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A detailed explanation of molecular models, focusing on lewis structures, electronic and molecular geometry, hybridization, bond angles, and polarity. It includes examples such as nh4+, bro4-, o3, n3-, brcl5, hno3, and pocl3. The document also covers resonance structures, formal charges, and vsepr theory, offering a comprehensive guide for understanding molecular shapes and bonding. It is designed to help students visualize and predict molecular properties based on electron distribution and repulsion, enhancing their understanding of chemical bonding principles and molecular geometry. Useful for students studying chemistry, especially those learning about molecular structure and bonding theories. It provides clear examples and explanations that aid in understanding complex concepts.
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Experiment 10- Dry lab - Molecular Models NAME(S) ______________Kellee Perry__________________ INSTRUCTIONS: (a) Total points for this Report/exercise = 10. All numbered problems are 1 pt each. (b) You will submit this by yourself or you may work with other students in the lab. However, there can be no more than three students and all students’ names must be written on this exercise. (c) Submit the completed exercise by the end of this week by email. (d) For all compounds draw the Lewis structures clearly and underneath the structure indicates various answers for questions 2-8. (e) Make sure you address all questions given and show all work to be eligible for the maximum points indicated for each question. Show and write all work only on the spaces provided after each question. Since this is a Word doc, you can add space between questions if you need more space. But under no circumstances are you to attach scratch sheets of paper! Afterwards, you are to upload this file as a PDF.
1. Define the following: a) Octet rule Octet rule refers to the tendency of atoms to achieve noble gas configuration by having eight electrons in their valence shell. b) Polar covalent bond A polar covalent bond forms when two atoms of different electronegativity share electrons with unequal contribution. Electronegativity difference around 2 refers to polar covalent bond. c) Resonance Resonance structures are Lewis structure that differ with each other with respect to the placement of valence electrons. Resonance describes the delocalization of electrons within molecule while satisfying the octet rule. d) Formal charge The formal charge of an atom is the charge that it would have in molecule if all atoms had same value of electronegativity. It is calculated as Formal charge = valence electrons – (bonding electrons / 2) – nonbonding electrons e) VSEPR Theory Valence shell electron pair repulsion (VSEPR) theory is used to predict the shape of molecules from the electron pairs (bonding as well as nonbonding) surrounding the central atom. According to it the electron pairs arrange themselves in such a way to minimize e- - e- repulsion. For questions 2- 8 , write the BEST Lewis dot structure for each molecule, being sure to give the electronic
geometry, molecular geometry, hybridization of the central element, polarity, and bond angle around the central element. 2. NH 4 + Total valence electrons = N+ 4H - 1 VT = 5 + 4(1) - 1 = 8 Total bond or lone pairs = 8 /2= 4 Central atom Electronic geometry Molecular geometry Hybridization Bond angle Polarity N tetrahedral tetrahedral sp 3 109.5 non-polar 3. BrO 4 - Total valence electrons = Br+ 4O + 1 VT = 7 + 4( 6 ) +1 = 32 Total bond or lone pairs = 32 /2= 16 Central atom Electronic geometry Molecular geometry Hybridization Bond angle Polarity Br Tetrahedral Tetrahedral sp^3 109.5 Non-polar 4. O 3 Total valence electrons = 3O VT = 3(6) = 18 Total bond or lone pairs = 18 /2= 9 Electron pair remaining after bonds = 9-2 = 7 Central atom Electronic geometry Molecular geometry Hybridization Bond angle Polarity O Trigonal planar Bent sp^2 <120 Polar 5. N 3 - Total valence electrons = 3 N + 1 VT = 3( 5 ) + 1 = 16 Total bond or lone pairs = 16 /2= 8 Electron pair remaining after bonds = 8 - 2 = 6
Central atom Electronic geometry Molecular geometry Hybridization Bond angle Polarity P Tetrahedral Tetrahedral sp^3 <109.5 polar 9. Assign formal charges to each atom in the three proposed resonance structures of NO 2 -. Which resonance structure is likely to contribute most to the correct structure of NO 2 -^ or are they equivalent resonance structures. Total valence electrons = N+ 2 O + 1 VT = 5 + 2( 6 ) + 1 = 18 Total bond or lone pairs = 18 /2= 9 Electron pair remaining after bonds = 9 - 2 = 7 Satisfying the octet of outer atoms Proposed resonance structures: Formal charge (FC) on Structure A: FC on left oxygen = valence electrons – (bonding electrons / 2) – nonbonding electrons FC = 6 – 2/ 2 – 6 = - 1 FC on nitrogen = valence electrons – (bonding electrons / 2) – nonbonding electrons FC = 5 – 4 / 2 – 2 = + 1 FC on right oxygen = valence electrons – (bonding electrons / 2) – nonbonding electrons FC = 6 – 2/ 2 – 6 = - 1 Formal charge (FC) on Structure B: FC on left oxygen = 6 – 2/ 2 – 6 = - 1 FC on nitrogen = 5 – 6 / 2 – 2 = 0 FC on right oxygen = 6 – 4 / 2 – 4 = 0 Formal charge (FC) on Structure B: FC on nitrogen = 5 – 6 / 2 – 2 = 0 FC on left oxygen = 6 – 4 / 2 – 4 = 0 FC on right oxygen = 6 – 2/ 2 – 6 = - 1
Out of three proposed structures, B and C are equivalent and contribute most to the structure. Structure A is least contributing as the octet of N is not completely filled and more charges are present compared to structure B and C. So, structure A is least contributing while structure B and C are equally contributing. 10. Draw the two best Lewis dot resonance structures of CNO- and identify which is the better structure. Total valence electrons = C + N +O + VT = 4 + 5 + 6 + 1 = 16 Total bond or lone pairs = 16 /2= 8 Electron pair remaining after bonds = 8 - 2 = 6 Completing the octet of central atom and drawing two best resonance structures. Calculating formal charge of structure, A: FC on Nitrogen = 5 – 6 / 2 – 2 = 0 FC on carbon = 4 – 8 / 2 – 0 = 0 FC on oxygen = 6 – 2 / 2 – 6 = - 1 Calculating formal charge of structure, B: FC on Nitrogen = 5 – 4 / 2 – 4 = - 1 FC on carbon = 4 – 8 / 2 – 0 = 0 FC on oxygen = 6 – 4 / 2 – 4 = 0 Structure A is better structure because the negative charge is present on more electronegative atom. In structure B the negative formal charge is present on nitrogen which is less electronegative than oxygen. Compared to structure A, structure B is less contributing.