Autumn 1999 2. The best available balances can weigh amounts as small as 10-5 g. If you were to count out water molecules at the rate of one per second, how long would it take to count a pile of molecules large enough to weigh 10-5 g?

Autumn 1999 2. The best available balances can weigh amounts as small as 10-5 g. If you were to count out water molecules at the rate of one per second, how long would it take to count a pile of molecules large enough to weigh 10-5 g? 1 molecule H2O has mass of 16 + 2 = 18 amu 1 mole H2O has mass of 18 g  6.022 x 1023 molecules

Autumn 1999 2. The best available balances can weigh amounts as small as 10-5 g. If you were to count out water molecules at the rate of one per second, how long would it take to count a pile of molecules large enough to weigh 10-5 g? 1 molecule H2O has mass of 16 + 2 = 18 amu 1 mole H2O has mass of 18 g  6.022 x 1023 molecules 10-5 g  10-5/18 moles = 5.6 x 10-7 moles

Autumn 1999 2. The best available balances can weigh amounts as small as 10-5 g. If you were to count out water molecules at the rate of one per second, how long would it take to count a pile of molecules large enough to weigh 10-5 g? 1 molecule H2O has mass of 16 + 2 = 18 amu 1 mole H2O has mass of 18 g  6.022 x 1023 molecules 10-5 g  10-5/18 moles = 5.6 x 10-7 moles  5.6 x 10-7 x 6.022 x 1023 molecules = 3.35 x 1017 molecules

Autumn 1999 2. The best available balances can weigh amounts as small as 10-5 g. If you were to count out water molecules at the rate of one per second, how long would it take to count a pile of molecules large enough to weigh 10-5 g? 1 molecule H2O has mass of 16 + 2 = 18 amu 1 mole H2O has mass of 18 g  6.022 x 1023 molecules 10-5 g  10-5/18 moles = 5.6 x 10-7 moles  5.6 x 10-7 x 6.022 x 1023 molecules = 3.35 x 1017 molecules 3.35 x 1017 s

Autumn 2000 13. Hemoglobin absorbs light of wavelength 407 nm. Calculate the energy (in J) of one millimole of photons of this light.

Autumn 2000 13. Hemoglobin absorbs light of wavelength 407 nm. Calculate the energy (in J) of one millimole of photons of this light. E = hn = hc/l = 6.626 x 10-34 x 3 x 108 /407 x 10-9 = J s m s-1 m-1

Autumn 2000 13. Hemoglobin absorbs light of wavelength 407 nm. Calculate the energy (in J) of one millimole of photons of this light. E = hn = hc/l = 6.626 x 10-34 x 3 x 108 /407 x 10-9 = J s m s-1 m-1 = 4.88 x 10-19 J

Autumn 2000 13. Hemoglobin absorbs light of wavelength 407 nm. Calculate the energy (in J) of one millimole of photons of this light. E = hn = hc/l = 6.626 x 10-34 x 3 x 108 /407 x 10-9 = J s m s-1 m-1 = 4.88 x 10-19 J 1 millimole = 10-3 mole = 6.022 x 1020 photons

Autumn 2000 13. Hemoglobin absorbs light of wavelength 407 nm. Calculate the energy (in J) of one millimole of photons of this light. E = hn = hc/l = 6.626 x 10-34 x 3 x 108 /407 x 10-9 = J s m s-1 m-1 = 4.88 x 10-19 J 1 millimole = 10-3 mole = 6.022 x 1020 photons energy of 1 millimole of photons  6.022 x 1020x 4.88 x 10-19 J

Autumn 2000 13. Hemoglobin absorbs light of wavelength 407 nm. Calculate the energy (in J) of one millimole of photons of this light. E = hn = hc/l = 6.626 x 10-34 x 3 x 108 /407 x 10-9 = J s m s-1 m-1 = 4.88 x 10-19 J 1 millimole = 10-3 mole = 6.022 x 1020 photons energy of 1 millimole of photons  6.022 x 1020x 4.88 x 10-19 J = 294 J

Quantization of e- energy levels by standing wave assumption Derived a mathematical description for e-s as 3-d standing waves in atoms: H = E , the wave function, is a function spatial position of e-; H, is a mathematical function which allows the calculation of the e- energy (Hamiltonian operator); and E, is the total energy of the electron

What is 2? " Amplitude probability function " e.g. for a linear standing wave: Amplitude = sin x , where x is distance along the wave

What is 2? " Amplitude probability function " e.g. for a linear standing wave: Amplitude = sin x , where x is distance along the wave Why amplitude probability function ? Heisenberg Uncertainty Principle (1927):

What is 2? " Amplitude probability function " e.g. for a linear standing wave: Amplitude = sin x , where x is distance along the wave Why amplitude probability function ? Heisenberg Uncertainty Principle (1927): "There is a fundamental limit to how precisely we can simultaneously determine both the position (x) and the momentum (m.v) of a particle"

What is 2? " Amplitude probability function " e.g. for a linear standing wave: Amplitude = sin x , where x is distance along the wave Why amplitude probability function ? Heisenberg Uncertainty Principle (1927): "There is a fundamental limit to how precisely we can simultaneously determine both the position (x) and the momentum (m.v) of a particle" x.mv > h/4 i.e. x.mv > 5.272(10-35) m2 kg s-1

Example: For a v of 0.1m/s, calculate x for (a) an electron (m=9.11 x 10-31kg) and (b) a football (m=0.4kg).

Example: For a v of 0.1m/s, calculate x for (a) an electron (m=9.11 x 10-31kg) and (b) a football (m=0.4kg). (a) x > 5.272(10-35)/9.11(10-31)(0.1) = 5.79(10-4) m

Example: For a v of 0.1m/s, calculate x for (a) an electron (m=9.11 x 10-31kg) and (b) a football (m=0.4kg). (a) x > 5.272(10-35)/9.11(10-31)(0.1) = 5.79(10-4) m x of e- is big c.f. size of an atom (2(10-10m))!

Example: For a v of 0.1m/s, calculate x for (a) an electron (m=9.11 x 10-31kg) and (b) a football (m=0.4kg). (a) x > 5.272(10-35)/9.11(10-31)(0.1) = 5.79(10-4) m x of e- is big c.f. size of an atom (2(10-10m))! (b) x > 5.272(10-35)/0.4(0.1) = 1.31(10-33) m

Example: For a v of 0.1m/s, calculate x for (a) an electron (m=9.11 x 10-31kg) and (b) a football (m=0.4kg). (a) x > 5.272(10-35)/9.11(10-31)(0.1) = 5.79(10-4) m x of e- is big c.f. size of an atom (2(10-10m))! (b) x > 5.272(10-35)/0.4(0.1) = 1.31(10-33) m x of ball is v. small c.f. its size (0.3m)

Example: For a v of 0.1m/s, calculate x for (a) an electron (m=9.11 x 10-31kg) and (b) a football (m=0.4kg). (a) x > 5.272(10-35)/9.11(10-31)(0.1) = 5.79(10-4) m x of e- is big c.f. size of an atom (2(10-10m))! (b) x > 5.272(10-35)/0.4(0.1) = 1.31(10-33) m x of ball is v. small c.f. its size (0.3m) Conclusion H.U.P. only important for v. small particles such as e-

Major implication of H.U.P.: It is not possible to find out the exact position & velocity of an e- in an atom

Major implication of H.U.P.: It is not possible to find out the exact position & velocity of an e- in an atom Can only define the shapes of electron "clouds" in terms of probability of finding electron at a given spot

Major implication of H.U.P.: It is not possible to find out the exact position & velocity of an e- in an atom Can only define the shapes of electron "clouds" in terms of probability of finding electron at a given spot Probability or e- density is given by the term 2

Major implication of H.U.P.: It is not possible to find out the exact position & velocity of an e- in an atom Can only define the shapes of electron "clouds" in terms of probability of finding electron at a given spot Probability or e- density is given by the term 2 Probability maps for a particular wavefunction are called electron orbitals

Major implication of H.U.P.: It is not possible to find out the exact position & velocity of an e- in an atom Can only define the shapes of electron "clouds" in terms of probability of finding electron at a given spot Probability or e- density is given by the term 2 Probability maps for a particular wavefunction are called electron orbitals Orbitals usually drawn as shape which encloses 90% of the total e- density for that wavefunction

Characterisation of Electron Orbitals Many different solutions to the Schroedinger Equation for an electron in an atom.

Characterisation of Electron Orbitals Many different solutions to the Schroedinger Equation for an electron in an atom. Each solution represented by an orbital

Characterisation of Electron Orbitals Many different solutions to the Schroedinger Equation for an electron in an atom. Each solution represented by an orbital A series of quantum numbers are used to describe the various properties of an orbital:

Characterisation of Electron Orbitals Many different solutions to the Schroedinger Equation for an electron in an atom. Each solution represented by an orbital A series of quantum numbers are used to describe the various properties of an orbital: Principle quantum number (n) has integral values (1,2,3, etc.) & describes orbital size and energy

Characterisation of Electron Orbitals Many different solutions to the Schroedinger Equation for an electron in an atom. Each solution represented by an orbital A series of quantum numbers are used to describe the various properties of an orbital: Principle quantum number (n) has integral values (1,2,3, etc.) & describes orbital size and energy Angular momentum quantum number (l) has integral values from 0 to n-1 for each value of n & describes the orbital shape

Magnetic quantum number (ml) has integral values between l and -l, including zero & describes the orientation in space of the orbital relative to the other orbitals in the atom

Magnetic quantum number (ml) has integral values between l and -l, including zero & describes the orientation in space of the orbital relative to the other orbitals in the atom Spin quantum number (ms) is either +1/2 or -1/2 for a given e- & describes the direction of spin of the e- on its axis

Magnetic quantum number (ml) has integral values between l and -l, including zero & describes the orientation in space of the orbital relative to the other orbitals in the atom Spin quantum number (ms) is either +1/2 or -1/2 for a given e- & describes the direction of spin of the e- on its axis Pauli Exclusion Principle: "no two electrons in an atom can have the same set of quantum numbers", or, only two electrons (of opposite spin) per orbital.

Some Quantum Nos. & Orbitals in the H Atom Sub-shell No. of n l Designation ml Orbitals No. of e-

Some Quantum Nos. & Orbitals in the H Atom Sub-shell No. of n l Designation ml Orbitals No. of e- 1 0 1s 0 1 2

Some Quantum Nos. & Orbitals in the H Atom Sub-shell No. of n l Designation ml Orbitals No. of e- 1 0 1s 0 1 2 2 0 2s 0 1 2 1 2p -1, 0, +1 3 6

Some Quantum Nos. & Orbitals in the H Atom Sub-shell No. of n l Designation ml Orbitals No. of e- 1 0 1s 0 1 2 2 0 2s 0 1 2 1 2p -1, 0, +1 3 6 3 0 3s 0 1 2 1 3p -1, 0, +1 3 6 2 3d -2, -1, 0, +1, +2 5 10 4...etc...

Which of the following are valid sets of quantum numbers? (a) n =1, l =1, ml = 0, ms = 1/2

Which of the following are valid sets of quantum numbers? (a) n =1, l =1, ml = 0, ms = 1/2 invalid l = 0 to n-1 (b) n =2, l =1, ml = 0, ms = 1/2 valid

Which of the following are valid sets of quantum numbers? (a) n =1, l =1, ml = 0, ms = 1/2 invalid l = 0 to n-1 (b) n =2, l =1, ml = 0, ms = 1/2 valid (c) n =2, l =0, ml = -1, ms = 1/2 invalid ml = -l to l

Which of the following are valid sets of quantum numbers? (a) n =1, l =1, ml = 0, ms = 1/2 invalid l = 0 to n-1 (b) n =2, l =1, ml = 0, ms = 1/2 valid (c) n =2, l =0, ml = -1, ms = 1/2 invalid ml = -l to l (d) n =3, l =1, ml = 0, ms = 0 invalid ms = -1/2 or +1/2

Which of the following are valid sets of quantum numbers? (a) n =1, l =1, ml = 0, ms = 1/2 invalid l = 0 to n-1 (b) n =2, l =1, ml = 0, ms = 1/2 valid (c) n =2, l =0, ml = -1, ms = 1/2 invalid ml = -l to l (d) n =3, l =1, ml = 0, ms = 0 invalid ms = -1/2 or +1/2 Write a valid set of quantum numbers for each of the following sub-shells: (a) 2s n = 2, l = 0, ml = 0, ms = 1/2

Which of the following are valid sets of quantum numbers? (a) n =1, l =1, ml = 0, ms = 1/2 invalid l = 0 to n-1 (b) n =2, l =1, ml = 0, ms = 1/2 valid (c) n =2, l =0, ml = -1, ms = 1/2 invalid ml = -l to l (d) n =3, l =1, ml = 0, ms = 0 invalid ms = -1/2 or +1/2 Write a valid set of quantum numbers for each of the following sub-shells: (a) 2s n = 2, l = 0, ml = 0, ms = 1/2 (b) 2p n = 2, l = 1, ml = -1, ms = 1/2

Which of the following are valid sets of quantum numbers? (a) n =1, l =1, ml = 0, ms = 1/2 invalid l = 0 to n-1 (b) n =2, l =1, ml = 0, ms = 1/2 valid (c) n =2, l =0, ml = -1, ms = 1/2 invalid ml = -l to l (d) n =3, l =1, ml = 0, ms = 0 invalid ms = -1/2 or +1/2 Write a valid set of quantum numbers for each of the following sub-shells: (a) 2s n = 2, l = 0, ml = 0, ms = 1/2 (b) 2p n = 2, l = 1, ml = -1, ms = 1/2 (c) 3d n = 3, l = 2, ml = -2, ms = 1/2

Principal quantum number specifies energy of e-

Principal quantum number specifies energy of e- For hydrogen, Schroedinger showed that E = - hR/n2 where R is the Rydberg constant = 3.29 X 1015 Hz

Principal quantum number specifies energy of e- For hydrogen, Schroedinger showed that E = - hR/n2 where R is the Rydberg constant = 3.29 X 1015 Hz Can calculate from theory the hydrogen spectrum

Principal quantum number specifies energy of e- For hydrogen, Schroedinger showed that E = - hR/n2 where R is the Rydberg constant = 3.29 X 1015 Hz Can calculate from theory the hydrogen spectrum For transition from n = 3 to n = 2 state

Principal quantum number specifies energy of e- For hydrogen, Schroedinger showed that E = - hR/n2 where R is the Rydberg constant = 3.29 X 1015 Hz Can calculate from theory the hydrogen spectrum For transition from n = 3 to n = 2 state DE = -hR[1/32 - 1/22] = - h x 3.29 x 1015[1/9 - 1/4]

Principal quantum number specifies energy of e- For hydrogen, Schroedinger showed that E = - hR/n2 where R is the Rydberg constant = 3.29 X 1015 Hz Can calculate from theory the hydrogen spectrum For transition from n = 3 to n = 2 state DE = -hR[1/32 - 1/22] = - h x 3.29 x 1015[1/9 - 1/4] = h x 4.57 x 1014 Hz = hn and n = 4.57 x 1014 Hz

Principal quantum number specifies energy of e- For hydrogen, Schroedinger showed that E = - hR/n2 where R is the Rydberg constant = 3.29 X 1015 Hz Can calculate from theory the hydrogen spectrum for transition from n = 3 to n = 2 state DE = -hR[1/32 - 1/22] = - h x 3.29 x 1015[1/9 - 1/4] = h x 4.57 x 1014 Hz = hn and n = 4.57 x 1014 Hz l = c/n = 3 x 108 ms-1/ 4.57 x 1014 s-1 = 656 nm

Line spectrum of hydrogen 400 550 700 nm

For hydrogen, E = - hR/n2 If H atom acquires enough energy e- can be promoted from one level to the next

For hydrogen, E = - hR/n2 If H atom acquires enough energy e- can be promoted from one level to the next To promote an e- from n =1 to n = 2 level DE = -hR[1/22 - 1/12] = - h x 3.29 x 1015[1/4 - 1] = 1.63 x 10-18 J

For hydrogen, E = - hR/n2 If H atom acquires enough energy e- can be promoted from one level to the next To promote an e- from n =1 to n = 2 level DE = -hR[1/22 - 1/12] = - h x 3.29 x 1015[1/4 - 1] = 1.63 x 10-18 J To promote an e- from n = 2 to n = 3 level DE = -hR[1/32 - 1/22] = - h x 3.29 x 1015[1/9 - 1/4] = 3.03 x 10-19 J

For hydrogen, E = - hR/n2 If H atom acquires enough energy e- can be promoted from one level to the next To promote an e- from n =1 to n = 2 level DE = -hR[1/22 - 1/12] = - h x 3.29 x 1015[1/4 - 1] = 1.63 x 10-18 J To promote an e- from n = 2 to n = 3 level DE = -hR[1/32 - 1/22] = - h x 3.29 x 1015[1/9 - 1/4] = 3.03 x 10-19 J To remove e- (ionise) from the atom DE = -hR[1/2 - 1/12] = hR = 2.18 x 10-18 J

For hydrogen, E = - hR/n2 If H atom acquires enough energy e- can be promoted from one level to the next To promote an e- from n =1 to n = 2 level DE = -hR[1/22 - 1/12] = - h x 3.29 x 1015[1/4 - 1] = 1.63 x 10-18 J To promote an e- from n = 2 to n = 3 level DE = -hR[1/32 - 1/22] = - h x 3.29 x 1015[1/9 - 1/4] = 3.03 x 10-19 J To remove e- (ionise) from the atom DE = -hR[1/2 - 1/12] = hR = 2.18 x 10-18 J To ionise one mole of hydrogen atoms DE = NAhR = 1.31 x 103 kJ H H+ + e-

Principle quantum number n = 1, 2, 3,….. describes orbital size and energy Angular momentum quantum number l = 0 to n-1 describes orbital shape Magnetic quantum number ml = l, l-1…-l describes orientation in space of the orbital relative to the other orbitals in the atom Spin quantum number ms = +1/2 or -1/2 describes the direction of spin of the e- on its axis Pauli Exclusion Principle: "no two electrons in an atom can have the same set of quantum numbers", or, only two electrons (of opposite spin) per orbital.

Write a valid set of quantum numbers for each of the following sub-shells: (a) 2 s n = 2, l = 0, ml = 0, ms = - 1/2 n = 2, l = 0, ml = 0, ms = ± 1/2 2 combinations

Write a valid set of quantum numbers for each of the following sub-shells: (a) 2 s n = 2, l = 0, ml = 0, ms = - 1/2 n = 2, l = 0, ml = 0, ms = ± 1/2 2 combinations (b) 2 p n = 2, l = 1, ml = -1, ms = - 1/2 n = 2, l = 1, ml = -1, 0 or 1, ms = ± 1/2 6 combinations

Write a valid set of quantum numbers for each of the following sub-shells: (a) 2 s n = 2, l = 0, ml = 0, ms = - 1/2 n = 2, l = 0, ml = 0, ms = ± 1/2 2 combinations (b) 2 p n = 2, l = 1, ml = -1, ms = - 1/2 n = 2, l = 1, ml = -1, 0 or 1, ms = ± 1/2 6 combinations (c) 3 d n = 3, l = 2, ml = -2, ms = - 1/2 n = 3, l = 2, ml = -2, -1, 0, 1, or 2, ms = ± 1/2 10 combinations

How many orbitals in a subshell? l = 1, px, py, pz 3 l = 2, dxy,, dxz,, dyz ,, dx2-y2, dz2 5

How many orbitals in a subshell? l = 1, px, py, pz 3 l = 2, dxy,, dxz,, dyz ,, dx2-y2, dz2 5 2 l + 1 orbitals per subshell

How many orbitals in a subshell? l = 1, px, py, pz 3 l = 2, dxy,, dxz,, dyz ,, dx2-y2, dz2 5 2 l + 1 orbitals per subshell How many orbitals in a shell? n = 1, 1s 1 n = 2, 2s, 2px, 2py, 2pz 4 n = 3, 3s, 3px, 3py, 3pz, 3dxy,, 3dxz,, 3dyz ,, 3dx2-y2, 3dz2 9

How many orbitals in a subshell? l = 1, px, py, pz 3 l = 2, dxy,, dxz,, dyz ,, dx2-y2, dz2 5 2 l + 1 orbitals per subshell How many orbitals in a shell? n = 1, 1s 1 n = 2, 2s, 2px, 2py, 2pz 4 n = 3, 3s, 3px, 3py, 3pz, 3dxy,, 3dxz,, 3dyz ,, 3dx2-y2, 3dz2 9 n2 orbitals per principal quantum level

Hydrogen atom- all orbitals within a shell have the same energy electrostatic interaction between e- and proton

Hydrogen atom- all orbitals within a shell have the same energy electrostatic interaction between e- and proton Multi-electron atoms- the energy level of an orbital depends not only on the shell but also on the subshell electrostatic interactions between e- and proton and other e-

Quantum Mechanical Model for Multi-electron Atoms electron repulsions He He+ + e- E = 2372 kJ mol-1 He has two electron which repel each other

Quantum Mechanical Model for Multi-electron Atoms electron repulsions He He+ + e- E = 2372 kJ mol-1 He has two electron which repel each other He+ He2+ + e- E = 5248 kJ mol-1 He+ has one electron, no electrostatic repulsion

Quantum Mechanical Model for Multi-electron Atoms electron repulsions He He+ + e- E = 2372 kJ mol-1 He has two electron which repel each other He+ He2+ + e- E = 5248 kJ mol-1 He+ has one electron, no electrostatic repulsion Less energy required to remove e- from He than from He+ Shielding of outer orbital electrons from +ve nuclear charge by inner orbital electrons => outer orbital electrons have higher energies

Quantum Mechanical Model for Multi-electron Atoms Penetration effect of outer orbitals within inner orbitals: ns > np > nd For a given n, energy of s < energy of p < energy of d

Quantum Mechanical Model for Multi-electron Atoms Penetration effect of outer orbitals within inner orbitals: ns > np > nd For a given n, energy of s < energy of p < energy of d Effective nuclear charge (Zeff) experienced by an electron is used to quantify these additional effects.

Quantum Mechanical Model for Multi-electron Atoms Penetration effect of outer orbitals within inner orbitals: ns > np > nd For a given n, energy of s < energy of p < energy of d Effective nuclear charge (Zeff) experienced by an electron is used to quantify these additional effects. Example: Sodium, Na, Z = 11 Na 1s e- : Zeff = 10.3 shielding effect is small Na 3s e- : Zeff = 1.84 large shielding effect by inner e-’s penetration effect counteracts this to a small extent

Orbital Energies 3dxy 3dxz 3dyz 3dx2-y2 3dz2 3px 3py 3pz 3s Energy 2px 2py 2pz 2s 1s

Electronic Configuration: Filling-in of Atomic Orbitals Rules: 1. Pauli Principle

Electronic Configuration: Filling-in of Atomic Orbitals Rules: 1. Pauli Principle 2. Fill in e-'s from lowest energy orbital upwards (Aufbau Principle)

Electronic Configuration: Filling-in of Atomic Orbitals Rules: 1. Pauli Principle 2. Fill in e-'s from lowest energy orbital upwards (Aufbau Principle) 3. Try to attain maximum number of unpaired e- spins in a given sub-shell (Hund's Rule)

Electronic Configuration: Filling-in of Atomic Orbitals Rules: 1. Pauli Principle 2. Fill in e-'s from lowest energy orbital upwards (Aufbau Principle) 3. Try to attain maximum number of unpaired e- spins in a given sub-shell (Hund's Rule) H (Z = 1) 1s1 2s 2p Energy 1s

Electronic Configuration: Filling-in of Atomic Orbitals Rules: 1. Pauli Principle 2. Fill in e-'s from lowest energy orbital upwards (Aufbau Principle) 3. Try to attain maximum number of unpaired e- spins in a given sub-shell (Hund's Rule) N (Z = 7) 1s2, 2s2, 2p3, 2p 2s Energy 1s

Electronic Configuration: Filling-in of Atomic Orbitals Rules: 1. Pauli Principle 2. Fill in e-'s from lowest energy orbital upwards (Aufbau Principle) 3. Try to attain maximum number of unpaired e- spins in a given sub-shell (Hund's Rule) B (Z = 5) 1s2, 2s2, 2p1 2p 2s Energy 1s

Electronic Configuration: Filling-in of Atomic Orbitals Rules: 1. Pauli Principle 2. Fill in e-'s from lowest energy orbital upwards (Aufbau Principle) 3. Try to attain maximum number of unpaired e- spins in a given sub-shell (Hund's Rule) F (Z = 9) 1s2, 2s2, 2p5 2p 2s Energy 1s

Hydrogen 2s 3s 4s 1s 2p 3p 4p 3d 4d 4f Multi-electron atoms 1s 2s 3s 4s 5 s 2p 3p 4p

1s 2s 2px 2py 2pz H 1s1 He 1s2 Li 1s2, 2s1 Be 1s2, 2s2 B 1s2, 2s2, 2px1 C 1s2, 2s2, 2px1, 2py1 N 1s2, 2s2, 2px1, 2py1, 2pz1 O 1s2, 2s2, 2px2, 2py1, 2pz1 F 1s2, 2s2, 2px2, 2py2, 2pz1 Ne 1s2, 2s2, 2px2, 2py2, 2pz2

H 1s1 He 1s2 Li [He], 2s1 Be [He], 2s2

H 1s1 He 1s2 Li [He], 2s1 Be [He], 2s2 B [He], 2s2, 2p1 Ne [He], 2s2, 2p6 Na [He], 2s2, 2p6, 3s1  [Ne], 3s1

H 1s1 He 1s2 Li [He], 2s1 Be [He], 2s2 B [He], 2s2, 2p1 Ne [He], 2s2, 2p6 Na [He], 2s2, 2p6, 3s1  [Ne], 3s1 Mg [He], 2s2, 2p6, 3s2  [Ne], 3s2 Al [Ne], 3s2, 3p1 Si [Ne], 3s2, 3p2

H 1s1 He 1s2 Li [He], 2s1 Be [He], 2s2 B [He], 2s2, 2p1 Ne [He], 2s2, 2p6 Na [He], 2s2, 2p6, 3s1  [Ne], 3s1 Mg [He], 2s2, 2p6, 3s2  [Ne], 3s2 Al [Ne], 3s2, 3p1 Si [Ne], 3s2, 3p2 P [Ne], 3s2, 3p3 S [Ne], 3s2, 3p4 Cl [Ne], 3s2, 3p5 Ar [Ne], 3s2, 3p6

H 1s1 He 1s2 Li [He], 2s1 Be [He], 2s2 B [He], 2s2, 2p1 Ne [He], 2s2, 2p6 Na [He], 2s2, 2p6, 3s1  [Ne], 3s1 Mg [He], 2s2, 2p6, 3s2  [Ne], 3s2 Al [Ne], 3s2, 3p1 Si [Ne], 3s2, 3p2 P [Ne], 3s2, 3p3 S [Ne], 3s2, 3p4 Cl [Ne], 3s2, 3p5 Ar [Ne], 3s2, 3p6 outermost shell - valence shell most loosely held electron and are the most important in determining an element’s properties

K [Ar], 4s1 Ca [Ar], 4s2 Sc [Ar], 4s2, 3d1 Ca [Ar], 4s2, 3d2

K [Ar], 4s1 Ca [Ar], 4s2 Sc [Ar], 4s2, 3d1 Ca [Ar], 4s2, 3d2 Zn [Ar], 4s2, 3d10 Ga [Ar], 4s2, 3d10, 3p1 Kr [Ar], 4s2, 3d10, 3p6

K [Ar], 4s1 Ca [Ar], 4s2 Sc [Ar], 4s2, 3d1 Ca [Ar], 4s2, 3d2 Zn [Ar], 4s2, 3d10 Ga [Ar], 4s2, 3d10, 3p1 Kr [Ar], 4s2, 3d10, 3p6 Anomalous electron configurations d5 and d10 are lower in energy than expected Cr [Ar], 4s1, 3d5 not [Ar], 4s2, 3d4 Cu [Ar], 4s1, 3d10 not [Ar], 4s2, 3d9

Electron Configuration of Ions Electrons lost from the highest energy occupied orbital of the donor and placed into the lowest unoccupied orbital of the acceptor (placed according to the Aufbau principle)

Electron Configuration of Ions Electrons lost from the highest energy occupied orbital of the donor and placed into the lowest unoccupied orbital of the acceptor (placed according to the Aufbau principle) Examples: Na [Ne], 3s1 Na+ [Ne] + e- Cl [Ne], 3s2, 3p5 + e- Cl- [Ne], 3s2, 3p6 Mg [Ne], 3s2 Mg2+ [Ne] O [He], 2s2, 2p4 O2- [He], 2s2, 2p6

Modern Theories of the Atom - Summary Wave-particle duality of light and matter Bohr theory Quantum (wave) mechanical model Orbital shapes and energies Quantum numbers Electronic configuration in atoms