λ = 650 nm = 6.50 x 10-7 m ν = c = 2.9979 x 108 m/s = 4.61 x 1014 Hz The brilliant red colors seen in fireworks are due to the emission of light with wavelengths around 650 nm when strontium salts such as Sr(NO3)2 and SrCO3 are heated. Calculate the frequency of red light of wavelength 6.50 × 102 nm. λ = 650 nm = 6.50 x 10-7 m ν = c = 2.9979 x 108 m/s = 4.61 x 1014 Hz λ 6.50 x 10-7 m

Ephoton = hc = (6.626 x 10-34 Js)(2.9979 x 108 m/s) λ 4.50 x 10-7 m The blue color in fireworks is often achieved by heating copper(I) chloride (CuCl) to about 1200°C. Then the compound emits blue light having a wavelength of 450 nm. What is the increment of energy (the quantum) that is emitted at 4.50 × 102 nm by CuCl? λ = 450 nm = 4.50 x 10-7 m Ephoton = hc = (6.626 x 10-34 Js)(2.9979 x 108 m/s) λ 4.50 x 10-7 m = 4.417 x 10-19 J

E = -2.178 x 10-18 J(Z2/n2) for H, Z = 1 so E = -2.178 x 10-18 J(1/n2) Calculate the energy required to excite the hydrogen electron from level n = 1 to level n = 2. Also calculate the wavelength of light that must be absorbed by a hydrogen atom in its ground state to reach this excited state. E = -2.178 x 10-18 J(Z2/n2) for H, Z = 1 so E = -2.178 x 10-18 J(1/n2) From n = 1 to n = 2 ΔE = -2.178 x 10-18 J( 1 𝑛2 2 - 1 𝑛1 2 ) = -2.178 x 10-18 J( 1 2 2 - 1 1 2 ) = 1.634 x 10-18 J λ = hc = (6.626 x 10-34 Js)(2.9979 x 108 m/s) Ephoton 1.634 x 10-18 J = 1.217 x 10-7 m

From n = 1 to n = ∞ since the electron is being removed Calculate the energy required to remove the electron from a hydrogen atom in its ground state. E = -2.178 x 10-18 J(Z2/n2) for H, Z = 1 so E = -2.178 x 10-18 J(1/n2) From n = 1 to n = ∞ since the electron is being removed ΔE = -2.178 x 10-18 J( 1 𝑛2 2 - 1 𝑛1 2 ) = -2.178 x 10-18 J( 1 ∞ 2 - 1 1 2 ) = -2.178 x 10-18 J(0 - 1) = 2.178 x 10-18 J

Exercise 5 Electron Subshells For principal quantum level n = 5, determine the number of allowed subshells (different values of ℓ), and give the designation of each. n = 5 so l = 0, 1, 2, 3, 4 or s, p, d, f, and then g l = 0  5s l = 1  5p l = 2  5d l = 3  5f l = 4  5g (theoretical, not actually observed)

Exercise 6 Give the full electron configuration and orbital diagram for each of the following elements. Sulfur Iron Lithium Aluminum Magnesium Silicon 1s22s22p63s23p4 1s22s22p63s23p64s23d6 1s22s1 1s22s22p63s23p1 1s22s22p63s2 1s22s22p63s23p2

Orbital Diagram—Sulfur Exercise 6 continued Orbital Diagram—Sulfur 1s 2s 2p 3s 3p

Orbital Diagram—Lithium Exercise 6 continued Orbital Diagram—Lithium 1s 2s

Orbital Diagram—Magnesium Exercise 6 continued Orbital Diagram—Magnesium 1s 2s 2p 3s

Exercise 6 continued Orbital Diagram—Iron 1s 2s 2p 3s 3p 4s 3d

Orbital Diagram—Aluminum Exercise 6 continued Orbital Diagram—Aluminum 1s 2s 2p 3s 3p

Orbital Diagram—Silicon Exercise 6 continued Orbital Diagram—Silicon 1s 2s 2p 3s 3p

Exercise 6 continued—other side of page Write the noble gas configuration for the following ions. S2-  Al3+  Mg2+ A sulfide ion has two more electrons than a sulfur atom. S2-  [Ne]3s23p6 isoelectronic with argon, Cl-, P3-, K+, Ca2+ [He]2s22p6 isoelectronic with neon, Mg2+, Na+, F-, O2-, N3- [He]2s22p6 isoelectronic with neon, Al3+, Na+, F-, O2-, N3-

Exercise 6 continued—other side of page Write the noble gas configuration for the following ions.  Fe3+  Br- [Ar]3d5 Electrons are lost from 4s first [Ar]4s24p6 Isoelectronic with Kr, Se2-, As3-, Rb+, Sr2+

Exercise 7 Determine the number of unpaired electrons in each element from Exercise 6.

Orbital Diagram—Sulfur Exercise 7 Orbital Diagram—Sulfur 1s 2s 2p 3s 3p

Orbital Diagram—Lithium Exercise 7 Orbital Diagram—Lithium 1s 2s

Orbital Diagram—Magnesium Exercise 7 Orbital Diagram—Magnesium No unpaired electrons 1s 2s 2p 3s

Exercise 7 Orbital Diagram—Iron 1s 2s 2p 3s 3p 4s 3d

Orbital Diagram—Aluminum Exercise 7 Orbital Diagram—Aluminum 1s 2s 2p 3s 3p

Orbital Diagram—Silicon Exercise 7 Orbital Diagram—Silicon 1s 2s 2p 3s 3p

Exercise 8 Trends in Ionization Energies The first ionization energy for phosphorus is 1060 kJ/mol, and the first ionization energy for sulfur is 1005 kJ/mol. Why? P  S  First IE = removing “last” electron For phosphorous, this means removing the electron that makes the 3p subshell half-full. A filled OR half-filled subshell is stable; the atom does not “want” to lose this stability. For sulfur, removing the last electron results in a half-filled 3p subshell. Since stability is gained, the first IE is lower for sulfur than for phosphorous.

Exercise 9 Ionization Energies Consider atoms with the following electron configurations: a. 1s22s22p6 b. 1s22s22p63s1 c. 1s22s22p63s2 Identify each atom. Which atom has the largest first ionization energy, and which one has the smallest second ionization energy? Explain your choices. Neon Largest first IE (noble gas, stable) Sodium Magnesium Smallest second IE because it gains a noble gas configuration when it loses 2 electrons

Exercise 10 Trends in Radii Predict the trend in radius for the following ions: Be2+, Mg2+, Ca2+, and Sr2+. Explain the trend as well. Be2+ < Mg2+ < Ca2+ < Sr2+ Beryllium has the smallest number of principal energy levels, so it is the smallest. As the number of principal energy levels (n) increases, the size of the atom increases.