Download Module 2 Atomic Bonds Lecture 2 Atomic Bonds

Module 2
Atomic Bonds
Lecture 2
Atomic Bonds
1
NPTEL Phase II : IIT Kharagpur : Prof. R. N. Ghosh, Dept of Metallurgical and Materials Engineering || |
Keywords: Electronic configuration of atoms & types of atomic bond, Metallic bond: Free electron models of metal, Electron mobility & conductivity, Limitation of free electron model, Band theory: conductor, semiconductor & insulator, Bond strength & stiffness. Introduction In this module we shall explore the nature of atomic bonds in solids and learn about its effect on properties. All solids are made of atoms which consist of subatomic particles; positively charged proton, neutron (neutral) and negatively charged electron. The entire mass of the atom is concentrated within a small nucleus consisting of proton and neutron whereas electrons having very little mass keep rotating about the nucleus in specific orbits. The sum total of proton and neutron gives the atomic mass number of an element; whereas the number of proton which is same as that of electron is its atomic number. The atom therefore is neutral. Hydrogen the lightest atom is made of one proton in the nucleus and an electron moving in a circular orbit known as k shell. The capacity of this orbit is 2. Helium is the next element with atomic number 2 and atomic mass 4. It has 2 protons & 2 neutrons in the nucleus and two electrons moving in a circular k shell in opposite directions. Since the capacity of this shell is 2 the outer orbit is full. Therefore even if two He atoms are brought close together there will be little interaction between the two. This is why He is a mono‐atomic inert gas having an extremely low boiling point. However if two H atoms are brought close together it would try to form a bond by sharing the outer most electrons. This is why it is a diatomic gas. Atoms having higher atomic numbers would need more numbers of shells (K, L, M, N ….) and sub‐shells (s, p, d, f ...) to accommodate orbiting electrons. The energy levels of these electrons are determined by four quantum numbers, principal quantum number (n) denoting shells, azimuthal quantum number (l) denoting sub‐shells, magnetic quantum number (m) denoting effect of magnetic field on orbiting electrons, spin quantum number (ms) representing direction of rotation about its axis. The electrons are so arranged in different orbits that no two of these have the same quantum numbers. Following this rule it is possible to find out the capacity of each shell and sub‐shell. You may refer to the following figures to get an idea about the shells, sub‐shells and their capacity. 2
Fig 1: Schematic diagram showing different Fig 2: Quantum numbers describing energy orbits. Electrons in inner shells are more levels of electrons orbiting around the nucleus NPTEL Phase II : IIT Kharagpur : Prof. R. N. Ghosh, Dept of Metallurgical and Materials Engineering || |
tightly bound. of an atom. Problem: How many sub‐shells would orbit N have? How many electrons can these accommodate? Answer: Principal quantum number for N orbit is 4. Look at the table in fig 2. This shows that l could have values from 0 to n‐1. Thus the possible numbers are 0, 1, 2, & 3. These are represented as sub‐
shell s, p, d & f. Magnetic quantum number of a sub‐shell can have values from –l to +l. The corresponding values of m for sub‐shell f are ‐3, ‐2, ‐1, 0, +1, +2, & +3. Each of these can accommodate 2 electrons with opposite spin. Its capacity is therefore 14. The total capacity of shell N = 2+6+10+14 = 32. Fig 3: Madelung Rule: arrows give the sequence in which sub‐shells in different orbits are filled up by electrons in an atom. The last row gives the capacity of respective sub‐shell Electrons occupying lower orbits are more tightly bound. These orbits are filled up depending on their energy level which is determined by the four quantum numbers. Fig 3 gives a table describing the sequence in which these are filled up. It is known as Madelung rule. Using this it is possible to write down electronic configuration of an atom if its atomic number is known. Example: How are electrons arranged in different orbits of Na (Z=11) & Cl (Z=17)? Answer: Na 1s2 2s2 2p63s1 Note that the superscript denotes the number of electrons in respective sub‐shells. This adds up to 11 which is its atomic number. Similarly the electronic configuration of Cl can be written as 1s22s22p63s22p5. Stability of atomic structure 3
The stability of atomic structure depends on its tendency to interact with its neighbor. This is essentially determined by the number of electrons in its outer orbit often called as valence band with respect to its capacity. The capacity of K shell is 2. He having atomic number 2 has completely filled outer shell. Therefore its atomic structure is very stable. It is one of the most stable inert gases which exist in nature in elemental form. It’s freezing and boiling points are extremely low. Apart from this all elements having 8 electrons in their valence band (outer orbit) have maximum stability with little chance for formation of bond. Inert gases (apart from He) like Ne, Ar, Kr, Xe, Rd NPTEL Phase II : IIT Kharagpur : Prof. R. N. Ghosh, Dept of Metallurgical and Materials Engineering || |
have 8 electrons in their outer orbit. I am sure all of you are familiar with these concept introduced in your earlier courses. The genesis of bond formation in elements as they solidify lies in their atomic configuration. As long as the atoms are far apart there is hardly any interaction between these. However when two atoms are brought very close to each other the outer orbits would overlap. Since no two electrons can have the same quantum numbers additional sub‐shells would have to form to accommodate more electrons. Depending on the way the outer shell electrons in neighboring atoms interact with each other there can be four different types of bonds in solids; metallic bond, ionic bond, covalent bond, and van der Waal bond. The properties of materials are determined by the nature of their bond. It also decides the type of crystal structure you expect. It is also possible to alter crystal structure by alloy addition and introduce defect structures thereby change properties of the material. Therefore it is important to know about atomic bond and crystal structure. Example: Here are the electronic configuration of inert gases He 1s2 ; Ne 1s2,2s2,2p6; Ar 1s2 2s2 2p6,3s2,3p6 These are stable & have low melting point, He: 1K, Ne: 24K, Ar: 84K, ….. Rn: 211K. As atoms become larger the melting point increases. This indicates that the tendency to form bonds increases. This is because most of the sub‐shells are not circular. The centre of all negatively charged electrons does not coincide with its nucleus where all the positive charges are concentrated. Therefore these tend to act as dipoles. This promotes formation of weak van der Waal bonds. Example: Why is C so stable that it is found in nature in its atomic form? Its atomic configuration is: C 1s2 2s2 2p2. Note that the number of electron in its outer orbit (L shell) is 4. This is half filled. Electrons in the outer orbit try to arrange themselves in such a fashion that its energy is the lowest (minimum). This happens if all the electrons have parallel spin. In such an event sp shell behaves as a hybridized orbit. This is known as Hund’s rule. 1s2,2s12sp3 (‐px,‐py,‐pz) electronic configuration of Diamond. This is schematically shown as follows: 1s 2s 2sp 2sp 2sp
Metallic bond When electrons in valence band are equally shared by all atoms it results in metallic bond. The range over which atomic forces are significant is very small. When two atoms are brought close enough the outer orbit or the valence band start overlapping, the force of attraction becomes significant. Since no two electrons can have identical set of quantum numbers the valence band must split to accommodate more electrons. This is schematically shown in the following sketches. 4
NPTEL Phase II : IIT Kharagpur : Prof. R. N. Ghosh, Dept of Metallurgical and Materials Engineering || |
Fig 4: Shows how the valence band splits to accommodate more number of electrons. The rectangle on the right hand indicates possible energy levels occupied by electrons. Note that in monovalent metal only half of the available shells are filled. Fig 5: Shows a pictorial representation of metallic bond. White spherical balls denoting inner part of a metal atom are immersed in a pool consisting of negatively charged valence electron. Electrons are loosely bound and free to move. These are equally shared by all atoms. Fig 6: Shows cases of metallic bond in elements having electrons in several shells. The difference between the size as well as energy level of two consecutive orbits keeps decreasing with increasing principal quantum numbers. The chance of having overlapping orbits increases as more shells (K, L, M, O, P, ‐‐
) are added. 5
This type of distribution makes metallic bond non‐directional. The strength of the bond depends on the number of valence electrons per atom, as if the electrons play the role of a glue to join layers of atoms. Look at the atomic configuration of elements in the fourth row of the periodic table. There are several transition elements (where the number of electrons in the outer most shell remains unchanged till the lower orbits are filled up) beyond Ca in this row. The inner orbits of Ca are similar to that of Ar. In addition it has 2 electrons in 4s. Its configuration is represented as Ca [Ar] 4s2. Electrons in atoms of subsequent elements tend to occupy 3d shell. For example the configuration of Ti is [Ar] 3d24s2. This trend continues till Zn: [Ar] 3d104s2 when 3d shell gets completely filled up. The valence band in these elements is made up of the electrons in overlapping 3d & 4s shells. Electrons having unpaired spin belong to the valence band. The number of such electrons per atom increases along the period (row). It reaches its peak at Cr and there after it decreases. This is the reason why the melting point (which is an indicator of the strength of the bond that develops between atoms) of the elements along this row keeps increasing reaches its peak at Cr and then decreases. Since in metals electrons are shared by all atoms and the electrons occupying the top most energy level in valence band (known as Fermi level) could easily move to the vacant higher energy levels; NPTEL Phase II : IIT Kharagpur : Prof. R. N. Ghosh, Dept of Metallurgical and Materials Engineering || |
they have high thermal & electrical conductivity, moderately high elastic modulus and are amenable to plastic deformation (ductile). Ionic Bond Ionic bond develops between two atoms having widely different electro‐negativity. For example such a bond can form between Na and Cl; two elements: one in group I and the other in group VII. Na has one loosely bound electron in its outer shell whereas Cl is short of one electron in its outer orbit to reach the stable octet configuration. Therefore if these two atoms come closer Na would lose one outer electron to form a stable octet and Cl would pick it up to attain a stable structure. Consequently Na & Cl atoms become charged particles (ions). The bond that forms between the two is due to the force of attraction between the two oppositely charged ions. Since the electric field is uniformly distributed in all direction such bonds do not exhibit any directionality. Like metallic bond ionic bond is also strong. Materials having such a bond have high elastic modulus. The electrons are tightly bound and are not free to move. Therefore its thermal & electrical conductivity are poor. However such materials exhibit ionic conductivity if molten or if present in aqueous solution. Ionic materials are brittle with little ductility. Covalent bond Covalent bond forms by sharing of electrons in the valence band. For example if two hydrogen atoms having one electron in their respective valence bands are brought close enough so that the bands overlap the two electrons having opposite spin fulfill the condition of a stable outer shell. They behave as if they belong to both atoms. This is shown pictorially in figure 7. This is why hydrogen exits in diatomic form in nature. Since the electrons are tightly bound to a pair of atoms they are not free to move as in metallic bond it has poor electrical conductivity. Likewise in the case of oxygen atom having 6 electrons in its outer orbit it is necessary to share two electrons from each atom. It has therefore 2 covalent bonds. Similarly nitrogen having 5 electrons in its outer orbit needs to share 3 electrons with its neighbor to form 3 covalent bonds. The distance between atoms is determined by the number of bonds and the number of orbits in the atom. The bond between two carbon atoms consists of four covalent bonds. This is because to fulfill the stability of the outer shell (octet) four electrons must be shared. The best way it can happen is by having four neighboring atoms. Tetrahedron having atoms at its centre and four corners is the best example one can think of. In fact this the way the atoms of carbon are arranged in diamond. This is illustrated in figure 8. Sharing of electrons gives it its directional characteristic. 6
NPTEL Phase II : IIT Kharagpur : Prof. R. N. Ghosh, Dept of Metallurgical and Materials Engineering || |
Figure 7: Schematic representation of four different types of atomic bonds and their characteristic features 7
Figure 8: Illustrates covalent bonds between carbon atoms in diamond. A tetrahedron is formed by joining four corners of a cube. Carbon atoms are located at each of these corners and at the centre of the tetrahedron. The thick line joining the corner with the central atom represents a covalent bond. Each atom has four such bonds. These are directed along specific direction. The angle between these is 109.5º. Van der Waal bond This forms between atoms that behave as dipoles. The tendency to form dipole increases as more orbits are added to accommodate electrons in an atom. The centre of negatively charged moving electrons often does not coincide with the positively charged nucleus. This leads to the formation of a dipole. As it comes near another dipole bond formation occurs as a result of attraction between opposite poles. However it is much weaker than the three primary bonds described above. It can form even between atoms of noble gases having stable outer shell with 8 electrons. The nature of bonds in water molecules is also similar. However precise measurement of bond strengths shows it is a little stronger than van der Waal bond. In order to distinguish the two it is often termed as hydrogen bond. Its origin nevertheless lies in dipolar characteristic of water molecules. Such bonds are also found in organic molecules. NPTEL Phase II : IIT Kharagpur : Prof. R. N. Ghosh, Dept of Metallurgical and Materials Engineering || |
Bond strength Bond formation takes place when two atoms are close enough. Under equilibrium there is no net force acting on these. However if you try to pull them apart a restoring attractive force acts on it. Likewise if you try to push them to bring the atoms closer there is a repulsive force which brings them back to the position of equilibrium. The net energy under such a condition should be the lowest and the net force should be zero. This is shown schematically in figure 9. 8
Figure 9: When atoms are far apart there is little interaction. As the distance decreases the force of attraction increases. It tries to bring the two atoms closer. This is why the force is negative. However if you push further there is a force of repulsion as well. This acts in the opposite direction. The net force is the sum total of the two. Under the position of equilibrium the net force is zero. This happens when the distance is r0. The net energy under such a condition is the lowest. The bond may therefore be visualized as an elastic spring. The stored energy is a function of its stiffness. This is given by equation 1. The negative term denotes the force of attraction and the positive term is the force of repulsion. The exponent m = 1 for ionic bond whereas m = 6 for weak van der Waal bond. Note that a lower exponent indicates a stronger bond. A B
Bond Energy  attraction  repulsion  U   m  n
(1) r
r
To find out the position of equilibrium equate the first differential of U with respect r to zero. On substitution of the same in equation 1 gives the expression for bond energy. These are given by the following equations. NPTEL Phase II : IIT Kharagpur : Prof. R. N. Ghosh, Dept of Metallurgical and Materials Engineering || |
For equilibrium :
dU
Am Bn
 Bn 
 m 1  n 1  0 or , r0  

dr r
r
 Am 
 m 


 nm 
nm
( 2)
 m   Am 
(3)
On substituing (2) in (1) U min  A  1 

 n   Bn 
The strength and the stiffness of solid are primarily determined by its atomic bond. Figure 10 gives the relation between stiffness and elastic modulus. The atoms in solid are arranged in a periodic fashion. The bond energy in the neighborhood of the mean position of an atom could be represented by a Taylor series. At equilibrium the first differential of U is zero. The change in energy due to an infinitesimal displacement of the atoms by x is approximately equal to the third term of the Taylor series (higher order terms can be neglected since x is small). Differentiating U with respect to displacement; x it is possible to find the relationship between elastic modulus and atomic bonds. The derivation is given in equation 4‐5. Figure 10: Derivation of the relation between stiffness S and elastic modulus. Note that the force is proportional to the displacement. The constant of proportionality is the stiffness of the bond. Displacement x over r0 is the elastic strain. 9
1  d 2U 
 dU 

U r0  x   U r0   
x
x 2  .... (4)
 2 

dr
2
dr

 r r 0

 r r 0
2
1 d U 
U r0  x   U r0    2 
x2
2  dr  r  r 0
dU  d 2U 


 2 
Force
F
x  Sx
dx
 dr  r  r 0
 d 2U 
Stiffness  S   2  at r  r0
(5)
 dr 
Using the expressions for U it is possible to find an equation for S in terms of the parameters A, B, m, & n NPTEL Phase II : IIT Kharagpur : Prof. R. N. Ghosh, Dept of Metallurgical and Materials Engineering || |
10
dU Am Bn


(6)
dr r m 1 r n 1
m  1Am  n  1Bn (7)
d 2U

2
dr
r m2
r n2
 1 


 Bn  n  m 
r0  
(8)

 Am 
 n2 


d 2U
 Am  n  m 
S  2  Bnn  m 
(9)

dr
 Bn 
Thus the elastic modulus is given by  n2 
 1 
 n 1 


Am   n  m   Bn   n  m 
Am   n  m 


E  S / r0  Bnn  m 
10
 Bnn  m 




 Am 
 Bn 
 Bn 
Table: Strengths of four major bonds found in common solids Bond Example
S, N/M
E, GPa
Covalent C‐C 180 1000 Metallic Al, Fe, Cu 75 300 Ionic Al2O3 24 96 Van der (C2H4)n 3 12 Waals wax 1 4 The above table gives an idea of the strengths of various types of bonds found in different solids. Note that carbon‐carbon bond in diamond is the strongest amongst all solids. Strength of most polymeric material is derived from van der Waal bonds between chains. It is the weakest of all the four. However strength of linear polymers along the chain could be substantially high because of covalent C‐C bond. In most materials except metals bonds may have mixed character. They might be partly ionic / covalent / van der Waal. Specific heat of metals: classical versus quantum mechanics Having established the nature of bonds in metals it should be possible to explain or guess few of its other properties as well. In metals atoms are arranged in a regular fashion with valence electrons being free to move as atoms in a gas. The atoms are not stationary. They keep moving about their NPTEL Phase II : IIT Kharagpur : Prof. R. N. Ghosh, Dept of Metallurgical and Materials Engineering || |
mean position of rest. On heating, the energy should be absorbed by both oscillating atoms as well as free electrons. The specific heat of metal would therefore be made up of two components as given by the following equation: Classical mechanics based on Maxwell‐Boltzmann (MB) statistics has been used to predict correctly specific heat of gases. Let us see what happens if the same is applied to metals. Assume atoms to be harmonic oscillator. It has three modes of vibrations along three different axes. Energy due to each mode has two components one due to kinetic energy and the other due to potential energy. According to MB statistics energy of each of these is equal to (kT/2). Thus the energy due to one mode of vibration is kT. Net energy / atom for 3 modes of vibration = 3kT. Therefore the total stored energy for one mole of metal E = 3kNT = 3RT. (Note that T is temperature in degree absolute, k is Boltzmann constant & N is Avogadro number, R is universal gas constant). Specific heat is therefore given by the following expression: 3
6
/
Molar specific heat of most metal is around 3R. This is the well‐known Dulong‐Petit rule. It is independent of temperature. The fact that it predicts specific correctly also indicates that the contribution of electrons towards specific heat is negligible. However if one applies MB statistics it should have significant contribution. In addition low temperature measurement has shown that specific heat decreases with decreasing temperature. It becomes 0 at 0º K. Figure 11 shows schematic variation of specific heat as a function of temperature. It suggests that classical mechanics based predictions are not valid when particles are small or if the temperature is low (which means that energy is low). Under such circumstances quantum mechanics based predictions are found to be accurate. 11
Figure 11: Molar specific heat of metals as function of normalized temperature. temperature D is Debye NPTEL Phase II : IIT Kharagpur : Prof. R. N. Ghosh, Dept of Metallurgical and Materials Engineering || |
Quantum mechanics imposes restrictions on energy levels a particle can have and whether a given level will be occupied by it. In contrast classical mechanics does not impose any restriction on energy levels. The probability that a given level is occupied is governed by MB statistics in the case of classical mechanics and by Fermi Dirac statistics in the case of quantum mechanics. Figure 12 – 13 presents the salient features of the two distributions. Figure 12: Maxwell – Boltzmann distribution: Figure 13: At absolute zero all discrete energy the chance of a lower energy level being levels up to E0, known as Fermi level are occupied is higher. Energy levels are occupied. With rising temperature only the continuous. particles near Fermi level move to higher level. Problem: Apply classical mechanics to estimate (Cv)electronic of mono‐valent metal. Answer: Each free electron has 3 degree of freedom (it could move along any 3 direction). According to MB statistics net energy for each degree of freedom is kT/2. One mole of metal has N number of free electrons. Therefore E = 3(kT/2)N &
3
3
/
. Note that this half of (Cv)atomic. Problem: Apply QM to estimate (Cv)electronic of mono‐valent metal. Answer: Only the electrons near the Fermi level can absorb thermal energy. The number of electrons near the Fermi surface is negligible in comparison to the total number of electrons. As an approximation one could assume that electrons occupying energy level E0‐kT moves over to E0+kT.Thus the net increase in energy = 2kT. This is very small in comparison to E0. This is why (Cv)electronic is negligible. Quantum mechanics based model for specific heat was first proposed by Einstein. According to this atoms assumed to be harmonic oscillator can have packets of energy called phonon. This is given by
. Where h is Planck’s constant, is frequency of oscillation and n is an integer representing quantum number. Even at 0ºK atoms can have small but finite energy. Using this 12
concept it can be shown that
∝
specific heat. This is given by
∝
. Later Debye derived more precise expression for where D is known as Debye temperature. Both the models predict that Cv becomes 0 at 0ºK although at low temperatures Debye model provides better predictions. NPTEL Phase II : IIT Kharagpur : Prof. R. N. Ghosh, Dept of Metallurgical and Materials Engineering || |
Band theory of metals Metals having excellent conductivity are mostly monovalent. Free electron model where electrons are free to move about can certainly explain its good electrical and thermal conductivity. However problem arises in explaining why contribution of electrons in absorbing thermal energy is negligible. In order to overcome this problem it was necessary to apply the concept of quantum mechanics. This states that when the valence bands overlap so that the outer shell electrons are equally shared by all the atoms the energy levels of these are governed by Pauli’s exclusion principle. No two electrons can have all quantum numbers same. Four quantum numbers are needed to describe the state of an electron. These are n1, n2, n3 (all are positive integers) and spin quantum number. The energy levels can therefore be visualized as a regular array of points in 3D in the form of a cubic lattice where each cell has one lattice point. Each point can have two electrons with opposite spin. Thus if there are n atoms there will be 2n number of sites to accommodate the valence electrons. Thus in a monovalent metal only half of the sites will be occupied whereas the rest are vacant. The distribution of electrons in the valence band follows Fermi Dirac statistics. At absolute zero all the levels up to a maximum value E0 (Fermi energy level) are occupied. This is schematically shown in figure 14. Since the capacity of this band is 2n only half of the band is filled up by electrons. Under the influence of electrical field the electrons near the Fermi level can move over to vacant sites having higher energy. In other words these are free to move. This is why these are excellent conductor of electricity. What happens if its valence is 2? In such a situation the valence band will be full. However if there are overlapping empty conductions bands it can still exhibit good conductivity. If there is a large gap between valence band and conduction band, it would behave as an insulator. There are elements where the gap between valence band and conduction band is small. In such a case either as a result of thermal excitation a few electrons can move over to conduction band and behave as semi‐conductor. Such a material exhibits higher conductivity with increasing temperature. 13
Figure 14: Schematic representation of band structures of conductor, semiconductor & insulator. In metals there are overlapping valence band & conduction
band. Horizontal lines represent energy levels occupied by electrons.
NPTEL Phase II : IIT Kharagpur : Prof. R. N. Ghosh, Dept of Metallurgical and Materials Engineering || |
Electrical conductivity Electrical conductivity of metal should be a function of the number of moving electrons and their average velocity. It is possible determine simple expression for conductivity from our current knowledge of structure of metal. Under the influence of electrical field (F) an electron would experience a force of magnitude eF. If m is the mass of the electron its acceleration should be given by eF/m. Let t be the relaxation time so that the average drift velocity of electron is given by (eF/m)t. If there are n free (mobile) electrons in unit volume the current density (J) should be given . In metals the kinetic energy . Therefore conductivity (s) should be
by
of electrons that are free to move could be assumed to be equal to its Fermi energy. Hence the average drift velocity (v) should be given by 14
and mean free path (l) of electrons is
. This helps us understand the effect . A more convenient expression for conductivity is
of temperature and alloy addition (impurity) on conductivity of metals. Mean free path of electrons moving through a periodic array of atoms oscillating about their mean position depends on their interaction. In terms of quantum mechanics this represents electron phonon interaction. As the temperature increases electron scattering too increases. This is why l is inversely proportional to temperature. However Fermi energy which determines average velocity of electron is a relatively weak function of temperature. Therefore conductivity can be assumed to vary inversely with temperature; whereas resistivity (which is reciprocal of conductivity) is directly proportional to temperature. Presence of impurity atoms increases phonon electron interaction. This results in reduction of mean free path (l) of electrons. Therefore conductivity decreases and resistivity increases as concentration of impurity atom increases. For a binary alloy resistivity ( ) is given by ∝ 1
where x is the mole fraction of the solute atom. Thermal conductivity Metals are good conductor of heat. In fact the ratio of thermal conductivity (Kth) over electrical conductivity is constant for all metals. It is known as Wiedemann – Franz law. This states 2.45 10 /°
. Like electrical conductivity thermal conductivity is also that
inversely proportional to temperature and impurity concentration. Truly thermal conductivity should depend on specific heat, mean free path of electrons and phonons. While mean free path of electrons is of the order of 100 atomic distances, mean free path of phonon is one atomic distance. Although the contribution of electron towards specific heat is negligible it plays a crucial role in heat conduction through metals. This is because most metals have half filled valence band and overlapping empty conduction band and the number of electrons near Fermi surface is fairly large. This determines thermal conductivity. NPTEL Phase II : IIT Kharagpur : Prof. R. N. Ghosh, Dept of Metallurgical and Materials Engineering || |
Problem: Estimate mean free path of electrons in gold assuming that all valence electrons are free to move. Given atomic weight = 197, density = 19.3g/cc, resistivity = 2.2x10‐8 m, e = 1.6x10‐19 Coulomb, m = 9.1x10‐31 kg, Avogrado number = 6.02x1023. Answer: Assuming that all valence electrons are free to move the number of electrons / unit .
volume
6.2 10
5.9 10
relaxation time
2.74
& 15
10 . Thus mean free path = vt = 1.0x108 x 2.74x10‐14 = 2.74x10‐6 cm. This is approximately equal to 100 atomic spacing. Summary Why bond forms in materials has been explained in terms of atomic configuration of various elements. The nature and main characteristics of three primary bonds (metallic, covalent and ionic bonds) have been introduced. The correlation between bond and certain properties of material like melting / boiling points, strength, stiffness and electrical & thermal conductivity have been illustrated with examples. An attempt has been made to give a little quantitative insight into the correlation between bond strength & elastic modulus of materials. Exercise: 1. What are the three primary bonds in materials? Which is the strongest? Why? 2. What is the electronic configuration of silicon atom? What type of bond do you expect here? 3. Inert gases have completely filled outer cell yet the boiling point of these increases with the atomic number. Explain why it is so. 4. Find out from Handbook the melting points of metals in the 4th row of the periodic table (K‐
Zn). Which has the highest melting point? Explain this in terms of their electronic configurations. 5. Stiffness of C‐C bond is around 200N/m. Estimate its elastic modulus assuming the distance between the atoms to be 2x10‐10m. Which form of carbon has such a modulus? Answer: 1. Three primary bonds are covalent, ionic and metallic. Boiling or melting points of material are indicators of the strength of the bond. C‐C bond (covalent) in diamond is possibly the strongest bond. Its melting point is ~3700°C. Its atomic diameter is small. It is verified by x‐
ray diffraction technique. 2. Atomic number of Si is 14. Its electronic configuration is 1s22s22p63s23p2. It has four atoms in the outer cell like carbon. It also forms covalent bond. Its atomic diameter is larger. The bond is not as strong as C‐C bond. Its melting pint is 1410°C. NPTEL Phase II : IIT Kharagpur : Prof. R. N. Ghosh, Dept of Metallurgical and Materials Engineering || |
3. Inert gases have completely filled outer cell. As you go down the group (from He to Xe) the diameter increases as more orbits are added. As the diameter increases the centre of orbital electrons will no longer coincide with the nucleus. The deviation increases with the increase in diameter. Such atoms will therefore behave like dipoles & promote bonding due to van der Waal force. Boiling point of He (Z=2) & Xe (Z=54) are 4.2°K & 165°K respectively. 4. The fourth row of periodic table has several metals with atomic numbers varying from 19 (K) to 30(Zn). Their melting points are given in the following table. Electronic configuration of Ar (18) is 1s22s22p63s23p6. K atom has one more electron. Its preferred site is 4s. The next atom Ca has 2 electron in 4s cell. Thereafter electrons occupy 3d cell. Unpaired 3d electrons too participate in bond formation. If the number of such electron increases the cohesive energy and hence melting point increases. (Mn: exception?). In Zn all 3d electrons are paired & do not take part in bonding. In this case number of electrons that take part in bond formation is less. Therefore its melting point is low. K 19 Ca Sc Ti 22 V 23 Cr 24 Mn25 Fe26 Co27 Ni28 Cu29 Zn30 20 21 64 833 1539 1668 1900 1875 1245 1539 1495 1453 1083 419
5. The bond is often visualized as a stiff spring which breaks without deformation. The relation between the force (F) needed to snap a bond and the stiffness (strength) of the bond (S) is given by F = S where is the minimum distance of separation at which the bond breaks. Note that the dimension of stiffness is N/m. The cross sectional area is of the order of a2. Therefore stress = F/a2 = (S/a) ( /a) = E where E is modulus & is strain. Thus E = S/a. If a is of the order of inter atomic distance (a~0.2nm). E = 200/ (0.2x10‐9) = 1012 N/m2 = 1000 GPa. 16
NPTEL Phase II : IIT Kharagpur : Prof. R. N. Ghosh, Dept of Metallurgical and Materials Engineering || |