Chem 121 Handout on E Diagram & Electron Configuration Page 1 Energy Diagram of Orbitals E

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Chem 121 Handout on E Diagram & Electron Configuration Page 1 Energy Diagram of Orbitals E 3d 5d 4f 4d 4p 4s 5p 5s 6s 2p Each line stands for an orbital, which can accommodate a maximum of 2 electrons, of different spins ( or ) If there is more than one orbital of the same energy, electrons will spread out into different orbitals before pairing up. RULES FOR RENTING APARTMENTS RULES FOR FILLING ELECTRONS INTO THE ENERGY DIAGRAM 1. Rent rooms with lowest rent first. 1. Electrons fill up levels with the lowest energy first. (Aufbau Principle) 2. Each room can take up to 2 occupants. 2. Each orbital can take up to 2 electrons. 3. The 2 occupants within a room must have different. 3. The 2 electrons within an orbital must have different spins. (Pauli Exclusion Principle) 4. If there are more than 1 room of the same rent, 4. If there are more than 1 orbital of the same energy, occupants will spread out first before pairing up electrons will spread out first before pairing up. (Hund s Rule) with roommates. The outershell is the E level with the highest n that contains electrons. Outershell electrons = electrons residing in the outershell (with the highest n). Valence electrons are electrons that participate in chemical reactions. For representative elements (A families), outershell electrons = valence electrons For transition elements (B families), valence electrons include the electrons in the d-orbitals as well. The ground state of an atom is the state where all the e are at the lowest E levels possible. There can only be one ground state. An excited state of an atom is a state where an e is boosted up to a higher E level. There can be more than one excited state. When an atom is in an excited state, it is unstable and will try to return to its ground state. You should be able to 1. Fill electrons into E diagrams for each element. 2. Write the electron configuration for each element. 3. Circle the outershell or valence electrons for each element.

Chem 121 Handout on E Diagram & Electron Configuration Page 2 Fill in the electrons for each of the elements, circle the valence electrons and then write the electron configuration. 23 V 33 As 24 Cr (an exception you must know) 4p 4p 4p 3d 3d 3d 4s 4s 4s 2p 2p 2p After writing the electron configuration, circle the valence electrons in the electron configuration. e config. for V = e config. for As = e config. for Cr = 29Cu (also an exception, how?) 16 S 2 - Fe 3+ 26 4p 4p 4p 3d 3d 3d 4s 4s 4s 2p 2p 2p After writing the electron configuration, circle the valence electrons in the electron configuration. e config. for Cu = e config. for S 2 = e config. for Fe 3+ =

Chem 121 Handout on E Diagram & Electron Configuration Page 3 How do the quantum numbers fit in? Principal quantum number (n) which apartment building? n = 1, 2, 3 Angular momentum quantum number (l) which floor? l = 0, 1, 2, 3 (n-1) = s, p, d, f MEMORIZE! Magnetic quantum number (ml) which room? ml = l 2, 1, 0, 1, 2 +l Spin quantum number (m s ) which roommate? m s = + ½, ½ 3d _ 4p 4s 1. The E diagram on the left is for the ground state of which element? 2. Give the quantum numbers of these 6 electrons. 3. Give the quantum numbers for these orbitals. 4. Give the quantum numbers for each of these 2 electrons. 5. What are the possible values of ml for electrons with n = 3, l = 1? 2p 6. Where would you find the electron with this set of quantum numbers: n = 3, l = 2, ml = 0 and m s = + ½ 7. What is wrong with this set of quantum numbers? n = 2, l = 2, ml = 0 and m s = + ½ 8. How many electrons can have n = 3, l = 2 and ml = 1? 9. What is the maximum number of electrons that can have n = 4, l = 3? 10. What is the maximum number of electrons that can have n =3? 11. What is the number of electrons that ground state phosphorus has with n = 3, l =1? n = 3, l =2? 12. For each of the following sets of quantum numbers, state whether it is allowed: a) n = 4, l = 4, ml = 0, m s = ½ b) n = 0, l = 0, ml = 0, m s = + ½ c) n = 3, l = 2, ml = 2, m s = + ½ d) n = 2, l = 0, ml = +1, m s = ½

Chem 121 Handout on E Diagram & Electron Configuration Page 4 Obtaining Electronic Configurations from the Periodic Table 1 H 1 s He He 2 3 4 5 6 7 2 s 3 s 4 s 5 s 6 s 7 s 3 d 4 d 5 d 6 d 2 p 3 p 4 p 5 p 6 p Lanthanide Series Actinide Series 4 f 5 f The diagram above will not be provided on an exam. A regular periodic table will be provided as usual. Commit to memory: 1. Locations of s, p, d, f 2. Numbering system of each section. Each square in the table above stands for an electron in an orbital. Each element has all the electrons of the previous element, plus one more. Give the electron configuration of each element below and circle the outershell electron(s). You may use only the periodic table. # of valence e Group # 6C 2 2 2p 2 7N 2 2 2p 3 8O 2 2 2p 4 9F 2 2 2p 5 10Ne 2 2 2p 6 e configuration using the core method: 17Cl 23V 34Se Cl V Se 24Cr 29Cu Fe 3+ You should be able to write electron configurations for all the elements (atoms and ions, and exceptions such as Cu and Cr).

Chem 121 Handout on E Diagram & Electron Configuration Page 5 1. Using the core method, give the e configuration of 50Sn 56Ba 15P 3 _ 26Fe 3+ _ 2. Fill in the electrons. Give the full e config. and circle the valence e. Na 3d Cl 3d 4s 4s 2p 2p Na + Cl 3. Give the e configuration of only the valence e : S S 2 K K + which resembles _ which resembles _ S 2, K +, Ar are said to be isoelectronic. 4. Name two other ions that are isoelectronic with them: and. 5. a) Are Br and Cl isoelectronic? Yes or No b) Are Ba 2+ and Cl isoelectronic? Yes or No An atom is said to be paramagnetic if it contains unpaired electrons. It is said to be diamagnetic if it does not contain any unpaired electrons. Using ONLY a periodic table, answer the following questions: 6. Is silicon paramagnetic or diamagnetic? 7. How many unpaired electrons does sulfur have? Ans. _ Is it paramagnetic or diamagnetic? Circle one. 8. How many unpaired electrons does the chromium atom have? Ans. _ 9. How many unpaired electrons does Fe 3+ have? Ans.