Chapter 3 Evens

Chapter 3 Evens - 3.6 I. + H :F:... 'F' F2 3. / . \Xe. .- /...

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Unformatted text preview: 3.6 I. + H :F:... 'F' F2 3. / . \Xe. .- / ' .F. (a) The shape of the XeF5+ ion is square pyramidal based upon an octahedral arrangement of electrons about the Xe atom. (b) There are two types of F~—Xe~F e bond angles. (c) The F——Xe—F bond angles are approximately 90° or 180°. 3.20 The Lewis structures are (a) (b) :61: \ . | B—F .gl__P: .- / :91: (c) iozsizofi (d) ’3" H/ \H Molecules (b) and ((1) will be polar; (a) and (c) will be nonpolar. 3.22 (a) (b) Ell T1 T T l Cl—T—Cl C1 H H H tetrachloromethane 2-propanol nonpolar polar (c) H O acetone polar . \ n . _ . /P—}§.. />§P F .13; I -'F :Br: " '. no lone pairs on central atom; two lone pairs on central atom; trigonal bipyramidyl; 90°, 120O T—shaped; 90°, 1800 (c) u + (d) F= ~ \‘1' a;- C F 0. O. .0 . ' '15: ..F../ I..F../ l O. : .13 1 no lone pairs on central atom; two lone pairs on central atom; trigonal bipyramidyl; 90°, 120O T—shaped; 90°, 1800 (e) 0 _ 2 (I): Q—llafi : 9: one lone pair on central atom trigonal pyramidyl; 1070 3.80 (a) The helium and hydrogen atoms have only their ls orbitals available for bonding. Combination of these two will lead to a 0' and a 0"“ pair of orbitals. The most stable species will be one in which only the 0' orbital is filled. Thus we need a species that has only two electrons. The charge on this species will be +1. (b) The maximum bond order will be 1. (c) Adding one electron will decrease the bond order by 1/2 because an antibonding orbital will be populated. Taking an electron away will also decrease the bond order by 1/2 because it will remove one bonding electron. ” ‘\ He ls 3.84 The three possible forms that maintain a valence of 4 at carbon are 3.86 cyclopropene, allene, and prOpyne, which have the structures H\ ,H H C l H H g H—c—-c—:c——H c_—c_—~c /C Q l / H H H H H The C—C—C bond angles in cyclopropene are restricted by the cyclic structure to be approximately 60°, which is far from the ideal value of 109.47° for an S193 hybridized carbon atom, or 120° for an 5192 hybridized C atom. Consequently, cyclopropene is a very strained molecule that is extremely unstable. The H—C—H bond angle at the CH2 group would be expected to be 109.470, but it is actually larger due to the narrow C—C—C bond angle. The carbon of the CH2 group is Sp3 hybridized, whereas the other two carbon atoms are sz. Likewise the C=C—H angles, which would normally be expected to be 120° due to sz hybridization at C, are somewhat larger. In allene, the middle carbon atom is Sp hybridized; the two end carbons are sz hybridized. The C—C—C bond angle is expected to be 180°; the H—C—H bond angles should be close to 120°. 1n propyne, one end carbon is Sp3 hybridized (109.5° angles); the other two C atoms are Sp hybridized, with 180° bond angles. The three structures are not resonance structures of each other as they have a different spatial arrangement of atoms. For two structures to be resonance forms of each other, only the positions of the electrons may be changed. These two compounds would be known as isomers (see section 18.3). (a) H H H H H H / \C:C< >C:C:C< >C:C:C:C H/ H H H H H (b) The hybridization at the atoms attached to two hydrogen atoms is sz, whereas that at the carbon atoms attached only to two other carbon atoms is Sp. (c) Double bonds connect all of the carbon atoms to each other. ((1) The HwC—H and C—C—H angles should all be ca. 120°. The C—C—C angles will all be 180°. 3.96 (a) The structure of caffeine and the hybridization of each non—hydro gen atom is shown below. For the sake of clarity all lone pairs of electrons have been left out; each N in the structure has one lone pair not shown, while each 0 has two lone pairs not shown: S 2 Sp 3 O 1? CH3 Sp 3 H / CH 2 3 Sp / C2\ 517/ Nsp3 N sp C \ SP2 I 2 H C — H SP2 CSP C // 0% \ /sp2\ N 2 SP3 I Sp CH3 (b) the bond angles around any atom marked Sp3 should be 1095" while the bond angles around any atom marked Sp2 should be 120°. 3.112 (a) The Lewis structure of hydrogen peroxide is H — — H ; no atoms have formal charges and the oxidation number of each 0 is —l . Oxidation number is more useful than formal charge in predicting the oxidizing ability of a compound as this will give us a clearer idea of whether a given atom wants to gain or lose an electron. (b) The bond angles should be about the same as that for water, ca. 104°. All atoms in this molecule are in the same plane but because of unrestricted rotation around the 0—0 bond it is expected that the hydro gens (and therefore the lone pairs of electrons on the oxygens) can rotate in and out of the plane; one of these rotations will lead to a polar structure (the different forms, arising from rotations around a single 0' bond, are called rotomers). Because of this, the molecule is expected to be polar, but slightly less so than water. .é/ .. H H H / I / \O/O\ H\O/O‘////: H\O/O‘\'u/H I , . . c 3 I . . planar, nonpolar rotomer 1313113r , p013r TOtomer nonplanar polar rotomer 7 (one of many) ...
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Chapter 3 Evens - 3.6 I. + H :F:... 'F' F2 3. / . \Xe. .- /...

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