Chem 103 exam 5 review questions and answers

Chem 103 exam 5 review questions and answers Score for this quiz: 92.5 out of 100 Submitted Apr 22 at 11:50pm This attempt took 72 minutes. Show the determination of the charge on the ion formed by the P15 atom. Your Answer: P15 = 1s2, 2s2, 2p6. 3s2,3p3 Gain 1e P15-1 P15 (nonmetal = gain electrons) 1s2 2s2 2p6 3s2 3p3 gain 3e → P-3 H = 2.1 I = 2.7 Li = 1.0 = 4.0 Be = 1.5 B = 2.0 C = 2.5 N = 3.0 O= 3.5 F Na = 1.0 = 3.0 Mg = 1.2 Al = 1.5 Si = 1.8 P = 2.1 S = 2.5 Cl K = 0.8 Br = 2.8 Ca = 1.0 Ga = 1.6 Ge = 1.8 As = 2.0 Se = 2.4 Using the electronegativities from the table above, show the determination of the polarity of each different type of bond in the following molecule Your Answer: H-0 3.5-2.1= 1.4 electronegativity difference ( bond is polar )= 1.6- 0.5 for polar range Br-0 2.8 -2.1= 0.7 electronegativity difference ( bond is polar ) = 1.6 - 0.5 for polar range H-O bond Polar electronegativity difference = 3.5 - 2.1 = 1.4 1.6 - 0.5 bond is Br-O bond Polar electronegativity difference = 3.5 - 2.8 = 0.7 1.6 - 0.5 bond is