EOC Chapter 11

# EOC Chapter 11 - CHAPTER 11 THEORIES OF COVALENT BONDING...

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CHAPTER 11 THEORIES OF COVALENT BONDING 11.1 Table 11.1 describes the types of shapes that form from a given set of hybrid orbitals. a) trigonal planar: three electron groups – three hybrid orbitals: sp 2 b) octahedral: six electron groups – six hybrid orbitals: sp 3 d 2 c) linear: two electron groups – two hybrid orbitals: sp d) tetrahedral: four electron groups – four hybrid orbitals: sp 3 e) trigonal bipyramidal: five electron groups – five hybrid orbitals: sp 3 d 11.2 a) sp 2 b) sp 3 c) sp 3 d d) sp 3 d 2 11.3 Carbon and silicon have the same number of valence electrons, but the outer level of electrons is n = 2 for carbon and n = 3 for silicon. Thus, silicon has 3 d orbitals in addition to 3 s and 3 p orbitals available for bonding in its outer level, to form up to 6 hybrid orbitals, whereas carbon has only 2 s and 2 p orbitals available in its outer level to form up to 4 hybrid orbitals. 11.4 Four , the same number of hybrid orbitals will form as the initial number of orbitals mixed. 11.5 The number of orbitals remains the same as the number of orbitals before hybridization. The type depends on the orbitals mixed. a) There are six unhybridized orbitals, and therefore six hybrid orbitals result. The type is sp 3 d 2 . b) Four sp 3 hybrid orbitals form from three p and one s atomic orbitals. 11.6 a) two sp orbitals b) five sp 3 d orbitals 11.7 To determine hybridization, draw the Lewis structure and count the number of electron groups. Hybridize that number of orbitals. a) The three electron groups (one double bond, one lone pair, and one unpaired electron) around nitrogen require three hybrid orbitals. The hybridization is sp 2 . NO b) The nitrogen has three electron groups (one single bond, one double bond, and one unpaired electron), requiring three hybrid orbitals so hybridization is sp 2 . O N O c) The nitrogen has three electron groups (one single bond, one double bond, and one lone pair so hybridization is sp 2 . O N O 11-1

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11.8 a) sp 2 O C O O 2 b) sp 2 CC O O O O 2 c) sp NC O 11.9 a) sp 3 The Cl has four electron groups (1 lone pair, 1 lone electron, and 2 double bonds) and therefore four hybrid orbitals are required; the hybridization is sp 3 . Note that in ClO 2 , the π bond is formed by the overlap of d -orbitals from chlorine with p -orbitals from oxygen. O Cl O b) sp 3 The Cl has four electron groups (1 lone pair and 3 bonds) and therefore four hybrid orbitals are required; the hybridization is sp 3 . OC l O O c) sp 3 The Cl has four electron groups (4 bonds) and therefore four hybrid orbitals are required; the hybridization is sp 3 . l O O O 11.10 a) sp 3 d FB r F F b) sp 3 Br O O 11-2
c) sp 3 d 2 F Br F F F F 11.11 a) Silicon has four electron groups (four bonds) requiring four hybrid orbitals; four sp 3 hybrid orbitals are made from one s and three p atomic orbitals .

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EOC Chapter 11 - CHAPTER 11 THEORIES OF COVALENT BONDING...

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