Question

Write the condensed electron configurations for the following atoms, using the appropriate noble-gas core abbreviations.

Part A Cs Express your answer in condensed form in order of increasing orbital energy as a string without blank space between orbitals. For example, [He]2s22p2 should be entered as [He]2s^22p^2. [Xe]6s^1 SubmitMy AnswersGive Up Correct Part B Ni Express your answer in condensed form in order of increasing orbital energy as a string without blank space between orbitals. For example, [He]2s22p2 should be entered as [He]2s^22p^2. [Ar]4s^23d^8 SubmitMy AnswersGive Up Correct Part C Se Express your answer in condensed form in order of increasing orbital energy as a string without blank space between orbitals. For example, [He]2s22p2 should be entered as [He]2s^22p^2. [Ar]4s^23d^104p^4 SubmitMy AnswersGive Up Correct Part D Cd Express your answer in condensed form in order of increasing orbital energy as a string without blank space between orbitals. For example, [He]2s22p2 should be entered as [He]2s^22p^2. [Kr]5s^24d^10 SubmitMy AnswersGive Up Correct Part E U Express your answer in condensed form in order of increasing orbital energy as a string without blank space between orbitals. For example, [He]2s22p2 should be entered as [He]2s^22p^2. SubmitMy AnswersGive Up Incorrect; Try Again; 4 attempts remaining Part F Pb Express your answer in condensed form in order of increasing orbital energy as a string without blank space between orbitals. For example, [He]2s22p2 should be entered as [He]2s^22p^2. SubmitMy AnswersGive Up Incorrect; Try Again; 4 attempts remaining

Answer 1

Part A Cs has an atomic number of 55. So, it has the configuration: The previous noble gas is Xe: 1s2 2s' 2pf 3s 3p*4s23d0 4p' 5s14d05p6 So the condensed configuration is [Xe]6s Part B Ni has an atomic number of 28. So, it has the configuration: The previous noble gas is Ar:1s 2s 2p6 3s2 3p So the condensed configuration is [Ar]4s 3d Part C Se has an atomic number of 34. So, it has the configuration: The previous noble gas is Ar: 1s 2s 2p6 3s2 3p So the condensed configuration is [Ar]4s 3d10 4p Part D Cd has an atomic number of 48. So, it has the configuration: ls2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s24d10 The previous noble gas is Kr: 1s2 2s2 2рб 3S2 3p6 4s2 3d10 4p6 So the condensed configuration is [Kr]5s 4d Part E U has an atomic number of 48. So, it has the configuration: The previous noble gas is Rn:1s2 2s2 2рб 3s2 3p6 4s2 3d10 4p 5s2 4d10 5p%" 4f14 5d10 6p So the condensed configuration is [Rn]7s2 5f3 6d Part F Pb has an atomic number of 82. So, it has the configuration The previous noble gas is Xe: 1s2 2s2 2p6 3s2 Зрб 4s2 3do 4p6 5s2 4d105p6 So the condensed configuration is [Xe]6s 4f1" sdO 6p