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The ratio of masses of one element that combine with a constant mass of another element.

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Presentation on theme: "The ratio of masses of one element that combine with a constant mass of another element."— Presentation transcript:

1 http://www.chem.uncc.edu/faculty/murphy/1251/slides/C12/sld003.htm The ratio of masses of one element that combine with a constant mass of another element can be expressed in ratios of small whole numbers.

2 Atomic models

3 Electrons Bohr’s atomic model (1913)- electrons revolve around the nucleus in distinct energy levels.

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5 How light is emitted... Excited state- electron(s) in higher energy levels than at ground state (lowest energy state). Excited state- electron(s) in higher energy levels than at ground state (lowest energy state). Energy is emitted when the electron(s) return to ground state. Energy is emitted when the electron(s) return to ground state. The frequency of the radiation emitted depends on the difference between the higher and lower energy levels. The frequency of the radiation emitted depends on the difference between the higher and lower energy levels.

6 Frequency-the number of wave cycles passing a point in a period of time. (C=3.0 x 10 8 m/s) As frequency increases, wavelength decreases

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8 Light of a particular wavelength (λ) has a particular frequency (v) and energy. Light of a particular wavelength (λ) has a particular frequency (v) and energy. E = h∙v and c = λ∙v E = h∙v and c = λ∙v If v, λ, or E are known, the other two can be calculated. If v, λ, or E are known, the other two can be calculated. c=3.0 x 10 8 m/s speed of light h=6.63 x 10 -34 joule-sec Planck’s constant E is the energy of one photon of light

9 Yellow light given off by a sodium vapor lamp has a wavelength of 589 nm. What is the frequency of this radiation? v = c/λ v = c/λ = 3.0 x 10 8 m/s x 10 9 nm = 5.09x10 14 s -1 = 3.0 x 10 8 m/s x 10 9 nm = 5.09x10 14 s -1 589nm 1 m 589nm 1 m

10 Calculate the energy of one photon of yellow light whose wavelength is 589nm. E = hv = (6.626 x 10 -34 J∙s)(5.09x10 14 s -1 ) = (6.626 x 10 -34 J∙s)(5.09x10 14 s -1 ) = 3.37 x 10 -19 J = 3.37 x 10 -19 J Substitute v= c/λ in E = h∙v and you get E = h∙c λ

11 Wave-particle duality of light Planck stated that energy is radiated in discrete packets called quanta. A photon is a quantum of light having the energy h∙v. Planck stated that energy is radiated in discrete packets called quanta. A photon is a quantum of light having the energy h∙v. Light’s particle nature is seen in its ability to eject electrons from a surface (photoelectric effect), and by the emission spectra of elements. Light’s particle nature is seen in its ability to eject electrons from a surface (photoelectric effect), and by the emission spectra of elements.

12 Principal Quantum Number (n) Tells its energy level and distance from the nucleus Tells its energy level and distance from the nucleus The maximum number of electrons at each energy level is 2n 2. The maximum number of electrons at each energy level is 2n 2. at n = 1, there can be 2(1) 2 =2 electrons at n = 2, there can be 2(2) 2 =8 electrons at n = 3, there can be 2(3) 2 =18 electrons

13 angular quantum number (l) Sublevel tells the shape of the electron cloud. tells the shape of the electron cloud. Sspherical (l = 0), P polar (l = 1), Dcloverleaf (l = 2) F (l = 3) (l) can be any integer between 0 and n - 1

14 magnetic quantum number (m) describes the orientation in space of a particular orbital describes the orientation in space of a particular orbital One pair of electrons can occupy each orbital One pair of electrons can occupy each orbital s sublevels have one orbital p sublevels have 3 orbitals d sublevels have 5 orbitals f sublevels have 7 orbitals m can be any integer between -l and +l m can be any integer between -l and +l

15 For p orbitals (l = 1), there are three possible orientations, so m can be -1, 0, or 1.

16 Spin The forth quantum number (s) indicates the direction of spin on the electron.   s = +½ or -½ Pauli Exclusion Principle states that no two electrons in an atom can have the same set of four quantum numbers. The two electrons in an orbital must have opposite spins. The two electrons in an orbital must have opposite spins.

17 Allowable Combinations of Quantum Numbers Shell n Subshell l Subshell Notation Orientation m Number of Orbitals 101s01 2 02s01 12p-1 0 +13 3 03s01 13p-1 0 +13 23d-2 -1 0 +1 +25 4 04s01 14p-1 0 +13 24d-2 -1 0 +1 +25 34f-3 -2 -1 0 +1 +2 +37

18 Example Quantum Number Problem List the quantum numbers of all 10 electrons in a neon atom. 1s : 1, 0, 0, ±½ 2s: 2, 0, 0, ±½ 2p: 2, 1, -1, ±½ 2, 1, 0, ±½ 2, 1, 1, ±½

19 Which of the following sets of quantum numbers are possible? 1, 0, 0, ½ 1, 3, 0, ½ Yes No, l must be less than n 4, 2, -2, -½ Yes 3, 1, 2, ½ No, -l <m < +l 3, 2, 1, 0No, spin is never an integer

20 Electrons occupy the lowest energy level available first (Aufbau Principle). Electron Configuration Notation H has a 1s electron H has a 1s electron He has two 1s electrons, or 1s 2 He has two 1s electrons, or 1s 2 Li has two 1s and a 2s electron, or 1s 2 2s 1 Li has two 1s and a 2s electron, or 1s 2 2s 1 Electron Dot Notation- show only the electrons in the outermost energy level H∙ He: Li∙ Be: Mg:

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22 1s 2s 2p 3s 3p 3d 4s 4p 4d 4f 5s 5p 5d 5f 6s 6p 6d 6f 7s 7p 7d 7f Sublevels fill in order of increasing energy. 1s2s2p3s3p4s3d4p5s4d5p6s4f5d6p7s5f6d7p

23 Predicting electron configurations from the periodic table.

24 Electron configuration notation Helium has 2 electrons, so its electron configuration would be 1s 2 Li 1s 2 2s 1 N 1s 2 2s 2 2p 3 Ne 1s 2 2s 2 2p 6 Na 1s 2 2s 2 2p 6 3s 1 or [Ne]3s 1

25 Orbital notation Hund’s rule- orbitals of equal energy are each occupied by one electron before any orbital is occupied by a second electron, and all electrons in singly occupied orbitals must have the same spin. 1s 2s 2p 1s 2s 2p N (7)  F (9) 

26 Paramagnetic- an atom having one or more unpaired electrons Paramagnetic- an atom having one or more unpaired electrons Diamagnetic- all electrons are paired Diamagnetic- all electrons are paired Which elements in period 2 are diamagnetic? Be and Ne

27 Ways to represent titanium 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 2 electron configuration             __ __ __ 1s 2s 2p 3s 3p 4s 3d Ti: 22 2 8 10 2

28 Ways of separating a mixture chromatography distillation


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