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CHEMISTRY 11 AP – ELECTRON CONFIGURATION WORKSHEET #1 1) Write the full electron configuration of the following: a) Co (1s2 2s2 2p6 3s2 3p6 4s2 3d7) b) Sr (1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2) c) Ar (1s2 2s2 2p6 3s2 3p6) d) Cd (1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10) e) Xe (1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p6) f) (1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p5) Br g) Cs (1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p6 6s1) h) Pb (1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p6 6s2 4f14 5d10 6p2) i) Ga (1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p1) j) Zr (1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d2) 2) Draw the orbital diagram of the following: a) H 1s 2s 2p 3s 3p 4s 3d 2p 3s 3p 4s 3d 3s 3p 4s 3d 3p 4s 3d é b) O 1s 2s éê éê éê é é c) Na 1s 2s éê éê 2p éê éê éê é d) Ti 1s 2s éê éê 2p éê éê 3s éê éê éê éê éê éê é é e) K 1s 2s éê éê 2p éê éê 3s éê éê 3p éê éê 4s éê é 3d f) Al 1s 2s éê éê 2p éê éê 3s éê éê 3p 4s 3d 3p 4s 3d 4s 3d é g) P 1s 2s éê éê 2p éê éê 3s éê éê é é é h) Ca 1s 2s éê éê i) éê éê 3s éê éê 3p éê éê éê éê Cl 1s 2s éê éê j) 2p 2p éê éê 3s éê éê 3p éê éê 4s 3d 4s 3d é Ni 1s 2s éê éê 2p éê éê 3s éê éê 3p éê éê éê éê éê éê éê é é Writing quantum numbers for a given electron is not required on the AP exam; however it is important to know what an appropriately formulated set of quantum numbers looks like. Use the following summary of quantum number rules to help you answer questions 3 and 4. Principal Quantum Number (n) – may be an integer value starting from 1. This represents the principal energy level of the atom in which the electron is located and is related to the average distance of the electron from the nucleus Angular Momentum Number (ℓ ) – may have any number from 0 up to n – 1. This designates the subshell of the electron and also represents the shape of the orbitals in the subshell. Magnetic Quantum Number (mℓ ) – may be any integer, including 0 from –ℓ to +ℓ . This designates the orientation of an orbital in space. Spin Quantum Number (ms) – may be either +½ or –½. This represents the “spin” of an electron. For electrons to pair up within an orbital, one electron must have a +½ value and the other a –½ value. Consider the orbital diagram of F: 1s 2s éê éê 2p éê éê é The last electron was the ê electron placed in the second p orbital therefore that electron has a n = 2 since it is in the second shell, a ℓ = 1 since it is a p subshell (all s = 0, p = 1, d = 2 and f = 3), a mℓ = 0 since it is in the second orbital of the 2p subshell (the first box is –1, the second box is 0 and the third box is +1) and a ms = –½ since the electron is pointed down 3) Complete the following table for the four quantum numbers for the last electron added to each of the first 20 elements: Element n ℓ mℓ ms H 1 0 0 +½ He 1 0 0 –½ Li 2 0 0 +½ Be 2 0 0 –½ B 2 1 –1 +½ C 2 1 0 +½ N 2 1 +1 +½ O 2 1 –1 –½ F 2 1 0 –½ Ne 2 1 +1 –½ Na 3 0 0 +½ Mg 3 0 0 –½ Al 3 1 –1 +½ Si 3 1 0 +½ P 3 1 +1 +½ S 3 1 –1 –½ Cl 3 1 0 –½ Ar 3 1 +1 –½ K 4 0 0 +½ Ca 4 0 0 –½ 4) Designate each of the following sets of quantum numbers as possible or impossible according to the rules of quantum numbers. a) 1, 0, 0, +½ d) 1, 3, 0, +½ g) 3, 2, 0, +½ j) 2, 2, 2, –½ b) 2, 0, 1, –½ e) 3, 1, -1, +½ h) 4, 2, -2, –½ k) 4, 4, 0, +½ c) 3, 2, 1, 0 f) 1, 1, 1, +½ i) 3, 1, 2, –½ l) 5, 3, –2, +½ a, e, g, h, i and l are valid. The others disobey one or more of the rules of quantum numbers as follows: d, e, f, g and k violate the angular momentum rule; b violates the magnetic quantum number rule and c violates the spin quantum number rule.