QUANTUM MECHANICAL MODEL OF THE ATOM

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Presentation transcript:

QUANTUM MECHANICAL MODEL OF THE ATOM

DEFINITION OF TERMS QUANTUM – UNIT OF ENERGY ORBITALS – space around a nucleus where the probability of finding an electron is greatest. QUANTUM NUMBERS – set of four numbers used to describe the electron’s energy level in terms of: a. distance from the nucleus b. shape of the orbital c. orientation in space (x, y, z) d. direction of electron spin (clockwise, counter-clockwise)

ENERGY LEVELS OF THE ATOMIC ORBITALS 1. MAIN ENERGY LEVEL (MEL). - represents the average distance of the electrons from the nucleus. - values 1, 2, 3, 4, 5, 6 and 7

Electrons in MEL closer to the nucleus have the lowest energy

2. SUBLEVELS or SUBSHELLS - define the shapes of the orbitals - has four possible shapes called s, p, d and f EACH SUBLEVEL HAS DIFFERENT NUMBERS OF ORBITALS AND EACH ORBITAL CAN ONLY HAVE A MAXIMUM OF 2 ELECTRONS

s p d f 1 orbital – 2 electrons 3 orbitals – 6 electrons

SUMMARY OF ENERGY LEVELS, ORBITALS AND ELECTRON RELATIONSHIPS MEL (n) MAXIMUM NUMBER OF ORBITALS (n2) MAXIMUM NUMBER OF ELECTRONS PER ENERGY LEVEL (2n2) 1 2 4 8 3 9 18 16 32

ELECTRONIC CONFIGURATION - Process of arranging the electrons around the nucleus The problem? The solution!

THE RESULT!!!

RULES GOVERNING ELECTRON CONFIGURATION AUFBAU PRINCIPLE – electrons enter orbitals of lowest energy first. PAULI’S EXCLUSION PRINCIPLE – only two electrons can occupy a single orbital. HUND’S RULE OF MAXIMUM MULTIPLICITY– within the same sublevel, electrons prefer to occupy empty rather than half-filled orbitals.

What is the electron configuration of Oxygen? 2s 2p 3s 3p 3d 4s 4p 4d 4f 5s 5p 5d 5f 6s 6p 6d 7s 7p Oxygen has 8 electrons. Its electron configuration is: 1s 2 2s 2 2p 4 2 + 2 + 4 = 8 electrons

ELECTRON CONFIGURATION OF IONS What is the electron configuration of S 2- ? SOLUTION: 1. Solve for the total number of electrons. e - = p+ - c = 16 – (-2) = 18 2. Use the electron configuration chart 1s2 2s2 2p6 3s2 3p6 3. Check the number of electrons used. 2 + 2 + 6 + 2 + 6 = 18

√ √ EXCEPTIONS TO THE RULE 24Cr 1s2 2s2 2p6 3s2 3p6 4s2 3d4 29Cu 1s2 2s2 2p6 3s2 3p6 4s2 3d9 1s2 2s2 2p6 3s2 3p6 4s1 3d10 √ √

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