+2 UNIT 1 PAGE-4
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.
BOND ORDER
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 ]
= ½
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 .
Fig 31 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 mol 1. 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 :
N22 KK(2s)2*(2s)2(2Px)2(2Py)2(2Pz)2*(2Px)1*(2Py)1
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 3
N2+ 2 ½
N2 2 ½
N22 2
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
O2+ 2 ½
O2 1 ½
O22 1
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.
Problems
32. Which of the following is expected to be coloured and why ?
Cu+ or Cr3+
33. Calculate the Bond Order for O2+ ion.
34. Give reasons for the following :
The bond in H2 is stronger than in H2+.
35. Why Helium exists as a monoatomic molecule ? Why He2 is not formed ?
36. Give the number of electrons , which occupy the bonding orbitals in H2+, H2 and He2.
37. Use the MO energy level diagram and show that N2 , would be expected to have a triple bond, F2 single and Ne2 no bond.
38. Calculate the bond orders in H2+ ion H2 and H2 molecule.
39. Arrange the following molecular species in increasing order of stability:
O2 , O2+, O2 and O22
40. Which of H2+, H2 and H22 have the same bond order. Justify their M.O diagrams.
41. 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+
42. The bond order in He2+ molecule is equal to the bond
order of H2 molecule. Justify the statement.
METALLIC BONDS
Metals constitute about three-fourth of all the known elements. They have characteristic properties such as bright lustre, high electrical conductivity, thermal conductivity , malleability and high tensile strength. The attractive force which binds various metal atoms together is called metallic bond. In order to understand the nature of bonding in metals, let us look into their general atomic properties. Metals are found to possess :
i) low ionisation energies.
ii) low electron affinities.
iii) large number of vacant orbitals in the valence shell.
iv) less number of valence electrons.
It may be noted that metal atoms cannot be bonded by ionic bond because such a bond cannot be formed between similar atoms. Metal atoms in solid state cannot be united by covalent bonds as well, because they may not be able to complete their octets by mutual sharing of electrons. Evidence in favour of above conclusions can be provided by the fact that both the ionic compounds as well as covalent compounds are bad conductors of electricity in the solid state, whereas metals are good conductors in solid state. Hence nature of bonding among metal atoms must be different than ionic or covalent bonds. Different theories have been put forward to explain the bonding in metals. These are :
i) Electron sea model or free electron model.
ii) Band model
1. Electron sea model for metallic bonding
The various features of electron gas model are given below
(i) The metal atom is supposed to consist of two parts, valence electrons and remaining part (nucleus and inner shells) which are called kernel.
(ii) The metallic crystal consists of closely packed metal atoms in three dimensions. The kernels of metal atoms occupy fixed positions called lattice sites, while the space between the kernels is occupied by valence electrons. The arrangement of kernels and valence electrons are shown in Fig.
Fig 37 Arrangement of metallic kernels
(iii) Due to smaller ionisation energy, the valence electrons of metal atoms are not held by the nucleus very firmly. Therefore, they can leave the field of influence of the other. The movement can take place through the vacant valence orbitals. Thus electrons are not localised but are in a state of continuous motion. The electrons are called mobile or delocalised electrons. As the movement of electrons in metallic crystal is just like gas molecules, hence the model is called Electron sea model.
(iv) The simultaneous force of attraction between the mobile electrons and the positive kernels is responsible for holding the metal atoms together is known as Metallic Bond.
The metallic bond is non-directional and is weaker than the covalent bond. The important points of differences between metallic bond and covalent bonds are given below :
Metallic bond Covalent bond
1. The simultaneous forces of attraction between mobile electrons and positive kernels constitute a metallic bond.
2. It is a weak bond because the mobile electrons are simultaneously attracted by a large number of kernels.
3. The bond is non-directional. 1. The mutual sharing of electrons between the combining atoms of the same or different elements is called a covalent bond.
2. It is a strong bond because the bonded pair is strongly attracted by two nuclei.
3. The bond is directional
Explanation of physical properties of metals
The metallic crystals consists of a three dimensional net work of positive kernels occupying the fixed crystal sites immersed in a sea of mobile electrons. The presence of these free electrons explains most of the characteristic properties of metals as discussed below :
1. Electrical conductivity : In a metal crystal, the mobile electrons are ordinarily flowing probably in all directions. But when the potential difference is applied across a metal by including it in an electrical circuit, the flow of electrons get directed along the applied field. Due to directed flow of electrons, the electric current is carried from one point to another and the metals are termed as good conductors.
Fig 38 Electrical conductivity of metals
2. Thermal conductivity : The thermal conductivity of the metals is also due to the presence of mobile electrons. On heating a part of the metal, the electrons in the region acquire high kinetic energy. These energised electrons rapidly move to the cooler parts of the metals and some of their energy is transferred to these parts. As a result, these parts also acquire higher temperature so that the heat gets conducted throughout the metal.
3. Metallic lustre : It is assumed that metallic lustre results from mobile electrons which have set in motion by photons of visible frequencies. In all metals , except copper and gold, the electrons absorb light of all wave lengths in spectrum. This energy in the form of light is momentarily accepted by mobile electrons and immediately re-emitted in the form of visible light.
This momentary exchange of light energy gives rise to the familiar metallic lustre.
4. Malleability and ductility : Metals are both malleable (made into sheets) and ductile (drawn into wires). These properties of metals can be explained by using the idea of metal structure by mobile electrons and positively charged ions. The mobile nature of electrons binding between atoms makes possible deformation of metal structure. When a metal changes shape, layers of atoms, acting as if they were rigid spheres, slide over each other from positions shown in Fig 38.
Fig 38 Displacement of metal ions in a metal crystal
Thus, the shape of metallic crystal can be changed or deformed without breaking.
5. Tensile strength : Metals have high tensile strength, i.e., they can resist stretching without breaking. Thus a large weight can be supported by even a wire of small cross-section. The high tensile strength is due to strong electrostatic attraction between positively charged kernels and mobile electrons surrounding them.
6. Hardness of metals : Metals are generally hard. The hardness of metal is due to the strength of the metallic bond. In general the strength of metallic bond depends upon the following factors :
(i) Number of valence electrons : Greater the number of valence electrons for delocalisation, stronger is the metallic bond.
(ii) Size of the kernel : Smaller the size of the kernel of the metal atom, greater is the attraction for the delocalised electrons. Consequently, the stronger is the metallic bond.
For example, alkali metals have only one valence electron and larger atomic kernels. Thus they form weak metallic bonds and are soft metals.
BAND MODEL : MOLECULAR ORBITAL THEORY OF BONDING IN METALS
The bonding in metals can be explained by extending the simple molecular orbital theory to metals. Let us consider an example of lithium metal. Ignoring inner 1s-electrons which do not involve in bonding, 2s-orbitals of lithium atoms may combine to form molecular orbitals. In the case of lithium Li2 (two atoms) 2s AOs on lithium atoms combine to give two MOs ; one bonding MO and one anti-bonding MO.
If we have three Lithium atoms, three 2s AOs would combine to form three MOs ; one bonding one anti-bonding and one non-bonding.
The energy of non-bonding MO is in between the bonding and anti-bonding MO , exactly at the same level as the energy of the AOs. Similarly in the case of four lithium atoms , four 2s AO’s would combine to form four MOs ; two bonding and two anti-bonding MOs.
Similarly, if we have N atoms ( N is Avogadro number, 6.023 x 1023 atoms), 2s AOs of N atoms would combine to give N MOs. As the number of molecular orbitals is large, the energy levels of MOs are so close together , that they may almost be treated as continuous.
Such group of energy levels is known as energy band and is responsible for the name band model for metals.
The energy levels in a band is equal to the number of atoms in the metal. The formation of energy band depends upon :
(i) The close proximity of large number of atoms.
(ii) The energy difference between the pure atomic orbitals.
Thus, 2s-atomic orbitals give rise to one energy band, the 2p-atomic orbitals also form a band of closely spaced energy levels.
The energy bands formed by the overlap of 1s, 2s and 2p AOs of lithium metal are shown in Fig 39.
The arrangement of electrons in different energy bands determine the characteristics of a metal. Now in the case of Li atom, the electronic configuration is 1s22s1. The 1s energy band will be completely filled. These electrons do not contribute towards bonding and may be ignored. In the case of Li2 , there will be 2 valence electrons and there are two MOs. Since each MO can accommodate two electrons (Pauli’s exclusion principle), only bonding molecular orbital will be completely filled. Similarly in case of Li4, the four valence electrons would occupy only the two lowest bonding MOs.
Thus, in these cases half of the molecular orbitals remain unoccupied. Similarly, N 2s atomic orbitals give N 2s molecular orbitals (energy band). Since Lithium atom has only one valence electron, only half of the molecular orbitals will be filled. In other words, only one half of the 2s band , will be filled in lithium metal. This is shown in Fig 40. ( The bands are displaced laterally for clarity).
Thus, there are many empty levels into which electrons can move. This movement of electrons from one level to another constitutes an electric current. Thus, lithium is a good conductor. At the same time, if we heat one end of metal, electrons at that end gain energy and move to empty energy bands where they can travel easily to another part of the metal. This accounts for thermal conduction of metals.
We can apply same arguments to beryllium. In beryllium, there are two valence electrons and therefore, N atoms of beryllium will have 2N electrons from 2s AOs. Therefore in beryllium metal, the 2s band is completely filled. There are no empty orbitals in 2s energy band into which electrons can move. However, 2p energy band is completely empty and it overlaps the 2s energy band. When some electric field is applied, electrons can move into empty band. The lowest lying empty band is called conduction band and the outermost filled band is called valence band. In the case of beryllium 2s band is valence band and 2p band is conduction band. Because of overlapping of 2s and 2p-bands, some of the electrons overflow into 2p-energy band. This is shown in Fig 41.
Thus some of the 2p energy band is occupied and some of the 2s band is empty. The electrons can , therefore , easily move and beryllium conducts electricity and behaves as a metal.
Similar explanation can be given for sodium ( one 3s-electron) and magnesium (two 3s electrons) . In the case of aluminium, there are three valence electrons per atom. Therefore the 3s band will be completely filled, while 3p band will be half-filled and the presence of vacant levels in 3p band are responsible for the conduction. (Fig 42).
In the case of Group 14 elements(C, Si, Ge and Sn) , there are four valence electrons per atom. The first two bands will be completely filled and next two bands do not overlap. The difference in energy between filled energy band (valence band) and next empty energy band (conduction band) varies from material to material. This will also determine the electrical conductivity of the material. The spaces between these bands represent energies forbidden to electrons and are called energy gaps. The valence band and conduction bands for metals, semi-metals and non-metals are shown in Fig 43.
(i) In metals, conduction band is close to valence band and therefore, the electrons can easily go into the conduction band. Therefore metals are good conductors.
(ii) In non-metals, the energy gap between valence band and conduction band is very large. Therefore empty bands are not accessible for conduction and therefore, electrons from valence band cannot move into conduction band.
(iii) Several solids have properties between metals and non-metals. These are called semi-metals and they have energy gap in between those of metals and non-metals. At room temperature, these are not good conductors of electricity. However, when the temperature is increased, some of the electrons move from valence band resulting in increase in conductivity and other metallic characteristics. For example, the energy gap in silicon is only 105 kJ mol1 and it behaves as semi-conductor or semi-metal, while energy gap in carbon , which is an insulator is 508 kJ mol1 . In Group 14 elements, the energy gap between conduction band and valence band in different elements is given below.
Element Energy gap (kJ mol1 )
Carbon 508
Silicon 105
Germanium 58
Tin 8
Therefore , carbon is an insulator, silicon and germanium are semi-conductors and tin is a metal
HYBRIDISATION
The phenomenon of intermixing of atomic orbitals of slightly different energies so as to redistribute their energies and to give an equal number of new orbitals of equivalent energy and shape is called hybridisation
Salient Features of hybridisation
1. The orbitals undergoing hybridisation should have almost equal energies.
2. The number of hybrid orbitals formed is equal to the number of atomic orbitals hybridised.
3. The hybrid orbitals formed are always equivalent in energy and identical in shape.
4. The hybrid orbitals form more strong bonds than the un-hybridised atomic orbitals due to better overlapping of hybid orbitals.
5. Both filled and half-filled orbitals can get hybridised.
6. Orbitals involved in pi-bond formation are not hybridised
Types of hybridisation
Hybridisation can be of different types depending on the number and type of atomic orbitals hybridised. The state of hybridisation of the central atom in a molecule determines the shape of the molecule. Different types of hybridisation and corresponding shapes of the molecules are shown in the following TABLE
SOME COMMON TYPES OF HYBRIDISATION AND THEIR CHARACTERISTICS
Type of Hybridisation Atomic
orbitals involved Orientation of Hybrid Orbitals Shape of the molecule Examples
sp
one s + one p
Linear
BeF2, BeCl2, BeH2, C2H2, HgCl2
sp2
one s + two p
Triangular planar
BF3, C2H4, NO3-, CO32-
sp3
one s + three p
Tetrahedral
CH4, CCl4, NH4+
dsp2
one d + one s
+ two p
square planar
[Ni(CN)4]2-
[PtCl4]2-
sp3d
one s + three p
+ one d
Trigonal bipyramidal
PF5, PCl5
sp3d2
one s + three p
+ two d
Octahedral
SF6, [Co(NH3)6]3+
HYBRIDISATION IN ELEMENTS OF THIRD PERIOD INVOLVING d-ORBITALS
The elements of the third period contain d-orbitals also in addition to ‘s’ and porbitals. These 3d-orbitals are comparable in energy to 3s and 3porbitals. These dorbitals are also involved in the hybridisation to explain the geometry of molecules of elements of third period. This results in covalencies of 5, 6 and 7, which are not known amongst the compounds of second period
elements. Due to availability of d-orbitals, P and S can exhibit covalence of 5 and 6 respectively whereas the corresponding elements Nitrogen and Oxygen of the second period cannot extend their octects. Let us discuss some common types of hybridisation involving d-orbitals and common examples of molecules.
Sp3d - Hybridisation
The hybridisation involves the mixing of one s , three p and one d-orbitals. These five orbitals hybridise to form five sp3d-hybrid orbitals. The mixing of five orbitals is shown in Fig 44 . These hybrid orbitals point towards the corners of a trigonal bipyramidal geometry. In this case , three orbitals forming a plane are directed towards the corners an equilateral triangle and the other two lying above and below it.
Fig 44 Formation of five sp3d type orbitals which
will adopt trigonal bipyamidal geometry
Formation of PCl5 Molecule
(sp3d-hybridisation)
The ground state electronic configuration of Phosphorus atom may be represented as :
In order to explain the pentavalency of phosphorus, it is assumed that under conditions of bond formation the 3s electrons get unpaired and one of it is promoted to to vacant 3dZ2 orbital. The excited configuration of phosphorus may represented as follows:
Now there are five orbitals , one s , three p and one d-orbital which are half-filled. These then hybridise to yield a set of five sp3d hybrid orbitals which point towards the five corners of a trigonal bipyramid ( Fig 45 )
Fig 45 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 Trigonal bipyramidal as shown in Fig 46 .
Fig 46 Trigonal bipyramidal geometry of PCl5 molecule
Formation of SF6 Molecule(sp3d2 hybridisation)
The ground and excited electronic configurations of sulphur are represented as :
In the excited state ie., under conditions of bond formation, the 3s and 3p electrons get unpaired and one electron out of each pair is promoted to vacant 3dx2 y2 and 3dZ2 orbitals. Now there are six orbitals ; one s, three p and two d-orbitals which are half-filled. These then hybridise to form six new sp3d2 hybrid orbitals which are projected towards the six corners of a regular octahedron as shown in Figure 47.
Fig 47 Formation of six sp3d2 hybrid orbitals
In SF6 six sp3d2 hybrid orbitals overlap with half filled orbitals of fluorine atoms to form six SF sigma
bonds. Because of sp3d2 hybridisation of sulphur , SF6 molecule adopts a regular octahedral geometry as shown in Fig 48.
Fig 48 Octahedral geometry of SF6 molecule
All the six SF bonds in SF6 have same bond length. Some other examples of sp3d2 are [CrF6]3, [Co(NH3)6]3+.
BOND DISTANCE
There is no fixed distance between any pair of atoms in a molecule, since atoms in a molecule are always vibrating with respect to each other. Thus , the average equilibrium distance between the centres of the nuclei of the bonded atoms in a molecule is known as bond length or bond distance. It is expressed in picometers (pm) and it can be determined experimentally by X-ray diffraction and other spectroscopic techniques.
Bond length depends on the nature of the atoms and the type of bonds between them. For example, the C to C bond length in ethane is 154 pm, in ethylene it is 134 pm and in acetylene it is 120 pm. This is because these compounds have different types of C to C bonds, i.e. single bond in ethane, double bond in ethylene and triple bond in acetylene. Thus the bond length decreases with increase in the multiplicity of bonds. The reason is that , greater the number of electrons shared by the two atoms, greater will be the attractive force between electrons and the nuclei. Bond lengths of some common bonds are given in TABLE.
Bond lengths and Bond energies of some bonds
Bond Bond energy kJ mol-1 Bond length
pm
HH 435 74
HC 414 110
HN 389 100
HO 464 97
HS 368 132
HF 569 101
HCl 431 136
HBr 368 151
HI 297 170
NN
N=N
NN
NO
N=O 163
418
949
230
590 145
123
109
136
115
CC 347 154
C=C 611 134
CC 837 120
CN 305 147
C=N 615 128
CN 891 116
CO 360 143
C=O 728 123
CCl 326 177
OO
O=O
FF
ClCl
BrBr
I I 142
498
159
243
192
151 145
121
128
199
288
266
Bond length is found to increase with increase in the size of the atom in the molecule. For example, the bond length between hydrogen and chlorine atoms in HCl molecule is 136 pm. The bond length between H and Br atoms in HBr is 151 pm.
It is found that the distance rAB between atoms A and B for a single bond A B is the average of the bond distances of AA and BB.
ie., rAB = ½ [ rAA+ rBB]
In the case of diatomic molecules (eg H2, Cl2 , HCl etc) bond length is the sum of atomic or covalent radii of the respective atoms, i.e.,
rAB = rA + rB
where rA and rB are single covalent radii of the atoms A and B respectively. Single covalent radius of an atom means half the bond length when the atoms are linked to another atom of the same type by a single covalent bond. Thus,
rA = ½ rA A and rB = ½ rB B
Problem
43. The HH and ClCl bond distance in hydrogen and chlorine
are 74 pm and 198 pm respectively. Calculate the bond
length in HCl
INTERMOLECULAR FORCES
In addition to covalent bond, ionic bond and co-ordinate bond, there are weak attractive intermolecular forces which occur in all kinds of molecular solids. These are present even in the case of non-polar molecules such as H2, O2 , Cl2, CH4 etc. in the solid and liquid states. It has been observed that even non-polar substances such as hydrogen, chlorine, methane etc., can be liquefied. Noble gases are monoatomic and their atoms do not combine to form molecules because they have stable filled electronic configurations. But these gases can also be liquefied. This means that some forces must be operating in their liquid state. These forces are however, very weak as indicated by their low melting and boiling points.
The existence of weak attractive forces among non-polar molecules in their liquid and solid states was first proposed by J.D. van der Waals and therefore , these are called van der Waals forces. Such forces come into existence due to the interplay of positive and negative charges between neighbouring atoms of non-polar molecules when they are very close to each other.
Van der Waals forces are composed of three types :
(i) Dipole-dipole ( or Keesom forces)
(ii) Dipole-induced dipole forces
(iii) Instantaneous dipole – instantaneous induced dipole forces ( or London dispersion forces)
(i) Dipole-dipole ( or Keesom forces)
These types of forces occur in molecules which have permanent electric dipole such as HCl, NH3, H2O etc. A polar molecule has separate centres of positive and negative charges. These forces arise due to electrostatic attraction between oppositely charged ends of the polar molecules. The positive end of one molecule attracts the negative end of the other molecule and vice-versa as shown in Fig 49.
Greater the dipole moment of the molecules, greater is the force of attraction. This effect was first studied by Keesom (1912) and is also referred to as orientation effect. These forces are also called Keesom forces and produce a net attractive influence. The interaction energy is inversely proportional to the sixth power of the distance between molecules, i.e.,
where k is a constant of proportionality and the negative sign indicates forces of attraction. This also shows that the forces are significant only at short distances of the order of 500 pm. The magnitude of the force also depends upon the nature of the molecule (its dipole moment).
(ii) Dipole induced – dipole forces
If the molecules of a substance are non-polar (such as noble gas, iodine molecule, etc.) , they do not possess any dipole. However, when these molecules come into the neighbourhood of other molecules or ions which are polarised, an induced dipole appears on these non-polar molecules.
This is because the electron cloud of non-polar atom or molecule gets deformed due to the electric field of the polar molecule. This causes a shift in the centre of gravity of the negative charge relative to the nuclear charge and results in the formation of an induced dipole moment. As the size of the atom or molecule increases, the influence of the electric dipole on it also increases.
The interaction energy is proportional to (1/r6) and depends upon the dipole moment of the polar molecule and polarizability of the non-polar atom or molecule. Molecules or atoms of larger size have more tendency to get polarised.
The forces existing between dipole and induced dipole are called dipole-induced dipole forces. The easier the molecules are deformed (easily polarizable), the greater are the attractive forces. This effect was studied by Debye (1920) and is known as induction effect.
(iii) Instantaneous dipole – Instantaneous induced dipole (London) forces
These forces arise in non-polar molecules such as noble gases. For noble gases , both Keesom forces and Debye forces do not arise because these are non-polar and their shells are rigid. However, the noble gases can be liquefied. This suggests that, there must be some other cause of intermolecular attractions.
In order to understand the origin of these forces, let us consider two noble gas atoms very close to each other. Each atom is uncharged because its electron cloud is symmetrically distributed around the nucleus (Fig 51 a)
However, due to the motion of electrons, we can imagine that for a fraction of time, the electron distribution is not symmetrical, i.e., the centres of positive and negative charges do not coincide. As a result of instantaneous asymmetrical distribution of electrons in the atom, there is a small temporary dipole known as instantaneous dipole (Fig 51 b above). This instantaneous dipole influences the electron distribution in the neighbouring atom and induces dipole known as instantaneous induced dipole (Fig 51 c above). This results into the attractive forces between the instantaneous dipole and instantaneous induced dipoles. These are called London forces.
The interaction energy due to these forces is also proportional to (1/r6) and these are significant only at short distances. These also depends upon the polarizability of the atom or molecule.
These forces are responsible for the condensation of noble gases and other gases such as H2. Cl2 , CH4 etc. which have no residual bonding capacity. In all cases, the solids and liquids are formed through these interactions.
The strength of these forces depend upon :
i) Size of the molecule
ii) Number of electrons present in the molecule.
iii) Molecular structure.
For example, large molecules have much diffused charged clouds. Hence, there are great chances of its distrotion and producing instantaneous dipoles. Therefore, the strength of the van der Waal’s forces increases with molecular size. This is evident from the boiling points of methane (CH4), silane (SiH4) and germane (GeH4) which increases from CH4 (112 K) to SiH4 (161 K) to GeH4 (183 K) due to increase in molecular size. Moreover, the larger the number of electrons in the molecules, the greater is the chances of distrotion of electron clouds. For example boiling points of monoatomic noble gases increase from helium to radon.
The above three types of forces are collectively called van der Waal’s forces. The magnitude of each types of interaction will vary depending upon the type of the molecules.
HYDROGEN BOND
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 more 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 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(about one-tenth of the strength of a covalent bond).
Comparison of bond energies (kJ/mol) of hydrogen bond with corresponding covalent bond
Bonds Bond energy (kJ/mol)
Hydrogen bond (……) Covalent bond ( )
H2NH….NH3 12.6 431
HOH….OH2 29.3 452
FH….FH 41.8 568
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
GROUP 15
Formula Molecular weight Boiling point (oC)
SbH3 125 17.0
AsH3 78 55.1
PH3 34 84.6
NH3 17 33.0
GROUP 16
H2Te 130 1.8
H2Se 81 42.0
H2S 33 59.6
H2O 18 + 100
GROUP 17
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 NH3 have abnormally high melting points in their groups.
Molecular mass
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 53 .
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.
Fig 53 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.
QUESTIONS
1. What are the dimensions of Plank constant ? What other physical quantity has the same dimension.
2. What experimental support is available for de Broglie concept ?
3. Two particles A and B are in motion. If the wavelength associated with particle A is 5 x 108 m, calculate the wave length associated with particle B if its momentum is half of A.
4. Calculate the wave length of 1000 kg rocket moving with a velocity of 3000 km per hour.
5. The sodium flame test has a characteristic yellow colour due to emissions of wavelength 589 nm. What is the mass equivalence of one photon of wavelength ?
6. Calculate the uncertainty in velocity of a wagon of mass 2000 kg whose position is known to an accuracy of 10 m.
7. Calculate the uncertainty in position of a dust particle with mass equal to 1 mg if the uncertainty in its velocity is 5.5 x 1020 ms1 .
8. On the basis of Heisenberg’s uncertainty principle, show that electron cannot exist witin the atomic nucleus( radius = 1015 m).
9. Why can’t we overcome the uncertainty predicted by building more precise devices to reduce the error in measurement below the h/4 limit ?
10. What physical meaning is attributed to the square of the absolute value of wave function 2 ?
11. Which of the four quantum numbers (n,ℓ , mℓ, ms) determine (a) energy of an electron in hydrogen tom and in many electron atom, (b) the size of the orbital , (c) shape of the orbital , (d) orientation of an orbital in space ?
12. Draw the sphaes (boundary surfaces) of the following orbitals : (a) 2Py (b) 3dz2 (c) 3dx2 y2. (show ordinate axes in your sketches)
13. Discuss the similarities and differences of 1s and a 2s-orbital.
14. For each of the following pair of hydrogen orbitals, indicate which is higher in energy :
(a) 1s, 2s (b) 2p, 3p (c) 3dxy, 3dyz
(d) 3s, 3d (e) 4f, 5s.
15. Which orbital in each of the following pairs is lower in energy in many electron atom ? (a) 2s, 2p (b) 3p, 3d (c) 3s, 4s (d) 4d, 5f
16. Explain the meaning of the symbol 4d6.
17. The ground state electron configurations listed here are incorrect. Explain what mistakes have been made in each and write the correct electron configurations.
Al : 1s22s22p43s23p3
B : 1s22s22p5
F : 1s22s22p6
18. Draw the energy level diagrams for atoms with the following electronic configuration :
(a) 1s22s22p5 (b) 1s22s22p63p3
(c) 1s22s22p63s23p64s23d7.
19. What do you understand by (i) Radial probability density , R2 and (ii) Radial probability function, 42R2 ? How do they vary for 1s, 2s and 2p atomic orbitals of hydrogen atom ?
20. 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.
21. Show the shapes of bonding and antibonding Mos formed by combination of (a) two s-orbitals (b) two p-orbitals (side to side).
22. 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 ?
23. Arrange the following species in order of increasing stability : Li2, Li2+, Li2. Justify your choice with a molecular orbital energy level diagram.
24. Use molecular orbital theory to explain why the Be2 molecule does not exist.
25. Explain why the bond order of N2 is greater than N2+ , but the bond order of O2 is less than O2+.
26. Compare the relative stability of the following species and indicate their magnetic properties (diamagnetic or paramagnetic) : O2, O2+, O2(super oxide) , O22 (peroxide ion).
27. Explain the significance of bond order . Can bond order is used for quatitative comparisons of the strengths of chemical bonds ?
28. What is the energy gap in band theory ? Compare its size in conductors, semiconductors and insulators.
29. Which of the following substances exhibit H-bonding ? Draw the H-bonds between two molecules of the substance where appropriate :
30. How can one nonpolar molecule induce a dipole in a nearby nonpolar molecule ?
31. 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