What are the other two reasons for choosing (b) in Example 9.7? Example 9.7 Formaldehyde (CH 2 O), a liquid with a disagreeable odor, traditionally has been used to preserve laboratory specimens. Draw the most likely Lewis structure for the compound. Strategy A plausible Lewis structure should satisfy the octet rule for all the elements, except H, and have the formal charges (if any) distributed according to electronegativity guidelines. Solution The two possible skeletal structures are First we draw the Lewis structures for each of these possibilities: To show the formal charges, we follow the procedure given in Example 9.6. In (a) the C atom has a total of five electrons (one lone pair plus three electrons from the breaking of a single and a double bond). Because C has four valence electrons, the formal charge on the atom is 4 − 5 = −1. The O atom has a total of five electrons (one lone pair and three electrons from the breaking of a single and a double bond). Because O has six valence electrons, the formal charge on the atom is 6 − 5 = +1. In (b) the C atom has a total of four electrons from the breaking of two single bonds and a double bond, so its formal charge is 4 − 4 = 0. The O atom has a total of six electrons (two lone pairs and two electrons from the breaking of the double bond). Therefore, the formal charge on the atom is 6 − 6 = 0. Although both structures satisfy the octet rule, (b) is the more likely structure because it carries no formal charges. Check In each case make sure that the total number of valence electrons is 12. Can you suggest two other reasons why (a) is less plausible?
What are the other two reasons for choosing (b) in Example 9.7? Example 9.7 Formaldehyde (CH 2 O), a liquid with a disagreeable odor, traditionally has been used to preserve laboratory specimens. Draw the most likely Lewis structure for the compound. Strategy A plausible Lewis structure should satisfy the octet rule for all the elements, except H, and have the formal charges (if any) distributed according to electronegativity guidelines. Solution The two possible skeletal structures are First we draw the Lewis structures for each of these possibilities: To show the formal charges, we follow the procedure given in Example 9.6. In (a) the C atom has a total of five electrons (one lone pair plus three electrons from the breaking of a single and a double bond). Because C has four valence electrons, the formal charge on the atom is 4 − 5 = −1. The O atom has a total of five electrons (one lone pair and three electrons from the breaking of a single and a double bond). Because O has six valence electrons, the formal charge on the atom is 6 − 5 = +1. In (b) the C atom has a total of four electrons from the breaking of two single bonds and a double bond, so its formal charge is 4 − 4 = 0. The O atom has a total of six electrons (two lone pairs and two electrons from the breaking of the double bond). Therefore, the formal charge on the atom is 6 − 6 = 0. Although both structures satisfy the octet rule, (b) is the more likely structure because it carries no formal charges. Check In each case make sure that the total number of valence electrons is 12. Can you suggest two other reasons why (a) is less plausible?
Solution Summary: The author explains the two Lewis structures for formaldehyde labelled as ‘a’ and ‘b’ in example 9.7 of the book.
What are the other two reasons for choosing (b) in Example 9.7?
Example 9.7
Formaldehyde (CH2O), a liquid with a disagreeable odor, traditionally has been used to preserve laboratory specimens. Draw the most likely Lewis structure for the compound.
Strategy A plausible Lewis structure should satisfy the octet rule for all the elements, except H, and have the formal charges (if any) distributed according to electronegativity guidelines.
Solution The two possible skeletal structures are
First we draw the Lewis structures for each of these possibilities:
To show the formal charges, we follow the procedure given in Example 9.6. In (a) the C atom has a total of five electrons (one lone pair plus three electrons from the breaking of a single and a double bond). Because C has four valence electrons, the formal charge on the atom is 4 − 5 = −1. The O atom has a total of five electrons (one lone pair and three electrons from the breaking of a single and a double bond). Because O has six valence electrons, the formal charge on the atom is 6 − 5 = +1. In (b) the C atom has a total of four electrons from the breaking of two single bonds and a double bond, so its formal charge is 4 − 4 = 0. The O atom has a total of six electrons (two lone pairs and two electrons from the breaking of the double bond). Therefore, the formal charge on the atom is 6 − 6 = 0. Although both structures satisfy the octet rule, (b) is the more likely structure because it carries no formal charges.
Check In each case make sure that the total number of valence electrons is 12. Can you suggest two other reasons why (a) is less plausible?
At 0oC and 1 atm, the viscosity of hydrogen (gas) is 8.55x10-5 P. Calculate the viscosity of a gas, if possible, consisting of deuterium. Assume that the molecular sizes are equal.
Indicate the correct option for the velocity distribution function of gas molecules:a) its velocity cannot be measured in any other way due to the small size of the gas moleculesb) it is only used to describe the velocity of particles if their density is very high.c) it describes the probability that a gas particle has a velocity in a given interval of velocitiesd) it describes other magnitudes, such as pressure, energy, etc., but not the velocity of the molecules
Indicate the correct option for the velocity distribution function of gas molecules:a) its velocity cannot be measured in any other way due to the small size of the gas moleculesb) it is only used to describe the velocity of particles if their density is very high.c) it describes the probability that a gas particle has a velocity in a given interval of velocitiesd) it describes other magnitudes, such as pressure, energy, etc., but not the velocity of the molecules
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Author:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell
Author:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell