Unit 3 Qz 1 (Chapter 10, Structure of Water)

The molecular formula for water is H2O. The first step to getting the structural formula is to find the central atom. The two criteria for finding the central atom are that it should be the least abundant atom, and the atom that can form the most number of bonds. The least abundant atom is _____ . From the periodic table, H can form only _____ bond, while O can form _____ bonds. Thus, the central atom in water is predicted to be _____ . The _____ are then placed around the central atom. a. Hydrogen b. Oxygen c. 0 d. 1 e. 2 f. 3 g. 4 h. 5

The molecular formula for water is H2O. The first step to getting the structural formula is to find the central atom. The two criteria for finding the central atom are that it should be the least abundant atom, and the atom that can form the most number of bonds. The least abundant atom is __b__ . From the periodic table, H can form only __d__ bond, while O can form __e__ bonds. Thus, the central atom in water is predicted to be __b__ . The __a__ are then placed around the central atom. Answer Key: b, d, e, b, a

Let's review what we learned about the Lewis structure for water. A water molecule has _____ Regions of Electron Density, REDs. There are _____ REDs that represent bonds to H atoms, and _____ REDS that represent non-bonding electron pairs. All 4 REDs repel/attract _____ each other. To get the REDs as far apart from one another as possible, water ends up as a tetrahedron with bond angles of _____ º. The x-ray diffractometer is designed to see atoms, and not electron clouds. Thus, the x-ray diffractometer see the water molecule as _____ . a. 1 b. 2 c. 3 d. 4 e. 5 f. 6 g. attract h. repel i. 45 j. 90 k. 109.5 l. 120 m. 180 n. diatomic o. bent p. planar trigonal q. tetrahedral r. pyramidal s. linear

Let's review what we learned about the Lewis structure for water. A water molecule has __d__ Regions of Electron Density, REDs. There are __b__ REDs that represent bonds to H atoms, and __b__ REDS that represent non-bonding electron pairs. All 4 REDs repel/attract ___h_ each other. To get the REDs as far apart from one another as possible, water ends up as a tetrahedron with bond angles of __k__ º. The x-ray diffractometer is designed to see atoms, and not electron clouds. Thus, the x-ray diffractometer see the water molecule as __o__ . Answer Key: d, b, b, h, k, o

Once the central atom is determined, and thus the bonding arrangement, the type of bonds must be determined. Since every atom in H2O is a nonmetal, _____ bonds are drawn between nonmetals. A covalent bond is designated by a _____ , while an ionic bond is designated by _____ . H2O has only _____ bonds. a. ionic b. covalent c. no bonds d. dash e. (+) and/or (-) charges

Once the central atom is determined, and thus the bonding arrangement, the type of bonds must be determined. Since every atom in H2O is a nonmetal, __b__ bonds are drawn between nonmetals. A covalent bond is designated by a __d__ , while an ionic bond is designated by __e__ . H2O has only __b__ bonds. Answer Key: b, d, e, b

The Lewis Electron Dot structure for water shows 4 Regions of Electron Density, REDs, around the central O atom. Two of the REDs come from bonding electron pairs shared between the O and the two H atoms. The other two REDs come from the nonbonding electron pairs on the O atom. Electron pairs repel one another and this is what determines a molecules shape. The perfect geometric shape containing 4 "points" would be a tetrahedron with 4 sides. The bond angles in the perfect tetrahedron would be 109.5º. Water is not perfect since the 4 points are not the same, but it gives us an approximation. The machine used to determine molecular shape is called the x-ray diffractometer. This machine sees atoms but not nonbonding electron pairs. The machine sees water as a bent shape with bond angles approximately 109.5º. Water has _____ REDs, _____ are from bonding electron pairs and _____ are from nonbonding electron pairs. The shape is _____ and the approximate bond angles are _____ º. a. 0 b. 1 c. 2 d. 3 e. 4 f. 5 g. 6 h. atoms i. diatomic j. planar trigonal k. tetrahedral l. pyramidal m. bent n. 109.5 o. 120 p. 90 q. 180 r. 60

The machine used to determine molecular shape is called the x-ray diffractometer. This machine sees atoms but not nonbonding electron pairs. The machine sees water as a bent shape with bond angles approximately 109.5º. Water has __e__ REDs, __c__ are from bonding electron pairs and __c__ are from nonbonding electron pairs. The shape is __m__ and the approximate bond angles are __n__ º. Answer Key: e, c, c, m, n

Using the first steps, a water molecule is shown as H - O - H. Next, the number of valence electrons is determined. All bonds are comprised of valence electrons. The goal of each atom in a molecule is to attain happiness. Small atoms try to attain the duet representing 1s2. Larger atoms try to attain the octet representing s2 p6. The total number of valence electrons in a water molecule is _____ . There are _____ valence electrons from the single O, and _____ valence electron from each of the two H atoms. When we distribute valence electrons into the molecule we put them first into the _____ , then onto the _____ , and finally onto the _____ . We see that this satisfies the octet rule for O and the duet rule for each H. a. 0 b. 1 c. 2 d. 3 e. 4 f. 5 g. 6 h. 7 i. 8 j. 9 k. 10 l. central O atom m. outer H atoms n. bonds

Using the first steps, a water molecule is shown as H - O - H. Next, the number of valence electrons is determined. All bonds are comprised of valence electrons. The goal of each atom in a molecule is to attain happiness. Small atoms try to attain the duet representing 1s2. Larger atoms try to attain the octet representing s2 p6. The total number of valence electrons in a water molecule is __i__ . There are __g__ valence electrons from the single O, and __b__ valence electron from each of the two H atoms. When we distribute valence electrons into the molecule we put them first into the __n__ , then onto the __m__ , and finally onto the __l__ . We see that this satisfies the octet rule for O and the duet rule for each H. Answer Key: i, g, b, n, m, l