CHEM 103MODULE 5 EXAM

ivy [COMPANY NAME] [Company address] CHEM 103MODULE 5 EXAM CHEM 103 MODULE 5 EXAM 2022 Question 1 Click this link to access the Periodic Table. This may be helpful throughout the exam. Show the determination of the charge on the ion formed by the Se34 atom. Se34 (nonmetal = gain electrons) 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 4 gain 2e → Se -2 Question 2 Click this link to access the Periodic Table. This may be helpful throughout the exam. H = 2.1 Li = 1.0 Be = 1.5 B = 2.0 C = 2.5 N = 3.0 O= 3.5 F = 4.0 3 3 Na = 1.0 2.1 Mg = 1.2 S = 2.5 Al = 1.5 Cl = 3.0 Si = 1.8 P = K = 0.8 2.0 Ca = 1.0 Se = 2.4 Ga = 1.6 Br = 2.8 Ge = 1.8 As = Using the electronegativities from the table above, show the determination of the polarity of each different type of bond in the following molecule H-C bond electronegativity difference = 2.5 - 2.1 = 0.4 0.5 bond is Nonpolar C-O bond electronegativity difference = 3.5 - 2.5 = 1.0 1.6 - 0.5 bond is Polar Question 3 Click this link to access the Periodic Table. This may be helpful throughout the exam. On a piece of scratch paper, draw the Lewis structure for the ClO -1 ion. Then choose the correct Lewis structure for ClO -1 from the options listed below. (B) Question 4 Click this link to access the Periodic Table. This may be helpful throughout the exam. On a piece of scratch paper, draw the Lewis structure for H2SO3. Then choose the correct Lewis structure for H2SO3 from the options listed below. (D) Question 5 Click this link to access the Periodic Table. This may be helpful throughout the exam. Determine the electron geometry and explain your answer for the Si atom in SiH4. The Si atom in SiH4 has 4 groups of electrons around it in its Lewis structure, therefore, its electron geometry would be tetrahedral. Question 6 Click this link to access the Periodic Table. This may be helpful throughout the exam. Determine the hybridization and explain your answer for the Si atom in SiH4. The Si atom in SiH4 has 4 groups of electrons around it in its Lewis structure, therefore, its hybridization would be sp 3 . Question 7 Click this link to access the Periodic Table. This may be helpful throughout the exam. Determine the shape and explain your answer for HCN. The C atom in HCN has 2 groups of electrons around it in its Lewis structure, therefore, its electron geometry would be linear and since there are 2 atoms around the central C atom, the shape would be linear. Question 8 Click this link to access the Periodic Table. This may be helpful throughout the exam. H = 2.1 Li = 1.0 3.0 Be = 1.5 B = 2.0 O= 3.5 F = 4.0 C = 2.5 N = Na = 1.0 2.1 Mg = 1.2 Al = 1.5 S = 2.5 Cl = 3.0 Si = 1.8 P = K = 0.8 2.0 Ca = 1.0 Ga = 1.6 Se = 2.4 Br = 2.8 Ge = 1.8 As = Use the electronegativities above and your knowledge of the shape of PH3 to determine the molecular polarity of PH3 explaining your answer in detail. The shape of PH3 is triangular pyramid and since the P-H bonds are all nonpolar, PH3 would be nonpolar since all the bonds are nonpolar. Question 9 Click this link to access the Periodic Table. This may be helpful throughout the exam. Is KNO3 Polar, Ionic or Nonpolar and List and Explain whether it is Soluble or Insoluble in Water? KNO3 is Ionic since it has Ionic bonds and since it is Ionic it is Soluble in water. Question 10 Click this link to access the Periodic Table. This may be helpful throughout the exam. Arrange the following compounds in a vertical list from highest boiling point (top) to lowest boiling point (bottom) and explain your answer on the basis of whether the substance is Polar, Nonpolar, Ionic, Metallic or Hydrogen bonding: Ar, NH3, Zn, HBr, NaBr Please note in this question you are not being asked to list BPs but the compounds in a list from highest to lowest BP on the basis of the type ofcompound. NaBr (ionic) = Zn (metallic)NH3 (Hydrogen Bonding) HBr (Polar) Ar (Nonpolar)