5.3 ELECTRON CONFIGURATIONS

5.3 ELECTRON CONFIGURATION Arrangement of electrons in the atom: Electron Hotel: 1) Stays in the cheapest (lowest energy) room possible. 2) No more than 2 (electrons) to a room. 3) Likes having own room. Will share if next room is more expensive. 4) “d” & “f” suites are especially expensive!

AUFBAU PRINCIPLE – Each electron will occupy the lowest energy orbital available Chart shows lowest to highest energy. Note all p, d, & f electrons in same energy level have same energy. “Energy overlap” of “d” & “f” electrons. Aufbau Hotel

PAULI EXCLUSION PRINCIPLE Each electron is unique: Unique by energy level, shape, orientation in space or spin If 2 electrons are in the same orbital, they must have opposite spins:  Aufbau Hotel electrons: 1 goes to football game ↑ 1 goes to the opera ↓

HUND’S RULE – Use when filling p, d, f orbitals: Must fill in single electron, same spin in each orbital before pairing up. Ex.    then   _ p p Hotel: Electrons like having own room, unless next available room is more expensive. Then they will share.

QUIZ 1) According to Aufbau, electrons fill the orbital having the ___ energy. 2) Pauli’s Exclusion Principle states that each electron is _____. If 2 electrons are in the same orbital they must have ____ spins. 3) Hund’s Rule states that when filling p, __, and f orbitals, you must first fill in single electron, same ___ in each orbital before pairing up.

Predicting last part of electron configuration Use periodic table: Period– tells what energy level being filled. (Ex. Na – 3rd energy level) Sublevels Groups s 1a – 2a (& He) p 3a – 8a d “B groups” f Lanthanide & Actinides (bottom 2 rows)

LOCATION OF s, p, d, f BLOCKS d block – 1 step behind f block - 2 steps behind

What is the ending for the following elements: 3) Li 4) Na 5) Mg 6) Al 7) Ar 8) K 9) Sc (“d” is 1 step behind) 10) U (“f” is 2 steps behind) 5f4

WRITING ORBITAL DIAGRAMS: Ex. O 1) Find atomic #. Gives #electrons (arrows). For O, atomic # = 8 2) Find what O ends in. 2 p4. 3) Draw orbitals (slots) from lowest (1s) to 2p4. __ __ __ __ __ Now draw in arrows. 1s 2s 2p Use Hotel rules:

FILL IN FROM LOWEST TO HIGHEST:

ELECTRON NOTATION Summarize arrows Called notation: 1s2 2s2 2p4

Write electron configuration (arrows) for Nitrogen. 1) N = #7 (7 electrons or arrows). 2) N ends in 2p3. 3) Draw orbitals up to 2p: (show all slots) __ __ __ __ __ Now draw in arrows. 1s 2s 2p Notation is 1s22s22p3

LEVEL 1 ASSIGNMENT: Write electron configuration (arrows) for: 1) F 2) Mg 3) O 4) P 1) Find atomic #. 2) Element ends in ____. 3) Draw orbitals up to ___: (show all slots) __ __ __ __ __ Now draw in arrows. 1s 2s 2p __ ___ ___ ___ 3s 3p 4) Show notation:

NOBLE GAS NOTATION: Short Cut: 1) Write previous noble gas in brackets. 2) Then write each block that is left over. Ex. for P, # 15. What is the previous noble gas? 1) Write previous noble gas: Ne This represents Neon’s 10 electrons. 2) Write what is left over (each block): Ne 3s23p3.

SHORTHAND NOTATION Find for:

WRITE IN SHORTHAND (NOBLE GAS) NOTATION FORM:: 1) Ca #20 (previous noble gas?) Ar Now, what’s left over? Ar 4s2. 2) S #16 Ne 3s23p4 3) Zr, #40 Kr 5s24d2

LEVEL 2 ASSIGNMENT: Write electron configuration (arrows) for: 1) K 2) V 3) Ge 4) Tc 1) Find atomic #. 2) Element ends in ____. 3) Draw orbitals up to ___: (show all slots) __ __ __ __ __ Now draw in arrows. 1s 2s 2p __ ___ ___ ___ 3s 3p 4) Show notation: 1s2…. 5) Show Shorthand notation: Noble gas Left over.

VALENCE ELECTRONS – Electrons in the atom’s outermost energy level Determine the chemical properties of the element. Include: s & p orbitals only d & f are in lower energy levels (not valence) Cl – 1s2 2s2 2p6 3s23p5 Outermost energy shell: n = 3 Valence electrons: 3s23p5 7 val. e-

PROBLEM: Find valence electrons for: Sodium (#11) 1 valence electron (n=3) Nitrogen (#7) 5 valence electrons (n=2) Argon (#18) 1s2 2s2 2p6 3s23p6 8 val. electrons (8 is stable!) n=3

FIND VALENCE ELECTRONS FOR: Valence electrons determine chemical properties! 2) Sr 3) Ra What do you notice? 4) N 5) As 6) Kr 7) V 8) La

NOTE: Valence electrons do not include “d” or “f” electrons!

NOBLE GASES – Have filled outer shells 8 valence electrons - stable octet! (He – has 2) Ar (#18) 1s22s22p6 3s23p6 (8 val.) He (#2 ) 1s2 (1st shell – full) Noble Gases end in “p6”

NOBLE GASES HAVE FILLED OUTER SHELLS! All end in: p6 8 val. electrons Except: He (#2 ) 1s2 2 val. Electrons (1st shell is full!)

EXCEPTIONS Half filled and filled sets of s & d orbitals very stable: Expected: Cr Ar 4s23d4 Actual: Ar 4s13d5 Ar ↑_ ↑_ ↑_ ↑_ ↑_ _↑_ 4s 3d Complex chemistry of transition elements.

EXCEPTIONS (cont.) Actual configuration: Cu #29 Predicted configuration: [Ar] 4s2 3d9 Actual configuration: [Ar] 4s1 3d10 Note: full “d” is stable.

CARBON FORMS 4 BONDS BECAUSE OF sp3 HYBRIDIZATION: Expected orbital configuration sp3 hybridization causes 4 like orbitals Carbon forms 4 bonds

BORON ALSO HYBRIDIZES: B forms 3 bonds!

THE END