£¨1£©HCOOH Ka=1.8?10-4 £¨2£©HAc Ka=1.76?10-5 £¨3£©NH3 Kb=1.76?10-5 £¨Ñ¡£¨1£©ºÏÊÊ£©
7.ijһԪÈõËáHAµÄŨ¶ÈΪ0.010mol/L£¬ÔÚ³£ÎÂϲâµÃÆäpHΪ4.0¡£Çó¸ÃÒ»ÔªÈõËáµÄ½âÀë³£ÊýºÍ½âÀë¶È¡££¨½âÀë³£ÊýºÍ½âÀë¶È·Ö±ðΪ1.0¡Á106£¬1%£©¡£
8.ÒªÅäÖÆ450mlpH=4.70µÄ»º³åÈÜÒº£¬ÇóÐè0.10mol/L HAcºÍ0.10mol/L NaOHÈÜÒºµÄÌå»ý¡£HAcµÄµçÀë³£ÊýΪ£º1.8¡Á10-5
£¨0.10mol/LHAcΪ306ml£¬0.10mol/LNaOHÈÜÒºµÄÌå»ýΪ144ml£©¡£
9.ÔÚѪҺÖÐH2CO3-NaHCO3»º³å¶ÔµÄ×÷ÓÃÖ®Ò»ÊÇ´Óϸ°û×éÖ¯ÖÐѸËÙ³ýÈ¥ÓÉÔ˶¯²úÉúµÄÈéËᣨ¼ò¼ÇΪHL£©¡£
-
£¨1£©ÇóHL+HCO3=H2CO3+L-µÄƽºâ³£Êý¡££¨3.3¡Á102£©
-3-2
£¨2£©ÈôѪҺÖÐ[H2CO3]=1.4¡Á10mol/L£¬[HCO3-]=2.7¡Á10mol/L£¬ÇóѪҺµÄpHÖµ¡££¨7.66£©
£¨3£©ÈôÏò1.0LѪҺÖмÓÈë5.0¡Á10-3mol HLºó£¬pHΪ¶à´ó£¿£¨6.91£© £¨ÒÑÖªH2CO3£ºKa1=4.2¡Á10-7£¬Ka2=5.6¡Á10-11£»HL£ºKa=1.4¡Á10-4£©
10£®¼ÆËãÏÂÁÐÈÜÒºµÄpH¡£
£¨1£©0.20mol/LHAcºÍ0.10mol/LNaOHµÈÌå»ý»ìºÏ£»£¨4.75£© £¨2£©0.20mol/LHAcºÍ0.20mol/LNaOHµÈÌå»ý»ìºÏ£»£¨8.72£© £¨3£©0.20mol/LNaACºÍ0.10mol/LHClµÈÌå»ý»ìºÏ£»£¨4.75£© 11£®ÊԱȽÏAgIÔÚ´¿Ë®ÖкÍÔÚ0.010mol/LKIÈÜÒºÖеÄÈܽâ¶È¡£¡²ÒÑÖªAgIµÄÈܶȻý=9.3¡Á10-17¡³£¨´¿Ë®ÖеÄÈܽâ¶ÈΪ9.6¡Á10-9£¬ÔÚ0.010mol/LKIÈÜÒºÖеÄÈܽâ¶ÈΪ9.3¡Á10-15£©¡£
12£®1gFeS¹ÌÌåÄÜ·ñÈÜÓÚ100ml1.0mol/LµÄÑÎËáÈÜÒºÖУ¿ÒÑÖªFeSµÄÈܶȻý³£ÊýΪ6.0¡Á10-18£¬H2SµÄ×ܵçÀëÆ½ºâ³£ÊýΪ9.23¡Á10-22£¬Ê½Á¿£¨FeS£©=87.9¡££¨ÄÜÈÜ£©
13.ÏÂÁÐ˵·¨ÊÇ·ñÕýÈ·£¿
£¨1£©¼ÈÓп¹Ëá³É·ÖÓÖÓп¹¼î³É·ÖµÄÈÜÒºÒ»¶¨ÊÇ»º³åÈÜÒº¡££¨´í£© £¨2£©Ksp´óµÄÄÑÈÜÎïµÄÈܽâ¶ÈÒ»¶¨´óÓÚKspСµÄÄÑÈÜÎï¡£ (´í) £¨3£©Í¬ÎÂÏÂZnSµÄKsp´óÓÚCuSµÄ,ËùÒÔͬÎÂÏÂZnSµÄÈܽâ¶È´óÓÚCuSµÄ¡£(¶Ô) £¨4£©ÔÚº¬ÓÐÂÈÀë×ÓµÄÈÜÒºÖмÓÈëÏõËáÒøÈÜÒº£¬Ò»¶¨ÓÐÂÈ»¯ÒøµÄ°×É«³Áµí²úÉú¡££¨´í£©
£¨5£©·Ó̪Óöµ½NaACÈÜÒº±äºì¡£ ʯÈïÓöµ½NH4ClÈÜÒº±äºì¡£ (¶Ô)
µÚÆßÕ Ô×ӽṹºÍÔªËØÖÜÆÚÂÉ
Ò»¡¢ÀýÌâ
1.¼ÆË㼤·¢Ì¬ÇâÔ×ӵĵç×Ó´ÓµÚÈýÄܼ¶²ãԾǨÖÁµÚ¶þÄܼ¶²ãʱËù·¢ÉäµÄ·øÉäÄܵįµÂÊ¡¢²¨³¤¼°ÄÜÁ¿¸÷ÊǶàÉÙ£¿
½â ¸ù¾ÝÀïµÂ±¤¹«Ê½£¬µÃËù·¢ÉäµÄ·øÉäÄÜÆµÂÊΪ£º
11 ?=R£¨2?2£©
n1n211 =3.289¡Á1015(2?2)
23 =4.568¡Á1014s-1
²¨³¤Îª
c2.998?108?109 ¦Ë===656.3nm 14?4.568?10Ëù·¢ÉäµÄÄÜÁ¿
E=h?=6.626¡Á10¡Á4.568¡Á1014
=3.027¡Á10-19J
×¢£ºcΪ¹âËÙ£¬2.998¡Á108m/s,1m=109nm
Z2±¾ÌâÒ²¿ÉÓɵç×ÓµÄÄÜÁ¿E=-13.62ev ¡÷E=h?½øÐмÆËã¡£
n2. Éè×Óµ¯µÄÖÊÁ¿Îª0.01kg £¬ËÙ¶ÈΪ1.0?103m/s.ÊÔͨ¹ý¼ÆËã˵Ã÷ºê¹ÛÎïÌåÖ÷Òª±íÏÖΪÁ£×ÓÐÔ£¬ÆäÔ˶¯·þ´Ó¾µäÁ¦Ñ§¹æÂÉ£¨Éè×Óµ¯ËٶȵIJ»È·¶¨³Ì¶ÈΪ10-3m/s£©
½â ¸ù¾ÝµÂ²¼ÂÞÒâ¹ØÏµÊ½£¬µÃ×Óµ¯Ô˶¯µÄ²¨³¤ h6.625?10?34-35-26
¦Ë===6.626¡Á10=6.626¡Á10nm 3mv0.01?1?10ÒÑÖª¦ÃÉäÏßÊÇ×î¶ÌµÄµç´Å²¨£¬Æä²¨³¤¦Ë=1¡Á10-5 nm;¿É¼û×Óµ¯µÄ²¨³¤Ì«Ð¡£¬¸ù±¾ÎÞ·¨²âÁ¿£¬¿ÉºöÂÔ¡£×Óµ¯µÈºê¹ÛÎïÌåÖ÷Òª±íÏÖΪÁ£×ÓÐÔ¡£
¸ù¾Ý²»È·¶¨¹ØÏµÊ½£¬×Óµ¯Î»ÖõIJ»È·¶¨Á¿
6.626?10?34hh?29¦¤x. ¦¤p¡Öh ¦¤x. ¦¤v¡Ö ¦¤x.?==6.626m ?10?3mm?V0.01?10λÖõIJ»È·¶¨³Ì¶ÈºÜС£¬¿ÉºöÂÔ¡£ËµÃ÷×Óµ¯µÄÔ˶¯ÓÐÈ·¶¨µÄ¹ìµÀ£¬Ô˶¯·þ´Ó¾µäÁ¦Ñ§¹æÂÉ¡£
3.ʲôÊÇÆÁ±ÎЧӦºÍ×괩ЧӦ£¿ÔõÑù½âÊÍͬһÖ÷²ãÖÐÄܼ¶·ÖÁѼ°²»Í¬Ö÷²ãÖеÄÄܼ¶½»´íÏÖÏó£¿
Z2´ð ¶ÔÓÚµ¥µç×ÓÔ×Ó£¬¸÷¹ìµÀµÄÄÜÁ¿ÎªE=¡ª13.62eV,µ¥µç×ÓÌåϵ£¨ÈçÇâÔ
n×Ó»òÀàÇâÀë×ÓLi?µÈ£©¹ìµÀµÄÄÜÁ¿Ö÷Ҫȡ¾öÓëÖ÷Á¿×ÓÊýn¡£¶ÔÓÚµ¥µç×ÓÔ×Ó»òÀë×Ó£¬Ö»ÒªnÏàͬ£¬¸÷¹ìµÀµÄÄÜÁ¿¾ÍÏàͬ£ºEns=Enp=End=E£¬ÕâЩÄÜÁ¿ÏàͬµÄ¹ìµÀ³Æ¼ò²¢¹ìµÀ¡£
ÔÚ¶àÔ×Ó»òÀë×ÓÖУ¬²»µ«ÓÐÔ×Ӻ˶Եç×ÓµÄÎüÒý£¬»¹Óеç×ÓÖ®¼äµÄÅų⡣ÔÚÌÖÂÛijһµç×ÓµÄÄÜÁ¿Ê±£¬½«ÄÚ²ãµç×ӶԸõç×ÓµÄÅųâ×÷Óùé½áΪ¶ÔºËµçºËµÄÆÁ±ÎºÍ²¿·ÖµÖÏû£¬´Ó¶øÊ¹ÓÐЧºËµçºÉ½µµÍ£¬Ï÷ÈõÁ˺˵çºÉ¶Ôµç×ÓµÄÒýÁ¦£¬ÕâÖÖ×÷ÓóÆÎªÆÁ±ÎЧӦ¡£ÕâÑùÒ»À´£¬¶àµç×ÓÌåϵ¹ìµÀµÄÄÜÁ¿Îª
2(Z-?)2Z* E= ¡ª13.62£¨ev£©= -13.6£¨ev£© 2nnʽÖУ¬z*ΪÓÐЧºËµçºÉ£¬?ΪÆÁ±Î³£Êý¡£Òò´Ë£¬¶àµç×ÓÔ×Ó»òÀë×Ó¹ìµÀµÄÄÜÁ¿³ýÓëÖ÷Á¿×ÓÊýnÓйØÍ⣬»¹ÓëÆÁ±Î³£Êý?ÖµÓйء£¶ø?ÖµÓë½ÇÁ¿×ÓÊýLµÈÐí¶àÒòËØÓйء£ÔÚnÏàͬµÄ¹ìµÀÖУ¬LֵԽСµÄµç×ÓÊÜÄÚ²ãµç×ӵįÁ±Î×÷ÓÃԽС£¬¹ÊÓйìµÀµÄÄÜÁ¿Ens©‚Enp©‚End©‚Enf£¬Ôì³É¡°Äܼ¶·ÖÁÑ¡±ÏÖÏó¡£ÎªÁ˽âÊÍ¡°Äܼ¶·ÖÁÑ¡±ÏÖÏ󣬱ØÐëÑо¿Ô×Ó¹ìµÀµÄ¾¶Ïò·Ö²¼º¯Êýͼ£¬¸ù¾ÝÔ×Ó¹ìµÀµÄ¾¶Ïò·Ö²¼Í¼¿ÉÖª£ºÖ÷Á¿×ÓÊýÏàͬµÄ3s£¬3p£¬3d¹ìµÀÖУ¬½ÇÁ¿×ÓÊý×îСµÄ3s¹ìµÀ²»½ö¾¶Ïò·Ö²¼·åµÄ¸öÊý×î¶à£¬¶øÇÒÔÚ¿¿½üºË¸½½üÓÐÒ»¸öС·å£¨¼´×êµÄÀëºËºÜ½ü£©£¬Òò´Ë3sµç×Ó
?34
±»ÄÚ²ãµç×ÓÆÁ±ÎµÄ×îÉÙ£¬Æ½¾ùÊܺ˵ÄÒýÁ¦½Ï´ó£¬ÆäÄÜÁ¿×îµÍ£¨ÖµµÃÖ¸³öµÄÊÇ£¬3s¹ìµÀ¾¶Ïò·Ö²¼Í¼ÖÐÀëºË½Ï½üµÄ·ìÃæ»ýËäС£¬µ«Õⲿ·Öµç×ÓÔÆ³öÏÖµÄÇøÓòÀëºË½Ï½ü£¬Òò¶øÓÐЧºËµçºÉ´ó£¬Ëü¶Ô½µµÍ¹ìµÀÄÜÁ¿½«ÆðÖØ´ó¹±Ï×£©£»¶ø3p¼°3dµç×Ó×êÈëÄÚ²ãµÄ³Ì¶ÈÒÀ´Î¼õÉÙ£¬ÄÚ²ãµç×Ó¶ÔÆäÆÁ±Î×÷ÓÃÖð½¥ÔöÇ¿£¬¹ÊËüÃǵÄÄÜÁ¿Ïà¼ÌÔö´ó¡£ÎÒÃǰÑÓÉÓÚÔ×Ó¹ìµÀµÄ¾¶Ïò·Ö²¼²»Í¬£¬µç×Ó´©¹ýÄÚ²ã×êµ½½üºËÄÜÁ¦²»Í¬¶øÒýÆðµç×ÓµÄÄÜÁ¿²»Í¬µÄÏÖÏó³ÆÎªµç×ÓµÄ×괩ЧӦ¡£¿É¼û×ê´©ÓëÆÁ±ÎÊÇ»¥ÏàÁªÏµµÄ¡£
×괩ЧӦµÄ½á¹û»¹»áÒýÆð¡°Äܼ¶½»´í¡±ÏÖÏó¡£ÀýÈçÔ×ÓÐòÊýΪ19ºÍ20µÈÔªËØµÄ¹ìµÀÄÜÁ¿E4s¤¯E3d¡£ÕâÊÇÒòΪ4sºÍ3d¹ìµÀµÄ¾¶Ïò·Ö²¼²»Í¬£¬ËäÈ»4sµÄ×î´ó·å±È3d×î´ó·åÀëºË½ÏÔ¶£¬µ«4sµÄС·å×êµ½¿¿ºËºÜ½üµÄÄڲ㣬Òò¶ø´ó´ó½µµÍÁË4sµç×ÓµÄÄÜÁ¿£¬ÒÔÖÂÓÚ4s±È3dµç×ÓÄÜÁ¿»¹µÍ¡£ 4. Çë½âÊÍÔÒò£º
(1) He+ÖÐ3sºÍ3p¹ìµÀµÄÄÜÁ¿ÏàµÈ£¬¶øÔÚAr+ÖÐ3sºÍ3p¹ìµÀµÄÄÜÁ¿²»ÏàµÈ¡£
(2) µÚÒ»µç×ÓÇ׺ÍÄÜΪCl>F£¬S>O£»¶ø²»ÊÇF>Cl , O>S¡£
½â£¨1£©He+ÖÐÖ»ÓÐÒ»¸öµç×Ó£¬Ã»ÓÐÆÁ±ÎЧӦ£¬¹ìµÀµÄÄÜÁ¿ÓÉÖ÷Á¿×ÓÊýn¾ö¶¨£¬nÏàͬµÄ¹ìµÀÄÜÁ¿Ïàͬ£¬Òò¶ø3sºÍ3p¹ìµÀµÄÄÜÁ¿Ïàͬ¡£¶øÔÚAr+ÖУ¬Óжà¸öµç×Ó´æÔÚ£»3s¹ìµÀµÄµç×ÓÓë3p¹ìµÀµÄµç×ÓÊܵ½µÄÆÁ±ÎЧӦ²»Í¬£¬¼´¹ìµÀµÄÄÜÁ¿²»½öºÍÖ÷Á¿×ÓÊýnÓйأ¬»¹ºÍ½ÇÁ¿×ÓÊýLÓйء£Òò´Ë£¬3sºÍ3p¹ìµÀµÄÄÜÁ¿²»Í¬¡£
£¨2£©Ò»°ãÀ´Ëµ£¬µç×ÓÇ׺ÍÄÜËæÔ×Ó°ë¾¶¼õС¶øÔö´ó£¬ÒòΪ°ë¾¶Ô½Ð¡£¬ºËµçºÉ¶Ôµç×ÓµÄÒýÁ¦Ô½´ó¡£Òò´Ë£¬µç×ÓÇ׺ÍÄÜÔÚͬ×åÖдÓÉϵ½Ï³ʼõÉÙµÄÇ÷ÊÆ¡£µ«µÚÒ»µç×ÓÇ׺ÍÄÜÈ´³öÏÖCl>F, S>OµÄ·´³£ÏÖÏó£¬ÕâÊÇÓÉÓÚOºÍF °ë¾¶¹ýС£¬µç×ÓÔÆÃܶȹý¸ß£¬µ±Ô×Ó½áºÏÒ»¸öµç×ÓÐγɸºÀë×Óʱ£¬ÓÉÓÚµç×Ó¼äµÄÅųâ×÷ÓýÏǿʹ·Å³öµÄÄÜÁ¿¼õÉÙ¡£
5. ¸ù¾ÝÔ×ÓÐòÊý¸ø³öÏÂÁÐÔªËØµÄ»ù̬Ô×ӵĺËÍâµç×Ó×é̬£º £¨a£©K (Z=19) £¨b£©Al (Z=13) £¨c£© Cl (Z=17) £¨d£©Ti£¨Z=22£© £¨e£©Zn£¨Z=30£© £¨f£©As£¨Z=33£©
½â £¨a£©[Ar]4s1£¨b£©[Ne]3s23p1£¨c£©[Ne]3s23p5£¨d£©[Ar]3d24s2£¨e£©[Ar] 3d104s2£¨f£©[Ar]3d104s24p3
6. ÒÔÏÂ+3¼ÛÀë×ÓÄÇЩ¾ßÓÐ8µç×ÓÍâ¿Ç£¿Al3+¡¢Ga3+¡¢Bi3+¡¢Mn3+¡¢Sc3+ ½â Al3+ºÍSc3+¾ßÓÐ8µç×ÓÍâ¿Ç¡£
7.ÅжÏÏÂÁи÷¶ÔÔªËØÄĸöÔªËØµÚÒ»µçÀëÄܴ󣬲¢ËµÃ÷ÔÒò¡£ SºÍP Al ºÍ Mg SrºÍRb CuºÍ Zn CsºÍAu
½â ÌâÖи÷¶ÔÔªËØÍ¬ÖÜÆÚÔªËØÒ»°ãÀ´Ëµ£¬ÔÚͬһÖÜÆÚÖУ¬´Ó×óµ½ÓÒËæ×ÅÓÐЧºËµçºÉµÄÔö¼Ó£¬°ë¾¶¼õС£¬µÚÒ»µçÀëÄÜ×ܵÄÇ÷ÊÆÊÇÔö´ó¡£µ«ÓÉÓÚµç×Ó¹¹ÐͶԵçÀëÄÜÓ°Ïì½Ï´ó£¬¿ÉÄÜ»áÔì³ÉijЩ·´³£ÏÖÏó¡£
23
P>S Òòpµç×Ó¹¹ÐÍΪ3s3p£¬3p¹ìµÀ°ë³äÂúÎȶ¨½á¹¹£¬´òµôµç×ÓÄÑ£»¶øSµÄµç×Ó¹¹ÐÍΪ3s23p4ʧȥһ¸öµç×ÓºóΪ°ë³äÂúµÄÎȶ¨½á¹¹£¬¹ÊP>S¡£
Mg>Al µÀÀíͬ£¨1£©Ò»Ñù¡£
Sr>Rb SrµÄºËµçºÉ±ÈRb¶à£¬°ë¾¶Ò²±ÈRbС¡£Æä´ÎSrµÄ5s2½ÏÎȶ¨¡£ Zn>Cu ZnµÄºËµçºÉ±ÈCu¶à¡£Í¬Ê±ZnµÄ3d¹ìµÀÈ«³äÂú£¬4s¹ìµÀÈ«³äÂú£»CuµÄ4s¹ìµÀ°ë³äÂú¡£Ê§È¥Ò»¸öµç×ÓºóΪ3d104s0Îȶ¨½á¹¹¡£
Au>Cs ¶þÕßΪͬÖÜÆÚµÄÔªËØ£¬AuΪIB×åÔªÊý£¬CsΪIA×åÔªËØ£¬½ðµÄÓÐЧºËµçºÉÊý½Ï蘆Ĵ󣬶ø°ë¾¶½Ï蘆ÄС£¬¶øÇÒï¤Ê§È¥Ò»¸öµç×Óºó±äΪ5s25p6Îȶ¨½á¹¹¡£
8. Åжϰ뾶´óС²¢ËµÃ÷ÔÒò£º
£¨1£©Sr ÓëBe £¨2£©Ca ÓëSc £¨3£©NiÓëCu £¨4£©ZrÓëHf £¨5£©S2- ÓëS £¨6£©Na+ÓëAl3+ £¨7£©Sn2+ÓëPb2+ £¨8£©Fe2+ÓëFe3+
½â £¨1£©Ba> Sr ͬ×åÔªËØBa±ÈSr¶àÒ»µç×Ӳ㡣 £¨2£©Ca>Sc ͬÖÜÆÚÔªËØ£¬ScºËµçºÉ¶à
£¨3£©Cu>Ni ͬÖÜÆÚÔªËØ£¬Cu´ÎÍâ²ãΪ18µç×Ó£¬Êܵ½ÆÁ±Î×÷Óôó£¬ÓÐЧºËµçºÉС£¬Íâ²ãµç×ÓÊÜÒýÁ¦Ð¡£¬
£¨4£©Zr?Hf ïçϵÊÕËõµÄ½á¹û£»
£¨5£©S2->S Í¬Ò»ÔªËØ£¬µç×ÓÊýÔ½¶à£¬°ë¾¶Ô½´ó£»
£¨6£©Na+>Al3+ ͬһÖÜÆÚÔªËØ£¬ÕýµçºÉÔ½¸ß£¬°ë¾¶Ô½Ð¡£»£»
2+2+
£¨7£© Pb>Snͬһ×åÔªËØµÄÀë×Ó£¬ÕýµçºÉÊýÏàͬ£¬µ«Pb2+±ÈSn2+¶àÒ»µç×Ӳ㣻
2+3+
£¨8£©Fe>Fe Í¬Ò»ÔªËØÀë×Ó£¬ÕýµçºÉÔ½¸ßÔò°ë¾¶Ô½Ð¡¡£ 9. Èç¹û·¢ÏÖ116ºÅÔªËØ£¬Çë¸ø³ö
£¨1£©ÄÆÑεĻ¯Ñ§Ê½£» £¨2£©¼òµ¥Ç⻯ÎïµÄ»¯Ñ§Ê½£» £¨3£©×î¸ß¼Û̬µÄÑõ»¯ÎïµÄ»¯Ñ§Ê½£»£¨4£©¸ÃÔªËØÊǽðÊô»¹ÊǷǽðÊô¡£
½â ÒòΪ¸ÃÔªËØµÄµç×Ó¹¹ÐÍΪ ?Rn?7s25f146d107p4¸ÃÔªËØÎªµÚÁùÖ÷×åµÄÔªËØ¡£Éè¸ÃÔªËØÎªRÔò(1)Na2R £¨2£©H2R £¨3£©RO3 £¨4£©¸ÃÔªËØÎª½ðÊôÔªËØ£© ¶þ¡¢Ï°Ìâ
1.ÇâÔ×ÓµÄÒ»¸öµç×Ó´ÓµÚ¶þÄܼ¶²ãԾǨÖÁµÚÒ»Äܼ¶²ãʱ·¢Éä³ö¹â×ӵIJ¨³¤ÊÇ 121.6nm£»µ±µç×Ó´ÓµÚÈýÄܼ¶²ãԾǨÖÁµÚ¶þÄܼ¶²ãʱ£¬·¢Éä³ö¹â×ӵIJ¨³¤ÊÇ656.3nm¡£ÎÊÄÄÒ»¸ö¹â×ÓµÄÄÜÁ¿´ó£¿
£¨´ÓµÚ¶þÄܼ¶²ãԾǨÖÁµÚÒ»Äܼ¶²ãʱ·¢Éä³ö¹â×ÓµÄÄÜÁ¿´ó£©¡£
2.ÒÑÖªµçÖÐÐԵĻù̬Ô×ӵļ۵ç×Ó²ãµç×Ó×é̬·Ö±ðΪ£º £¨a£©3s3p5£¨b£©3d64s2£¨c£©5s2£¨d£©4f96s2£¨e£©5d106s1
ÊÔ¸ù¾ÝÕâ¸öÐÅϢȷ¶¨ËüÃÇÔÚÖÜÆÚ±íÖÐÊôÓÚÄǸöÇø¡¢Äĸö×å¡¢ÄĸöÖÜÆÚ¡£ £¨£¨a£©pÇø£¬¢÷A×壬µÚÈýÖÜÆÚ £¨b£©dÇø£¬¢ø×壬µÚËÄÖÜÆÚ £¨c£©sÇø£¬¢òA×壬µÚÎåÖÜÆÚ £¨d£©fÇø£¬¢óB×壬µÚÁùÖÜÆÚ £¨e£©dsÇø£¬¢ñB×壬µÚÁùÖÜÆÚ £©
3. ¸ù¾ÝTi¡¢Ge¡¢Ag¡¢Rb¡¢NeÔÚÖÜÆÚ±íÖеÄλÖã¬ÍƳöËüÃǵĻù̬Ô×ӵĵç×Ó×é̬¡£
(TiλÓÚµÚËÄÖÜÆÚ¢ôB×壬ËüµÄ»ù̬Ô×ӵĵç×Ó×é̬Ϊ[Ar]3d24s2£» GeλÓÚµÚËÄÖÜÆÚ¢ôA×壬ËüµÄ»ù̬Ô×ӵĵç×Ó×é̬Ϊ[Ar]3d104s24p2£» AgλÓÚµÚÎåÖÜÆÚ¢ñB×壬ËüµÄ»ù̬Ô×ӵĵç×Ó×é̬Ϊ[Kr] 4d105s1£» RbλÓÚµÚÎåÖÜÆÚ¢ñA×壬ËüµÄ»ù̬Ô×ӵĵç×Ó×é̬Ϊ[Kr] 5s1£» NeλÓÚµÚ¶þÖÜÆÚ0×壬ËüµÄ»ù̬Ô×ӵĵç×Ó×é̬Ϊ[He] 2s22p6¡£)
4.Ä³ÔªËØµÄ»ù̬¼Û²ãµç×Ó¹¹ÐÍΪ5d26s2£¬¸ø³ö±È¸ÃÔªËØµÄÔ×ÓÐòÊýС5µÄÔªËØµÄ»ù̬Ô×Óµç×Ó×é̬¡£(¸ÃÔªËØµÄ»ù̬Ô×Óµç×Ó×é̬Ϊ[Xe] 4f116s2)
5.Ä³ÔªËØ»ù̬Ô×Ó×îÍâ²ãΪ5s2£¬×î¸ßÑõ»¯Ì¬Îª+4£¬ËüλÓÚÖÜÆÚ±íÄĸöÇø£¿Êǵڼ¸ÖÜÆÚµÚ¼¸×åÔªËØ£¿Ð´³öËüµÄ+4Ñõ»¯Ì¬Àë×ӵĵç×Ó¹¹ÐÍ¡£ÈôÓÃA´ú±íËüµÄÔªËØ·ûºÅ£¬Ð´³öÏàÓ¦Ñõ»¯ÎïµÄ»¯Ñ§Ê½¡££¨¸ÃÔªËØµÄ»ù̬Ô×Óµç×Ó×é̬Ϊ[Kr] 4d25s2£¬¼´µÚ40ºÅÔªËØï¯£¨Zr£©¡£ËüλÓÚdÇø£¬µÚÎåÖÜÆÚ¢ôB×壬+4Ñõ»¯Ì¬Àë×ӵĵç×Ó¹¹ÐÍΪ[Kr]£¬¼´1s22s22p63s23p63d104s24p6£¬ÏàÓ¦Ñõ»¯ÎïΪAO2¡££©
51
6.Ò»ÔªËØµÄ¼Ûµç×Ӳ㹹ÐÍΪ3d4s,Ö¸³öÆäÔÚÖÜÆÚ±íÖеÄλÖã¨Çø¡¢ÖÜÆÚºÍ
2