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Published byLenard Dorsey Modified over 9 years ago
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By Lauren and Joe
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Electromagnetic Spectrum From www.lcse.umn.edu/specs/labs/images/spectrum.gif
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Electromagnetic Spectrum Frequency and wavelength are inversely related, as demonstrated by the equation c=λν c-speed of light constant-2.9979x10 8 m/s λ-wavelength(m) ν-frequency(Hz or s -1 )
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Quantum Theory Einstein—Light behaves as if it consists of quantized energy packets, meaning that energy can have only certain allowed values given by the equation E photon =hν E photon -(J) h-Planck’s constant-6.626x10 -34 J-sec ν-frequency (Hz or s -1 )
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Another Equation to Remember E photon =E remove electron/threshold + E kinetic
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Quantum Numbers n-principal quantum number (shell) i.e. 3s, n=3 l-azimuthal quantum number (sub-shell) The value of l corresponds to the sub-shell of the orbital s=0, p=1, d=2, f=3 i.e. if n=3, can have 3s, 3p, 3d. Accordingly, you can have l values 0, 1, and 2.
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More Quantum Numbers! m l -magnetic quantum number Each orbital has number from –L to +L i.e. 3p 4, occupies 1 st orbital, m l = -1 m s -magnetic spin quantum number Value is ±1/2 If electron points up, +1/2 If electron points down, -1/2 i.e. 3p 4, points down, m s = -1/2
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DeBroglie Wavelength Matter has a characteristic wavelength that depends on its momentum, mv λ=h/mv λ-wavelength(m) h-Planck’s constant-6.626x10 -34 J-s m-mass of particle(kg) (e - =9.11x10 -31 kg) v-velocity (m/s)
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Bohr’s Model A model of the hydrogen atom that explains its line spectrum Light emitted when the electron drops from a higher energy state to a lower energy state Light must be absorbed to excite the electron from a lower energy state to a higher energy state
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Electron Configuration Use the periodic table to write electron configurations Core electron configuration—use largest noble gas that is smaller than atom/ion, then write additional electrons Remember that each orbital can hold 2 electrons each Place 1 electron in each orbital before putting a second one Electron configurations are most stable when the orbitals are full or half-full
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Electron Configuration D Block (Transition metals) -5 orbitals @ 2 electrons each=10 electrons D block 1 behind s/p block F Block 7 orbitals @2 electrons each=14 electrons F block 1 behind D block, 2 behind s/p block
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Electron Configuration of Ions When determining configurations for cations, remove electrons first from the orbital with the largest quantum number n For example, Sn=[Ar] 4s 2 3d 10 4p 2 Sn 3+= =[Ar]4s 1 3d 10
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Hybridization Mixing of s, p, and d orbitals to form hybrid orbitals A particular mode of hybridization corresponds with each of the five common electron-domain geometries note: electron domain geometry is arrangement of electron domains around a central atom. Each bond, whether it is single, double, or triple, and each lone pair is one electron domain.
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Hybridization Linear-2 electron domains-sp hybridization Trigonal planar-3 electron domains- sp 2 Tetrahedral-4 domains, sp 3 Trigonal bipyramidal- 5 domains, sp 3 d Octahedral- 6 domains, sp 3 d 2
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Valence Bond Theory Bonds form when atomic orbitals overlap between two atoms The greater the overlap between two orbitals, the stronger the bond Sigma Bond Covalent bonds formed from end to end overlap of s orbitals Pi Bond Bond formed from the sideways overlap of p orbitals
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Molecular Orbital Theory Electrons exist in allowed energy states called molecular orbitals (MOs) Like an atomic orbital, an MO can hold two electrons of opposite spin Occupation of bonding MOs favors bond formation Occupation of antibonding MOs (denoted with an *) is unfavorable
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Molecular Orbital Theory Bond Order Bond Order = ½(# of electrons bonding - # of electrons anti-bonding) The principle of anti-bonding sets molecular orbital theory apart from valence bond theory
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Paramagnetism and Diamagnetism paramagnetism—an attraction of a molecule by a magnetic field due to unpaired electrons diamagnetism—a weak repulsion from a magnetic field by paired electrons
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Question 1 What is the core electron configuration of Pb 3+ ?
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Answer [Xe]6s 1 4f 14 5d 10
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Question 2 What type of hybridization does the central atom in the following compounds assume? a. NH3 b. SF 6 c. ClF 3
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Answers a. Sp 3 b. Sp 3 d 2 c. Sp 3 d
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Final Question! What score will you get on the AP Chem Exam after seeing this presentation?
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ANSWER 5
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