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《结构化学》第三章习题答案

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《结构化学》第三章习题答案

《结构化学》第三章习题答案3001(A,C)3002Hab=∫a[-2--+]bd=EHSab+Sab-∫abd=EHSab+K因EH=-13.6eV,Sab为正值,故第一项为负值;在分子的核间距条件下,K为负值。所以Hab为负值。3003∫gud=(4-4S2)-1/2∫(+)((-)d=(4-4S2)-1/2∫[2-2]d=(4-4S2)-1/2[1-1]=0故相互正交。3004(C)3006描述分子中单个电子空间运动状态的波函数叫分子轨道。两个近似(1)波恩-奥本海默近似(核质量>>电子
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导读《结构化学》第三章习题答案3001(A,C)3002Hab=∫a[-2--+]bd=EHSab+Sab-∫abd=EHSab+K因EH=-13.6eV,Sab为正值,故第一项为负值;在分子的核间距条件下,K为负值。所以Hab为负值。3003∫gud=(4-4S2)-1/2∫(+)((-)d=(4-4S2)-1/2∫[2-2]d=(4-4S2)-1/2[1-1]=0故相互正交。3004(C)3006描述分子中单个电子空间运动状态的波函数叫分子轨道。两个近似(1)波恩-奥本海默近似(核质量>>电子
《结构化学》第三章习题答案

3001    ( A, C )                                                          

3002    Hab =∫a[-2-  -  + ] bd                     

           =EHSab +  Sab - ∫ab d                        

           = EHSab + K                                             

        因 EH= -13.6eV, Sab 为正值,故第一项为负值; 在分子的核间距条件下, K为负值。 所以 Hab为负值。                                     

3003    ∫gud=(4 - 4S2)-1/2∫(+)((-)d 

                 = (4 - 4S2)-1/2∫[2 -2 ] d                   

                 = (4 - 4S2)-1/2 [ 1 - 1 ] = 0                           

     故相互正交。                                                      

3004    ( C )                                                             

3006    描述分子中单个电子空间运动状态的波函数叫分子轨道。                 

     两个近似 (1) 波恩 - 奥本海默近似 ( 核质量 >> 电子质量 )           

              (2) 单电子近似 (定态)                                    

3007    单个电子                                                          

3008    (B)                                                               

3009   (1) 能级高低相近                                                   

       (2) 对称性匹配                                                     

       (3) 轨道最大重叠                                                   

3010    不正确                                                            

3011    (B)                                                               

3012    = (0.8)1/2A + (0.2)1/2B                                     

3013    能量相近, 对称性匹配, 最大重叠                                    

     >  ,  <  或  <  ,  >                                              

3014    正确                                                              

3015    不正确                                                            

3016                                                             

3017 

px

pz

dxy

dxz

px

//
pz

///
dxy

///
dxz

//
3018    z                                                                

3019    (C)                                                               

3020                                                                    

3021    轨道: s-s, s-pz , s-dz,  pz –pz ,  pz -,  -  ,        

 轨道px –px ,px –dxz  ,py –py ,py –dyz  ,dyz –dyz  ,dxz –dxz         

 轨道:dxy-dxy, d- d 

3022                               不能         不能               

3023    (B)                                                               

3024            原子轨道对                    分子轨道                 

                pz-dxy                           ×       

                px-dxz                            

                d- d                    

                -                          

                px –px                            

3025    12221432 ,   3 ,    反磁                              

3026    dxy ,                                                     

3027    py,    dxy                                           

3028    C2 ( 1g)2( 1u)2( 1u)2+2  s-p 混杂显著.

     因1u为弱反键,而1g 和1u 均为强成键,故键级在2-3之间.

3029    N2:  (1g)2(1u)2(1u)4(2g)2              

           O2: 2s22s2pz22px22py22px*2py*1  

           或 ( 1g)2(1u)22g 2(1u )4(1g)2

3030    ( 1g)2( 1u)2( 1u)4( 2g)2            

     的三重键为 1 个键 (1g)2,2个键 (1u)4,键级为 3  

( 1u)2和(2g)2分别具有弱反键和弱成键性质, 实际上成为参加成键作用很小的两对孤对电子,可记为 :N≡N: 。因此N2的键长特别短,键能特别大, 是惰性较大的分子。

3031    O2[KK(g) 2 (u*) 2 (g2p*)2 (u)4 (g2*)1 (g2*)1 ]        顺磁性                                                            

     C2 [KK(g) 2 (u*) 2(g2)2 (g2)2]                     反磁性                                                            

3032    KK( 1g)2(1u)2 (1u )3        约  3/2                    

        [1222321442]5224              1      

3033    (1) 12223214       1         反                       

        (2) 1s21s2 2s22s22pz22py22pz22py*22px*1        1.5     顺                                                    

3034    3py ,  3pz ;    3px      

3035    CN-( 1)2(2) 2(1)2+2(3)2                           

     键级: 3                                                           

3036    CF KK-( 1)2(2) 2(3)2 (1)4(2)1                         

不论 s-p 混杂是否明显, 最高占据的 MO 为(2)1 , 它是反键轨道。故(C-F)+键强些, 短些。                                               

3037    Cl2: 3s23s*3px23py23pz23py*22pz*2              反磁性                                                      

     O2+:  2s22s*22px22py22pz22py*`                                顺磁性.                                               

        CN-:     -( 1)2(2) 2 (1)4(3)2                                    反磁性.                                                     

3038    (A), (C)                                                          

3039    (C)                                                               

3040    NF+( 1)2(2) 2(3)2 (1)4(2)1                              

         键级  2.5                                                     

         磁性  顺磁性 ( 有一个不成对电子)                              

3041    (A)                                                               

3042    OF+> OF > OF-                                         

3043    E N2 > E N2+                                            

     E O2+ > E O2                                       

     E OF > E OF-                                        

     E CF+> E CF                                        

     E Cl2+ > E Cl2                                    

3044    是一个极性较小的分子, 偶极矩方向由氧原子指向碳原子。              

3045    OH 的 HOMO 是 1轨道. 这是个非键轨道, 基本上是 O 原子的 2p 轨道.因此,    OH 的第一电离能与 O 原子的 2p 轨道的电离能非常接近。            

        HF 的结构与 OH 类似, 它的 HOMO 是 1 轨道, 也是个非键轨道, 基本上是 F 的 2p 轨道。因此,HF 的第一电离能与 F 原子 2p 轨道的电离能非常接近。

3046    (1) OH 基的电子结构为:                                            

        ( 1)2(2)2(3)2(12py)2(12pz)1    

        (2) 未成对电子占据轨道                                        

        (3) 1 轨道是非键轨道, 仍保持 O 原子的 2p 轨道的特性            

        (4) OH-的最低的电子跃迁的能量比 OH 基的要高                      

3047    H 和 F 以键结合;                                              

           HF = N[H (1)  F (2) +  F (1)  H (2)] [H (1)  F (2) -  F (1) H (2)]                                 

           N 为归一化系数                                              

3048    (1)  ( 1)2(2)2(3)2(1)4                              

        (2)  HF 分子的键长                                                

          r HF= r H + r F - 0.09 ,     r H , r F  是共价半径.     

             =( 37 + 71 - 0.09×1.9)pm  ,  = 4.0 - 2.1 = 1.9          

             = 107.8 pm                                             

3049    H2+;                                                              

     He2+;                                                             

     C2+;                                                              

     NO                                                                

3050     分  子         N22-    O22-    F22-    N2    O2    F2                

     电子数         16     18     20     14    16    18                

     净成键电子数    4      2      0      6     4     2                

     F22- 净成键电子数为 0 , 不能稳定存在                               

     N22- 和 O22- 能稳定存在                                             

        r N22- > rN2                                     

        r O22- > rO2                                     

3051     分子            N2+   O2+   F2+    N2    O2    F2                 

     电子数          13    15    17    14    16    18                  

     净成键电子数     5     5     3     6     4     2                  

        r N22- > rN2                                     

        r O22- > rO2                                     

        r F2+ > r F2                                           

3052     分子    电子数    最高占有分子轨道    磁性                        

     N2      14       ↑↓ 3g             反                   

     O2      16       ↑ ↑ 2g            顺             

     F2      18       ↑↓ ↑↓ 2g        反             

     O22+     14      ↑↓ 3g             反                   

     F2+     17       ↑↓ ↑ 2g          顺             

3053    CO   KK (1)2(2)2(1)4(3)2                        

     NO   KK (1)2(2)2(1)4(32)(2)1                    

     NO 在高能反键轨道上有一电子, I1较低。                           

3054    [Kr] (1)2(2)2(32) (1)4                              

3055    N2+< N2;    NO+> NO;    O2+> O2;    C2+< C2                   

        F2+> F2;    CN+< CN;    CO+< CO                                

3056    (1) O2 电子结构为                                                 

        KK(2s2) (2s*2) (32pz2) (2py2) (2pz2) (2py*1) (2px*1)          

        顺磁性分子                                                        

     (2) O22-> O2-> O2 > O2+                                         

     (3)  1865 cm-1                                                    

3057    比 (AB)+的键能小: O2, NO,                                      

        比 (AB)-的键能小: C2, CN                                         

3058    Li2> Li2+                                                        

     C2 > C2+                                                          

     O2 < O2+                                                          

     F2 < F2+                                                          

3059   小                                                                 

3060    H2+ 比 H2 在成键轨道 (1s)上少一个电子, H2+的键级为 0.5, H2 的键级为 1  。                                                     

     O2+比 O2 在反键轨道 (2p* )上少一个电子, O2+的键级为 2.5; O2 的键级为 2.0  。                                              

3061    (C)                                                               

3062    成键轨道                                                          

    1s= N(e + e)                   

    在 M 点, ra= 40 pm,  rb= 106 pm + 40 pm = 146 pm              

    M2= N2(e-40/52.9 + e-146/52.9)2            

    在 N 点, ra= 40 pm, rb= 66 pm                                 

    N2= N2(e-40/52.9 +e-66/52.9)2              

3062    M,N 两点概率密度之比为 : = 0.4958                          

3063    反键1s * 轨道                                      

     = N(e + e)                               

     在 M 点, ra= 106 pm - 40 pm = 66 pm,    rb=40 pm              

     M 2= N2(e-66/52.9- e-40/52.9)2             

     在 N 点, ra= 146 pm, rb= 40 pm                                

     N2= N2(e-146/52.9 - e-40/52.9)2            

     M,N 两点概率密度之比为:                                           

                   = 0.2004                                     

30    已知 MN = 10.0, AM = 22.4, 故                                     

     AN = ( 22.42-10.02)1/2 = 20.0                                    

     BN = 106 - 20 = 86.0                                              

     BM = ( 86.02+10.02)1/2 = 86.6                                    

     从而得 M 点之 rA= 22.4 pm,   rB= 86.6 pm                      

    M = 8.210×10-4(e-22.4/52.9 + e-86.6/52.9)    

        = 6.98×10-4 pm-3/2                                           

    M 2= 4.87×10-7 pm-3                                      

3065    H2 分子体系:                                                      

       空间波函数s= a(1) b(2) + a(2) b(1)                

                  A= a(1) b(2) - a(2) b(1)                

       自旋波函数 1=  (1)  (2),  2= (1) (2)               

                  3= (1) (2),  4= (1)  (2)               

                  5= -3+ 4= (1) (2) + (1)  (2)           

                  6= 3- 4=  (1) (2) - (1)  (2)            

      完全波函数                                                       

          单重态Ⅰ=s6                                   

          三重态Ⅱ=A1                                   

                 Ⅲ=A2                                   

                 Ⅳ=s5                                   

3066    (A.)                                                                

3067   =(2+2Sab2)-1/2 [a(1) b(2)+ a(2) b(1)]×× [ (1) (2)- (2) (1)]                                    

3068    MO=(2+2S)-1/2 [a(1) +b(1)][ a(2) +b(2)]        

        VB=(2+2S)-1/2 [a(1) b(2)+ a(2) b(1)]             

     简单分子轨道理论将电子 (1) 和 (2) 安放在分子轨道 (a+b)上, 分子轨道是基函数; 价键理论将电子 (1) 和 (2) 安放在原子轨道a和b上, 原子轨道是基函数                                                  

     MO中包含共价项和离子项 ( 各占 50% )                      

     VB中只含共价项                                           

3069    [-(12+22)- -  + ] -  +  +] (x1y1z1x2y2z2)           

    = E (x1y1z1x2y2z2)                                  

    式中 ra1 ,r a2 分别为 He 核与电子 1,2 间距离       

         rb1 , r b2分别为 H 核与电子 1,2 间距离        

         r12 为 1,2 电子间距离                                 

         rab为 He, H 核间距离                                  

3070    转动, 振动和电子运动                                              

     转动                                                              

     振动--转动                                                        

     电子--振动--转动                                                  

3071    (D)                                                               

3072    能级差增大的有: 一维势箱, 刚性转子                                

     能级差减小的有: 非谐振子, H 原子                                  

     能级等间隔排列的有: 谐振子                                        

3073    4,   3,   2                                                       

3074    对称伸缩, 不对称伸缩, 面内弯曲和面外弯曲                          

     对称伸缩                                                          

     不对称伸缩和弯曲振动                                              

3075    H2, CH4                                                           

3076    = 2B = h/(42Ic)                                        

     相邻转动能级差的递增值相等, 选律J = ±1  。                   

3077    H2,                                                             

3078    =  ( J+1 ) →∝                           

     HCl = DCl  = 1.95 DCl               

3079    由下到上能级间隔愈来愈大, 按 2B 增加。                           

3080    (B)                                                               

3081    (A)                                                               

3082    (D)                                                               

3083    (B)                                                               

3084    (A)                                                               

3085    (C), (D)                                                          

3086    (1) X-Y-X 型                                                      

     (2) 反对称伸缩 2349 cm-1, 弯曲振动 667 cm-1                       

     (3) 测定偶极矩, 为 0 ;  测拉曼, 只有一种基本振动频率              

3087          S           S              S                       

                                                       

          O     O    O      O       O      O                  

            1151cm-1    1361cm-1         519cm-1                            

     均为红外活性和 Raman 活性。                                       

3088    因空间位阻的缘故。                                                

30    (D)  b-a=2B                                                    

        a=2B(J+1)                                                  

        J= (2a-b)/( b-a)                                                 

3090    (D)                                                               

     3091       因为电子从基态向激发态跃迁, 平衡核间距来不及变化,所以跃迁概率最大的  跃迁是那些与核间距保持不变的状态对应的态态间的跃迁。                

3092    kCN/k CN+ = 1.714                                                      

3093    2B = 16.94 cm-1   B = 8.470 cm-1                                

    Re = 142.0 pm                                                    

3094    (1) =  ()1/2 = 1.394                                 

    (2) E0=  h0                                               

        ∴  =  = 1.394                                    

3095    (B)                                                               

3096    (1) k = 312 N·m-1                                               

  (2) D0= De- E0= 4.02 eV                                     

3097    r = 113.1 pm                                                     

     '=  =× 3.842 = 3.673 (cm-1)          

3098    = c= 6.509×1013 s-1                                   

      h= 12.99 kJ.mol-1                                           

3099    (1) k=1900 N·m-1                                                

        (2) E0=1.298×104J·mol-1                                      

3100    (1) B=8.5 cm-1                                                   

     (2) I=3.291×10-47 kg·m2                                 

        re=141.6 pm                                                  

     (3) Er=1.22×10-20 J                                      

3101    k = 312 N·m-1                                                   

     E0= 2.295×10-20 J                                        

     I = 4.27×10-47 kg·m2                                    

     re=161 pm                                                       

3102    h= 2.130×10-20 J                                      

        k = 513 N·m-1                                                   

3103    k = 516 N·m-1                                                   

3104    (1) x = 0.01736                                                  

        0= 29.7 cm-1                                              

     (2) k = 512.1 N·m-1                                             

     (3) E0= 2.946×10-20 J                                    

3105    I = 2.726×10-47 kg·m2                                   

        r = 129.9 pm                                                     

3106    De = 5.90 eV                                                     

        E0= 0.254 eV                                                    

3107    r = 126.9 pm                                                     

3108    k = ( 2C )2 = 604.6 N·m-1                                

        '=  (  /')1/2 = 2401 cm-1                                  

3109     这些光谱是由 HCl、HBr 产生的。                                   

        rH--Cl = 128.5 pm                                          

        rH--Br = 142.4 pm                                       

3110    I = 1. × 10-46 kg·m2                                  

3111    121.5 pm                                                          

3112    (1) k = 71.43 N·m-1                                             

        (2) = 310.6 cm-1                                              

3113    HD = ()1/2H2 = 3603 cm-1                     

        D2 = ()1/2 H2 = 2942 cm-1                   

3114    k(HF) = 961.27 N·m-1                                            

     k(H35Cl) = 512.43 N·m-1                                         

     k(H81Br) = 409.18 N·m-1                                   

     k(H127I) = 312.23 N·m-1                                   

3115    D0= 941.48 kJ·mol-1                                            

3118    2.295×10-20 J                                              

3119    激发态 R = 1.309×10-8 cm , 比基态 1.2×10-8 cm 长。              

3120    (a) 由分子 M 的 V=0 的状态到分子离子 M+的 V=0 的状态所需能量;  

        (b) 由分子 M 基态到分子离子 M+, 不改变核间距离的跃迁概率最高的振动态所需能量。                                                

3122    (D)        

3123    (C)                                                               

3124    (B)                                                               

3125    (D)                                                               

3126    (D)                                                               

3127    (C)                                                               

3128    (A)                                                               

3136    对称的伸缩振动, 为非红外活性。                                    

  不对称的伸缩振动 , 为 2349 cm-1,                                    

    弯曲振动为 667 cm-1 。                                            

3138    不正确。                                                          

3140    (B)                                                               

3141    (B)                                                               

3142    (D)                                                               

3143    参看 《 结构化学基础 》 ( 周公度编著 ) ,p.153 ,图 3.5-4 。       

3144    [-(12+22)- (+  ++) +(+)]=E                      

     假定两核 H 和 D 相对位置不变.                                     

3145    NO+和 PS+键级为 3, 最稳定;                                      

        SCl-,ArCl+键级为 1, 最不稳定。                                  

3146    NO+                                                              

3147    如 O2-                                                            

3148    沿键轴平行分布无节面的轨道。                                      

3149    沿键轴平行分布有一个节面的轨道。                                  

3150    沿键轴平行分布有二个节面的轨道。                                  

3151    由 AO 组成 MO 时,能级低于 AO 者。                                 

3152    由 AO 组成 MO 时,能级高于 AO 者。                                 

3153    由 AO 组成 MO 时,能级等于 AO 者。                                 

3154    碳原子端。                                                        

3155                                                                    

3156                                                                    

3157                                                                  

3158                                                                    

3159                                                                   

3160    不对                                                              

3161    不对                                                              

3162    E = ( v+)h ,   =                        

3163    E =  J(J+1)                                 

31    (D)                                                               

3165    (D)                                                               

3166    (D)                                                               

3167    (D)                                                               

3168    (A)                                                               

3169    (C)                                                               

3170    (D)                                                               

3171    (A)                                                               

3172    (1) =2B                                                   

    所以  B =8.5cm-1                                                

        (2) EJ=hcBJ(J+1)                                     

             =6.626×10-34×3×108×8.5×102×10×11J          

             =1.859×10-20J                                   

(3) (J)= =2B(J+1)=169cm-1                          

             J+1=169/(2×8.5)=10                                                 

    故为E9与E10之间的跃迁产生的。                                      

3173    3,3,3。                                                              

3174    4,4。                                                                

3175    (1)   (2s2)2()2(2p )4                                

 (2)   rHF = rH +rF -9(x F -x H )=90.9pm      

3176    (1) =                                              

    k=422m=42×(2.4)2×0.045N/m=10.23N/m                    

        (2)振子能量为                                                         

     E=kA2=×10.23×(0.04)2 J=8.18×10-3J                      

    若用量子力学处理                                                   

    Ev=(+v)h                                                  

    V=-=(-)J ≈ 5.1×1030 J     

3177    设键轴方向为z轴,                                                  

        原子轨道s与s,s与pz,pz与pz组合,得到的分子轨道是圆柱对称的,称为轨道;                                                        

     原子轨道px与px,py与py组合,得到的分子轨道通过键轴有一个节面,这种分子轨道称为轨道;                                            

        原子轨道dx2-y2与dx2-y2,dxy与dxy组合,得到的分子轨道通过键轴有两个节面,这种分子轨道称为轨道;                

3178    N─N─O                                                              

3179    =R0  平衡核间距(键长)                                  

        cd=D   平衡解离能                                          

3180    包括核排斥能,因中包括核排斥项1/R                               

     E1=(Haa+Hab)/(1+Sab)                       

     E2=( Haa-Hab)/(1-Sab)                       

     求Haa, Hab 时都要用到,Haa, Hab , Sab都是核间距R的函数。                                       

3181    (1) 自旋-轨道偶合所产生的两个UPS峰的相对强度为                

            (2J1+1):(2J2+1)=(l+1):l                      

     由题意知,(l+1):l=2:1,所以l=1,即从3p轨道击出电子。                

     (2) 1S0;2P3/2;2P1/2          

     (3) Ar(1S0)→Ar+(2P3/2)+e  I=15.759eV   

           Ar(1S0)→Ar+(2P1/2)+e  I=15.937eV   

3182    (B)                                                               

3183    分子1的UV-vis与苯相似。                                          

    分子1中甲烯烃与苯环没有共轭,2中形成大键体系1010,所以,后者的UV-vis与苯不一样。                                          

3184    (1).KK 12 -22 -14 -32,        3,     C∞v             

        (2).丙酮中最长,CO其次,CO2中的C--O键长最短。因CO2中有2个34, CO中有-, -,-配键,而丙酮中只-,- 键。                           

     (3).Cr(CO)6    正八面体                                       

        Fe(CO)5    三角双锥                                       

        Ni(CO)4   正四面体                                       

     (4). =c=6.509×1013s-1                                    

                   2B=h/(42r2)                                

                   r =[h/(82B)]1/2                       

3185    N2中1s轨道近似能量等于N原子1s轨道能量                  

          E1s ≈E1s=-(Z2R)/n2=-49R=-666.4eV      

3186    光电子动能变小,变化值均为18.87eV .                                 

3187    B=                                                

     同位素取代不改变re                                        

    B HD =BH2= BH2=44.48cm-1  

    B D2= BH2= BH2=29.66cm-1 

3188    12223213 , 2                                       

31    122232421452,1∑                               

3190    12223214,  1∑                                       

3191    (1s)2(1s *)1,   2∑g                   

3192    1s 2 ,   ∧=0  ,  S=0   ,    1∑g                

3193    (1g)2(1u)2 (1u )4 ,     1∑             

3194    激发线频率                                                        

      0=c/=(3×108)/(435.8×10-9) s-1=6.884×1014s-1        

     斯托克斯线                                                        

     =0-e                                          

       =6.884×1014-8.667×1013 s-1                          

       =6.017×1014s-1                                         

     斯托克斯线波长                                                    

     =c/=(3×108)/(6.017×1014)m=4.986×10-7m             

       =498.6nm                                                        

3195    均可发生跃迁                                                      

     2∑g←→ 2∑u                                       

     2∏←→ 2∑                                                     

3196    谱线间距为4B                                                     

      B=0.9752/4 cm-1=0.2438cm-1                                         

      =[(35×35)/(35+35)]×1.6606×10-27kg                      

        =2.9061×10-26kg                                       

      R=[h/(82cB)]1/2                                          

       =()1/2 m            

       =1.987×10-10m                                          

       =198.7pm                                                      

3197    C2H2有3×4-5=7种简正振动,其中,Q3,Q6,Q7是红外活性的,                                                               

     所列简正振动为其全部简正振动,其中Q4,Q5简并,Q6, Q7简并,所以有5个简正振动频率,其中有2个红外活性的简正振动频率。  

3198    2906.2=e+2B                                              

        2925.9=e+4B                                              

     解之得                                                            

     e=2886.5cm-1                                              

     B=9.85cm-1                                                       

     =1.6145×10-27kg                                          

     re=[h/(82Bc)]1/2                                    

      =1.326×10-10m                                      

      =132.6pm                                                  

      k=(2ce)2                                          

       =477.95N·m-1                                                   

3199    前两条谱线为P(1)和P(2)                                            

       2865.1cm-1=e -2B                                        

       2843.6cm-1=e -4B                                        

     解之得e =2886.6cm-1                                       

             B=10.75cm-1                                              

          re =[h/(82Bc)]1/2                                     

            =()1/2 m        

            =1.270×10-10m                                     

            =127.0pm                                                 

      k=(2ce)2                                           

       =(2×3×108×2886.6×102)2×1.6151×10-27 N·m-1           

       =478.2N·m-1                                                  

3200    (1) HCl                                                           

         E0=e /2 - xee /4                         

           =1331.8cm-1=0.165eV                                    

         D0=De-E0=(5.33-0.165)eV=5.165eV                      

        (2) DCl                                                           

          DCl=(HCl/DCl)1/2HCl           

             =×26.7 cm-1                    

             =[37.5/(2×36.5)]1/2×26.7 cm-1=1927.8cm-1                  

         由于在势能曲线中                                              

       a2=(4c0x)/     DCl与HCl之a相同,故                   

       0x∝ 1/                                                   

       xe (DCl)=·xe (HCl)                     

                =[37.5/(2×36.5)]×52.05 cm-1                               

                =26.97cm-1                                           

        E0=×1927.8-×26.97 cm-1=957.1cm-1=0.1187eV                   

        D0=(5.33-0.12)eV=5.21eV                                          

3201      =[(mamb)/(ma+ mb)]×1.6606×10-27 kg           

       e =(k/)1/2 / (2c)                                     

       k = (2ce)2                                         

       e’/e=(’/)1/2                                     

                    HF        HCl        HBr        HI        

     /(kg×10-27)    1.52    1.6266     1.6529     1.6603          

                    DF        DCl        DBr        DI             

     /(kg×10-27)    3.0238    3.1623     3.2632     3.2922          

     e/cm-1        3002.3    2143.7     1885.9     10.1          

     k/N·m-1        967.1     515.6      411.8      314.2          

3202    2B平均值                                                         

     (60.81408-24.32592)/3 GHz =12.16272 GHz                                

     B=6.08136 GHz                                              

     I=h/(82B)                                                   

      =(6.626×10-34)/(82×6.08136×109) kg·m2                     

      =1.3799×10-45kg·m2                                    

     不能测键长,因16O12C32S中有二个键长,仅由转动惯量不能确定这两个键长。                                                  

3203    B=13.10/2=6.55cm-1                                               

      =[(1×126.9)/(1+126.9)]×1.6606×10-27 kg                   

       =1.76×10-27kg                                        

     r=[h/(82 Bc)]1/2                                           

      =1.610×10-10m                                             

      =161.0pm                                                         

3204    参见周公度编著,《结构化学基础》,第132页。                         

3205    选极化率各向异性分子                                              

    H2,HCl,CH3Cl,CH2Cl2,CH3CH3,H2O                                   

3206    (1) 1.6145×10-27kg;  2.6224×10-47kg·m2            

     (2) 3.1417×10-27kg;  5.1032×10-47kg·m2            

     (3) 1.6169×10-27kg;  2.62×10-47kg·m2            

3207    12C16O:  =1.1387×10-26kg,                     

              B=1.930cm-1,第一条谱线位于3.860cm-1;                    

        13C16O:  =1.1910×10-26kg,                     

              B=1.845cm-1,第一条谱线位于3.690cm-1;                    

     所以,分辩率至少应为0.1cm-1。                                      

3208    HCl: B=10.225cm-1                                                

     DCl: B=10.225×( HCl/ DCl) cm-1=5.27cm-1                

     前4条谱线波数为:                                                  

        10.54 cm-1,21.08 cm-1,31.62 cm-1,42.16cm-1                                    

3209    De=D0+E0                                               

          =[1.9+he(0+1/2)]eV                                          

          =(1.9+0.097)eV=1.997eV                                            

3210    基本光带    1=e(1-2x)                                      

        第一泛音带    2=2e(1-3x)                                   

              2143.0=e-2xe                                         

           4260.0=2e-6xe                                        

    解之得                                                            

          e=2169.0cm-1                                              

          x=0.00599                                                   

       De=e/(4xe)=1.81×105cm-1                           

     亦可                                                              

       De=(hce)/(4xe)=3.60×10-20J                      

3211    =(k/ )1/2 / (2c)                                              

     DCl/HCl=( HCl/ DCl )1/2               

     DCl=( HCl/ DCl)1/2 ×HCl             

         =2143.4cm-1                                                  

3212    零点能                                                            

      E0=hec=×6.626×10-34×1580.0×102×3×108 J   

        =1.57×10-20J                                        

      D0=De-E0=8.00×10-19J                                

3213    =(k/ )1/2 / (2)                                                

          =9.627×1012s-1                                           

    E0=h=3.1×10-21J                                   

     (或E0==/(2c)=1.605×104m-1=160.5cm-1)                 

3214    =(k/ )1/2/(2c)                                              

      k=(2c)2                                                 

       =(2×3×108×3.780×104)2×[(39×35)/(39+35)]              

           ×1.6606×10-27 N·m-1                                      

       =155.5N·m-1                                                  

3215    间隔=2B=h/(42r2)=6.350×1011s-1                    

       =[(1×35)/(1+35)]×1.6605×10-27 kg                       

         =1.6144×10-27kg                                      

       r =[(6.626×10-34)/(42×1.6144×10-27×6.350×1011)]1/2 m                                    

         =1.28×10-10m                                         

         =128pm                                                      

3216   =[(1×127)/(1+127)]×1.6605×10-27 kg=1.75×10-27kg

       B=h/(82r2c)                                           

        =6.5993cm-1                                                    

3217     B=h/(82Ic)                                                

       =1.482cm-1                                                      

     按钢体模型,H12C14N的微波谱是由一系列等间隔谱线组成,其第一条谱线及相邻两谱线间的距离均为2.9cm-1                         

3218    合频: 2+3=(517.69+1361.76) cm-1=1879.45 cm-1≈1875.55cm-1             

     倍频: 21=(2×1151.38) cm-1=2302.76 cm-1≈2295.88cm-1                     

     合频: 1+3=(1151.38+1361.76) cm-1=2512. cm-1≈2499.55cm-1            

3219    基本光带1-0=e-2xe                                 

                      =(29.7-2×45.21) cm-1=2559.3cm-1                      

     第一泛音带2-0=2e-6xe                              

                      =(2×29.7-6×45.21) cm-1=5028.1cm-1                   

     第二泛音带3-0=3e-12xe                             

                      =(3×29.7-12×45.21 )cm-1=7406.6cm-1                  

     第三泛音带4-0=4e-20xe                             

                      =(4×29.7-20×45.21 )cm-1=9694.6cm-1                  

3220    CH2CCl2,CCl4,HCOOH,Cl2CO,BF3,C2N2,HCN             

3221    CH2CCl2,HCOOH,Cl2CO,HCN                                    

3222    N2,CH2CCl2,HCOOH,Cl2CO,BF3,C2N2,HCN               

3223    PCl3,SO2,H2S2,H2CO,N2O                             

3224    PCl3,SO2,H2S2,CH2CH2,H2CO,N2O,Ni(CO)4             

3225    O2,PCl3,SO2,H2S2,CH2CH2,H2CO,N2O                  

3226    =[(6×19)/(6+19)]×1.6606×10-27kg=7.5723×10-27kg  

     B=h/(82cr2)                                             

      =1.518×102m-1=1.518cm-1                                        

     e=(k/)1/2/(2c)=9.639×104m-1=963.9cm-1                   

      [P(T)]= e-2BJ=963.9-3.04J                                   

      [R(T)]= e+2B(J+1)=963.9+3.04(J+1)                          

     P支前三条  960.9 cm-1  957.8 cm-1  954.8 cm-1                              

     R支前三条  967.6 cm-1  970.7 cm-1  973.7cm-1                              

3227     [P(T)]= e-2BJ   J=1,2,...                                  

      [R(T)]= e+2B(J+1)   J=0,1,...                              

     B=(56000×106)/(3×108) m-1=186.7m-1=1.867cm-1                     

      [P(T)]=(2143.0-3.734J ) cm-1   J=1,2,...                    

      [R(T)]=[2143.0+3.734(J+1)] cm-1    J=0,1,...               

     P支前三条谱线为  2139.3cm-1  2135.5cm-1  3132.0cm-1               

     P支前三条谱线为  2146.7cm-1  2150.5cm-1  2154.2cm-1               

3228    B=×3634×106=1817×106                                      

     =[(39×127)/(39+127)]×1.6605×10-27kg                    

       =4.955×10-26kg                                         

     r=[h/(82B)]1/2                                             

      =3.05×10-10m                                              

      =305pm                                                           

3229    本质相同:                                                         

     但取坐标参考零点不同,前一个取双原子分子中原子无限远处时E=0;后一个取V=0。                                                              

3230    1=2B1(J+1)=[h(J+1)]/(42cI1)     I1=1r12       

        2=2B2(J+1)=[h(J+1)]/(42cI2)     I2=2r22       

     对HCl及DCl, r1=r2                                             

     =1-2=[h/(42c)](J+1)(1/I1-1/I2)                

         =1(1-I1/I2)                                            

         =1(1-1/2)                                            

3232    不一定,  根据坐标选取及相位取向而定。                             

     键因为相位相同,负值表示反键,与键不同。                        

3233    (1) *为g                                                       

     (2) 分子无对称中心,无g,u性质                                      

     (3) 为g                                                         

     (4) *为u                                                       

3234    NO的价电子组态为: 1222143221,键级为2.5,当NO失去一个电子成为NO+时,失去的是反键电子(21),键级增至为3,所以NO易被氧化为NO+。                                                         

3235    羰基中,C,O之间是+键,由于C,O电负性不同,所以羰基C=0中的极性很大。而一氧化碳中,C,O之间是++键,三重键中一对电子完全由氧原子提供,这就抵消了由于O电负性大而使电荷集中在O端.由于这种配键的存在,CO的极性很小。                                                              

3236    (A)                                                               

3237    (D)                                                               

3238    (1) (1g)2(1u)2(1u )42g 1,2.5  

     (2) (1)2(2)2(1)4(32),3.0                              

     (3) (2s )2(2s *)2(2pz )2 (2p y)2 (2p x )2 (2p y *)1,2.5         

3239    (B)                                                               

     若知是第几条谱线,一条即可计算.                                    

3240    ①非活性,  ②活性,  ③非活性,  ④活性,                            

3241    De=D0+h=D0+hc                                    

          =255.9+×6.626×10-34×2.9979×108×2233×102×6.023×1023×10-3 kJ·mol-1                                    

          =(255.9+13.55) kJ·mol-1=269.5kJ·mol-1                                  

3242    G()=e(v+)- exe(v+)2                           

     分数==            

     由e=29.7, exe=52.05得xe=0.01741                        

             v                分数                                    

             0                0.00878                                  

             5                0.1059                                   

             10               0.2237                                   

3243   G()=e(v+)- exe(v+)2                         

     1331.84=e-xe                                           

     3917.44=3/2 e-9/4 xe                                       

     解之得                                                            

         e=26.7cm-1                                               

         xee=52.04                                                

         xe=0.01935                                                  

         De=e/4x=3.475×104cm-1                                 

     亦可 De=(hce)/(4x)=4.308eV                                

          D0=De-E0=(3.475×104-1331.8) cm-1=3.3418×104cm-1         

     亦可  D0=4.143eV                                                

           k=(2ce)2=(2×3×108×26.7×102)2           

                  ×[(1×35.5)/(1+35.5)]×1.6605×10-27 N·m-1        

            =415.2N·m-1                                               

3244    G()=e(v+)- exe(v+)2                         

     142.81=e-xe                                           

     427.31=3/2 e-9/4 xe                                        

     解之得                                                            

         e=285.99cm-1                                               

         xee=0.74                                                 

         xe=0.00259                                                  

         =[(23×127)/(23+127)]×1.6606×10-27 kg                 

           =3.2337×10-26kg                                    

         k=(2ce)2=(2×3×108×285.99×102)2×3.2337×10-26 N·m-1                                  

          =9.397×102N·m-1=93.97N·m-1                           

         De=e/4x=27605m-1                                        

     亦可 De=(hce)/(4x)=5.484×10-19J=3.422eV            

          D0=De-E0=(27605-142.81) cm-1=27462cm-1=5.455×10-19J

3245    ②结构式合理,                                                     

            O                                                           

                                                                      

         ─ S ─中的S原子对应于E k 较小的峰                        

                                                                      

            O                                                           

3246    CO2有不对称伸缩和弯曲二个红外活性的简正振动频率,伸缩振动频率一般大于相应的弯曲振动,所以图中波数较小,靠近6.0×104m-1的那个峰来自于CO2的弯曲振动,另一个波数较大的峰为其不对称伸缩振动。               

3247    键级是表示化学键的相对强度的,对于定域共价键,键级=(成键电子数-反键电子数).对于离域键.相邻两原子i和j之间的键键级为                 

        Pij=nkckickj         

     式中nk为第k个分子轨道中的电子数, cki和ckj分别为第k个分子轨道中第i个和第j个原子轨道前的系数。                  

3248    有两个包含键轴的节面的分子轨道称为轨道。                        

    或:                                                                

     角动量沿键轴方向分量为2·(h/2)的分子轨道称为轨道。           

     轨道上的电子称为电子。                                        

3249    有一个包含键轴的节面的分子轨道称为轨道。                        

    或:                                                                

        角动量沿键轴方向分量为1·(h/2)的分子轨道称为轨道。           

     轨道上的电子称为电子。                                        

     由成键电子形成的共价键称为键。                                

3250    没有包含键轴的节面的分子轨道称为轨道。                          

    或:                                                                

     角动量沿键轴方向分量为0的分子轨道称为轨道。                     

     轨道上的电子称为电子。                                        

     由电子形成的共价键称为键。                                    

3251    (C)                                                               

                                                                          

3171    (A)                                                               

3254

(1)

(2)

3255

     

3256

     

3257

     

     

3258

     键轴内

     

     键轴外

     

3259

     

     在此平面上任一点,距两原子核距离相等,ra=rb,所以

3260

     

                                                                 

                                                                

              

            对称伸缩                     弯曲振动                 不对称伸缩

            1157cm-1                       519cm-1                             1361cm-1

                   Raman活性                    红外活性                 红外活性

3261

     

     

     

3262

        J=0,1,2….

     kg

     

         

     

3263

(1)

(2)

(3)

32

(1) 

(2) 

(3) 

3265

         

     N(E)

                                              I

        左边第一个峰即是。

3266

     

N(E)               *        

                                               I

        打星号者即是。

3267

     描述分子中电子的轨道运动的单电子波函数。

3268

     

     :分子中内层电子i的化学位移;

     :分子中内层电子i的结合能;

     :自由原子中内层电子i的结合能。

3269

     

     电子的动能算符;

     电子受a原子核的吸引势能算符;

     电子受b原子核的吸引势能算符;

     a,b两原子核的排斥势能算符。

3270

                      

       2.5      2.0     1.5      1.0

3271

                      

       121      126     149      112    (pm)

3272

                     

      626.1   493.5    392.9    138.1    (kJ/mol)

文档

《结构化学》第三章习题答案

《结构化学》第三章习题答案3001(A,C)3002Hab=∫a[-2--+]bd=EHSab+Sab-∫abd=EHSab+K因EH=-13.6eV,Sab为正值,故第一项为负值;在分子的核间距条件下,K为负值。所以Hab为负值。3003∫gud=(4-4S2)-1/2∫(+)((-)d=(4-4S2)-1/2∫[2-2]d=(4-4S2)-1/2[1-1]=0故相互正交。3004(C)3006描述分子中单个电子空间运动状态的波函数叫分子轨道。两个近似(1)波恩-奥本海默近似(核质量>>电子
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