Rank the bonds from most polar to least polar. a. C-O, C-F, C-N b. C-a, C-I, C-B
ID: 1023904 • Letter: R
Question
Rank the bonds from most polar to least polar. a. C-O, C-F, C-N b. C-a, C-I, C-Br c. H-O, H-N, H-C d. C-H, C-C, C-N Draw a Lewis structure for each of the following: a. CH_3 CHO b. CH_3 OCH_3 c. CH_3 COOH Draw a skeletal structure for each of the compounds in Problem 58. What is the hybridization of the indicated atom in each of the following? a. CH_3 CH-CH_2 b. CH_3 CCH_3 c. CH_3 CH_2 Oh d. CH_3 C = N e. CH_3 CH = NCH_3 f. CH_3 OCH_2 CH_3 Predict the approximate bond angles for the following: a. the C-N-H bond angle in CH_3 CH_2 NH_2 b. the F-B-F bond angle in^- BF_4 c. the C-C-N bond angle in CH_3 C = N d. the C-C-N bond angle in CH_3 CH_2 NH_2Explanation / Answer
Q57.
Polarity --> Tendency of a molecule to form poles, and, in overall, a positive charged pole + negativeley charged pole. Therefore the substance is called "polar". This poles are created by electron lone pairs dsitributions in the geometry of the structure.
a)
Carbon has an electronegativity value of 2.5
Oxygen has a 3.5 value
Fluorine has 4.0
Nitrogen 3.0
Therefore, the most polar bondings are given by the HIGHEST difference in electronegativity
F-C = 4.0-2.5 = 1.5 (most polar)
O-C = 3.5-2.5 = 1.0 (second most polar)
N-C = 3.0-2.5 = 0.5 (least polar)
b.
Similar to a, find electronegativity values
Cl = 3.0
I = 2.5
Br = 2.8
Then,
Cl > Br > I; and this makes sense since the smaller the halogen, the stronger the electronegativity.
c.
Similar case as in A), but now we compare vs. H
H = 2.1 electronegativity value
so
C-H = 2.5-2.1 = 0.4 (least polar)
O-H = 3.5-2.1 = 1.0 (most polar)
N-H = 3.0-2.1 = 0.5 (second most polar)
d.
Once again, compare electronegativity differences
C-H = 2.5-2.1 = 0.4 (second most polar)
C-C = 2.5-2.5 (not even polar, this is nonpolar)
C-N = 3.0-2.5 = 0.5 (most polar)
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