e skill 1.15 For each of the following compounds, identify any polar covalent bonds by drawing 8+ and 8- symbols in the appropriate locations: H HH H H-C-O-C C C-H (a) H H H (b) H:O: H C C -Ö-H (d) HHH :F: H-C-CI: HOH H-C-C-C-H (e) H H H-C-Mg-Br: (c) H :CI: :-C-C: H (f) :CI:
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- 3-35 Classify each bond as nonpolar covalent, polar covalent, or ionic. (a) CCl (b) CLi (c) CN3-79 Answer true or false. (a) The letters VSEPR stand for valence-shell electron-pair repulsion. (b) In predicting bond angles about a central atom in a covalent molecule, the VSEPR model considers only shared electron pairs (electron pairs involved in forming covalent bonds). (c) The VSEPR model treats the two electron pairs of a double bond as one region of electron density and the three electron pairs of a triple bond as one region of electron density. (d) In carbon dioxide, OCO, carbon is surrounded by four pairs of electrons and the VSEPR model predicts 109.5° for the OCO bond angle. (e) For a central atom surrounded by three regions of electron density, the VSEPR model predicts bond angles of 120°. (f) The geometry about a carbon atom surrounded by three regions of electron density is described as trigonal planar. (g) For a central atom surrounded by four regions of electron density, the VSEPR model predicts bond angles of 360°/4 = 90°. (h) For the ammonia molecule, NH3, the VSEPR model predicts HNH bond angles of 109.5°. (i) For the ammonium ion, NH4, the VSEPR model predicts HNH bond angles of 109.5°. (j) The VSEPR model applies equally well to covalent compounds of carbon, nitrogen, and oxygen. (k) In water, HOH, the oxygen atom forms covalent bonds to two other atoms, and therefore, the VSEPR model predicts an HOH bond angle of 180°. (l) If you fail to consider unshared pairs of valence electrons when you use the VSEPR model, you will arrive at an incorrect prediction. (m) Given the assumptions of the VSEPR model, the only bond angles it predicts for compounds of carbon, nitrogen, and oxygen are 109.5°, 120°, and 180°.3-109 Until several years ago, the two chlorofluorocarbons (CFCs) most widely used as heat transfer media in refrigeration systems were Freon-li (trichloro fluoromethane, CC13F) and Freon-12 (dichiorodi fluoromethane, CCl2F2). Draw a three-dimensional representation of each molecule and indicate the Direction of it.s polarity.
- 1.35. Arrange the single covalent bonds within each set in order of increasing polarity. (а) С-Н, О—н, N— H (b) С—Н, В —Н, О —Н (с) С-Н, С —СІ, С —І (d) С S, C —О, С —N (е) С—Li, C —В, С -MgProblem 1.17 Draw a second resonance structure for each species in parts (a), (b), and (c). Draw two additional resonance structures for the ion in part (d). a. b. CH3 d. E H H E c. H₂N C I H CH3 Η Η H I H :0: || C C :O: || H + CH3 C I H :0: || CH31.65 Write an equation for the Brønsted-Lowry acid-base reaction that occurs when each of the following acids reacts with water. Show all unshared electron pairs and formal charges, and use curved arrows to track electron movement. + :0-Н Н-С%3DN: N. H3C- H. H :0-CH3 .. (а) (b) (c) +
- Determine a molecular formula. Question 103. Are the following bonds polar or nonpolar? Tip: You should use electronegativity values to back up your answers. Consult your textbook if necessary. (a) C-H (b) N-H (c) C-O (d) C-NProblem 1.37 Draw a Lewis structure for each of the following species: Part A Part B CO²- Draw the molecule by placing atoms on the canvas and connecting them with bonds. Include all lone pairs of electrons. [] i + P F Br CI More C H O N S I X Press [SPACE] to restore all settings to the default. Press [TAB] to move to the next option. Press [ALT+A) to get to the bonds and electrons editor. Press [ALT+P] to get to the elements editor. Press (ALT+Y) to get to the elements on the canvas. Press
- Which of the bonds, shown by the dash, has the greatest polarity? O H-CI O H-CH, H-OH H-SH O H-NH,Use the first image to help with the questionIdentify the correct dipole for the C-O bond in the Lewis structure of methanol shown using the symbols d* and õ. :0 A) (6*)C – O(5) B) (6)С — О(6') Н—с—Н C) (6)С - 0(6) D) (õ')C – O(5*) н