1 The Quantum theory and the electronic structure of atoms

2 - Light energy is transmitted by waves, each wave has 2 properties. 1- Wave length (λ) : -It is the distance between 2 identical points in successive waves. Its unit is nm = 10 A o = m = cm Therefore 1A o = cm = m i.e cm ( x 107) nm 2- Frequency (ν) : - It is the number of waves per second. - Its unit is sec -1 or hertz (Hz). - Since C (the speed of light)= ν λ = 3 x 10 8 m/s. Therefore ν = C/ λ Amplitude : It is the vertical distance from midline to peak or trough. Amplitude Direction of propagation

3 Electromagnetic radiation : - Electromagnetic waves has electric and magnetic field transmitted perpendicular on each other and have, λ and C = 3 x 10 8 m/s. i.e. γ (Gamma) rays have the higher frequency and the shorter wave length, while radio waves have the lower frequency and the longer wave length. γ raysX raysU.V. raysVisible raysI.R.raysMicro waves Radio waves nm nm 800 nm  m λ increases, E and decrease

4 Planck’s quantum theory :- - The atom or molecule can emit or absorb a discrete quantity of energy i.e fixed amount of energy called “Quanta”. - The energy of a single quantum is proportional to the frequency i.e. E α therefore E = h Where h is Plank’s constant = 6.63 x J.s Since C = λ therefore E = h C/ λ Example - What is the energy emitted at 450 nm by CuCl 2 ? Solution E = h = h C/ λ = (6.63 x J.s) (3 x 10 8 m/s) / 450 x m = 4.4 x J

5 Bohr’s theory of hydrogen atom : 1- The atom consists of a + ve center (nucleus) and a number of negative electrons equal to the number of + ve charges in the nucleus. 2- A centrifugal force appears when electrons rotate around the nucleus which is balanced by the attraction force of the nucleus. 3- Electrons rotate in definite energy levels. 4- Each electron has a definite amount of energy depending on the distance between its energy level and the nucleus. 5- Each energy level is expressed by a whole number called principle quantum number KLM N O PQ

6 6- The electron remains in the lowest energy level (ground state). If it acquires an amount of energy by heating or electricity, the electron becomes excited and jumps to a higher energy level depending on the amount of energy absorbed, the excited electron (in the higher level) is now unstable and soon returns to its original level losing the same amount (quantum or photon) of energy, but in the form of light. n=1 n=2 photon P initial

7 7- The energy of electron is calculated from the equation : E n = - R H x 1/n 2 - Where R H is Rydberg constant = 2.18 x J, n is an integer called principle quantum number, n=1, 2, 3 (no. of energy level occupied by the electron) -The negative sign in the equation means that the energy of the electron in the hydrogen atom is always lower than that when it is in the free state i.e when n = ∞ (where E= 0). -Thus as n increase E will increase and the most negative energy value (lowest energy) is obtained when n = 1 (the lowest energy level)

8 8- The difference between the energies of the initial and the final is given by ΔE where ΔE = E 2 – E 1 E 2 = -R H x 1/n 2 2 and E 1 = -R H x 1/n 1 2 therefore ΔE =-R H x 1/n ( -R H x 1/n 1 2 ) ΔE = R H (1/n /n 2 2 ) such that n 2 > n 1 N.B. If ΔE is + ve therefore energy is absorbed and if – ve therefore energy is emitted Since ΔE = h C/ λ therefore λ = h C /ΔE nfnf nini EE E initial E final

9 Example - Calculate the energy required to excite the hydrogen electron from level n=1 to level n=2. Also calculate the wavelength of light that must be absorbed by a hydrogen atom in its ground state to reach this excited state Solution Use the equation ΔE = R H (1/n /n 2 2 ) ΔE =2.18 x J (1/1 2 -1/2 2 ) = 1.63 x J - The positive sign of ΔE indicates that the system has gained energy. λ = h C /ΔE = (6.626 x J.s)(3 x 10 8 m/s)/1.63 x J = 1.21 x m

10 9- The radius of each circular orbit is expressed as r =0.529 x n 2 A o Example :- Calculate radius of H atom ( n = 1) Solution : r = x (1) 2 = A o (Bohr radius)

11 Quantum numbers -Quantum numbers describe the distribution of electrons in the atom. 1-The principle quantum number (n) n is an integer = 1, 2, 3, 4,……7 It determines : 1- The energy of an orbit (principle energy level) (E= -R H x 1/n 2 ) 2- The size of the orbit ( r = n 2 ) 3- The maximum no. of electrons that can occupy the orbit 2(n 2 ) i.e. n 1 contains 2 electrons, n 2 contains 8 electrons and n 3 contains 18 electrons.

12 2- The angular momentum quantum number (l):- - It tells us the no. and shape of subshells in the orbit. The value of l depends on n where l is an integer from 0 to (n-1) l hgfdpssubshell i.e when n = 1 l = 0 1s n = 2 l = 0, 1 2s 2p n = 3 l = 0, 1, 2 3s 3p 3d n = 4 l = 0, 1, 2, 3 4s 4p 4d 4f

13 3- The magnetic quantum number (m l ) :- - It tells us the number of orbitals with in the subshell. m l = 2l + 1 (this eq. gives no. of orbitals) where value of m l from –l to +l including 0 i.e - For l = 0 (s-subshell) m l = 2 x = 1 therefore s has 1 orbital therefore s-orbital is divided into one orbital its ml = 0 - For l = 1 (p-subshell) m l = 2 x = 3 therefore p has 3 orbitals i.e. p-orbital is divided into three orbitals their ml = -1,0, +1 or px, py, pz For l = 2 (d-subshell) m l = 2 x = 5 therefore p has 5 orbitals i.e. d-orbital is divided into five orbitals their m l = -2, -1, 0, +1, +2

14 4- The electron spin quantum number (m s ) :- - It tells the electron magnetic properties due to its spin. Any electron can rotate clock wise (+1/2) or anticlock wise (-1/2( Therefore m s has only two values +1/2 and -1/2. Example - For principle quantum level n = 5 determine the number of subshells (l) and give the designation of each ? Solution n = 5 therefore l = n – 1 = 0, 1, 2, 3, 4 (5s 5p 5d 5f 5g)

15 The shape of orbitals: s- orbital :- All s- orbital are spherical in shape. They differ in size, as the principle quantum no. (n) increase as the size increase. p- orbitals : are three orbitals p x, p y, p z identical in size, shape (two lopes) and energy. They increase in in size from 2p to 3p to 4p. 1s 2s3s p z -orbital

16 d- orbitals : are five orbitals dxy, dyz, dxz, dx 2 -y 2, dz 2 each one has four lopes. The energy of orbitals :- - It was assumed that the energy of orbitals increase in the order : 1s < 2s = 2p < 3s =3p =3d < 4s =4p =4d =4f <…. But this was wrong, due to orbital overlap the energy of orbitals depends not only on the principle quantum no. (n) but also on the angular quantum no. The order is: 1s< 2s<2p< 3s<3p< 4s<3d<4p< 5s<4d<5p < 6s<4f<5d<6p< 7s<5f<6d<7p, where 1s has the lower energy. y zz y x x z d xy d yz d xz d x 2 - y 2 d z 2

17 Electron configuration : - Any electron can be described in an orbital by knowing the four quantum numbers (n, l, m l, m s ) Eg. (1, 0, 0, +1/2) or (1, 0, 0, -1/2) represent an electron in 1s-orbital. Pauli exclusion principle : ‘ No two electrons in an atom can have the same four quantum numbers ‘ eg. He 2 : 1s 2 E,gHOr1s 1 Denotes the principal quantum number n denotes the number of electrons in the orbital 1s 1 Denotes the angular momentum quantum number l

18 Hund’s rule: “ The most stable arrangement of electrons in subshell is the one with the greatest number of parallel spin” i.e. no pairing is required. Eg C 6 electronic configuration : 1s 2 2s 2 2p 2 -If the two electrons are in the same 2p x orbital, great repulsion will be present than when they occupy two separate p-orbitals. 1s 2 2s 2 2p 2

19 Examples : - Which of the following are possible sets of quantum numbers for an electron ? 1.n = 1, l = 0, m l = 1, m s = +1/2 2.n = 7, l = 3, m l = -3, m s = -1/2 3.n = 2, l = 1, m l = 0, m s = 0 4.n = 1, l = 1, m l = 1, m s = +1/2 5.n = 3, l = 2, m l = 3, m s = +1/2 6.n = 4, l = 0, m l = 0, m s = -1/2 Solution :Remember: n is an integer from 1 to ∞, l is an integer= (n-1), m l from – l to +l and m s has only two values +1/2 and -1/2 1- False, for l = 0 ml = 0 2- True, it represents an electron in 7f 1 3- False, m s has only two values +1/2 and -1/2 4- False, for n = 1 l = 0 5- False for l = 2 m l = -2, -1, 0, 1, 2 6 -True, it represents an electron in 4s 1

20 Paramagnetic atoms : - When an atom contains unpaired electrons in its shell it is called paramagnetic. Eg. N :z = 7 C :z = 6 Diamagnetic atoms : - If it has no unpaired electrons it is called diamagnetic. eg. Ne :z = 10 Ar :z =18 N.B. No more than two electrons can be placed in each orbital. 1s2s2p   odd n o 1s2s2p  Even n o 1s2s2p3s3p  1s2s2p   odd n o