Chemistry by Steven Zumdahl EOC Solutions -- Chapter 13
Chemistry by Steven Zumdahl EOC Solutions -- Chapter 13 Chemistry 142
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Popular in Chemistry
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Reviews for Chemistry by Steven Zumdahl EOC Solutions -- Chapter 13
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d n o b c i n o i n A . s m o t a o w t n e e r a l o p A . d n o b a n i ) s ( d n a 1 + f d o i r e p a s s o r c a t h g i r o t t f p u o r g i S < e G < n S . c O < B < a G . f a d e t c i d e r p s a e m a s ) 5 . 3 ( e m a s ) 0 . 3 ( l t n e r e f f i d ) 8 . 1 ( n S t n e r e f f i d ) 9 . 0 ( a r e t c a r a h e h t d n a s n o t o r p f o r e b m u n e h . s a g e l b o n a f o s n o r t c e l e f s e c r o f e h t f I . r e h t o h c a e _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ : d e m r o f s d n o B : n e k o r b s d n H l C : d e m r o f s d n o B : n e k o r b s d n N H : d e m r o f s d n o B : n e k o r b s d n C H N 2 N N F 532 44. . 45. C 1 : n e k o r b s d n o B 46. : d e m r o f s d n o B : n e k o r b s d n o C 4 2 ) l o m / J k 5 9 4 ( O = O 2 47. : d e m r o f s d n o B : n e k o r b s d n o C 2 C 1 ) l o m / J k 5 9 4 ( O = O 2 / 5 48. 49. : d e m r o f s d n o B : n e : ) * ( d e m r o f s d n o B : ) * ( n e k : d emr o fs dn:n e ko r b C 3 × N N 1 : d e m r o f s d n o B : n e k o r b s d n o C 3 × N 3 × O : d e m r o f s d n o B : n e k o r b s d n C C C 1 _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ m r o f e l b a t s t s o m e h t t o n e r : d e m r o f s d n e 4 s e s u e . s n o r t c e l e e c n e l a v 8 = 7 + e r u t c u r t s s i w e L e r u t c u r t s e h t e v a h o t n o s a e r o n s i e r 540 CHAPTER 13 BONDING: GENERAL CONCEPTS ▯ N C O . b has 6 + 4 + 5 + 1 = 16 valence electrons. We can draw three resonance structures for OCN▯. OCN OCN OCN ▯ N C S has 6 + 4 + 5 + 1 = 16 valence electrons. Three resonance structures can be drawn. SCN SCN SCN N ▯ has3(5)+1=16valenceelectrons. AswithOCN -andSCN , three different 3 resonance structures can be drawn. NN N NN N NN N 61. Ozone: O3has 3(6) = 18 valence electrons. Two resonance structures can be drawn. OOO OO O Sulfur dioxide: 2Ohas6+2(6)=18valenceelectr ons. Two resonance structures are possible. OS O OS O Sulfur trioxide:3SOs6+3(6)=24valenceelectrons. Threeresonancestructuresare possible. O O O S S S O O O O O O 62. PAN (H C NO ) has 3(1) + 2(4) + 5 + 5(6) = 46 valence electrons. 3 2 5 H O This is the skeletal structure with complete octets O about oxygen atoms (46 electrons used). HC CO O N O H CHAPTER 13 BONDING: GENERAL CONCEPTS 541 This structure has used all 46 electrons, but there are only six electrons around one of the carbon atoms and the nitrogen atom. Two unshared pairs must become shared; i.e., we must form double bonds. H O H O O O HC CO O N HC C O O N H O H O H O O HC CO O N (last form not important) H O 63. CH3NCO has 4 + 3(1) + 5 + 4 + 6 = 22 valence electrons. The order of the elements in the formula give the skeletal structure. H H H HC NCO HC NC O HC NCO H H H 64. Resonance occurs when more than one valid Lewis structure can be drawn for a particular molecule. A common characteristic of resonancestructures is a multiple bond(s) that moves from one position to another. We say the electrons in the multiple bond(s) are delocalized in the molecule. This helps us rationalize why the bonds in a molecule that exhibit resonance are all equivalent in length and strength. Any one of thes indicates different types of bonds within that molecule. This is not correct, hence none of the individual resonance structures are correct. We think of the actual structure as an average of all the resonance structures; again, this helps explain the equivalent bonds within the molecule that experiment tells us we have. 65. Benzene has 6(4) + 6(1) = 30 valence electrons. Two resonance structures can be drawn for benzene. The actual structure of benzene is anaverage of these two resonance structures; i.e., all carbon-carbon bonds are equivalent with a bond length and bond strength somewhere between a single and a double bond. H H H H H H C C C C C C C C C C C C H H H H H H 542 CHAPTER 13 BONDING: GENERAL CONCEPTS 66. We will use a hexagon to represent the six-membered carbon ring, and we will omit the four hydrogen atoms and the three lone pairs of electrons on each chlorine. If no resonance exists, we could draw four different molecules: Cl Cl Cl Cl Cl Cl Cl Cl If the double bonds in the benzene ring exhibit resonance, then we can draw only three different dichlorobenzenes. The circle in the hexagon in the following illustrations represent the delocalization of the three double bonds in the benzene ring (see Exercise 13.65). Cl Cl Cl Cl Cl Cl With resonance, all carbon-carbon bonds are equivalent. We can’t distinguish between a single and double bond between adjacent carbons that have a chlorine attached. That only three isomers are observed supports the concept of resonance. 67. Borazine (3N3H6) has 3(3) + 3(5) + 6(1) = 30 valence electrons. The possible resonance structures are similar to those of benzene in Exercise 13.65. H H H H H H B B N N N N B B B B N N H H H H H H O N , d n o b e l b u o d a n a h t r e g n h t , n w a r d e b n a c s e r u t c u r t s CHAPTER 13 BONDING: GENERAL CONCEPTS 545 F l C , 7 + 3(7) = 28 e rB , 3(7) + 1 = 22 e 3 3 F Br Br Br Cl F F Row 3 and heavier nonmetals can have more than 8 electrons around them when they have to. Row 3 and heavier elements have empty d orbitals that are close in energy to valence s and p orbitals. These empty d orbitals can accept extra electrons. For example, P in P5 has its five valence electrons in the 3s and 3p orbitals. These s and p orbitals have room for three more electrons, and if it has to, P can use the empty 3d orbitals for any electrons above 8. 72. SF6, 6 + 6(7) = 48 e F lC 5, 7 + 5(7) = 42 e F F F F F F S Cl F F F F F F e X , 8 + 4(7) = 36 e 4 F F Xe F F 73. CO 3▯ has 4 + 3(6) + 2 = 24 valence electrons. 2- 2- 2- O O O C C C O O O O OO Three resonance structures can be drawn for C32. The actual structure for CO3▯isan average of these three resonance structures. That is, the three Cond lengths are all equivalent, with a length somewhere between a single and a double bond. The actual bond length of 136 pm is consistent with this resonance view o. CO 3 O d n a C n e e w t e b d n o b 3 / 1 1 t s e t r o h s m o r f r e d r o e h t , s e 548 CHAPTER 13 BONDING: GENERAL CONCEPTS NN O NN O Assigning formal charges for all three resonance forms: NN O NN O NN O -1 +1 0 0 +1 -1 -2 +1 +1 For: N , FC = 5 - 4 - 1/2(4) = -1 N , FC = 5 - 1/2(8) = +1 , Same for N and N , N FC = 5 - 6 - 1/2(2) = -2 ;N, FC = 5 - 2 - 1/2(6) = 0 , , O FC = 6 - 4 - 1/2(4) =;0 O FC = 6 - 6 - 1/2(2) = -1 O , FC = 6 - 2 - 1/2(6) = +1 We should eliminate N ‒N ≡O since it has a formal charge of +1 on the most electronegative element (O). This is consistent with the observation that ‒N bond is between a double and triple bond, and that the Nbond is between a single and double bond. 2▯ ▯ 3▯ 79. See Exercise 13.58a for the Lewis structures of P3ClO4 , ClO4and PO 4 . All of these compounds/ions have similar Lewis structures to those of2SO2Cl and X4O shown below. 2▯ l C O P . a 3: P, FC = 5 ▯ 1/2(8) = +1 b. SO 4 : S, FC = 6 ▯ 1/2(8) = +2 ▯ 3▯ O l C . c 4: Cl, FC = 7 ▯ 1/2(8) = +3 d. PO 4 : P, FC = 5 ▯ 1/2(8) = +1 O S . e Cl , 6 + 2(6) + 2(7) = 32 e .fOX , 8 + 4(6) = 32 e 2 2 4 O O Cl S Cl OeXO O O 6 = C F , S ▯ 1/2(8) = +2 Xe, FC = 8 ▯ 1/2(8) = +4 e z i m i . S f o e g r a h c l a m r o f s i e c o e o t P o t s d n o b e v i f m r o f t s u . e g r a h c l a m r o f e z i m i n : N r o f e g r a h c l a m r o f o r e z a n d l a r d e h a r t e . e l o p i d t n e n a m r e p o n s a H . b . e l o p i d t n e n a m r e p a s a H . d . e l o p i d t n e n a m r e p o n s a H . f H P . b ° 0 8 1 , r a e n i l : N ° 5 . 9 0 1 ; l a r d e h a r t e t e r a l ° 5 . 9 0 1 < ; d i m a r y p l a n o g i r t ° 5 . 9 0 1 < ; d e p a h s - V CHAPTER 13 BONDING: GENERAL CONCEPTS 553 c. PCl 3as 5 + 3(7) = d. SCl h2s 6 + 2(7) = 26 valence electrons. 20 valence electrons P Cl Cl S Cl Cl Cl Trigonal pyramid; all angles are <109.5°. V-shaped; angle is <109.5°. e. SiF 4has 4 + 4(7) = 32 valence electrons. F Si Tetrahedral; all angles are 109.5°. F F F Note: In PCl , SCl , and SiF , there are four pairs of electrons about the central atom in each 3 2 4 case. All the structures are based on a tetrahedral geometry, but only SiF 4 hasatetrahedral structure. We consider only the relative positions of the atoms when describing the molecular structure. 89. From the Lewis structures (see Exercise 13.71), Br 3▯would have a linear molecular structure, ClF 3 would have a T-shaped molecular structure, and SF wo4ldhaveasee-saw molecular structure. For example, consider ClF 3 (28 valence electrons): F The central Cl atom is surrounded by five electron pairs, which requires a trigonal bipyramid geometry. Since there are three bonded atoms and two lone pairs of electrons about Cl, we describe the Cl F molecular structure of ClF 3asT-shapedwithpredictedbondangles F of about 90°. The actual bond angles will be slightly less than90° due to the stronger repulsive effect of the lone pair electrons as compared to the bonding electrons. 90. From the Lewis structures (see Exercise 13.72), XeF would have a square planar molecular 4 structure and ClF5 would have a square pyramid molecular structure. 91. a. XeCl has 8 + 2(7) = 22 valence electrons. 2 Cl Xe Cl 180 o There are five pairs of electrons about the central Xe atom. The structure will be based on a trigonal bipyramid geometry. The most stable arrangement of the atoms in XeCl 2 is a linear molecular structure with a 180° bond angle. . s n o r t c e l e e c n e l a v . s n o r t , d i CHAPTER 13 BONDING: GENERAL CONCEPTS 555 93. Let us consider the molecules with three pairs of electrons around the central atom first; these molecules are SeO and SeO , and both have a trigonal planar arrangement of electron 3 2 pairs. Both these molecules have polar bonds, but only SeO 2hasanoverallnetdipole moment. The net effect of the three bond dipoles from the three polar Se ‒O bonds in SeO 3 will be to cancel each other out when summed together. Hence SeO 3isnonpolarsincethe overall molecule has no resulting dipole moment. In SeO , the two Se‒O bond dipoles do not 2 cancel when summed together; hence SeO hasanet2ipolemoment(ispolar). SinceOis more electronegative than Se, the negative endof the dipole moment is between the two O atoms, and the positive end is around the Se atom. The arrow in the following illustration represents the overall dipole moment in SeO .2Note that to predict polarity for SeO , 2ither of the two resonance structures can be used. Se O O The other molecules in Exercise 13.88 (PCl 3 SCl 2 and SiF 4 have a tetrahedral arrangement of electron pairs. All have polar bonds; in SiF 4 theindividualbonddipolescancelwhen summed together, and in PCl 3and SCl 2he individual bond dipoles do not cancel. Therefore, SiF4 has no net dipole moment (is nonpolar), and PCl 3 and SCl 2ave net dipole moments (are polar). For PCl , the negative end of the dipole moment is between the more electronegative 3 chlorine atoms, and the positive end is around P. For SCl 2, the negative end is between the more electronegative Cl atoms, and the positiv e end of the dipole moment is around S. 94. The molecules in Exercise 13.91 (XeCl 2, ICl3, TeF4, and PCl 5 all have a trigonal bipyramid arrangement of electron pairs. All of these molecules have polar bonds, but only TeF 4and ICl3 have dipole moments. The bond dipoles from the five P ‒Cl bonds in PCl ca5cel each other when summed together, so PCl has 5o dipole moment. The bond dipoles in XeCl also 2 cancel: Cl Xe Cl Since the bond dipoles from the two Xe ‒Cl bonds are equal in magnitude but point in opposite directions, they cancel each other, and XeCl 2hasnodipolemoment(isnonpolar). For TeF and ICl , the arrangement of these molecules is such that the individual bond dipoles 4 3 do not all cancel, so each has an overall net dipole moment. The molecules in Exercise 13.92 (ICl 5, XeCl 4 and SeCl )6all have an octahedral arrangement of electron pairs. All of these molecules have polar bonds, but only ICl has an overall 5 dipole moment. The six bond dipoles in SeCl 6all cancel each other, so SeCl 6as no dipole moment. The same is true for XeCl 4 Cl Cl Xe Cl Cl l e c n a c t o n o d s d n o b r a l o p e ° 0 CHAPTER 13 BONDING: GENERAL CONCEPTS O C 2 and COS both have linear molecular structures with a 180 because the individual bond dipoles cancel each other out, but COS is polar. By replacing an O with a less electronegative S atom, the molecule is not symmetric any more. The individual bond dipoles do not cancel because the C ▯ S bond dipole is smaller than the C resulting in a polar molecule. 96. a. XeCl 4, 8 + 4(7) = 36 e XeCl Cl Cl Xe Cl Cl r a l opnon , ° 081 , r a enr iaLl opnon , ° 09 , r ana l p e r auqS Both compounds have a central Xe atom and terminal Cl atoms, and both compounds do not satisfy the octet rule. In addition, both are nonpolar because the Xe and lone pairs around Xe are arranged in such a manner that they cancel each other out. The last item in common is that both have 180° bond angles. Although we haven’t emphasized this, the bond angle between the Cl atoms on the diagonal in XeCl apart from each other. b. We didn’t draw the Lewis structures, but all are polar covalent compounds. The bond dipoles do not cancel out each other when summed together. The reason the bond dipoles are not symmetrically arranged in these compounds is that they all have at least one lone pair of electrons on the central atom, which disrupts the symmetry. Note that there are molecules that have lone pairs and are nonpolar, e.g., XeCl problem. A lone pair on a central atom does not guarantee a polar molecule. 97. Only statement c is true. The bond dipoles in CF4and KrF are arranged in a manner that they all cancel each other out, making them nonpolar molecules (CF structure, whereas KrF 4has a square planar molecular structure). In SeF this see-saw molecule do not cancel each other out, so SeF molecules have either a trigonal planar geomerty or a trigonal bipyramid geometry, both of which have 120° bond angles. However, XeCl 2has three lone pairs and two bonded chlorine atoms around it. XeCl h2salinearmolecularstructurewitha180°bondangle.Withthree lone pairs, we no longer have a 120° bond angle in XeCl shaped molecular structure with a bond angle of about 120°. CS angle and SCl 2isV-shapedbutwithanapproximate109.5°bondangle.Thethreecom- pounds do not have the same bond angle. For statement d, central atoms adopt a geometry to minimize electron repulsions, not maximize them. 98. EO 3▯ is the formula of the ion. The Lewis structure has 26 valence electrons. Let number of valence electrons of element E. 26 = x + 3(6) + 1, x= 7 valence electrons l C O ; r a l o e m o m e l o p i d g n i t l u s e r e h T . g n i w a r d e h t ; r a l o p , d i m a r y p l a n o g . l e c n a c t o n o d s e l o p i d d n o b ; r a l o p , . l e c n a c t o n o d s e l ; r a l o p n o n , d i m a r y p i b l a n o g i . l e c n a c s e l o p i d d n o b . a l a u q e e v a h d n a h g u o n e e g r a l e r a e 562 2▯ 105. CO 3 has 4 + 3(6) + 2 = 24 valence electrons. 2- O C O O ▯ OCH 3 has 1 + 4 + 3(6) + 1 = 24 valence electrons. - H O C O O H 2CO 3as 2(1) + 4 + 3(6) = 24 valence electrons. O C O O H H : e r as t cudo r pdna s t na t c a eeh t r o f s e r u t cu r ts iweL ehT O C O O H H : demr o fs d:neko r b s dnoB C 2 ‒O (358 kJ/mol) O 1 ‒H (467 kJ/mol) ΔH = 2(358) + 467799 + 467) = two carbon-oxygen single bonds; hence CO 106. TeF5has 6 + 5(7) + 1 = 42 valence electrons. - F F F Te F F CHAPTER 13 BONDING: GENERAL CONCEPTS The lone pair of electrons around Te exerts a stronger repulsion than the bonding pairs, pushing the four square-planar F's away from the lone pair and thus reducing the bond angles between the axial F atom and the square-planar F atoms. 107. As the halogen atoms get larger, it becomes more difficult to fit three halogen atoms around the small nitrogen atom, and the NX3 108. XeF 2l 2 8 + 2(7) + 2(7) = 36 e Cl F Cl Xe Xe Cl F F Polar Nonpolar The two possible structures for XeF2 apart from each other, and the Cl atoms are also 90° apart. The individual bond dipoles would not cancel in this molecule, so this molecule is polar. In the second possible structure the F atoms are 180° apart, as are the Cl atoms. Here, the bond dipoles are symmetrically arranged, so they do cancel out each other, and this molecule is nonpolar. Therefore, measurement of the dipole moment would differentiate between the two compounds. These are different compounds and not resonance structures. 109. The stable species are: r B a N n I : r B a N . a 2, the sodium ion would have a 2+ charge, assuming that each bromine has a 1▯ charge. Sodium doesn’t form stable Na compounds. ▯ O l C . b 4 : ClO4has 31 valence electrons, so it is impossible to satisfy the octet rule for all atoms in ClO4. The extra electron from the 1 plete octets for all atoms. O e X . c 4 We can’t draw a Lewis structure that obeys the octet rule for SO (30 electrons), unlike with Xe4 (32 electrons). F e S . d 4: Both compounds require the central atom to expand its octet. O is too small and doesn’t have low-energy d orbitals to expand its octet (which is true for all Row 2 elements). 110. If we can draw resonance forms for the anion after loss of H stability of the anion causes the proton to be more readily lost, i.e., makes the compound a better acid. . a - - O HCO 564 CHAPTER 13 BONDING: GENERAL CONCEPTS b. - O O O O CH CHCCH C 3 3 CH 3 CHCCHC 3 . c O O O O O In all three cases, extra resonance forms canbe drawn for the anion that are not possible when the H +is present, which leads to enhanced stability. + ▯ ▯ + 111. a. Radius: N < N < N ; IE: N < N < N + ▯ N has the fewest electrons held by tseven protons in the nucleus whereas N has the most electrons held by the seven protons. The seven protons in the nucleus will hold the + ▯ + electrons most tightly in Nnd least tightl▯ in N . Therefore, N has the smallest radius with the largest ionization energy (IE), and Nthe largest species with the smallest IE. + ▯ ▯ + b. Radius: Cl < Cl < Se < Se ; IE: Se < Se < Cl < Cl The general trends tell us that Cl has a smaller radius than Se and a larger IE than Se. + Cl , with fewer electron-electron repulsions than Cl,willbesmallerthanClandhavea larger IE. Se , with more electron-electron repulsions than Se, will be larger than Se and have a smaller IE. c. Radius: Sr < Rb < Br ; IE: Br < Rb < Sr 2+ 2+ These ions are isoelectronic. The species with the most protons (Sr ) will hold the electrons most tightly and will have the smallest radius and largest IE. The ion with the ▯ fewest protons (Br ) will hold the electrons least tightly andwillhavethelargestradius and smallest IE. CHAPTER 13 BONDING: GENERAL CONCEPTS 112. This molecule has 30 valence electrons. The only C–N bond that can possibly have a double bond character is the N bound to the C with O attached. Double bonds to the other two C–N bonds would require carbon in each case to have 10valenceelectrons(whichcarbonnever does). O H O H HCNCH HCNCH H H H CH H CH H H 1molF 113. Assuming 100.00 g of compound: 42.81 g F = 2.253 mol F 19.00g F 1molX FX n i X f o s e l om f o r ebmun ehT 5is: 2.253 mol F × = 0.4506 mol X 5molF This number of moles of X has a mass of 57.19 g (= 100.00 g – 42.81 g). The molar mass of X is: 57.19g X = 126.9 g/mol; This is element I. 0.4506 molX F I , 7 + 5(7) = 42 e 5 F F F The molecular structure is square pyramid. I F F 114. For carbon atoms to have a formal charge of zero, each C atom must satisfy the octet rule by forming four bonds (with no lone pairs). For nitrogen atoms to have a formal charge of zero, each N atom must satisfy the octet rule by forming three bonds and have one lone pair of electrons. For oxygen atoms to have a formal charge of zero, each O atom must satisfy the octet rule by forming two bonds and have two lone pairs of electrons. With these bonding requirements in mind, then the Lewis structure of histidine, where all atoms have a formal charge of zero, is: HC N CH C NH 2 H H C 1 H O H CN C H H O n e e w e l g n a d n o b ° 0 2 1 e l g n a d n s m o t a O n e e w t e b s m o t a O a r i a . CHAPTER 13 Challenge Problems 117. KrF2, 8 + 2(7) = 22 e ; from the Lewis structure, we have a trigonal bipyramid arrangement of electron pairs with a linear molecular structure. Hyperconjugation assumes that the overall bonding in KrF is a combination of covalent and ionic contributions (see Section 13.12 of the text for discussion of hyperconjugation). Using hyperconjugation, two resonance structures are possible that keep the linear structure. 118. The skeletal structure of caffeine is: : e r a s t neme r i uqe rgn i dnob eh t , smo t a l l a no o r e zf o eg r ahc l amr o f a r oF a. Four bonds and no lone pairs for each carbon atom mo t a nego r t i n hc a e r o f r i ap eno l eno dna s dnob e e r hT . b eno l owt dna s dnob owT . c d. One bond and no lone pairs for each hydrogen atom ) g ( i L 2 ) s ( i L 2 ) g ( lCH 2 e 2 + ) g ( lC 2 ) g (H 2 _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ ) g ( lCH 2 + : e r a s e r u t c u r t s y l e k i l t s o m . y l e k i l t s o m e r a . ° 0 2 1 ( e v i t a g e e v i t a g e n , E L : r e h t e g o t e m o c o . ) c i m r e h t o x e ( e v i t a g e n e b t _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ N l o m 2 1 . 3 = g 1 0 . 4 1 / l o m 1 O l o m 2 1 . 3 = g 0 0 . 6 1 / l o m 1 H l o m 5 2 . 6 = g 8 0 0 . 1 / l o m 1 = 574 CHAPTER 13 BONDING: GENERAL CONCEPTS For a correct molar mass, the molecular formula of compound2D2i4 N O H4or2NH NO . A Lewis structure is: + - H N HN H O O H Note: One more resonance structure for2NO can be drawn. Compound E: A basic solution (part g) that is commercially availaMl(part c) is ammonium hydroxide (NH 4H). This is also consistent with the information given in parts b and d. The Lewis structure for NH OH is: 4 + H - HN H OH H
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