Determine what elements are denoted by the following electron configurations: 14) \( 1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{4} \) 15) \( 1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 4 s^{2} 3 d^{10} 4 p^{6} 5 s^{1} \) 16) \( \left[\mathrm{Krr}^{1}\right] 5 s^{2} 4 d^{10} 5 p^{3} \) 17) \( [\mathrm{Ke}] 6 s^{2} 4 f^{14} 5 d^{6} \)

We need to identify the element corresponding to each electron configuration by counting its electrons (or otherwise recognizing the configuration). 14) 1s² 2s² 2p⁶ 3s² 3p⁴ • The inner shells (1s, 2s, 2p) account for 2 + 2 + 6 = 10 electrons. • The third shell (3s² and 3p⁴) provides an additional 2 + 4 = 6 electrons. • Total electrons = 10 + 6 = 16. Element 16 is sulfur (S). 15) 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s¹ • Adding up:   1s² = 2 electrons,   2s² 2p⁶ gives 2 + 6 = 8 electrons,   3s² 3p⁶ gives 2 + 6 = 8 electrons,   4s² gives 2 electrons,   3d¹⁰ gives 10 electrons,   4p⁶ gives 6 electrons,   5s¹ gives 1 electron. • Total = 2 + 8 + 8 + 2 + 10 + 6 + 1 = 37 electrons. Element 37 is rubidium (Rb). 16) [Krr¹] 5s² 4d¹⁰ 5p³ The shorthand “[Krr¹]” appears to stand in for a noble gas core. Recognize that the standard configuration for antimony is given as:   [Xe] or [Kr] plus the remaining electrons. Here the breakdown is:   Assume the core represents krypton ([Kr]) which has 36 electrons.   5s² gives 2 electrons (36 + 2 = 38),   4d¹⁰ adds 10 (38 + 10 = 48),   5p³ adds 3 (48 + 3 = 51). Element 51 is antimony (Sb). 17) [Ke] 6s² 4f¹⁴ 5d⁶ The core “[Ke]” is likely a misprint or shorthand for the xenon core ([Xe]), which has 54 electrons. Then:   6s² adds 2 electrons (54 + 2 = 56),   4f¹⁴ adds 14 electrons (56 + 14 = 70),   5d⁶ adds 6 electrons (70 + 6 = 76). Element 76 is osmium (Os). Final Answers: 14) Sulfur (S) 15) Rubidium (Rb) 16) Antimony (Sb) 17) Osmium (Os)