1559Tch0101-1710/22/051:48Page6 6.chapter 1 STRUCTURE AND BONDNG IN ORGANIC MOLECULES 4 (group for C)-13 (half of shared c in bonds)+2 (unsharcd c)l=-1 (d)Proceed as for part(a).by considering simpler.analogous species.Suitable examples are ammo ds,an H-NHH-N-HH-g-H 3 (group for B)-[4 (half of shared e in bonds)]=-1 Therefore.we can arrive at the solution (below,center): H- H-0-H B-0 H-B-H H-6 (f)The oxygens are the same as in water,no problem there.The nitrogen is unusual:With two bonds and one lone pair it has no familiar analogs.Let's do the math: 5(group for N)-[2 (half of shared e-in bonds)+2(unsharede]=+1 The answer is H-6---H. 25.(a)(i)and (ii)Don't move any atoms!Resonance forms differ only in the location of the electron 0 gives .gives HO○o: HOO: favorable as a conributor to the hybrid.The first and third forms have only one charged atom and are the major contributors -H 沙1 (b)Construct on able Lewis structure first:HH All the(c) A new system, but not complex. You do not have a simpler species for comparison, so just do the calculation. 4 (group # for C) [3 (half of shared e in bonds)
2 (unshared e)]  1 It is a carbon anion, or carbanion. It is isoelectronic—it has the same number of valence electrons (5)—with neutral ammonia and positive hydronium ion. (d) Proceed as for part (a), by considering simpler, analogous species. Suitable examples are ammo￾nium ion for nitrogen with four bonds, and water for oxygen with two bonds and two lone pairs. Thus, we can write the answer (below, center): The species is called hydroxylammonium ion. (e) All three oxygen atoms are “normal”: two bonds and two lone pairs, and therefore all three are neutral. What about boron? We have seen two relevant examples: borane, BH3, neutral boron with three bonds, and borohydride, BH4 – , boron with four bonds and a negative charge, based upon the calculation 3 (group # for B) [4 (half of shared e in bonds)]  1 Therefore, we can arrive at the solution (below, center): (f) The oxygens are the same as in water, no problem there. The nitrogen is unusual: With two bonds and one lone pair it has no familiar analogs. Let’s do the math: 5 (group # for N) [2 (half of shared e in bonds)
2 (unshared e)] 
1 The answer is . 25. (a) (i) and (ii) Don’t move any atoms! Resonance forms differ only in the location of the electrons. The forms shown place the negative charge on two of the oxygen atoms. Continue with a Lewis structure that places the charge on the third oxygen: (iii) All three Lewis structures have octets on every large atom, but the middle structure has three charged atoms and two instances of plus–minus charge separations, making it relatively less favorable as a contributor to the hybrid. The first and third forms have only one charged atom and are the major contributors. (b) Construct one reasonable Lewis structure first: . All the atoms are neutral, so we can move electron pairs in a couple of ways to see what we get. Let’s begin by moving a pair of N H H O H C O O O O HO O HO
O HO gives gives H O N O H
H O H O H O H O B H H H B H
H H H H N H
H H H N O H O H 6 • Chapter 1 STRUCTURE AND BONDING IN ORGANIC MOLECULES 1559T_ch01_01-17 10/22/05 1:48 Page 6