Portage Learning - CHEM 103 MODULE 5
Exam
1. As33 (nonmetal = gain electrons) 1s2 2s2 What will be the charge of
the
2p6 3s2 3p6 4s2 3d10 4p3 gain 3 ion formed from each of these
’ As-3 atoms? Explain your
answer.
Ge32 (metal = lose electrons) 1s2 2s2 2p6
3s2 3p6 4s2 3d10 4p2 As33, Ge32
lose 4 ’ Ge+4
2. 1) Cl-O bond electronegativity
difference Using the electronegativity val-
= 3.5 - 3.0 = 0.5 ues from the equation bank,
(1.6 - 0.5 bond is Polar) show the determination of
H-O bond electronegativity difference = the polarity of each
3.5 - 2.1 = 1.4 different type of bond in the
(1.6 - 0.5 bond is Polar) following molecules:
2) CaN bond electronegativity difference =
3.0 - 2.5 = 0.5
(1.6 - 0.5 bond is Polar)
H-C bond electronegativity difference =
2.5 - 2.1 = 0.4
(< 0.5 bond is Nonpolar)
3. Your Answer: H
NH3 ..
N = 5, H = 3, 8 total H:N:H
1/
9
, .. Write Lewis dot structures
SO2 for the following molecules:
S = 6, O = 12, total = 18
.. .. .. NH3
: O : S :: O SO2
.. .. CH3O
H
CH3OH HNO2
C = 4, H = 4, O = 6, total = N2
14 H CH2O
.. ..
H:C:O:H
2/
9
Exam
1. As33 (nonmetal = gain electrons) 1s2 2s2 What will be the charge of
the
2p6 3s2 3p6 4s2 3d10 4p3 gain 3 ion formed from each of these
’ As-3 atoms? Explain your
answer.
Ge32 (metal = lose electrons) 1s2 2s2 2p6
3s2 3p6 4s2 3d10 4p2 As33, Ge32
lose 4 ’ Ge+4
2. 1) Cl-O bond electronegativity
difference Using the electronegativity val-
= 3.5 - 3.0 = 0.5 ues from the equation bank,
(1.6 - 0.5 bond is Polar) show the determination of
H-O bond electronegativity difference = the polarity of each
3.5 - 2.1 = 1.4 different type of bond in the
(1.6 - 0.5 bond is Polar) following molecules:
2) CaN bond electronegativity difference =
3.0 - 2.5 = 0.5
(1.6 - 0.5 bond is Polar)
H-C bond electronegativity difference =
2.5 - 2.1 = 0.4
(< 0.5 bond is Nonpolar)
3. Your Answer: H
NH3 ..
N = 5, H = 3, 8 total H:N:H
1/
9
, .. Write Lewis dot structures
SO2 for the following molecules:
S = 6, O = 12, total = 18
.. .. .. NH3
: O : S :: O SO2
.. .. CH3O
H
CH3OH HNO2
C = 4, H = 4, O = 6, total = N2
14 H CH2O
.. ..
H:C:O:H
2/
9