regents review,may 31, june 1,2 2008 caiafa chem is fun d. hh the ionic character is 0.0, the bond...

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D. OO STEP 1 OO STEPSTEP 2 OO DRAW LEWIS VALENCE DOT STRUCTURE, CONNECT ONLY SINGLE ELECTRONS, 2 FOR EACH OXYGEN EACH O SHARES THE 4 e-(DOUBLE BOND) ELECTRONS IT NEEDS, IT NOW HAS THE OCTET CONFIGURATION OF NEON Ne O2O2 Important: ALL OF THIS SERIES ( H 2,O 2,N 2,Cl 2,Br 2,I 2,F 2 ) ARE NON POLAR FOR THE FOLLOWING REASONS: 1)THEY HAVE IONIC CHARACTER OF 0.0 2)THEY ARE LINEAR (SYMMETRICAL) – ONE BOND 3)THE BONDS ARE NON POLAR. 4)ALL (EXCEPT I AND Br) ARE GASSES AT STP. 5)ALL ARE ELEMENTAL MOLECULAES, OXIDATION STATE 0!

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REGENTS REVIEW,MAY 31, JUNE 1, CAIAFA CHEM IS FUN D. HH THE IONIC CHARACTER IS 0.0, THE BOND IS NON-POLAR COVALENT STEP 1 HH DRAW LEWIS VALENCE DOT STRUCTURE, CONNECT ONLY SINGLE ELECTRONS DRAW LEWIS VALENCE DOT STRUCTURE, EACH SINGLE BOND IS 2 e- STEP 2 HH DRAW THE DOT STRUCTURES OF H 2,O 2,N 2,Cl 2,Br 2,I 2, F 2. 1) How would you estimate the relative melting and boiling points of these? 2) Put them in ascending boiling point order and indicate how you do this. 3) Indicate the type of molecule and type of bond. 4) List the characteristics and explain them with the bonding and type of attraction. THIS H SHARES THE 2 ELECTRONS IT NEEDS, H IS NOT AN OCTET ELEMENT THIS H SHARES THE 2 ELECTRONS IT NEEDS, IT NOW HAS THE CONFIGURATIONOF He : DRAW LEWIS VALENCE DOT STRUCTURE, EACH ELECTRON PAIR IS A DASH- LINE! H2H2 = UNBONDED e- PAIRS ON CENTRAL ATOM BONDS ON CENTRAL ATOM GEOMETRY EXAMPLES 0 4 TETRAHEDRAL CH4CH4 13 PYRAMIDAL NH3NH3 22 BENT (V SHAPE) H2OH2O NO CENTRAL, ONLY 2 ATOMS 1(in molecule)LINEARH2H2 03 TRIGONAL PLANAR BF3BF3 VSEPR (valence shell electron pair repulsion) 1)Gives geometry to assess polarity etc. 2)Assume central atom is LEAST electronegative. 3)Hydrogen cannot be the central atom as it makes only one bond. D. OO STEP 1 OO STEPSTEP 2 OO DRAW LEWIS VALENCE DOT STRUCTURE, CONNECT ONLY SINGLE ELECTRONS, 2 FOR EACH OXYGEN EACH O SHARES THE 4 e-(DOUBLE BOND) ELECTRONS IT NEEDS, IT NOW HAS THE OCTET CONFIGURATION OF NEON Ne O2O2 Important: ALL OF THIS SERIES ( H 2,O 2,N 2,Cl 2,Br 2,I 2,F 2 ) ARE NON POLAR FOR THE FOLLOWING REASONS: 1)THEY HAVE IONIC CHARACTER OF 0.0 2)THEY ARE LINEAR (SYMMETRICAL) ONE BOND 3)THE BONDS ARE NON POLAR. 4)ALL (EXCEPT I AND Br) ARE GASSES AT STP. 5)ALL ARE ELEMENTAL MOLECULAES, OXIDATION STATE 0! NN STEP 1 NN STEP 2 NN Connect only single electrons for covalent (molecular) substances. Each nitrogen has an octet, isoelectronic to Ne Each nitrogen shares 6 electrons, triple bond. RELATIVE ASCENDING BOILING AND MELTING SERIES H 2 (2g/mol), O 2 (32), N 2 (28), Cl 2 (70), Br 2 (161), I 2 (254 ), F 2 (38) H 2 (2g/mol), N 2 (28), O 2 (32), F 2 (38), Cl 2 (70), Br 2 (161), I 2 (254 ) 1-LOWEST HIGHEST Van der Waals increase with molar mass (gfm), use gfm for nonpolar Liquid at STP Solid at STP The electron configuration of nonmetals will resemble the noble gas of their own period DRAW THE DOT STRUCTURES FOR H 2 O, H 2 S, SiO 2, CO 2, CCl 3 F, CH 4, HF. a. - WHICH AR POLAR, WHICH ARE NOT. b. - INDICATE THE TYPE OF BOND AND THE TYPE OF MOLECULE FOR EACH. c. - PUT IN ASCENDING BOILING POINTS, INDICATE HOW YOU SO THIS. d. - List the characteristics of each and explain why. OH H Connect single electrons STEP 1 O HH Each H shares the 2 e- achieve He : configuration O shares the 8 e- achieve Ne configuration Ne STEP 2 HOH Next slide IONIC CHARACTER = 2.1 3.5 = 1.4, POLAR COV. H O H THE BONDING IS CORRECT, BUT THE GEOMETRY IS NOT! WITH 2 BONDS AND 2 UNSHARED PAIRS OF e-, THE GEOMETRY IS BENT. HOH H O H THE MOST ELECTRONEGATIVE ATOM IS THE NEGATIVE POLE AND GETE THIS NEGATIVE POLE SYMBOL - THE LEAST ELECTRONEGATIVE ATOM IS THE POSITIVE POLE AND GETS THIS POSITIVE POLE SYMBOL YELLOW ARROWS ARE BOND POLARITY VECTORS, PONT TO MOST ELECTRONEGATIVE ELEMENT, NEGATIVE POLE. THE 2 YELLOW VECTOR ARROWS ADD UP TO THIS BIG ONE, WHICH REPRESENTS MOLECULAR POLARITY, POINTS TO NEGATIVE POLE. + H O H HH S IONIC CHARACTER(EACH BOND) =| 2.1 2.6| = 0.5, WEAKLY POLAR COVALENT. AS IONIC CHARACTER INCREASES causes BOND POLARITY INCREASES causes MOLECULAR POLARITY INCREASES ( IF ASSYMETRICAL) H 2 O HAS GREATER IONIC CARACTER THAN H 2 S causes H 2 O HAS GREATER BOND POLARITY THAN H 2 S causes WATER IS MORE POLAR HAS HIGHER MELTING AND BOILING POINTS + = Si O O OO EACH OXYGEN HAS SHARED 4 e- TO ACHIEVE AN OCTET EACH THE Si HAS SHARED 4 e- TO ACHIEVE AN OCTET STEP 1 STEP 2 OOSi SiO 2 (silicon dioxide) is NONPOLAR, it is LINEAR ( 2 bonds) and symmetrical. The BONDS ARE polar but the molecule is not! NOTICE THE VECTOR ARROWS CANCEL = 0 IONIC CHARACTER = |3.5 1.9| = 1.6, STRONGLY POLAR COV. Note: THE Si IS HYBRID, NOT GROUND STATE. Si SI d OOC CO 2 (carbon dioxide) is NONPOLAR, it is LINEAR ( 2 bonds) and symmetrical. The BONDS ARE polar but the molecule is not! = 0.0 NOTICE THE VECTOR ARROWS CANCEL IONIC CHARACTER(EACH BOND) =| 2.1 2.6| = 0.5, WEAKLY POLAR COV. C H HH H H HF H C POLAR BONDS,SYMMETRICAL NON POLAR MOLLECULESYMMETRICAL POLAR BONDS, A-SYMMETRICAL POLAR MOLLECULE + Note: THE C IS HYBRID, NOT GROUND STATE. C Mg+ Mg IS AN ACTIVE METAL THAT NEEDS TO LOSE 2 e- IONIC CHARACTER = |1.3 3.2| = 1.9, IONIC Cl Cl IS AN ACTIVE NON-METAL THAT NEEDS TO GAIN 1 e- STEP 1 Mg +2 Cl [ [ - [ [ - Mg 2+ Cl 1- MgCl 2 Note: Mg+ 2 has a Ne electron configuration. Note: Cl- has an Ar electron configuration. The electron configuration of metals will resemble the noble gas of the period before the period of the metal Al 3+ F 1- AlF 3 CHARACTERISTICS OF IONIC SOLIDS 1)Form crystal lattice solid with very strong electrostatic attractions between ions. 2)Ions are fixed in place, rendering the crystal non-conductive, brittle and hard. 3)The electrons cannot leave the ions, cannot conduct. 4)The strong electrostatic attractions make it difficult to separate the ions, and this requires great amounts of energy, thus the melting and boiling points are very high. CHARACTERISTICS OF IONIC liquids and solutions (aq) 1)Soluble ionic compounds (table F) form conductive solutions mobile ions. 2)The melt of an ionic substance is always conductive. 3)Remember: FOR CONDUCTIVITY MUST HAVE MOBILITY! (OF IONS OR ELECTRONS) COO CARBON DIOXIDE EACH OXYGEN HAS 2 SINGLE ELECTRONS, NEEDS TO SHARE TWO MORE FOR AN OCTET, AND TO RESEMBLE NEON. EACH HYBRID CARBON HAS 4 SINGLE ELECTRONS, NEEDS TO SHARE FOUR MORE FOR AN OCTET, AND TO RESEMBLE NEON. COO CARBON DIOXIDE, VENN DIAGRAM THE YELLOW CIRCLE REPRESENTS THE OXYGENSET WITH AN OCTET THE BLUE CIRCLE REPRESENTS THE CARBONSET WITH AN OCTET THE GREEN INTERSECTION AREAS REPRESENT THE 4 SHARED ELECTRONS OF A DOUBLE BOND COO CARBON DIOXIDE, VENN DIAGRAM CO 2 HAS AN IONIC CHARACTER OF 0.9, WHICH IS CONALENT POLAR EACH DOUBLE BOND IS 4 SHARED ELECTRONS, IS SHORTER AND STRONGER THAN THE SINGLE BOND. AS WIH ALL COVALENT COMPOUNDS, CO 2 HAS WEAK ATTRACTIONS, LOW M.P. AND B.P.AND IS NEVER CONDUCTIVE. Covalent bonds are formed by sharing at least one pair of electrons. Every covalent bond has a characteristic length that leads to maximum stability. This is the bond length. Energy required to break a covalent bond in an isolated gaseous molecule is called the bond dissociation energy.