General and Inorganic Chemistry Lecture 2 PDF

General and Inorganic Chemistry (Chem101-WS24) Lecture 2 QUANTUM NUMBERS Prof. Dr. Rasha Hanafi  Did you contact Your supporter? List is in front of B5-109 and B5-103! Take advantage of t...

General and Inorganic Chemistry (Chem101-WS24) Lecture 2 QUANTUM NUMBERS Prof. Dr. Rasha Hanafi  Did you contact Your supporter? List is in front of B5-109 and B5-103! Take advantage of this Unique Support !!! FACULTY OF PHARMACY AND BIOTECHNOLOGY FIRST YEAR SUPPORT HEY FRESHMEN STUDENTS! Under the umbrella of “Chairman First Year Initiative”, the faculty of Pharmacy and Biotechnology offers support to freshmen students by our “First Year Supporters” Each 5 students have 1 supporter they can visit weekly within 2hr to talk about any aspect related to improvement of university experience and academic performance. You will find the list of supporters you are assigned to on the board in front of B5-109. Go and check the name and location of your supporter and take the first step and pass by him/her. In case the office hours does not suit your schedule, feel free to modify it with your supporter. For any inquiry, contact Vice Dean of Student Affairs: [email protected] (Office B5-103)  Key Elements By the end of this session, the student should be able to: Domain 1: Fundamental Knowledge 1-1- Competency Key Elements 1-1-1- Demonstrate understanding of basic principles of the quantum theory. 1-1-2 - Identify the quantum numbers of electrons 1-1-3 Derive the electronic configuration of elements and ions by applying Pauli Exclusion principle and Hund’s rule Lecture 2, CHEM101-WS21 3 BOHR’S MODEL FOR THE HYDROGEN ATOM Bohr found that 1. the e- of H moves in a circular orbit. 2. “the electrostatic force of attraction by the p+ for the e- is balanced with the centrifugal force due to circular motion of the e-.” 3. Energies that an e- in H atom can occupy are given by En = -RH/n2 Where, En : energy of e- RH: Rydberg constant (2.180 x 10-18 J) Energy levels of H and various emission series depend on n is an integer: the principal quantum excitation of e- followed by its number (n= 1, 2, 3…..etc) relaxation via emission of BUT the theory was only valid for 2,HCHEM101-WS21 Lecture atom photons of light at specific4 wavelengths  THE QUANTUM MECHANICAL MODEL Based on the postulates of the dual nature of light, De Broglie hypothesized that also an e- (which Bohr treated only as a particle) could behave as a standing wave. Schrödinger, 2, is directly formulated an proportional to the equation (Wave probability of finding functions, ) that the e- in any point. describes the the depth of colour is behavior and proportional to the energies of probability of finding submicroscopic the e- at a given point. particles in general The closer to the (analogous to nucleus, the higher the Newton’s When laws equation Schrödinger of was solved,probability many wave of functions finding motion (orbitals) the eof for to satisfy it. Each were found - is.these orbitals is macroscopic by a series of numbers called quantum characterized objects). which describes various properties of the orbital. numbers, Lecture 2, CHEM101-WS21 5  QUANTUM NUMBERS ANSWER … How Where is does the the e-? e- What is behave? the e-? The 4 quantum numbers (q.n.) inform about the : 1. Distance from the nucleus (Principal q.n. n). 2. Shape of the orbital (Angular Momentum q.n. ℓ). 3. Position of the e- in the orbital (Magnetic Orbital q.n. mℓ). Lecture 2, CHEM101-WS21 6 4. Spin of e (Magnetic Spin q.n. ms) - THE PRINCIPAL QUANTUM NUMBER (n) n = integral value 1, 2, 3, …. The value of n determines the energy of an orbital. relates to the average distance of the electron from the nucleus in a particular orbital, and is less tightly bound to the nucleus. The larger n is, the greater the average distance of an electron in the orbital from the nucleus and therefore the larger the orbital. n=1  first principal level Lecture 2, CHEM101-WS21 7  THE ANGULAR MOMENTUM QUANTUM NUMBER (ℓ) The angular momentum quantum number (ℓ) describes the “shape” of the orbitals. Value of ℓ depends on value of the principal quantum number, n, where for a given value of n, ℓ has possible integral values from [0 ( n -1)] If n= 1  ℓ = [0 1-1]In the first principal level there is only 1 sublevel, ℓ= 0: (s) If n=2, ℓ= 0 , 1: (s, p) If n= 3, ℓ = 0, 1, 2: (s, p, d) If n=4, ℓ = 0, 1, 2, 3 : (s, p, d, f) l 0 1 2 3 4 5 In general, in the nth principal level, there are n different sublevels. Name of s p d f g h orbital Lecture 2, CHEM101-WS21 8  THE MAGNETIC ORBITAL QUANTUM NUMBER (mℓ) Has integral values between ℓ and -ℓ, including zero. mℓ = ℓ,….,+1, 0, -1,…., - ℓ The value of mℓ is related to the orientation of the orbital in space relative to the other orbitals in the atom.  For ℓ=0 or s , mℓ can have only one value 0. This means that an s sublevel contains only 1 orbital.  For ℓ=1 or p, mℓ =1, 0, -1. This means that the p sublevel has 3 orbitals.  For ℓ=2 or d, mℓ = 2, 1, 0, -1, -2 (5 orbitals)  For ℓ=3 or f, mℓ = 3, 2, 1, 0, -1, -2, -3 (7 orbitals) Lecture 2, CHEM101-WS21 9  ATOMIC ORBITALS, SHAPES AND SIZES s orbitals 3p orbitals 5 d orbitals 7f orbitals Lecture 2, CHEM101-WS21 10  MAGNETIC SPIN QUANTUM NUMBER (ms) To describe e- spin. e- have magnetic properties that correspond to those of a charged particle spinning on its axis clockwise or counter clockwise. ms can only have 1 of 2 values ms = +1/2 or –1/2 e- that have the same value of ms have parallel spin and those of different values of ms have opposed spins. Lecture 2, CHEM101-WS21 11 Lecture 2, CHEM101-WS21 12  PAULI EXCLUSION PRINCIPAL No 2 electrons in an atom can have the same set of 4 quantum numbers.  only 2 e- can fit into an orbital and they must have opposed spins. (Otherwise they would have had the same set of 4 quantum numbers). Example: Can the following set of quantum numbers (n, ℓ, mℓ, ms)= (3, 2, 0, +1/2) be valid for 1 e-? If yes, identify the orbital involved. Answer: n=3, so ℓ can have values from 02. So ℓ =2 is possible and has the name of d. Orientation can be +1/2 or - 1/2. 3, 2, 0, +1/2 valid, 3d So this orbital is valid2,and is -1/2 2, 0, 3d. not valid, ℓ cannot equal n 2, 0, 0, +1/2 valid, 2s 4, 3, 2, valid, 4f +1/2 Lecture 2, CHEM101-WS21 13 2, 1, 0, 0 not valid, ms can not be  ELECTRONIC CONFIGURATION IN ATOMS e- enter the available sublevels in order of increasing sublevel energy. Ordinarily, a sublevel is filled to capacity before the next one starts to fill. The order of filling orbitals: 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 4f, 5d, 5f,….. etc. Lecture 2, CHEM101-WS21 14 HUND’S RULE OF ORBITAL FILLING “The most stable arrangement of electrons in subshells is the one with the greatest number of parallel spins”  When several orbitals of equal energy are available, e- enter singly with parallel spins. NO PAIRING until all orbitals are half filled Lecture 2, CHEM101-WS21 15  Test yourself Write the electronic configuration of the following atoms with reference to the nearest noble gas 5B 1s22s22p1 [He] 2s2 2p1 1s22s22p2 [He] 2s2 2p2 6C 1s22s22p3 [He] 2s2 2p3 7N 1s22s22p4 [He] 2s2 2p4 8O 1s22s22p5 [He] 2s2 2p5 9F 1s22s22p6 10Ne 1s22s22p63s2 12Mg [Ne]3s2 18Ar 1s22s22p63s23p6 1s22s22p63s23p64s2 20Ca [Ar]4s2 30Zn 1s22s2Lecture 2p63s 2 3p64s23d10 [Ar] 36Kr 2, CHEM101-WS21 16 2 10  Exceptions to rules of electronic configuration 24 Cr 1s2 2s2 2p6 3s2 3p6 4s2 3d4 1s2 2s2 2p6 3s2 3p6 4s1 3d5 Orbitals are more stable when they are empty half filled completely filled 29Cu 1s2 2s2 2p6 3s2 3p6 4s2 3d9 1s2 2s2 2p6 3s2 3p6 4s1 3d10 Lecture 2, CHEM101-WS21 17  MAGNETISM OF ELEMENTS Based on Hund’s and Pauli’s rules, it is possible to determine the number of unpaired e- in an atom. PARAMAGNETIC elements are those that contain net unpaired spins and are attracted by a magnet. DIAMAGNETIC elements do not contain net unpaired spins and are slightly repelled by a magnet, because when e- spins are paired (antiparallel to each Lecture 2, CHEM101-WS21 18 other), magnetic effects ELECTRONIC CONFIGURATION OF IONS In general, when an ion is formed from its atom, e- are added to or removed from sublevels of the highest energy level. Anions Cations 7 N + 3e -  7 N 3- 11 Na  11 Na + (1s22s22p3) (1s22s22p63s1)  (1s22s22p6) (1s22s22p6) + e- N has now the Na acquired Ne noble gas configuration of the structure by losing e-to nearest noble gas Ne form cations. by gaining e-. Lecture 2, CHEM101-WS21 19  Test yourself How many electrons are gained/lost by the following atoms? Write the electronic configuration of the atom and the ion. Oxygen, fluorine, Magnesium, Aluminum (use the periodic table to get their atomic numbers). O (1s2 2s2 2p4 ) + 2e-  O2- (1s2 2s These are 8, 9, 12, and 13 respectively. 8 8 2 2p6 ) 9 F(1s2 2s2 2p5 ) + e-  9 F- (1s2 2s2 2p6 ) 12 Mg (1s22s22p63s2) 12Mg2+ (1s22s22p6) + 2e- 13 Al (1s 2 2s2 2p6 3s2 3p Lecture 1 2, )  13 Al CHEM101-WS21 3+ (1s2 2s2 2p6 ) + 20 - CLASSIFICATION OF GROUPS OF ELEMENTS IN THE PERIODIC TABLE ACCORDING TO THE TYPE OF SUBSHELL BEING FILLED WITH e- Lecture 2, CHEM101-WS21 21  REFERENCES 1. Chemistry, 10th ed., Raymond Chang, ISBN 978-0-07-017264-7, McGraw Hill. Chapter 7. Lecture 2, CHEM101-WS21 22