferred direction to have minimum repulsion between electron pairs and thus a stable arrangement. Therefore the type of hybridization indicates the geometry of molecules.
TYPES OF HYBRIDISATION IN CARBON COMPOUNDS
Carbon compounds involve three main types of hybridisations, viz., sp3, sp2 and sp.
(i) Tetrahedral hybridisation(sp3)
(ii) Triogonal hybridisation(sp2)
(iii) Diagonal hybridisation(sp)
1. Tetrahedral hybridisation (sp3) : In tetrahedral hybridisation , one s- and three p-orbitals are combined to form four equivalent orbitals called sp3 hybrid orbitals the hybridisation is known as tetrahedral hybridisation.
These hybrid orbitals point towards, the four corners of a regular tetrahedron with carbon atom in the centre. The hybridisation is therefore, also known as tetrahedral hybridisation. The newly formed sp3 hybrid orbital consists of two lobes one of which is big while the other one is small.
Orientation of sp3 hybrid orbitals
1. Formation of methane molecule : In methane, the carbon atom acquires sp3 hybrid state as follows :
One orbital of 2s-subshell and three orbitals of 2p-subshell of the excited carbon atom get hybridised to form four sp3 hybrid orbitals. The process involving promotion of 2s-electron followed by hybridisation is shown in Fig (above). The sp3 hybrid orbitals of carbon atom are directed towards the corners of a regular tetrahedron. Each of the sp3 hybrid orbitals of carbon atom are directed towards the corners of a regular tetrahedron. Each of the sp3 hybrid orbitals overlap axially with half-filled 1s-orbital of the hydrogen atom constituting a sigma bond. The HCH bond angle is 109028’.
Formation of Methane molecule
(a) excitation of carbon atom and sp3 hybridisation
(b) overlapping of hybrid orbitals with 1s orbitals of H atoms
(c) CH4 molecule
2. Ethane molecule C2H6 : In the formation of ethane molecule, both the carbon atoms assume sp3 hybrid state. One of the hybrid orbitals of carbon atom overlaps axially with similar orbitals of the other carbon atoms to form sp3 sp3 sigma bond. The other three hybrid orbitals of each carbon atom are used in forming
sp3 s sigma bonds with hydrogen as described below :
Formation of C2H6 molecule
(a) Two excited carbon atoms undergoing sp3 hybridisation separately
(b) Overlapping of hybrid orbitals
(c) Ethane molecule
Each C H bond in ethane is sp3-s sigma bond with bond length 108 pm. The C C bond is sp3sp3 sigma bond with bond length 154 pm.
Sp2 Hybridisation : In trigonal hybridisation one s and two p-orbitals of carbon are combined to form three hybrid orbitals and the hybridisation is known as trigonal hybridisation.
Orientation of Three sp2 hybrid orbitals
The unhybridised 2pz orbital of carbon is oriented in a plane perpendicular to the plane containing three hybridised orbitals.
Each sp2 hybrid orbital has one-third
s-character and two-third p-character.
Formation of ethylene molecule, C2H4
In ethylene both the carbon atoms present in the molecule undergoes sp2 hybridisation. One of the three sp2 orbitals of each carbon atom mutually overlap to form a bond between two carbon atoms. The other two hybridised orbitals of each carbon atom form bonds with four hydrogen atoms. Thus all the six atoms of ethylene lie in the same plane and the bond angle between any bonds is 1200. The remaining unhybridised 2p-orbitals of the carbon atoms overlap to a small extent( in the form of electron cloud) at a plane perpendicular to the plane of six atoms to form another CC bond. It is however weak and is termed as bond. Thus the two bonds comprising the double bond between carbon atoms are not equally strong ; the bond being stronger than the bond.
Formation of ethene molecule
In ethene molecule, the C=C bond consists of one sp2-sp2 sigma bond and one pi-bond. Its bond length is 134 pm. CH bond is sp2-s sigma bond with bond length 108 pm. The HCH angle is 117.6 , while that of HCC angle is 1210.
Sp-Hybridisation or Diagonal hybridisation
In diagonal hybridisation, one s and one p-orbital are combined to form two equivalent orbitals called sp-hybrid orbitals and the hybridisation is known as diagonal hybridisation. These sp-hybridised orbitals are oriented in space at angle of 1800. Each sp hybrid orbital has equal s and p character, i.e., 50% s-character and 50% p-character. The molecules in which the central atom is sp hybridised and is linked to two or other atoms directly have linear shape.
Formation of sp-hybrid orbitals
Formation of acetylene C2H2
Both carbon atoms in acetylene are sp-hybridised. The two sp-hybrid orbitals of carbon atom are linear and are directed to an angle of 1800, whereas the unhybridised p-orbitals are perpendicular to sp-hybrid orbitals and also perpendicular to each other as shown in Fig.
Two sp-hybrid orbitals and two p-orbitals
In the formation of acetylene , carbon atom uses its one of the sp-hybrid orbital for overlapping with similar orbital of the other carbon to form CC sigma bond. The other sp-hybrid orbital of each carbon atom overlaps axially with 1s orbital of H atom to form CH sigma bond. Each of the two unhybridised orbitals of both carbon atoms overlaps sidewise to form -bonds. The electron clouds of one -bond lie above and below the internuclear axis, whereas those of the other -bond lie in front and back of the internuclear axis. The overlapping of orbitals has been shown in Fig.
Formation of acetylene molecule
Thus four --electron clouds so formed further merge into another to form a single cylindrical electron cloud around the inter-nuclear axis representing CC sigma bond. It has been shown in Fig.
Cylindrical electron cloud in acetylene
Thus in acetylene molecule C C bond length is 120 pm. CH bond is sp-s sigma bond. The HCC angle is 1800 .i.e., the molecule is linear.
HYBRIDISATION IN BERYLLIUM COMPOUNDS
The electronic configuration of Be is 1s22s2. This means that beryllium has no half-filled orbital and it would be inert. Actually it is not so. Beryllium forms compounds such as BeH2, BeF2 etc. in which it is divalent. To be divalent, it is assumed that Be is raised to an excited state in which two 2s electrons are unpaired and one of them is promoted to the empty 2px orbital or the configuration of excited beryllium atom is 1s22s12px1. Now there are two half-filled orbitals available for bond formation or in other words, beryllium is divalent. It is evident that two electrons available for bond formation belong to different types of orbitals( s and p ) , hence the two bonds will not be identical in nature. But actually both the bonds of beryllium are equivalent. Therefore, in order to get two equivalent bonds, it is postulated that these two orbitals must be mixed or hybridised to form two identical orbitals each having 50% s and 50% p character.
When one s- and one p-orbitals are mixed to form two equivalent orbitals, the hybridisation is known as diagonal hybridisation.
Hence beryllium forms two sp-type orbitals. The two orbitals lie in one axis in a straight line and the bond angle between the orbitals is 1800. Each sp-hybrid orbital consists of one large and one small lobe. The large lobe is much bigger than that of the original p-orbital. This can give better overlapping with an orbital of the other atom. The resulting bond is therefore stronger.
Diagonal hybridisation of Be showing two sp-orbitals
Beryllium fluoride, BeCl2
Diagonal hybridisation of beryllium is illustrated by the orbital picture of beryllium fluoride. In BeCl2 two sp-hybrid orbitals of Be overlap with two half-filled p-orbitals of two fluorine atoms to form two bonds and the molecule is linear, bond angle ClBeCl being 1800.
Linear Beryllium fluoride molecule formed by
HYBRIDISATION IN BORON COMPOUNDS
The electronic configuration of B is 1s22s22px12py02pz0. Boron should therefore form only one bond with other atoms. Boron forms compounds such as BF3 in which it is trivalent. To be trivalent, it is assumed that the boron atom is raised to empty Py orbital or configuration of the excited boron atom is :
B (excited) : 1s22s12px12py12pz0.
Now, there are three half-filled orbitals available for bond formation or in other words, boron is trivalent. The three equivalent orbitals are obtained by mixing one s and two p-orbitals, which are called sp2 type hybrid orbitals. This type of hybridisation is called trigonal hybridisation.
Formation of sp2 hybrids and BCl3 molecule
The shape of the trigonal boron atom is planar i.e., the three orbitals lie in the same plane. The bond orbitals are directed towards the corners of an equilateral triangle and the angle between the bond orbitals is 1200. This shown in the Fig (above). Each of the sp2 hybrid orbitals of boron overlaps axially with half-filled orbital of chlorine atom to form three BCl sigma bonds as shown in Fig.
Hybridisation of Elements involving d orbitals
The elements in the third period contain d-orbitals in addition to s and p orbitals. The energy of 3d orbitals are also comparable to those of 4s and 4p orbitals. As a consequence the hybridization involving either 3s , 3p and 3d or 3d , 4s and 4p is possible. However, since difference in energies of 3p and 4s orbitals is significant , no hybridization involving 3p, 3d and 4s orbitals is possible.
The important hybridization schemes involving s, p and d-orbitals are summarized below :
Shape of molecule / ions Hybridisation type Atomic orbitals Examples
square planar dsp2 d + s + p(2) [Ni(CN)4]2
Trigonal bipyramidal sp3d s + p(3)+d PF5 , PCl5
Square pyramidal dsp3 d + s + p(3) BrF5 , XeOF4
d(2)+s+p(3) SF6, [CrF6]3
Formation of PCl5 (sp3d hybridization)
The ground state and excited state outer electronic configurations of phoshorus ( Z = 15) are represented below.
Now the five orbitals (ie., one s , three p and one d orbitals) are available for hybridization to yield a st of five sp3d hybrid orbitals which are directed towards the five corners of a trigonal bipyramidal as shown in Fig.
Five sp3d hybrid orbitals pointing towards the five corners of a trigonal bipyramid
Three of hybrid orbitals are oriented towards the three corners of an equilateral triangle making an angle of 120 between them. The remaining two hybrid orbitals are oriented at right angles to these orbitals. In PCl5 , these five sp3d orbitals of phosphorus overlap with half-filled orbitals of chlorine atoms to form five PCl sigma bonds. Three PCl bonds lie in one plane and make an angle of 120 with each other. These bonds are called equatorial bonds. The remaining two PCl bonds are perpendicular to the plane of the equatorial bonds, i.e., one above the plane and other below the plane. These bonds are called axial bonds. As the axial bond pair suffer more repulsive interactions from the equatorial bond pairs, therefore, to minimise them the axial bonds slightly elongate. Now as the axial bonds are slightly longer, therefore, these are weaker than equatorial bonds. Because of weaker axial bonds, PCl5 molecule is more reactive.
Trigonal bipyramidal geometry of PCl5 molecule
Formation of SF6 (sp3d2 hybridization)
In SF6 the central sulphur atom has the the central sulphur atom has the ground state outer electronic configuration 3s23p4. In the excited state the available six orbitals i.e., one s , three p and two d are singly occupied by electrons. These orbitals hybridise to form six new sp3d2 hybrid orbitals , which are projected towards the six corners of a regular octahedron in SF6. These six sp3d2 hybrid orbitals overlap with singly occupied orbitals of fluorine atoms to form six SF sigma bonds. Thus SF6 molecule has a regular geometry as shown in Fig.
Octahedral geometry of SF6 molecule
24. Describe the change in hybridization (if any) of the Al atom in the following reaction :
AlCl3 + Cl AlCl4
25. Considering X-axis as the internuclear axis which out of the following will not form a sigma bond and why ?
(a) 1s and 1s (b) 1s and 2Px
(c) 2Py and 2Py (d) 1s and 2s
26. Which hybrid orbitals are used by carbon atoms in the following molecules ?
(a) CH3CH3 (b) CH3CH=CH2
(c) CH3CH2OH (d) CH3CHO
27. The bond angle in NH4+ and CH4 are same but NH3 has different bond angle. Why ?
28. Predict which out of the following species are planar.
(i) NH4+ (ii) CH3+ (iii) SF4 (iv) OF2 (v) H2O
29. Why MgCl2 molecule is linear whereas the molecule of SnCl2 has angular shape ?
30. The hybridization state of oxygen in both water and diethyl ether molecules is the same but they differ in their bond angles. Explain.
MOLECULAR ORBITAL THEORY
Molecular orbital theory was put forward by F. Hund, R.S. Mullikan, Leonard Jones and Charles Coulson (1932) . The theory is modern and rational. It assumes that in molecules atomic orbitals loose their identity and the electrons in molecules are present in new orbitals called molecular orbitals which are not associated with a particular atom but belong to the molecule as a whole.
The salient features of this theory are given below :
i) In molecules , electrons are present in new orbitals called molecular orbitals . Molecular orbitals like atomic orbitals are characterised by a set of quantum numbers.
ii) Molecular orbitals are formed by combination of atomic orbitals of nearly same energies.
iii) Molecular orbitals are not associated with a particular atom but belong to nuclei of all the atoms constituting the molecule. Nuclei of different atoms in the molecule behave as polycentric nucleus.
iv) The number of molecular orbitals formed is equal to the number of atomic orbitals undergoing combination. Among new molecular orbitals formed, half are of lower energy than combining atomic orbitals (bonding molecular orbitals ) and the other half are of energy greater than combining atomic orbitals ( anti-bonding molecular orbitals ).
v) The shapes of molecular orbitals depend upon the shapes of combining atomic orbitals.
vi) The molecular orbitals are filled in the increasing order of energies , starting with orbital of least energy (Aufbau principle ).
vii) A molecular orbital like atomic orbitals can accommodate only two electrons and these electrons must have opposite spins (Pauli’s exclusion principle)
viii) While filling molecular orbitals of equal energy pairing of electrons does not take place until all such orbitals are singly occupied (Hund's Rule).
LINEAR COMBINATION OF ATOMIC ORBITALS (LCAO METHOD)
Molecular orbitals are formed by the combination of atomic orbitals of bonded atoms. In wave mechanics atomic orbitals are expressed by wave functions (). The wave functions are obtained as the solutions of Schrodinger wave equation. Just like atomic orbitals, Schrodinger wave equation can be written to describe the behaviour of the electron in molecules also. However, because of the complex nature of the Schrodinger wave equation, it may not be easy to solve it for molecules. Thus, in view of it, for the sake of convenience an approximate technique to obtain the wave functions for molecular orbitals was applied. This approximate method is known as Linear Combination of Atomic Orbitals Method ( LCAO ).
Let us apply this theory to homonuclear diatomic molecules such as hydrogen molecule. Let us consider two atoms of hydrogen in the molecule as A and B. Each hydrogen has one electron in 1s-orbital in the ground state . These atomic orbitals may be represented by wave functions A and B. Then according to LCAO method , the molecular orbitals in H2 molecule are given by linear combination ( addition or subtraction of wave functions of individual atoms ) of A and B as shown below :
MO = A B
b = A + B
a = A B
The molecular orbital b formed by the addition overlap ( constructive interference of waves ) of atomic orbitals is called bonding molecular orbitals and the molecular orbital a formed by subtraction overlap (destructive interference of waves) of atomic orbitals is called anti-bonding molecular orbital .
Formation of bonding( ) and antibonding (*) molecular orbitals by the linear combination of atomic orbitals A and B centred on two atoms A and B respectively.
The formation of Bonding and Anti-bonding orbitals can be interpreted in terms of sign of wave functions of the orbitals which interact. Such a wave possesses a crest and a trough , therefore the positive and negative signs are arbitrarily assigned to the crest and trough respectively.
Now if the crest of one wave overlaps with the crest of the other , the two waves interact in a constructive interference and therefore the new resulting wave is reinforced i.e., add up or there is in-phase overlap , hence bonding orbitals result by overlap of atomic orbitals with the same sign. On the contrary, if crest of one overlaps with the trough of the other two waves interact in a destructive manner or out of phase overlap or subtraction overlap and thus the resulting wave gets weakened. Thus the anti-bonding orbitals result form atomic orbitals with opposite sign.
The combination of 1s orbitals of hydrogen atoms to form molecular orbitals has been shown in Fig 20.
1s 1s 1s Bonding MO
(A) (B) (A + B)
1s 1s *1s Antibonding MO
(A) (B) (A B)
Molecular orbitals formed by combination of
As is clear from the figure, in the bonding molecular orbital the electron density is mainly concentrated in between the nuclei , so the electrons feel greater force of attraction in the orbital. Hence bonding molecular orbital is of lower energy as compared to atomic orbitals. On the other hand , in anti-bonding molecular orbital the electron density is mainly concentrated away from the nuclei ; in between the nuclei there is a node. So the electrons feel less force of attraction in this orbital and hence anti-bonding molecular orbital is of higher energy as compared to atomic orbitals. Relative energies of atomic orbitals and molecular orbitals are shown in Fig 21.
Fig 21 Relative energies of bonding and
anti-bonding molecular orbitals.
It may be noted that bonding molecular orbital is stabilised almost to the same extent as the anti-bonding molecular orbital is destabilised relative to atomic orbitals.
Differences between bonding and anti-bonding
Bonding Molecular Orbital Anti-bondingMolecular Orbital
1. Bonding molecular orbital is formed by the addition overlap of atomic orbitals, 1. Anti-bonding molecular
orbital is formed by the subtraction overlap of atomic orbitals.
2. It may or may not have a node
2. It always has a node in between the nuclei of bonded atoms.
3. It has lower energy than
the AOs from which it is
formed. 3. It has higher energy than
the AOs from which it is
4 The electron charge density in between the nuclei is high and hence the repulsion between the nuclei is very low. This results in stabilisation of BMO , in other words, the electrons in the BMO favour stable bond formation. 4.The electron charge density in between the nuclei is less and hence the repulsion between the nuclei is high. This results in de-stabilisation of anti-bonding MO. In other words, the electron in the ABMO oppose bond formation.
CONDITIONS FOR COMBINATION OF ATOMIC ORBITALS
For atomic orbitals to combine, resulting in the formation of molecular orbitals , the main conditions are :
i) The combining atomic orbitals should have almost the same energies. For example, in the case of diatomic molecules, 1sorbital of one atom can combine with 1s - orbital of the other atom, but 1sorbital of one atom cannot combine with 2sorbital of the other atom.
ii) The extent of overlap between the atomic orbitals of the two atoms should be large.
iii) The combining atomic orbitals should have the same symmetry about the molecular axis. For example, 2Px orbital of one atom can combine with 2Px orbital of the other atom but not with 2Pz orbital .
Note : It may be noted that Z-axis is taken as the inter-nuclear axis according to modern conventions.
Designations of Molecular Orbitals
Just as atomic orbitals are designated as s, p, d, f etc molecular orbitals of diatoimic molecules are named as (sigma) , (pi ) , (delta ) etc.
The molecular orbitals which are cylindrically symmetrical around inter-nuclear axis are called - molecular orbitals. The molecular orbital formed by the addition of 1s orbitals is designated as 1s and the molecular orbital formed by subtraction of 1s orbitals is designated as *1s (Fig 20) . Similarly combination of 2s orbital results in the formation of two - molecular orbitals designated as 2s and *2s as illustrated in Fig.
2s 2s 2s Bonding MO
(A) (B) (A + B)
2s 2s *2s Antibonding MO
(A) (B) (A B)
Molecular orbitals formed by
addition and subtraction of 2s-orbitals
They differ from 1s and *1s MOs with regard to their size only,
Combination of p-atomic orbitals
There are three p-atomic orbitals represented as Px, Py, and Pz. As a convention, the Z-axis is taken as the internuclear axis. The X- and Y-axes would then be perpendicular to the nuclear line. The combination of two Pz atomic orbitals belonging to different nuclei would give a sigma bonding molecular orbital, represented as 2Pz and sigma anti-bonding molecular orbital *2Pz. The combination of two Px atomic orbitals or two Py atomic orbitals on the other hand would give rise to pi-Bonding Molecular orbitals represented as : 2Px and 2Py and pi-antibonding Molecular Orbitals represented as : *2Px and *2Py. The region of overlapping of pi MOs is at right angles to the molecular axis ie., Z-axis.
The formation of 2Pz BMO by linear additive combination of two Pz atomic orbitals is represented in the Fig 23. The thick dots represent the nuclei.
Formation of (2Pz) BMO by the linear additive combination of two Pz atomic orbitals.
In this combination, the positive lobes of two AOs overlap so that the electron charge density between the two nuclei is enhanced. The nuclei are thus well shielded from each other since the repulsion between the nuclei in such case is minimum, the energy of the MO is lower than the energy of any atomic orbitals which have gone into its formation.
The corresponding ABMO is obtained by the linear subtractive combination of two Pz atomic orbitals , as represented in the following figure (Fig 24).
Overlapping of two Pz atomic orbitals by the linear subtractive combination to form an *(2Pz) ABMO orbital.
COMBINATION OF Px AND Py ORBITALS
Suppose a Px orbital of an atom overlaps with a Px orbital of another atom. The overlap would be positive , but it would be side to side and not end to end as in the case of Pz orbitals. The resulting molecular orbitals will , thus be called pi-molecular orbitals. The formation of the BMO designated as (2Px) and is shown in Fig 25.
Overlapping of two Px AO’s formed by
linear additive combination to give 2Px BMO
The corresponding ABMO formed by linear subtractive combination of Px AOs designated as *2Px is shown in Fig.
Overlapping of two Px AO's by linear subtractive combination to give *2Px ABMO.
The energy of the anti-bonding molecular orbital would be high because the similarly charged nuclei are not effectively screened by the electronic charge and therefore repel each other considerably.
The formation of 2Py BMO and *2Py ABMOs is similar to the formation of 2Px and *2Px MOs. The only difference is that the AOs which combine and MOs which are formed now lie perpendicular to the plane of the paper.
ENERGY LEVEL DIAGRAM FOR MOLECULES
The combination of two 1s atomic orbitals of atoms form two new molecular orbitals designated as 1s and *1s. In the same manner , the 2s and 2p atomic orbitals ie., eight atomic orbitals of two atoms give rise to eight new molecular orbitals viz.,
2s , *2s; 2Px, *2Px ; 2Py , *2Py ; 2Pz ; *2Pz.
The increasing order of energies of various molecular orbitals are given below.
1s , *1s ; 2s , *2s; 2Pz ; 2Px = 2Py ; *2Px =*2Py ; *2Pz.
The molecular orbital energy diagram for homonuclear diatomic molecules like H2, H2+, He2, He2+, Li2, Be2, O2 , F2 and Ne2 is given in .
Molecular Energy Level Diagram for
Diatomic Molecules such as O2, F2 etc.
However, the given sequence of energy levels of molecular orbitals is not correct for all molecules. For instance , it has been observed experimentally that in some diatomic molecules such as B2 , C2 , N2 etc, the molecular orbital energy level diagram shown in Fig 28 is followed.
Fig 28 Molecular Orbital Energy Level Diagram for Diatomic Homonuclear molecules such as B2, C2, N2 etc.
The order for writing molecular orbital configurations of B2 , C2 and N2 molecules can be understood in terms of electron-electron or orbital-orbital interactions which come into play during overlap of orbitals. Thus, owing to these interactions the energy level diagram is modified. Due to the close proximity of 2s and 2Pz orbitals , 2s , *2s and *2Pz. orbitals undergo mixing interactions in view of which the energy of 2Pz orbitals is raised and it becomes greater than 2Px and 2Py which do not experience these intermixing interactions.
It may be noted that following N2, the next molecules O2 and F2 do not exhibit these interactions. It is presumably because of large difference of energy between their 2s and 2Pz orbitals ,such intermixing is insignificant. The fact is also supported by spectroscopic measurements.
RULES FOR ADDING ELECTRONS TO MOs
The molecules are built up by adding electrons to molecular orbitals in the same way as the atoms are built up by adding electrons to atomic orbitals. The principle involved are the same in both cases. These may be summed up as follows :
i) The molecular orbital with lowest energy is filled first.
ii) The maximum number of electrons in a MO cannot exceed two and the two electrons must have opposite spins.
iii) If there are two or more orbitals at the same energy level, pairing of electrons will occur only after each orbital of same energy has one electron.
ELECTRONIC CONFIGURATION OF A MOLECULE
The distribution of electrons among various molecular orbitals is called electronic configuration of the molecule. It can give us very important information about the molecule. From electronic configuration, it is possible to find out the number of electrons in bonding molecular orbitals (Nb) and number of electrons in anti-bonding molecular orbitals (Na).
i) The molecule is stable if Nb Na
ii) The molecule is unstable if Nb Na
iii) The molecule is unstable if Nb = Na
It may be noted that even if number of electrons in bonding MO and number of electrons in anti-bonding MO are equal , the atoms do not combine to form molecule. This is because of the fact that effect of anti-bonding electrons is slightly more than that of bonding electrons.
The stability of the molecule can be determined from the parameter called Bond Order. Bond order may be defined as half the difference between number of electrons in bonding molecular orbitals and number of electrons in antibonding molecular orbitals.
INFORMATIONS CONVEYED BY BOND ORDER
The Bond Order conveys the important informations :
i) If the value of bond order is positive , it indicates a stable molecule and if the value of bond order is negative or zero, it means that the molecule is unstable and is not formed.
ii) Dissociation energy of the molecule is directly proportional to the bond order of the molecule. In other words, greater the bond order, the greater is the bond dissociation energy.
iii) Bond length of the molecule is inversely proportional to the bond order of the molecule. In other words , greater the bond order shorter will be the bond length.
iv) Knowing the bond order, the number of covalent bonds between the atoms in the molecule can be predicted. Bond order of a molecule is equal to the number of covalent bonds between the atoms in the molecule.
v) The magnetic behaviour of molecules can also be predicted, i.e., if all the electrons in a molecule are paired, the substance is diamagnetic and in case there are unpaired electrons in a molecule, the substance is paramagnetic. It may be noted that if the bond order is fractional, the molecule will be definitely paramagnetic.
BONDING IN SOME DIATOMIC MOLECULES
1. Hydrogen molecule (H2)
It is formed by the combination of two hydrogen atoms . Each hydrogen atom has one electron in 1s orbital. Therefore in all there are two electrons in hydrogen molecule which are present in 1s molecular orbital. So the electronic configuration of hydrogen molecule is :
H2 : (1s)2
The molecular orbital diagram of hydrogen molecule is given in Fig 29. The bond order of H2 can be calculated as given below :
Bond order = ½ [Nb - Na]
= ½ [2 0 ] = 1
This means that the two hydrogen atoms are bonded together by a single covalent bond.
Fig 29 Molecular orbital diagram for H2 molecule
The bond dissociation energy of hydrogen molecule has been found to be 458kJ mol1 and bond length equal to 74 pm. Since there are no unpaired electrons in hydrogen molecule, therefore, it is diamagnetic.
2. Hydrogen molecule ion (H2+ )
This is formed by the combination of hydrogen atom containing one electron and hydrogen ion having no electron. This electron is present in the outer most MO ie 1s . So the electronic configuration of H2+ is :
H2+ : (1s)1
The molecular orbital diagram for H2+ is given in Fig 30.
Fig 30 Molecular Orbital Diagram For H2+ ion.
The Bond Order for H2+ ion can be calculated as given below :
Bond Order = ½ [Nb Na]
= ½ [1 0 ]
Since there is an unpaired electron in H2+ion so that it is expected to be paramagnetic. This fact has been confirmed experimentally. The bond length of H2+ is found to be 104 pm and its bond dissociation energy is found to be 269 kJ mol1. Bond lenth of H2+ ion is longer than that of H2 molecule. This is due to its smaller bond order than H2 molecule.
3. HYDROGEN MOLECULE NEGATIVE ION ( H2 )
It is formed by the combination of hydrogen atom containing one electron and hydrogen negative ion containing two electrons. Accordingly, the MO configuration of the ion is :
H2 : (1s )2 (*1s)1
The Bond Order of the ion H2 is :
Bond Order = ½ [ Nb Na] = ½ [2 1 ] = ½
Like H2+ ion its bond order is also positive but this ion H2 is less stable than H2+ion. This is attributable to the presence of one electron in the anti-bonding orbital due to which the destabilising effect is more and hence the stability is less. The presence of one unpaired electron in it , makes it paramagnetic.
4. HELIUM MOLECULE (He2)
The electronic configuration of Helium atom is 1s2. Each Helium atom contains 2 electrons, therefore, in He2 molecule there would be 4 electrons. The electrons will be accommodated in 1s and *1s molecular orbitals. Therefore the electronic configuration of He2 would be :
He2 : (1s )2 (*1s)2
The Molecular orbital diagram for He2 is given in Fig .
Molecular Orbital Diagram of He2 molecule.
Bond Order would be :
Bond order = ½ [ Nb Na]
= ½ [2 2 ] = 0
Since Bond Order for He2 comes out to be zero, therefore this molecule is unstable and does not exist.
5. Helium molecule ion (He2+ )
This ion is formed by the combination of He atom and He+ ion. Thus there are three electrons to be filled into the molecular orbitals. The configuration of the ion is :
He2+ : (1s )2 (*1s)1
Bond Order would be :
Bond Order = ½ [ Nb Na]
= ½ [2 1 ] = ½
The bond order is positive. Therefore , this ion is formed. Its bond dissociation energy is found to be 244 kJ mol1. Since there is one unpaired electron in it, the ion is paramagnetic in nature.
Diatomic molecules in the second row of elements
In these elements, the 1s orbitals are completely filled. In the formation of molecular orbitals, the electrons in the inner shells ( i.e., 1s electrons) of each atom remain essentially unperturbed in their respective atomic orbitals and may be kept out of consideration. In the formation of electronic configurations of these molecules, the letters KK are generally used for denoting the fully filled shells (K shells) in the two atoms.
6. Lithium Molecule Li2
The electronic configuration of lithium ( Z = 3 ) is 1s22s1. There are six electrons in lithium molecule. The four electrons are present in K-shell and there are only two electrons to be accommodated in molecular orbitals. These two electrons go into the 2s BMO which has lower energy and the * 2s ABMO remains empty.
Fig 32 Molecular orbital energy level diagram for Li2 molecule
The electronic configuration of Li2 molecule is thus :
Li2 : KK (2s)2
Bond order = ½ [ 2 0 ] = 1
Thus there is one Li Li sigma bond. The bond dissociation energy of the molecule is quite low, being 105 kJ mol1. The bond length of the molecule is 267 pm.
7. Beryllium molecule Be2
The electronic configuration of Be (Z = 4 ) is 1s22s2. In the formation of a diatomic molecule, two electrons of each atom, i.e., four in all , have to be accommodated in molecular orbitals. Two of these (one pair) will go into 2s BMO, while the other two (second pair) will have to go into * 2s ABMO.
Fig 33 Molecular energy level diagram for Be2 molecule
The electronic configuration of Be2 molecule is, thus :
Be2 : KK (2s)2 *(2s)2
Bond order = ½ [ 2 2 ] = 0
The zero bond order suggests that the existence of stable Be2 molecule is not possible.
8. Boron molecule B2
The electronic configuration of boron (Z= 5) is 1s22s22p1. The outer shell of each contains 3 electrons. In the formation of B2 molecule , there will be six electrons to be accommodated in the molecular orbitals of B2.
Fig 34 Molecular orbital energy level diagram for B2 molecule
The electronic configuration of B2 molecule is :
B2 : KK (2s)2 *(2s)2 (2px)1(2py)1
Bond order = ½ [ 4 2 ] = 1
The molecule has only one bond. The molecule is expected to stable. Since orbital contains a single electron, i.e., unpaired , the molecule B2 is paramagnetic.
9. Carbon molecule C2
The electronic configuration of carbon is 1s22s22p2. The outer shell of each atom contains four electrons. In the formation of C2 molecule, there will be evidently, 8 electrons to be accommodated in the molecular orbitals of C2.
The electronic configuration of C2 is :
C2 : KK (2s)2 *(2s) 2 (2px) 2(2py) 2
Bond order = ½ [ 6 2 ] = 2
Since the molecule does not have any unpaired electron, it is diamagnetic, as observed experimentally. The bond dissociation energy of C2 molecule has found to be 627.9 kJ mol1 and the bond length is equal to 131 pm.
10. Nitrogen molecule N2
The electronic configuration of nitrogen ( Z = 7 ) is 1s22s22p3. The outer shell in this case contains 5 electrons. Thus there are 10 electrons to be accommodated in the molecular orbitals of N2. The electronic configuration of nitrogen molecule is :
N2 : KK(2s) 2*(2s) 2(2Px) 2(2Py) 2(2Pz) 2
Bond order = ½ [ 8 2 ] = 3
Thus , nitrogen molecule contains a triple bond.
It is evident that in the formation of nitrogen molecule, only one anti-bonding orbital is involved ( the remaining four being all bonding orbitals). Hence , nitrogen molecule is highly stable. This is confirmed by its high dissociation energy , viz.., 945.6 kJ mole-1 and small bond length equal to 110 pm. Since there are no unpaired electrons in any orbital, N2 molecule is diamagnetic.
11. Nitrogen Molecule ion (N2+)
This ion is formed by the removal of one electron from N2 molecule.
N2 e N2+
From the electronic configuration of N2 , it is clear that this electron would be removed from (2Pz) bonding molecular orbital. Therefore the electron configuration of N2+ ion is :
N2+ : KK(2s)2*(2s)2(2Px)2(2Py)2(2Pz)1
Bond Order = ½ [ Nb Na] = ½ [7 2 ] =2.5
Since the bond order of N2+ is smaller than N2, therefore , it will have longer bond length and smaller bond dissociation energy than N2 molecule.
12. Nitrogen Molecule ion (N2 )
N2 ion is formed by the gain of one electron by N2 molecule. The electron will enter into *(2Py) orbital. Hence electronic configuration of N2 ion will be :
N2 : KK(2s)2*(2s)2(2Px)2(2Py)2(2Pz)2*(2Px)1
Nb = 8 : Na = 3
Bond order = ½ [ 8 3] = 5/2
This has one unpaired electron in the *(2Px) orbital , therefore it is paramagnetic.
12. The N22 ion
The ion N22 is formed by gaining two electrons. These two electrons will be added to *(2Px) and *(2Py) molecular orbitals one in each. Hence electronic configuration of N22 ion will be :
Nb = 8 : Na = 4
Bond order = ½ [ 8 4] = 2
As it contains two unpaired electrons , it is paramagnetic.
Relative stabilities , Bond dissociation energies and Bond lengths of N2 , N2+ , N2 and N22 species
The bond orders of the species are :
Species Bond order
N2+ 2 ½
N2 2 ½
As bond energies are directly proportional to the bond orders, therefore the dissociation energies of these molecular species are in the order :
N2 > N2+ > N2 > N22
As greater the bond dissociation energy, greater is the stability, the stability of these species is also in the above order.
As bond length is inversely proportional to the bond order , therefore there bond lengths will be in the order :
N22 > N2 > N2+ > N2
If bond order is considered, N2 and N2+ have B.O = 2.5 , but N2 contains more electron in antibonding MO, so it is considered less stable. The results are summed up in the following TABLE.
Species Bond order Stability Magnetic character
N2 3 More stable Diamagnetic
N2+ 2 ½ Less stable Paramagnetic
N2 2 ½ Less stable Paramagnetic
N22 2 Least stable Paramagnetic
13. Oxygen Molecule (O2)
The electronic configuration of oxygen is 1s2 2s22p4. In the formation of O2 molecule, 12 outer electrons are to be accommodated in MOs. Four of these electrons fill the (2s) and * (2s) orbitals. We are now left with 8 electrons. Six of them ( i.e., 3 pairs) go into the three MOs , viz., (2 Pz) , (2Px) and (2Py). Since all the bonding MOs are now filled, the remaining two electrons go into the antibonding MOs. The lowest ones are *(2Px) and *(2Py).
The electronic configuration of oxygen molecule is thus :
O2 : KK (2s)2 , *(2s)2; (2Pz)2 ; (2Px)2 , (2Py)2 ;
*(2Px)1, *(2Py )1
Bond Order = ½ [ Nb Na] = ½ [8 - 4 ] = 2
So in oxygen molecule , atoms are held by a double covalent bond . The bond dissociation energy is 495 kJ mol1 and bond length is 121 pm. Moreover, from molecular orbital diagram of O2 molecule , it may be noted that it contains 2 unpaired electrons in *(2Px) and *(2Py) molecular orbitals. Therefore, O2 molecule has paramagnetic nature. In this way, the MO theory successfully explains the paramagnetic nature of oxygen. The Molecular Orbital Diagram of oxygen molecule is shown in Fig 35.
Fig 35 M.O diagram for O2 molecule
1. O2+, O2 and O22 ions
O2+, is formed by the loss of one electron from O2 molecule. The electron will be lost from *(2Px) or *(2Py ) molecular orbital. Hence the electronic configuration of O2+ will be :
O2+ : KK (2s)2 , *(2s)2; (2Pz)2 ; (2Px)2 , (2Py)2 ;
*(2Px)1, *(2Py )0
Nb = 8 : Na = 3
Bond order = ½ [ 8 3 = 5/2 = 2 ½
Since O2+ has has one unpaired electron in the *(2Px) orbital, therefore it is paramagnetic in nature.
O2 ion (superoxide ion) is formed by the gain of one electron by O2 molecule. This electron will be added up in the *(2Px) or *(2Py ) molecular orbital. Hence electronic configuration of O2 will be :
O2 : KK (2s)2 , *(2s)2; (2Pz)2 ; (2Px)2 , (2Py)2 ;
*(2Px)2, *(2Py )1
Nb = 8 : Na = 6
Bond order = ½ [ 8 5] = 3/2 = 1 ½
Since it has one unpaired electron in (2Py) molecular orbital, therefore it is paramagnetic.
O22 ion (peroxide ion) is formed when O2 gains two electrons. These two electrons will be added to the *(2Px) and *(2Py ) molecular orbitals, one in each. Hence the electronic configuration of O22 ion will be :
O22 : KK (2s)2 , *(2s)2; (2Pz)2 ; (2Px)2 , (2Py)2 ;
*(2Px)2, *(2Py )2
Nb = 8 : Na = 6
Bond order = ½ [ 8 6] = 1
Since it contains no unpaired electron, therefore it is diamagnetic in nature.
Relative stabilities , Bond dissociation energies and Bond lengths of O2+, O2 and O22 ions
The Bond order of these species are :
Species Bond order
O2+ 2 ½
O2 1 ½
As the bond dissociation energies are directly proportional to the bond orders, therefore dissociation energies of these molecular species are in the order :
O2+ > O2 > O2 > O22
As greater the bond dissociation energy, greater is the stability, the stability of these species is also in the above order.
As bond length is inversely proportional to bond order, therefore , there bond lengths will be in the order :
O22 > O2 > O2 > O2+
15. Fluorine molecule (F2)
The electronic configuration of fluorine ( Z = 9 ) is 1s2 2s22p5. In the formation of F2 molecule by combination of two fluorine atoms , there would be 14 outer electrons to be accounted in molecular orbitals. The electronic configuration of fluorine molecule is thus,
F2 : KK (2s)2 , *(2s)2; (2Pz)2 ; (2Px)2 , (2Py)2 ;
*(2Px)2 *(2Py )2
The bond order of the molecule is :
Bond Order = ½ [ Nb Na]
= ½ [8 6 ] =1
The fluorine molecule , thus contains a single bond. Because of the presence of as many as 6 electrons in the ABMOs, the bonding in F2 is weaker than that in O2. Therefore, bond dissociation energy of F2 is very low being 155 kJ mol1 and bond length is 142 pm.
13. Neon molecule (Ne2)
The electronic configuration of Neon ( Z = 10 ) is 1s2 2s22p6. In the formation of Ne2 , 16 outer electrons have to be accommodated in the molecular orbitals. The electronic configuration of hypothetical neon molecule is :
Ne2 : KK (2s)2 , *(2s)2; (2Pz)2 ; (2Px)2 , (2Py)2 ;
*(2Px)2 ; *(2Py )2; *(2Pz)2
The Bond Order of the molecule is :
Bond Order = ½ [ Nb Na]
= ½ [8 8] = 0
Since antibonding MOs slightly dominate over the bonding MOs , Ne2 molecule would be incapable of existence.
31. Calculate the Bond Order for O2+ ion.
32. Give reasons for the following :
33. The bond in H2 is stronger than in H2+.
34. Why Helium exists as a monoatomic molecule ? Why He2 is not formed ?
35. Give the number of electrons , which occupy the bonding orbitals in H2+, H2 and He2.
36. Use the MO energy level diagram and show that N2 , would be expected to have a triple bond, F2 single and Ne2 no bond.
37. Calculate the bond orders in H2+ ion H2 and H2 molecule.
38. Arrange the following molecular species in increasing order of stability:
O2 , O2+, O2 and O22
39. Which of H2+, H2 and H22 have the same bond order. Justify their M.O diagrams.
40. Using M.O. diagrams and occupancy of electrons in orbitals, arrange the following in the order of their stabilities.
(i) H2 ii) H2 iii) H2+
41. The bond order in He2+ molecule is equal to the bond
order of H2 molecule. Justify the statement.
In compounds of hydrogen with strongly electronegative elements, such as fluorine, oxygen and nitrogen, electron pair shared between the two atoms lie far away from the hydrogen atom. As a result, the hydrogen atom becomes highly electropositive with respect to the other atom. This phenomenon of charge separation in the case of hydrogen fluoride is represented as . Such a molecule is said to be polar. The molecule behaves as a dipole because one end carries a positive charge and the other end a negative charge. The electrostatic force of attraction between such molecules should be very strong. This is because the positive end of one molecule is attracted by the negative end of the other molecule. Thus, two or molecules may associate together to form larger cluster of molecules. This is illustrated below for the association of several molecules of hydrogen fluoride.
The cluster of HF molecules may be described as (HF)n.
It may be noted that hydrogen atom is bonded to fluorine atom by a covalent bond in one molecule and by electrostatic force or by hydrogen bond to the fluorine atom in the adjacent molecule. Hydrogen atom is thus seen to act as a bridge between the two fluorine atoms.
The attractive force which binds the atom of one molecule with an electronegative atom (such as fluorine, oxygen or nitrogen) of another molecule, generally of the same substance , is known as hydrogen bond.
The hydrogen bond is represented by a dotted line, as shown above. The solid lines represent the original(covalent ) bond present in the molecule.
Chlorine, bromine and iodine are not as highly electronegative as fluorine and therefore, the shared pair of electrons in the case of HCl , HBr and HI do not lie as far away from hydrogen as in the case of HF. The tendency to form hydrogen bond in these cases is therefore less.
Water molecule, because of its bent structure, is also a dipole, oxygen end carrying a negative charge and hydrogen end carrying a positive charge. Hydrogen bond taking place in this case as well, as represented below:
The cluster of water molecules may be described as (H2O)n
The nature of hydrogen bond : The hydrogen bond is a class in itself. It arises from electrostatic forces between positive end (pole) of one molecule and the negative end(pole) of the other molecule generally of the same substance. The strength of hydrogen bond has been has been found to vary between 1040 kJ/mol (i.e., 6.02 x 1023 bonds) while that of a covalent bond has been found to be of the order of 400 kJ per mole. Thus a hydrogen bond is very much weaker than a covalent bond. Consequently, the length of hydrogen bond is bigger than the length of a covalent bond.
In the case of hydrogen fluoride, for instance, while the length of the covalent bond between F and H atoms is 100 pm, the length of hydrogen bond between F and H atoms of neighbouring molecules is 155 pm.
Types of hydrogen bonding : Hydrogen bonding may be classified into two types :
(a) Intermolecular hydrogen bonding : This type of hydrogen bonding involves electrostatic forces of attraction between hydrogen and electronegative element of two different molecules of the substance. Hydrogen bonding in molecules of HF, NH3, H2O etc are examples of intermolecular hydrogen bonding.
(b) Intramolecular hydrogen bonding : This type of bonding involves electrostatic forces of attraction between hydrogen and electronegative element both present in the same molecule of the substance. Examples o-nitrophenol and salicylaldehyde.
p-Nitrophenol , on account of large distance between two groups , does not show any intramolecular hydrogen bonding. On the other hand, it shows the usual inter molecular hydrogen bonding , as illustrated below:
As a result of intermolecular hydrogen bonding, the para derivative undergoes association, resulting in an increase in molar mass and hence an increase in boiling point. In ortho derivative, on account of intramolecular hydrogen bonding, no such association is possible. Consequently, the ortho derivative is more volatile than the para derivative. Thus, while ortho nitrophenol is readily volatile in steam , para nitrophenol is completely non-volatile. The two derivatives can thus be separated from each other by steam distillation.
Consequences of Hydrogen Bonding
1. Boiling points of Binary Hydrogen compounds Consider boiling points of compounds of hydrogen with various elements of Group 15, 16 and 17, are shown in the TABLE
Formula Molecular weight Boiling point(oC)
SbH3 125 17.0
AsH3 78 55.1
PH3 34 84.6
NH3 17 33.0
H2Te 130 1.8
H2Se 81 42.0
H2S 33 59.6
H2O 18 + 100
HI 128 3.5
HBr 81 67.1
HCl 36.5 85.0
HF 20 + 19.4
It is seen that although the boiling point in each group decreases with decrease in molecular mass, there is a sudden reversal in the case of ammonia, water and hydrogen fluoride in Groups 15, 16 and 17 respectively. The unusually high boiling point of each compound , is a consequence of strong intermolecular forces due to hydrogen bonding.
Thus, while H2S, in which there is no hydrogen bonding, is a gas, H2O , in which there is considerable hydrogen bonding, is a high boiling liquid.
Similarly, it has been observed that the three compounds HF, H2O and N H3 have abnormally high melting points in their groups.
Relative boiling points of hydrides of Group 15, 16 and 17
2. Association of molecules : Due to hydrogen bonding, many compounds exist as aggregates of two or more molecules. For example , formic acid exists as a dimer, (HCOOH)2, as shown below :
3. Solubility : Hydrogen bonding also accounts for the solubility of certain compounds in water. For example, lower alcohols are freely miscible with water because their molecules form hydrogen bonds with water molcules.
Some Unique Properties of Water : Water has some unique properties . Two of these are :
1. Density in solid state(ice) is less than that in liquid state. This is some what unusual because in most substances density in solid is more than that in liquid state.
2. Water contracts when heated between 00C and 40C. This is again unusual because most substances expand when heated in all temperature ranges.
Both these peculiar features are due to hydrogen bonding, as discussed below :
(i) In ice, hydrogen bonding between H2O molecules is more extensive than in liquid water. A substance in solid state has a definite structure and the molecules are more rigidly fixed relative to one another than in the liquid state. In ice, the H2O molecules are tetrahedrally oriented with respect to one another. This has been shown in Fig .
At the same time , each oxygen atom is surrounded tetrahedrally by four hydrogen atoms, two of these are bonded covalently and the other two by hydrogen bonds.
The tetrahedral open cage-like crystal structure of ice. The central oxygen atom A is surrounded tetrahedrally by the oxygen atoms marked 1,2, 3 and 4.
The hydrogen bonds are weaker and therefore, longer than covalent bonds. This arrangement gives rise to an open cage-like structure, as shown in the Fig. There are evidently , a number of ‘holes’ or open spaces. These holes are formed because the hydrogen bonds holding the H2O molecules in ice are directed in certain definite angles. In liquid water such hydrogen bonds are fewer in number. Therefore, as ice melts, a large number of hydrogen bonds are broken. The molecules, therefore, move into the ‘hole’ or open spaces and come closer to one another than they were in the solid state. This results in a sharp increase in density. The density of liquid water is, therefore higher than that of ice.
(ii) As liquid water is heated from 00C to 40C, hydrogen bonds continue to be broken and the molecules come closer and closer together. This leads to contraction. However, there is some expansion of water also due to rise in temperature as in other liquids. It appears that up to 40C, the former effect predominates and hence the volume increases as the temperature rises.
Importance of Hydrogen Bonding in Sustaining Life
It can be easily realised that without hydrogen bonding , water would have existed as a gas like hydrogen sulphide. In that case no life would have been possible on this globe.
Hydrogen bonding also exists in all living organisms, whether of animal or of vegetable kingdom. Thus, it exists in various tissues, organs, blood, skin and bones in animal life. It plays an important role in determining structure of proteins which are so essential for life.
Hydrogen bonding plays an important role in making wood fibres more rigid and thus makes it an article of great utility. The cotton, silk or synthetic fibres owe their rigidity and tensile strength to hydrogen bonding. Thus hydrogen bonding is of vital importance for our clothing as well. Most of our food materials also consists of hydrogen bonded molecules. Sugars and carbohydrates , for example, have many -OH groups. The oxygen of one such group in one molecule is bonded with -OH group of another molecule through hydrogen bonding. Hydrogen bonding is thus a phenomenon of great importance in every day life.
42. KHF2 exists while KHCl2 does not. Explain.
1. What is meant by Lewis structure of an atom or ion ? Give some examples.
2. Draw the Lewis symbols for the following elements :
Na, Ca, B, Br, Xe, As, Ge
3. Write the electronic configurations of inert gases. What conclusion can be drawn from them ?
4. What is Lewis and Kossel concept of chemical bonding
5. Why do elements combine with each other ?
6. What is an ionic bond ? Give an example to illustrate ionic bond formation.
7. What are the conditions for formation of ionic bonds ?
8. Discuss the empirical formula and draw the Lewis structures for ionic compounds formed by the following pairs of elements.
i) Na, O v) Al, F
ii) K , S vi) Ca, O
iii) Na, P vii) Li, S
iv) Mg, Br
9. What is a covalent bond ? Give some examples to
10. Depict the Lewis structures for the formation of : (I) Cl2 (ii) H2O (iii) H2S (iv) NH3 (v) CH4 (vi) C2H6 (vii) C2H4 (viii) C2H2
11. Draw Lewis structures for each of the following molecules and ions. F2, PH3, H2S, SiCl4, C3H8, F2O, Na+, Br
12. Three elements having the following Lewis symbols :
(i) Place the elements in the appropriate group of the periodic table.
(ii) Which elements in the appropriate group of the periodic table.
(iii) Write the formulae and Lewis structures of the covalent compounds formed between :
(a) A and B and (ii) A and C
13. Is He2 molecule possible ?
14. What is a co-ordinate bond ? Give a few examples to illustrate it.
15. Give the points of difference between electrovalent and covalent compounds.
16. What are cations and anions ?
17. In the formation of a compound XY2, atom X gives one electron to each Y atom. What is the nature of bonds in XY2 ? Give two properties of XY2.
18. State and explain the differences between an atom and the corresponding ion.
19. In the formation of a compound AB, atom of ‘A’ lost one electron each ; while atoms of ’B’ gained one electron each.
(i) What is the nature of bond in AB ?
(ii) Does solution of AB in water allow electric current to pass or not ?
(iii) Write electron dot structure of AB.
20. Explain how the VB theory differs from Lewis concepts.
21. Draw diagrams showing the formation of a double bond and a triple bond between carbon atoms in the C2H4 and C2H2 molecules.
22. From the atomic structures of A,B,C,D and E identify the atoms or ion
Element/ion Number of protons Number of electrons
(i) A 1 0
(ii) B 6 6
(iii) 9 10
(iv) D 11 10
(v) E 12 12
23. The electron arrangement of atoms of three elements A, B and C are : A(2,8,1) : B(2, 8, 6) : C (2, 8, 7 )
(i) Write down the formula of a molecule of B and its electron dot diagram. Mention the type of bonding.
(ii) Write down the formula of the compound formed between A and C and the type of bonding.
(iii) Classify the elements A, B and C as metal and non-metal.
(iv) Which element is likely to be a good conductor of electricity and why ?
24. Explain why Na+ is not reactive , but Na metal is very reactive.
25. A solution of ionic compound behaves as an electrolyte; while that of a covalent compound is a non-electrolyte. Explain why ?
26. Explain why covalent compounds generally low melting ?
27. What are the characteristics of covalent compounds ?
28. Which of the bonding is likely in a binary compound, when both elements are (I) non-metals (ii) ion forming with opposite electric charges ?
29. For three elements X, Y and Z, following data are given :
Element/ion Mass number Number ofneutrons
X 35 18
Y 23 12
Z 24 12
Given the chemical formula and nature of the
compounds(electrovalent/covalent) formed between : (I) X
and X (ii) X and Y (iiii) Z and X
30. Which type of bonding is likely in a binary compound, when both elements are (I) non-metals (ii) ion forming with opposite electric charges.
31. What type of bonding would you expect between :
(i) a metal and a non-metal ?
(ii) a metal and another metal
(iii) a non-metal and another non-metal ?
32. Identify the types of bonds would you expect to find in the molecule of each of the following substances : (a) water (b) ethylene (c) methane and (d) sodium chloride.
33. You are given the electronic configurations of five neutral atoms, A, B, C D and E.
A : 1s2,2s2,2p6,3s2
B : 1s2,2s2,2p6,3s1
C ; 1s2,2s2,2p1
D : 1s2,2s2,2p5
E : 1s2,2s2,2p6
Write the empirical formula for substances containing ; (a) A
and D ; (b) B and D ; (c ) only D ; (d) only E
34. Explain the terms : (I) non-polar bond (ii) polar bonds.
35. In general, what conditions cause two atoms to combine to form each of the following :
(i) A bond that is covalent.
(ii) A bond that is ionic
(iii) A polar molecule
36. Which of the following hydrogen halides possesses the most polar molecule and why ?
37. Write a note on ionic character of covalent bond.
38. What is meant by dipole moment ? What is its utility ?
39. Dipole moment of a molecule provides valuable information regarding its structure. Illustrate this fact by giving examples of some polyatomic molecules.
40. The molecule of SO2 has a dipole moment.. Is the molecule is linear or bent. Explain your reasoning.
41. Predict the dipole moment of :
(i) A molecule of the type AX4, having a square planar geometry.
(ii) A molecule of the type AX5, having a trigonal bipyramidal geometry.
(iii) A molecule of the type AX6, having an octahedral geometry.
42. What information regarding structure of molecule can be obtained from the fact that :
(i) CCl4 has no dipole moment, but CHCl3 has ?
(ii) BF3 has no dipole moment, but NF3 has ?
43. What is hydrogen bond ? Explain the formation of hydrogen bond in HF molecule.
44. What requirements should a molecule fulfil for the formation of a hydrogen bond ?
45. Give reason why H2O is liquid ; while H2S is a gas at ordinary temperature.
46. Define the terms : (I) intermolecular hydrogen bond (ii) intramolecular hydrogen bond. Illustrate with suitable example.
47. HF is more polar than HI. Why ?
48. Chlorine forms a more polar hydride than iodine. Why ?
49. Which of the hydrogen halides has most polar molecule and why ?
50. How do you account for the high boiling point and high viscosity of sulphuric acid ?
51. Concentrated sulphuric acid is highly viscous. Explain.
52. Methane(mol wt = 16) and water (mol wt = 18) have similar molecular mass, yet methane has boiling point(=112 K) and water(=372 K). Explain.
53. Boiling point of water is higher than that of hydrogen fluoride, even though fluorine is more electronegative and smaller in size than oxygen. Explain why ?
54. Why glycerol is more viscous ethanol ?
55. Explain why boiling point of ammonia is higher than that of phosphine.
56. In what respect is hydrogen bonding different from ionic and covalent bonding.
57. What is meant by the term resonance ? Give examples to illustrate.
58. What are the conditions of resonance ?
59. Explain the concept of metallic bond.
60. Why is electrical conductivity of metals much higher than those of ionic liquids ?
61. Identify the type of bonding in the following molecules . (I) water (ii) methane (iii) silicon (iv) aluminium.
62. What type of bonding would you expect between :
(i) metal and non-metal
(ii) a metal and other metal
(iii) a non-metal and another non-metal.
63. Explain the formation of a covalent bond in terms of electrostatic forces of attraction and repulsion.
64. What are van der Waal’s forces ? What is their origin ?
65. What are essentials of overlap theory of covalent bond formation ?
66. Define : (a) a sigma bond and (b) pi-bond.
67. Differentiate between a sigma and pi-bond.
68. Sketch the shapes of M.O orbitals formed by the overlap of :
(i) two s-orbitals (ii) a and p-orbitals (iii) End on overlapping of two p-orbitals (iv) side-on overlap of two p-orbitals.
69. Describe the model of : (I) Hydrogen molecule (ii) hydrogen fluoride molecule (iii) Fluorine molecule (iv) Nitrogen molecule (v) Nitrogen molecule (vi) Oxygen molecule (vii) water molecule (vii) Ammonia molecule.
70. What do you understand by hybridisation of orbitals ?
71. Write notes on : (I) sp3 hybridisation (ii) sp2 hybridisation and (iii) sp-hybridisation.
72. Draw the shapes of the following hybrid orbitals (I) sp, (ii) sp2 and (iii) sp3.
73. Making use of the concept of hybridisation, predict the shape of (I)methane (ii) Ethane (iii) Ethene (iv) Ethyne.
74. Discuss VSEPR theory.
75. Explain the geometry of (I) Beryllium fluoride molecule (ii) Carbon dioxide molecule (iii) Boron trifluoride molecule (iv) Ammonia molecule.
76. Based on VSEPR theory, give the ideal geometry and shapes for molecules with 2, 3, 4, 5 and 6 electron pairs about the central atom. Give one example each.
77. Predict the shapes of the following molecules using VSEPR theory.
BeCl2, SiCl4, H2S, HgB2, PH3, GeF2.
78. Amongst LiF and LiI, which has more covalent character and why ?
79. Explain how the valence bond theory accounts for : (i) carbon-carbon double bond (ii) Existence of cis-trans isomers.
80. Explain why a sigma-bond is stronger than a pi-bond ?
81. Explain in terms of valence bond theory the following observations . HSH angle in H2S is closer to 900 than HOH angle in water.
82. Explain the formation and difference between a sigma bond and a pi bond.
83. Explain how the valence bond theory accounts for :
(i) a carbon-carbon double bond.
(ii) A carbon-carbon triple bond.
84. Drawthe Lewis structures for H2CO3, SF6, PF5, IF7 and CS2. Is the octet rule obeyed in these cases ?
85. Define lattice enthalpy. How is it related to stability of an ionic compound ?
86. How can lattice enthalpy of an ionic compound like NaCl can be determined by using Born Haber cycle ?
87. Define electronegativity. How does it differ from electron affinity ?
88. What is a polar covalent bond ? Give two examples of compounds containing polar covalent compounds ?
89. Arrange the the bonds in order of increasing ionic character in molecules : LiF, K2O , N2, SO2 and ClF3.
90. Arrange the following bonds in the order of increasing ionic character : C H, F H, Br H, Na I, K F and LiCl.
91. The skeletal structure of CH3COOH as shown below is correct, but some of the bonds are wrongly shown. Write the correct Lewis structure for acetic acid.
92. Define bond length, resonance and resonance structures.
93. Explain why BeH2 molecule has a zero dipole moment although BeH bonds are polar.
94. Sketch the bond moments and resultant dipole moments in the following molecules
(i) H2O (ii) PCl3 (iii) NH3 (iv) NF3
95. What is meant by the term bond order in Lewis concept ? Calculate the bond order of : N2, O2 and CO.
96. Write the formal charges of the atoms in the carbonate ion.
97. Write the formal charges of the atoms in the nitrite ion.
98. Which out of the two molecules OCS and CS2 has a higher dipole moment and why ?
99. Sketch the bond moments and resultant dipole moments in SO2, cis- and trans forms of C2H2Cl2.
100. The dipole moment of hydrogen halides decreases from HF to HI. Explain this trend.
101. Two p orbitals from one atom and two p-orbitals form another atom are combined to form molecular orbitals. How many MOs will result from this combination ? Explain.
102. Show the shapes of bonding and antibonding Mos formed by combination of (a) two s-orbitals (b) two p-orbitals (side to side).
103. How do the bonding and antibonding Mos formed from a given pair of AOs compare to each other with respect to (a) energy (b) presence of nodes (internuclear electron density ?
104. Arrange the following species in order of increasing stability : Li2, Li2+, Li2. Justify your choice with a molecular orbital energy level diagram.
105. Use molecular orbital theory to explain why the Be2 molecule does not exist.
106. Explain why the bond order of N2 is greater than N2+ , but the bond order of O2 is less than O2+.
107. Compare the relative stability of the following species and indicate their magnetic properties (diamagnetic or paramagnetic) : O2, O2+, O2(super oxide) , O22 (peroxide ion).
108. Explain the significance of bond order . Can bond order is used for quatitative comparisons of the strengths of chemical bonds ?
109. What is the energy gap in band theory ? Compare its size in conductors, semiconductors and insulators.
110. Which of the following substances exhibit H-bonding ? Draw the H-bonds between two molecules of the substance where appropriate :
111. How can one nonpolar molecule induce a dipole in a nearby nonpolar molecule ?
112. What types of intermolecular forces exist between the following pairs ?
(a) HBr and H2S (b) Cl2 and CBr2
(c) I2 and NO3 (d) NH3 and C6H6