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Chemistry. Chemistry Chemistry is the study of the composition, properties, and transformations of matter.

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Presentation on theme: "Chemistry. Chemistry Chemistry is the study of the composition, properties, and transformations of matter."— Presentation transcript:

1 Chemistry


3 Chemistry Chemistry is the study of the composition, properties, and transformations of matter.

4 2N CO 2 + 2 N H 2 O + PHOTONS → 2(CH 2 O) N + 2N O 2 CARBON DIOXIDE + WATE R + LIGHT ENERGY → CARBOHYDRATE + OXYGENPHOTONS2(CH 2 O) N C6H12O6 + 6O2 → 6H 2O +6CO2+energy

5 Text Book : Genelral Chemistry, R. H. Petrucci, W.S. Harwood, Prentice Hall International, Inc., 2002, 8th Ed. Chemistry a molecular approach Nivaldo J Tro Pearson Education 2008 and all other chemistry books 1 st Midterm : 7 Nov. 2009 Saturday 10-12 (Topic: 1,2,3,4,5,6,7,9,10) 2 nd Midterm : 19 Dec. 2009 Saturday 10-12 (Topic: 11,12,13,14 ) Final : All topics Grading: 1 st Midterm 25%, 2 nd Midterm %25, and Final 50 % KIM 101E General Chemistry Mustafa Özcan

6 WEEKDATETOPICS 11 Oct. 2009Electronic Structure of Atom (Chp:1-2-9) 2/08 Oct. 2009Periodic Table, Chemical Compounds (Chp:10 ve 3) 315 Oct. 2009Chemical Reactions, Reactions in Aqueous Solutions (Chp:4 ve 5) 422 Oct. 2009Gases (Chp:6) 529 Oct. 2009 605 Nov. 2009Thermochemistry (Chp:7) 7 Nov. 20091 st Midterm (Topic: 1,2,3,4,5,6,7,9,10) 712 Nov2009Chemical Bonding –I (Chp:11) 819 Nov 2009Chemical Bonding –II (Chp:12) 926 Nov 2009 1003 Dec. 2009Liquids, Solids, and Intermolecular Forces (Chp:13) 1110 Dec. 2009Liquids, Solids, and Intermolecular Forces (Chp:13) 1217 Dec. 2009Solutions and Their Physical Properties (Chp:14) 19 Dec. 20092 nd Midterm (Topic: 11,12,13,14) 1324 Dec. 2009Chemical Equilibrium (Chp:16) 1431 Dec. 2009Acids and Bases (Chp:17)

7 MATTER is anything that occupies space and has mass All matter is formed from one or more of 114 presently known elements—fundamental substances



10 Early Chemical Discoveries Antonie Lavosier:  LAW OF MASS CONSERVATION Mass is neither created nor destroyed in chemical reactions. French chemist Joseph Proust:  LAW OF DEFINITE PROPORTIONS Different samples of a pure chemical substance always contain the same proportion of elements by mass John Dalton (1766–1844), proposed a new theory of matter as follows:  Elements are made of tiny particles called atoms.  Each element is characterized by the mass of its atoms. Atoms of the same element have the same mass, but atoms of different elements have different masses.  Chemical combination of elements to make different substances occurs when atoms join together in small whole-number ratios.  Chemical reactions only rearrange the way that atoms are combined; the atoms themselves are unchanged.

11 Subatomic particles.  1897 J.J Thomson proposed that cathode rays must consist of tiny negatively charged particles, which we now call electrons. Thomson was able to calculate the ratio of the electron’s electric charge to its mass—its charge-to-mass ratio, e/ m = 1.758 820 x 10 8 C/g  Robert Millikan found mass of electron as 9.109 382 x 10 -28 g by oil drop experiment  Ernest Rutherford found that atoms also contain positively charged particles. in 1911  Rutherford proposed that most of the mass and positive charges of an atom are concentrated in a tiny central core that he called the nucleus.  1932 James Chadwick discovered some subatomic particles other than protons named neutrons.

12 Particle name SymbolElectric Charge Coulomb C Relative Electric Charge Mass gAtomic mass unit Location Protonp+p+ +1.602 x 10 -19 +11.673x10 -24 1.0073Inside nucleus Neutronn001.675x10 -24 1.0087Inside nucleus Electrone-e- -1.602 x 10 -19 9.109x10 -28 0.00055Outside nucleus Subatomic particles.

13 A X Z Atomic number Z = Number of protons in atom’s nucleus= Number of electrons around atom’s nucleus A = Z + N. Mass number (A) = Number of protons (Z) + Number of neutrons (N) İsotopes : atom with the same number of protons but different numbers of electrons are called isotope 1 H 1 2 H 1 3 H 1

14 A  (p-e) X Z Ion: an atom or group of atoms which either positevely charged or negatively charged As a result of the loss or gain of electrons


16 Electrons in Atom Lights has many characteristics in common with electrons

17 Light is a kind of electromagnetic wave carrying energy. ( Waves can be identified as a motion carrying energy) The wave nature of light The particle nature of light

18 The wavelength ( ) is the distance between any two identical points in consecutive cycles. The frequency ( ) is the number of wave cycles that pass a point each second. Its unit is 1/second (Hertz). The amplitude of a wave is its height UnitSymbol (m) AngstromÅ10 -10 Nanomet re nm10 -9 Mi k rometr e m10 -6 Millimet re mm10 -3 s entimetr e cm10 -2 Met re m1 the velocity of a wave is the product of its frequency and its wavelength, for light c =.

19 Elektromagetic spectra

20 question: calculate the frequency, in s -1, of an X- ray that has a wavelength of 8.21nm

21 Particlelike Properties of Electromagnetic Radiation: Quantum theory: (1900 The Planck Equation ) E= h.. = h c / h=Planck constant = 6.626 x 10- 34 J.s Photoelectric effect : ( 1905 Albert Einstein) When a beam of light shines on certain surfaces particularly certain metals, e beam of electrons is produced. This phenomenon is called the photoelectric, Einstein considered that electromagnetic radiation has particle like characteristics and that particles of light called photons, posses a characteristic energy, given by planck’s equation.

22 calculate the energy of X ray having 8,21nm vawelength calculate the energy of the violet light that has 6,15x10 14 s -1 frequency

23 Arrange the followings in order to increasing vawelength and energy per photon a- X-Ray b- UV c- Visible light d- Microwave e- Infrared

24 A nitrogen gas laser pulse with wavelength of 337nm contains 3.83mJ of energy. How many photons does it contain?

25 Bohr Atom Theory En = - R h / n 2 Rh= 2.179 x 10 –18 J

26 En = - R h / n 2 Rh= 2.179 x 10 –18 J



29 = R H ( ni2ni2 1 nf2nf2 – 1 ) = h = hc/λ ΔE = E f – E i = -R H nf2nf2 ni2ni2 –

30 Calculate the energy change in electron transition from Hirojen n=5 to n=3 in hydrogen atom Determine the wavelength of light emitted when an electron in a hydrogen atom Hiydrogen atom makes a transition from an orbital n=6 to n=5.

31 Bohr Atom Theory the electrons in an atom move at a certain distance from nucleus and their motions are stable. Each stationery state has a definite energy. Electrons move in each stationary energy state in a circular orbital. These circular orbitals are called energy levels or shells. The possible states for the electron are numbered, n=1, 2, 3 and so on. When an electron is in a stationary state, the atom does not emit light. However when an electron falls back to a lower energy level from a higher one, it emits a quantum of light that is equal to the energy differences between these two energy levels.

32 the Heisenberg Uncertainty Principle imagine what would happen if we tried to determine the position of an electron at a given moment. For us to “see” the electron, light photons of an appropriate frequency would have to interact with and bounce off the electron. But such an interaction would transfer energy from the photon to the electron, thereby increasing the energy of the electron and making it move faster. Thus, the very act of determining the electron’s position would make that position change. In mathematical terms, Heisenberg’s principle states that the uncertainty in the electron’s position, times the uncertainty in its momentum, is equal to or greater than the quantity h/4p: Δx Δp ≥ h 4π4π h = ------------- DE BROGLIE EQUATION m.

33 the mass of an electron is 9,109x10 -28 g and its velocity 5,97x10 6 m/s, what is the wavelength of this electron?

34 Quantum Numbers The principal quantum number (n) is a positive integer on which the size and energy level of the orbital primarily depend. n= 1, 2, 3,4, 5,6,7 The angular-momentum quantum number (l) defines the three-dimensional shape of the orbital. l can have any integral value from 0 to n-1 If n = 1, then l = 0 2 If n = 2, then l = 0 or 1 If n = 3, then l = 0, 1, or 2 The magnetic quantum number (ml)defines the spatial orientation of the orbital with respect to a standard set of coordinate axes. Ml can have any integral value ml = from -l to + l. spin quantum number electrons behave as if they were spinning around an axis,much as the earth spins daily. Unlike the earth, though, electrons are free to spin in either a clockwise or a counterclockwise direction. This spinning charge gives rise to a tiny magnetic field and to a spin quantum number (which can have either of two values, -1/2 or +1/2)







41 s orbitals p orbitals d orbitals f orbitals l= 0 I =1 l =2 l =3 ml= 0 ml= -1,0,+1 ml =-2,-1,0,+1,+2 ml= -3,-2, 1,0,+1,+2,+3 _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ s p d f max e: 2 max electron: 6 max electron:10 max electron: 14

42 Electron Configurations of Multielectron Atoms - Lower-energy orbitals fill before higher-energy orbitals. 1s -2s -2p -3s- 3p –4 s-3d -4p -5s -4d -5p –6s-4f –5d -6p -7s -5f - 6d (Aufbau), no two electrons in an atom can have the same four quantum numbers. In other words, the set of four quantum numbers associated with an electron acts as a unique “address” for that electron in an atom, and no two electrons can have the same address.(Pauili) HUND’S RULE If two or more orbitals with the same energy are available, one electron goes in each until all are half-full. The electrons in the half-filled orbitals all have the same value of their spin quantum number.

43 Diamagetic: the state of an atom or ion that contains only paired electrons Paramagnetic: the state of an atom or ion that contains only unpaired electrons İsoelektronic: atoms or ions which have the same numbers of electron are isoelectronic. (same electron configuration)

44 Question: which one shows the quantum numbers of last electron of Neutral Na atom has the elecron configuration as 1s 2 2s 2 2p 6 3s 1 a- n=1 l=0 ml=0 ms= +½ b- n=3 l=1 ml=0 ms= +½ c- n=3 l=0 ml=1 ms= +½ d- n=1 l=1 ml=1 ms= +½ e- n=3 l=0 ml=0 ms= +½

45 Soru: which one has the paramagnetic properties a- 2 He b- 18 Ar c- 20 Ca d- 30 Zn e- 40 Zr

46 explain the following quantum numbers are possible or not? a.n=3 l=2 ml=-1 b.n=3 l=3 ml=-3 c.n=2 l=3 ml=-1 d.n=5 l=2 ml=-1

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