# Unit 2: Chemical Bonding and Organic Chemistry

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1 Chemistry AP Unit : Chemical Bonding and Organic Chemistry Unit : Chemical Bonding and Organic Chemistry Chapter 7: Atomic Structure and Periodicity 7.1: Electromagnetic Radiation Electromagnetic (EM) Radiation: - energy that travels at the speed of light in a form of perpendicular waves. This includes everything from cosmic rays to radio waves. Wavelength (λ): - the length of a wave (from crest to crest). Frequency (v): - the number of wave in one second; measures in Hertz (Hz) or s 1. (1 Hz = 1 s 1 ) 1 wavelength (λ) Electric Waves Frequency Wavelength c = λv Magnetic Waves c = speed of light ( m/s) λ = wavelength (m) lambda v = frequency (Hz or s 1 ) nu EM Wave Frequency (Hz) Wavelength (m) EM Wave Frequency (Hz) Wavelength (m) Cosmic Wave Infrared Gamma Wave Microwaves X-Ray FM Radio 10 8 or (100 MHz) 1 to 10 Ultraviolet Shortwave Radio 10 6 (1 MHz) 10 Visible (7.5 to 4.3) (blue to red) (4 to 7) nm to 700 nm (blue to red) AM Radio 10 4 (10 khz) 10 4 Example 1: One of the spectral lines of hydrogen has a wavelength of 410 nm. Determine the corresponding frequency. c = m/s λ = 410 nm = m v =? c = λv 8 c v = = ( m/s) λ 9 ( m) v = s 1 (Hz) Example : What is the wavelength being broadcast by a radio station with a frequency of 10.1 MHz? c = m/s v = 10.1 MHz = s 1 λ =? c = λv 8 c λ = = ( m/s) ν 6 1 ( s ) λ =.94 m Copyrighted by Gabriel Tang B.Ed., B.Sc. Page 51.

2 Unit : Chemical Bonding and Organic Chemistry Chemistry AP 7.: The Nature of Matter Photoelectric Effect: - Einstein s experiment with photoelectric tube found that electrons can only be emitted from a metal surface in a cathode ray tube when a threshold (minimum) frequency is directed at it. (Intensity or Brightness of the light is not a factor) - the slope of amount of energy of the emitted electrons per frequency used is called the Planck s constant (h) = J s (in honour of Max Planck who first hypothesized this relationship). Max KE of photoelectrons (x J) Check out the Photoelectric Effect website at: Example 1: Calculate the energy change (in J and ev) associated with the longest spectral line of hydrogen at 656 nm. Page 5. Visible light Threshold Frequencies (v min ) (Lowest Frequency required for emitting e of a particular metal) Ultra violet c = m/s h = J s λ = 656 nm = m E =? potassium magnesium Frequency (x Hz) Work Functions (W = hv min ) (Minimum Energy required for emitting e ) E = hv = λ hc c = speed of light ( m/s) h = Planck s constant = J s λ = wavelength (m) v = frequency (Hz or s 1 ) Electron Volts (ev): - energy of an accelerated electron through 1 V of potential difference hc E = = ( J s)( m/s) λ 9 ( m) 1 ev = J E = J E = J E = 1.89 ev 1eV Copyrighted by Gabriel Tang B.Ed., B.Sc. 19 J

3 Chemistry AP Unit : Chemical Bonding and Organic Chemistry Example : What is the minimum frequency needed if it takes 450 kj to remove one mole of electrons from a metal surface of a photoelectric tube? h = J s E min = W = 450 kj/mol (negative because of energy input) v min =? E min = mol J/mol = J e 1 W W = hv min v min ( ) = = J h 34 ( J s) v min = s 1 Duality of Light: - EM Radiation has characteristics of wave (reflection, refraction, and diffraction) and particles (momentum and kinetic energy as demonstrated by Einstein). These light particles are called photons. - Photons have no mass. But its mass equivalence can be calculated as follows. E = mc (Einstein s Mass Energy Conservation) E m = c hc m = λ hc hc = (Substitute E = and simplify) c λc λ Mass Equivalence of a Photon m = h λc (m is in kg) de Broglie Wavelength: - since light can behave like particles; particles can have wave properties. - rearranging the mass equivalence of a photon formula above and replacing c with speed of the actual particle (nothing that has mass can travel at the speed of light Einstein Theory of Relativity), we have the following de Broglie Wavelength h λ = mυ υ = speed of particle (m/s) m = mass of particle (kg) Example 3: Calculate the de Broglie wavelength of a proton with a mass of kg moving at 5.0% of the speed of light. h = J s υ = 0.5 ( m/s) = 74,947,500 m/s m = kg λ =? λ = 34 h = ( J s) mυ 7 ( kg)( m/s) λ = m (1 J = 1 kg m /s ) Copyrighted by Gabriel Tang B.Ed., B.Sc. Page 53.

4 Unit : Chemical Bonding and Organic Chemistry Chemistry AP Example 4: What is the velocity of an alpha particle ( protons) that has a wavelength of m as it is travelling towards a piece of gold foil in the Rutherford experiment? h = J s m = ( kg) = kg λ = m υ =? λ = υ = h mυ 34 h = ( J s) 7 14 ( kg)( m) mλ υ = m/s (1 J = 1 kg m /s ) 7.3: The Atomic Spectrum of Hydrogen Quantized Energy: - because electrons can only exist at a particular orbital at one time, and never between orbitals, it can only emit a certain wavelength as it comes down from an excited state to the ground state. Diffraction: - as wave encountered an opening (slit), its wavelength is separated into as the wave is scattered. Diffraction Pattern: - the pattern of bright and dark spots due to the interference of the wave front because a wave has travelled through a slit. Continuous Spectrum: - as white light travelled through a prism, it is diffracted and the visible light spectrum (red to violet) is observed. Line Spectrum: - when light from an excitation of a particular atom, the quantized energy is emitted as specific wavelength as it travels from higher excited orbitals to the lower orbitals nm nm nm 410. nm Hydrogen Spectrum at Visible range Assignment 7.1 pg. 337 #39, pg #41 to 50 Page 54. Copyrighted by Gabriel Tang B.Ed., B.Sc.

5 Chemistry AP Unit : Chemical Bonding and Organic Chemistry 7.4: The Bohr Model Quantum Model: - proposed by Niels Bohr, electron in hydrogen atom moves around the nucleus in specific circular orbits, which have certain energies. - Energy needs to be input ( E > 0) to move up the orbitals. - Energy is given off as frequency ( E < 0) to move down the orbitals. E = 0 J n = n = 5 n = 4 n = 3 n = n = 1 n = 6 E 1 = J Ground State Bohr s Hydrogen Energy Level E n = n Example 1: Verify that the blue line (434.0 nm) of the visible part of the hydrogen spectrum is indeed from the emission of energy of an electron moving from the 5 th orbital to the nd orbital. E = E f E i hc E J 18 5 = J λ E = n f n i hc λ = E E = J J s m/s n f n i 19 ( J) E 5 = J (Using net energy change) 5 E 5 = J ( E < 0 means energy is released) λ = m λ = nm Example : Determine the ionization energy (from ground state to infinity) of hydrogen in J and in ev. Explain the meaning of the sign in the calculated value. 18 J E n = Energy Level of a specific nth hydrogen orbital n = the number of orbital from the nucleus R H = Rydberg Constant for H Atom ( J) λ = ( )( ) E = E f E i E = n f 18 J n i E 1 = J 1 E 1 = J (0 1) 18 J = J 1 (Note: 0) 1 1 n f n i E 1 = J ( E > 0 means energy is absorbed) E 1 = J 1eV E 1 = ev J Copyrighted by Gabriel Tang B.Ed., B.Sc. Page 55.

6 Unit : Chemical Bonding and Organic Chemistry Chemistry AP 7.5: The Quantum Mechanical Model of the Atom Schrödinger Wave Equation: - Schrödinger recognized that since electron can behave like waves (de Broglie), he proposed that electron can have quantized energy when it achieved a standing wave (wave that appears standing because it contains nodes which signifies exactly half a wavelength). Standing Wave ( nodes 1 half-λ) Standing Wave (3 nodes half-λ) Standing Wave (4 nodes 3 half-λ) Schrödinger Wave Equation Ĥψ = Eψ Ĥ = Hamiltonian Operator (This is not a variable, but a combination of calculation procedures like integrals, ψ (x, y, z) d (x, y, z), and derivatives dψ(x, y, z)/d(x, y, z).) ψ = Wave Function E = Total Energy (Electric Potential and Kinetic Energies) Note: If the wave function (ψ) remains the same after the operation by the Hamiltonian Operator (Ĥ), then the coefficient of the final wave function is the total energy of the atom (E). The wave function can now be used to calculate the probability distribution of the electron around the nucleus. Heisenberg Uncertainty Principle: - at very high velocity, we cannot know precisely the position and momentum (mass times velocity) of a particle at a given time. Heisenberg Uncertainty Principle x (mυ) h 4π x = Uncertainty in particle s position (mυ) = Uncertainty in particle s momentum As x (more certain in particle s position), (mυ) (less certain in particle s momentum) Probability Distribution (R ): - the probability of the likely position of an electron around a nucleus. - the probability distribution is equalled to the square of the wave function. Probability Distribution R = [ψ (x, y, z)] ψ (x, y, z) = Wave Function in x, y, and z directions Page 56. Copyrighted by Gabriel Tang B.Ed., B.Sc.

7 Chemistry AP Unit : Chemical Bonding and Organic Chemistry Electron Density (Probability) Map: - a diagram that shows the probability distribution of an electron around the nucleus; also called atomic orbital. Electron Density or Probability (R ) Distance from Nucleus (r) Radial Probability Distribution: - when total probability (Surface Area of a Sphere Probability) is plotted against distance from the nucleus. - at small radius, probability is big but surface area is small (yields low total probability). - at large radius, probability is small but surface area is big (yields low total probability) - at optimal radius, surface area probability is large (maximum total probability = Atomic Radius). Surface Area = 4πr Radial Probability (4πr R ) Hydrogen s Maximum Radial Probability at nm = 0.59 Å (Atomic Radius of Hydrogen) 1 Å (angstrom) = m 7.6: Quantum Numbers Distance from Nucleus (r) Quantum Numbers: - a series of number that demonstrates the properties of different atomic orbitals. a. Principal Quantum Number (n): - natural number values {1,, 3, } are used to describe the energy and size of the orbital. b. Angular Momentum Quantum Number (l): - whole numbers {0 l (n 1)} are used to indicate the shape of the atomic orbitals. l = 0 (s orbital) l = 1 (p orbital) l = (d orbital) l = 3 (f orbital) l = 4 (g orbital) sharp principal diffuse fundamental (follows the alphabet after f) Copyrighted by Gabriel Tang B.Ed., B.Sc. Page 57.

8 Unit : Chemical Bonding and Organic Chemistry Chemistry AP c. Magnetic Quantum Number (m l ): - integral numbers { l m l l} are used to show the orientation of the orbital in space relative to the other orbitals in the atom. - each m l value represents an atomic orbital. Check out the websites at: 1. Schrödinger Equation in Three Dimensions: Hydrogen Schrödinger Equation: Subshell: - orbital designation as indicated by Principal Quantum Number (n) and the letter used for Angular Momentum Quantum Number (l). n l Subshell m l Number of Orbitals Total Number of Orbitals 1 0 1s s p 1, 0, s p 1, 0, 1 3 3d, 1, 0, 1, s p 1, 0, 1 3 4d, 1, 0, 1, 5 3 4f 3,, 1, 0, 1,, s p 1, 0, 1 3 5d, 1, 0, 1, 5 3 5f 3,, 1, 0, 1,, g 4, 3,, 1, 0, 1,, 3, 4 9 Example 1: Explain why the following sets of quantum numbers or subshells are incorrect. a. 3f b. n = 3, l = 3, m l = 5 n = 3 l = 0, 1, or (s, p, or d subshells) Since n = 3, we cannot have a f subshell. n = 3 l = 0, 1, or m l =, 1, 0, 1 or Since n = 3, we cannot have a l = 3. c. 1d d. n = 5, l = 4, m l = 5 Page 58. n = 1 l = 0 (s subshell only) Since n = 1, we cannot have a d subshell. Assignment 7.4 pg. 338 #51 to pg. 339 # pg. 338 #61 to 64 n = 5 l = 0, 1,, 3 or 4 m l = 4, 3,, 1, 0, 1,, 3 or 4 Since n = 5, and the maximum l = 4, we cannot have a m l = 5. Copyrighted by Gabriel Tang B.Ed., B.Sc.

9 Chemistry AP Unit : Chemical Bonding and Organic Chemistry 7.7: Orbital Shapes and Energies Orbital Shapes: - from the Schrödinger wave functions, the graph the radial probability distributions in 3 dimensions forms the orbital shapes. Nodal Surfaces (Nodes): - the areas between orbitals where there is zero probability of electron distribution. Nodes (n 1) 3s orbital s orbital 1s orbital m l = 0 m l = 1 m l = 1 m l = m l = m l = 1 m l = 1 m l = 0 m l = 0 Check out the Hydrogen Atom Orbital Viewer at Degenerate: - in the case of hydrogen atom where orbitals of the same n value shares the same energy level. E ns = E np = E nd = E nf E 3s 3p 3d s p 1s Copyrighted by Gabriel Tang B.Ed., B.Sc. Page 59.

10 Unit : Chemical Bonding and Organic Chemistry Chemistry AP 7.8: Electron Spin and the Pauli Principle Electron Spin: - when electron spins clockwise, it creates a magnetic north pole in the upward direction. Conversely, when electron spins counter-clockwise, it creates a magnetic north pole in the downward direction. N e e Electron Spin Quantum Number (m s ): - values of 1 ± to denote the electron spin direction. Pauli Exclusion Principle: - in a given atom, no two electrons can have exactly the same set of four quantum numbers (n, l, m l, and m s ). - an orbital can only hold two electrons with opposing spins. Example 1: In the 1s orbital, the four quantum numbers for the two electrons are: N n = 1 l = 0 m l = 0 m s = 1 1s orbital n = 1 l = 0 m l = 0 1 m s = 7.9: Polyelectronic Atoms Polyelectronic Atoms: - atoms with more than one electron (all non-hydrogen atoms). Electron Correction Problem: - due to the mathematical difficulty in calculating the repulsion of electrons in polyelectronic atoms (since Heisenberg Uncertainty Principle rules out predicting the exact location and momentum of an electron), an assumption needs to be made that the valence electrons are screened or shielded from the nuclear charge which composes of the inner electrons and the protons in the nucleus. Page 60. Copyrighted by Gabriel Tang B.Ed., B.Sc.

11 Chemistry AP Unit : Chemical Bonding and Organic Chemistry Penetration (Tunnelling) Effect: - the result where electrons in higher s or p orbitals have a other maximum radial probabilities in closer proximity to the nucleus than d orbital at the same energy level. Radial Probability Tunnelling of 3s Penetration of 3p 3s 3d 3p Distance from Nucleus E ns < E np < E nd < E nf E 3s 3p 3d s p 1s Check out the websites at: 1. Electron Configurations: Whizzy Periodic Table: Assignment 7.7 pg. 339 #67 and pg. 339 #69 and 79 Copyrighted by Gabriel Tang B.Ed., B.Sc. Page 61.

12 Unit : Chemical Bonding and Organic Chemistry Chemistry AP 7.10: The History of the Periodic Table Johann Dobereiner: - first to discover a pattern of a group of elements like Cl, Br, and I (called triads). John Newland: - suggested elements should be arranged in octaves because they repeat their properties for every eighth elements. Dmitri Mendeleev: - conceived the first modern periodic table of elements. (independently worked on by Julius Lothar Meyer). - insisted certain spots of the table be left blank until the actual element is found that matched the predicted properties. This was done to preserve the elements with similar properties called groups or families. 7.11: The Aufbau Principle and the Periodic Table Aufbau Principle: - Aufbau German for building up - for each element, electrons are added up into the hydrogen like quantum orbitals as protons are being added to the nucleus. Orbital Diagram: - a diagram that shows the arrangements of electrons in quantum subshells. Hund s Rule: - for subshells that can have more than 1 orbital (as in p, d, f ), the lowest energy can be achieved when the electrons are arranged so that there are a maximum number of unpaired electrons. These unpaired electrons are drawn spinning up ( ) in the orbital diagram. Electron Configuration: - the arrangement of electrons in atomic subshells. Example 1: Draw the quantum orbital diagrams and state the electron configurations for the following atoms. Atomic # Atom Electron Configuration 1 H 1s 1 He 1s 3 Li 1s s 1 4 Be 1s s 5 B 1s s p 1 6 C 1s s p 7 N 1s s p 3 8 O 1s s p 4 9 F 1s s p 5 10 Ne 1s s p 6 Orbital Diagram 1s s p Page 6. Copyrighted by Gabriel Tang B.Ed., B.Sc.

13 Chemistry AP Unit : Chemical Bonding and Organic Chemistry Core Electrons: - inner electrons that have completed a row in the Periodic Table of Elements. Valance Electrons: - electrons in the outermost principal quantum number of an atom. - elements in the same group or family contains the same valence electron configuration. Example : Draw the quantum orbital diagrams and state the electron configurations for the following atoms. Atomic # Atom Electron Configuration 19 K [Ar] 4s 1 [Ar] 0 Ca [Ar] 4s [Ar] 1 Sc [Ar] 4s 3d 1 [Ar] Ti [Ar] 4s 3d [Ar] 3 V [Ar] 4s 3d 3 [Ar] 4 *Cr [Ar] 4s 1 3d 5 [Ar] 5 Mn [Ar] 4s 3d 5 [Ar] 6 Fe [Ar] 4s 3d 6 [Ar] 7 Co [Ar] 4s 3d 7 [Ar] 8 Ni [Ar] 4s 3d 8 [Ar] 9 *Cu [Ar] 4s 1 3d 10 [Ar] 30 Zn [Ar] 4s 3d 10 [Ar] 31 Ga [Ar] 4s 3d 10 4p 1 [Ar] 3 Ge [Ar] 4s 3d 10 4p [Ar] 33 As [Ar] 4s 3d 10 4p 3 [Ar] 34 Se [Ar] 4s 3d 10 4p 4 [Ar] 35 Br [Ar] 4s 3d 10 4p 5 [Ar] 36 Kr [Ar] 4s 3d 10 4p 6 [Ar] Orbital Diagram 4s 3d 4p *From Hund s Rule, Cr and Cu can achieved lowest energy if the 4s e was moved to the 3d 5 or 3d 10. 1s Order of appearance for Subshell s p 3s 3p 3d 4s 4p 4d 4f 5s 5p 5d 5f 6s 6p 6d 7s 7p Increasing Energy Copyrighted by Gabriel Tang B.Ed., B.Sc. Page 63.

14 Unit : Chemical Bonding and Organic Chemistry Chemistry AP Some Exceptions in Electron Configurations 1. Chromium ( 4 Cr): [Ar] 4s 1 3d 5 5. Palladium ( 46 Pd): [Kr] 4d 10. Copper ( 9 Cu): [Ar] 4s 1 3d Silver ( 47 Ag): [Kr] 5s 1 4d Molybdenum ( 4 Mo): [Kr] 5s 1 4d 5 7. Platinum ( 78 Pt): [Xe] 6s 1 4f 14 5d 9 4. Technetium ( 43 Tc): [Kr] 5s 1 4d 6 8. Gold ( 79 Au): [Xe] 6s 1 4f 14 5d Lanthanum ( 57 La): [Xe] 6s 5d 1 (Lanthanide Series: beginning of 4f subshell) 10. Actinium ( 89 Ac): [Rn] 7s 6d 1 (Actinide Series: beginning of 5f subshell) Representative Elements (Main Groups) Electron Configurations in the Periodic Table of Elements Representative Elements (Main Groups) Transition Metals Lanthanide Series Actinium Series Assignment 7.11 pg. 339 #71 to 75, 77, 81, 83 Page 64. Copyrighted by Gabriel Tang B.Ed., B.Sc.

15 Chemistry AP Unit : Chemical Bonding and Organic Chemistry 7.1: Periodic Trends in Atomic Properties There are many different trends regarding the physical and chemical properties of the elements in the Periodic Table. However, we will limit to three atomic properties. They are ionization energy, electron affinity, and atomic size. Shielding Effect: - the outer electrons are pushed away because of the repulsion between them and the core electrons. The net result is that the protons in the nucleus cannot hold on to these outer electrons as tightly as they would for the core electrons. Effective Nuclear Charge (Z eff ): - the net nuclear charge actually experienced by an electron (the difference between the number of protons, Z, and the number of shielded core electrons). - the higher it is for Z eff, the less shielding effect the outer electrons will experience. Z eff = Z Shield Core Electrons Example 1: Calculate the effective nuclear charge of Na and Ar (first and last elements of period 3). Sodium (Na): Z = 11 protons Z eff = Argon (Ar): Z = 18 protons Shield Core e = 10 (e in the first two shells) Z eff = 1 for Na Shield Core e = 10 (e in the first two shells) Z eff = Z eff = 8 for Ar (Ar has experience LESS shielding effect than Na) Ionization Energy: - the energy needed to completely remove an electron from a gaseous atom or gaseous ion (plasma). First Ionization Energy: - the ionization energy required to remove the highest-energy electron from an atom. X (g) X + (g) + e Na (g) Na + (g) + e I 1 = 495 kj/mol (energy is required) Second Ionization Energy: - the ionization energy required to remove the second highest-energy electron from the ion. X + (g) X + (g) + e Na + (g) Na + (g) + e I = 4560 kj/mol Copyrighted by Gabriel Tang B.Ed., B.Sc. Page 65.

16 Unit : Chemical Bonding and Organic Chemistry Chemistry AP Successive Ionization Energies (kj/mol) for Elements in Row 3 of the Periodic Table First Ionization Energies generally INCREASE within a Period. (INa < IMg > IAl < ISi < IP > IS < ICl < IAr) Elements I 1 I I 3 I 4 I 5 I 6 I 7 Na Mg Core Electrons Al Si P S Cl Ar Successive Ionization Energies INCREASES within each element. (I 1 < I < I 3 < ) Several Notes on Trends in Ionization Energies 1. There is an Increase in Successive Ionization Energies because each successive electron has to jump from a lower level. Besides, these successive electrons are bind more tightly with the nucleus because they are closer to the protons.. Ionization Energies Decrease Down a Group. This is due to the fact as the atom has more orbitals, it is increasing in size. It is easier (takes less energy) to take away a valence electron because the protons are having a more difficult time to hold on to the electron. 3. In general, Ionization Energies Increase as one move from Left to the Right of a Period. This is because the increases in protons in the same row increase the effective nuclear charge (the protons in the nucleus have more pull on the outer electrons, decreasing shielding), thus requiring more energy to ionize them. 4. The notable exceptions to the above statement is between Mg and Al along with P and S. In the case between magnesium and aluminum, the I 1 decreases because of the change in subshells ([Ne] 3s and [Ne] 3s 3p 1 ). It is easier to remove a 3p 1 electron than an electron in 3s. Between P and S, the electron configuration changes between [Ne] 3s 3p 1 x 3p 1 y 3p 1 z to [Ne] 3s 3p x 3p 1 y 3p 1 z. Again, it is easier to remove an electron from a 3p x orbital because of Hund s Rule (the lowest energy state is achieved when all electrons in the subshell is unpaired). Page 66. Copyrighted by Gabriel Tang B.Ed., B.Sc.

17 Chemistry AP Unit : Chemical Bonding and Organic Chemistry Electron Affinity: - the change in energy associates with an addition of an electron to a gaseous atom. - the larger negative electron affinity, the more stable the anion formed. X (g) + e X (g) Cl (g) + e Cl (g) EA = kj/mol (348.7 kj/mol is released) Several Notes on Trends in Electron Affinities 1. In general, Electron Affinity DECRESES Down a Group (less energy is released). This is due to the fact there are more orbitals as the number of row increases. Since the protons in the nucleus cannot attract another electron as effectively due to the increase distance involved, less energy is released. The trend holds except for row. The reason is because of the small size of the p orbitals. Electron repulsion cause smaller values of electron affinities than expected for row.. In most cases, Electron affinity INCRESASES (becomes more negative) across the Period from Left to Right up to the Halogen group (more energy is released). Metals have EA > 0 because they like to form cations (low ionization energy). Non-metals have EA < 0 because they like to form anions in order to form a stable octet. 3. Exceptions can be found in the A and 5A Groups. With the A Alkaline Earth Group, it is more difficult to add an electron into the s orbital such that it will be filled to ns np 1. This is because the A Group already has an electron configuration of ns (subshell is already filled). Therefore, they have a more positive (less negative) electron affinity. In the case of the 5A Group, the atoms find themselves difficult to add another electron because this will mean a np 4 configuration, which contains a paired electron from three unpaired electrons in np 3. Again, Hund s Rule frowns on such arrangement because it is not of the lowest energy state. Thus, Group 5A tends to have a more positive (less negative) electron affinity. Copyrighted by Gabriel Tang B.Ed., B.Sc. Page 67.

18 Unit : Chemical Bonding and Organic Chemistry Chemistry AP Atomic Radius: - the size of an atom as measured by the distances between atoms in chemical compound. r Cl Molecule r = 13 Å 13 Å Cl Atom r = 66 Å Several Notes on Trends in Atomic Radii 1. In general, Atomic Radii decrease as one move to the right of a period. This is because the increases in protons in the same row increase the effective nuclear charge (the protons in the nucleus have more pull on the outer electrons, decreasing shielding), thus drawing these outer electrons closer to the nucleus, decreasing in sizes as the result.. Atomic Radii INCRESES Down a Group. This is due to the fact there are more orbitals as the number of row increases. The outer electrons are, of course, further away from the nucleus Assignment 7.1 pg #85 to 91, 93 to 99, 101 Page 68. Copyrighted by Gabriel Tang B.Ed., B.Sc.

### Copyrighted by Gabriel Tang B.Ed., B.Sc.

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