s (2 families) p (6 families) d (10 families) f (14 families)

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1 Electron Configuration Hint Sheet The chart below shows the regions of the periodic table in which the outermost electron is in s,p,d and f orbitals. It requires 2 e- s to fill an s orbital, 6 e- s to fill a p orbital, 10 e- s to fill a d orbital and 14 e- s to fill an f orbital. Each row and region in the periodic table is associated with a different quantum number. For s and p orbitals, the row number is the principle quantum number. For d orbitals, the row number minus one is the principle quantum number. For f orbitals, the row number minus two is the principle quantum number. 1 2 s (2 families) p (6 families) d (10 families) f (14 families) Total Electron configuration of Neutral Atoms Example: Se 1) Locate the element on the Periodic Table and determine what row it is in. Ah, it s in the 4 th row, in the p block. 2) Go up one row and all the way over to the noble gases. You can save some writing by putting the noble gas in brackets to indicate everything the noble gas has. OK, Argon is the noble gas in the 3 rd row, so write [Ar]. 3) Now go back to the row the element is in, and start counting over. It s going to take 2 e- s to fill up the s-block, so that s 4s 2. Page 1 7/3/03 L.M. Petrovich

2 It s going to take 10 e s to fill up the d-block, so that s 3d 10. No it s not a mistake. Remember it s (row #-1) for d s. Se is 4e s over into the p-block, so that s 4p 4. Example: Ba Total electron configuration of Se is [Ar] 4s 2 3d 10 4p 4 1) Locate the element on the Periodic Table and determine what row it is in. Ah, it s in the 6 th row, in the s block. 2) Go up one row and all the way over to the noble gases. You can save some writing by putting the noble gas in brackets to indicate everything the noble gas has. OK, Xenon is the noble gas in the 5th row, so write [Xe]. 3) Now go back to the row the element is in, and start counting over. It s going to take 2 e- s to fill up the s-block to where Ba is, so that s 6s 2. Example: Pd Total electron configuration of Ba is [Xe] 6s 2 1) Locate the element on the Periodic Table and determine what row it is in. Ah, it s in the 5 th row, in the d block. 2) Go up one row and all the way over to the noble gases. You can save some writing by putting the noble gas in brackets to indicate everything the noble gas has. OK, Kryton is the noble gas in the 4th row, so write [Kr]. 3) Now go back to the row the element is in, and start counting over. It s going to take 2 e- s to fill up the s-block, so that s 5s 2. Pd is 8e s over into the p-block, so that s 4d 8. Total electron configuration of Pd is [Kr] 5s 2 4d 8 Examples for you to try: 1a) Sn 1b) Fe 1c) Na 1d) N 1e) Ti 1f) Br Page 2 7/3/03 L.M. Petrovich

3 Valence Electron configuration of Neutral Atoms The VALENCE electron configuration of neutral atoms is simply the highest energy s orbital with electrons, plus any other energy levels that are not full of electrons. The valence electrons are the electrons the atom uses to participate in chemical reactions. Example: Se 1) Figure out the total electron configuration The total electron configuration of Se is [Ar] 4s 2 3d 10 4p 4, from the previous example. 2) Remove the noble gas core and any filled sublevels, other than the highest filled s. To get valence, we just throw out the core, [Ar] and the filled 3d sublevel. Valence electron configuration of Se is 4s 2 4p 4 Example: Ba 1) Figure out the total electron configuration Total electron configuration of Ba is [Xe] 6s 2 2) Remove the noble gas core and any filled sublevels, other than the highest filled s. To get valence, we just throw out the core, [Xe]. Valence electron configuration of Ba is 6s 2 Example: Pd 3) Figure out the total electron configuration Total electron configuration of Pd is [Kr] 5s 2 4d 8 4) Remove the noble gas core and any filled sublevels, other than the highest filled s. To get valence, we just throw out the core, [Kr]. Valence electron configuration of Pd is 5s 2 4d 8 Examples for you to try: 2a) Sn 2b) Fe 2c) Na 2d) N 2e) Ti 2f) Br Page 3 7/3/03 L.M. Petrovich

4 Total Electron configuration of Ions Example: Se 2- The total electron configuration of Se is [Ar] 4s 2 3d 10 4p 4, from the previous example. Se 2- is an anion of-2, so add two electrons to the next available slot: the 4p orbital. The total electron configuration of Se 2- is [Ar] 4s 2 3d 10 4p 6, or more simply [Kr] Example: Ba 2+ The total electron configuration of Ba is [Xe] 6s 2 Ba 2+ is a cation of +2, so remove two electrons from the outermost s orbital, since the highest energy p is part of the core. The total electron configuration of Ba 2+ is [Xe] Example: Pd 2+ The total electron configuration of Pd is [Kr] 5s 2 4d 8 Pd 2+ is a cation of +2, so remove two electrons from the outermost s orbital, since the highest energy p is part of the core. Remember, s leave before d. The total electron configuration of Pd 2+ is [Kr] 4d 8 Page 4 7/3/03 L.M. Petrovich

5 Example: Tl 1+ The total electron configuration of Tl is [Xe] 6s 2 4f 14 5d 10 6p 1 Tl 1+ is a cation of +1, so remove two electrons from the outermost p orbital. Remember, p leave first, then s, then d. The total electron configuration of Tl 1+ is [Xe] 6s 2 4f 14 5d 10 Examples for you to try: 3a) Sn 4+ 3b) Fe 2+ 3c) Na 1+ 3d) N 3-3e) Ti 2+ 3f) Br 1- Page 5 7/3/03 L.M. Petrovich

6 Answers: 1a) Sn: [Kr] 5s 2 4d 10 5p 2 1b) Fe: [Ar] 4s 2 3d 6 1c) Na: [Ne] 3s 1 1d) N: [He] 2s 2 2p 3 1e) Ti: [Ar] 4s 2 3d 2 1f) Br: [Ar] 4s 2 3d 10 4p 5 2a) Sn: 5s 2 5p 2 2b) Fe: 4s 2 3d 6 2c) Na: 3s 1 2d) N: 2s 2 2p 3 2e) Ti: 4s 2 3d 2 2f) Br: 4s 2 3d 10 4p 5 3a) Sn 4+ : [Kr] 4d 10 3b) Fe 2+ : [Ar] 3d 6 3c) Na 1+ :: [Ne] 3d) N 3- :: [Ne] 3e) Ti 2+ :: [Ar] 3d 2 3f) Br 1- :: [Kr] Page 6 7/3/03 L.M. Petrovich