ÌâÄ¿ÄÚÈÝ
£¨12·Ö£©A¡¢B¡¢C¡¢D¡¢E·Ö±ð´ú±íÎåÖÖ¶ÌÖÜÆÚÔªËØ£¬ÇÒÔ×ÓÐòÊýÒÀ´ÎÔö´ó£¬ÒÑÖª£ºBµÄ×îÍâ²ãµç×ÓÅŲ¼ÊÇnsnnpn£«1£»CµÄpÄܼ¶ÉÏδ³É¶ÔµÄµç×Ó±ÈBÉÙÒ»¸ö£»DµÄ¶þ¼ÛÑôÀë×ÓÓëCµÄÒõÀë×Ó¾ßÓÐÏàͬµÄµç×Ó²ã½á¹¹£»EÓëDͬÖÜÆÚÇÒEÔÚ¸ÃÖÜÆÚÖÐÔ×Ӱ뾶×îС£»BÓëAµÄµ¥ÖÊÄÜÉú³É¾ßÓд̼¤ÐÔÆøζµÄÆøÌ壬¸ÃÆøÌ弫Ò×ÈÜÓÚË®¡£
(1)BÔ×ÓºËÍâµç×ÓÅŲ¼Ê½Îª____________________________¡£
(2)AÓëE»¯ºÏʱÔ×Ó¼äÒÔ_____________¼üÏà½áºÏ£¬DÓëC»¯ºÏʱÔ×Ó¼äÒÔ_____________¼üÏà½áºÏ¡£
(3)д³öA¡¢CµÄµ¥ÖÊÖ±½Ó»¯ºÏÐγɵĻ¯ºÏÎïÓëEµ¥ÖÊ·´Ó¦µÄÀë×Ó·½³Ìʽ£º________________________________¡£
(4)AÓëBÐγɻ¯ºÏÎïʱ£¬ÖÐÐÄÔ×Ó²ÉÈ¡________ÔÓ»¯³É¼ü£¬ÆäÁ¢Ìå½á¹¹Îª________£¬ÊôÓÚ________(Ìî¡°¼«ÐÔ¡±»ò¡°·Ç¼«ÐÔ¡±)·Ö×Ó¡£
(5) ½«°×É«µÄÎÞË®CuSO4ÈܽâÓÚA2CÖУ¬ÈÜÒº³ÊÀ¶É«£¬ÊÇÒòΪÉú³ÉÁËÒ»ÖÖ³ÊÀ¶É«µÄÅäºÏÀë×Ó¡£Çëд³öÉú³É´ËÅäºÏÀë×ÓµÄÀë×Ó·½³Ìʽ£º ¡£
£¨6£©ÎåÖÖÔªËØÖе縺ÐÔ×î´óµÄÊÇ £¨ÌîÔªËØÃû³Æ£©¡£
(1)BÔ×ÓºËÍâµç×ÓÅŲ¼Ê½Îª____________________________¡£
(2)AÓëE»¯ºÏʱÔ×Ó¼äÒÔ_____________¼üÏà½áºÏ£¬DÓëC»¯ºÏʱÔ×Ó¼äÒÔ_____________¼üÏà½áºÏ¡£
(3)д³öA¡¢CµÄµ¥ÖÊÖ±½Ó»¯ºÏÐγɵĻ¯ºÏÎïÓëEµ¥ÖÊ·´Ó¦µÄÀë×Ó·½³Ìʽ£º________________________________¡£
(4)AÓëBÐγɻ¯ºÏÎïʱ£¬ÖÐÐÄÔ×Ó²ÉÈ¡________ÔÓ»¯³É¼ü£¬ÆäÁ¢Ìå½á¹¹Îª________£¬ÊôÓÚ________(Ìî¡°¼«ÐÔ¡±»ò¡°·Ç¼«ÐÔ¡±)·Ö×Ó¡£
(5) ½«°×É«µÄÎÞË®CuSO4ÈܽâÓÚA2CÖУ¬ÈÜÒº³ÊÀ¶É«£¬ÊÇÒòΪÉú³ÉÁËÒ»ÖÖ³ÊÀ¶É«µÄÅäºÏÀë×Ó¡£Çëд³öÉú³É´ËÅäºÏÀë×ÓµÄÀë×Ó·½³Ìʽ£º ¡£
£¨6£©ÎåÖÖÔªËØÖе縺ÐÔ×î´óµÄÊÇ £¨ÌîÔªËØÃû³Æ£©¡£
£¨¹²12·Ö£¬³ý·½³Ìʽ2·Ö£¬ÆäÓàÿ¿Õ1·Ö£©
(1)1s22s22p3¡¡ 2·Ö (2)¹²¼Û¡¡Àë×Ó 2·Ö
(3)Cl2£«H2O===H£«£«Cl££«HClO¡¡2·Ö (4)sp3¡¡Èý½Ç׶ÐΡ¡¼«ÐÔ 3·Ö
£¨5£©Cu2++4H2O=[Cu(H2O)4]2+ 2·Ö £¨6£©Ñõ 1·Ö
(1)1s22s22p3¡¡ 2·Ö (2)¹²¼Û¡¡Àë×Ó 2·Ö
(3)Cl2£«H2O===H£«£«Cl££«HClO¡¡2·Ö (4)sp3¡¡Èý½Ç׶ÐΡ¡¼«ÐÔ 3·Ö
£¨5£©Cu2++4H2O=[Cu(H2O)4]2+ 2·Ö £¨6£©Ñõ 1·Ö
¸ù¾ÝÔªËصĽṹ¼°ÓйØÐÔÖÊ¿ÉÖª£¬A¡¢B¡¢C¡¢D¡¢E·Ö±ðÊÇH¡¢N¡¢O¡¢Mg¡¢Cl¡£
£¨1£©¸ù¾Ý¹¹ÔìÔÀí¿ÉÖª£¬µªÔ×ӵĺËÍâµç×ÓÅŲ¼Ê½Îª1s22s22p3¡£
£¨2£©HºÍClÔ×Ó¶¼ÊǷǽðÊô£¬ËùÒÔ¶þÕßµÄÐγɵĻ¯Ñ§¼üÊǼ«ÐÔ¼ü£»ÑõÔªËØÊÇ»îÆõķǽðÊô£¬Ã¾ÊÇ»îÆõĽðÊô£¬¶þÕßÐγɵĻ¯Ñ§¼üÊÇÀë×Ó¼ü¡£
£¨3£©A¡¢CµÄµ¥ÖÊÖ±½Ó»¯ºÏÐγɵĻ¯ºÏÎïÊÇË®£¬ËùÒÔºÍÂÈÆø·´Ó¦µÄÀë×Ó·½³ÌʽÊÇCl2£«H2O===H£«£«Cl££«HClO¡£
£¨4£©¸ù¾Ý¼Û²ãµç×Ó¶Ô»¥³âÀíÂÛ¿ÉÖª£¬ÔÚ°±Æø·Ö×ÓÖеªÔªËغ¬ÓеŶԵç×Ó¶ÔÊýÊÇ£¨5£1¡Á3£©¡Â2£½1£¬ËùÒÔ°±ÆøÊÇÈý½Ç׶Ðνṹ£¬µªÔ×ÓÊÇsp3ÔÓ»¯£¬ÓÉÓÚÕý¸ºµçºÉ²»ÄÜÖغϣ¬ËùÒÔ°±ÆøÊôÓÚ¼«ÐÔ·Ö×Ó¡£
£¨5£©ÁòËáÍÖеçÀë³öµÄÌúÀë×ÓÄܺÍË®·Ö×ÓÐγÉÅäλ¼ü£¬ÆäÖÐË®ÊÇÅäÌ壬ÅäλÊýÊÇ4£¬ËùÒÔ·´Ó¦µÄÀë×Ó·½³ÌʽÊÇCu2++4H2O=[Cu(H2O)4]2+ ¡£
£¨6£©·Ç½ðÊôÐÔԽǿ£¬µç¸ºÐÔÔ½´ó¡£ÓÉÓÚÑõÔªËصķǽðÊôÐÔ×îÇ¿£¬ËùÒÔÑõÔªËصĵ縺ÐÔ×î´ó¡£
£¨1£©¸ù¾Ý¹¹ÔìÔÀí¿ÉÖª£¬µªÔ×ӵĺËÍâµç×ÓÅŲ¼Ê½Îª1s22s22p3¡£
£¨2£©HºÍClÔ×Ó¶¼ÊǷǽðÊô£¬ËùÒÔ¶þÕßµÄÐγɵĻ¯Ñ§¼üÊǼ«ÐÔ¼ü£»ÑõÔªËØÊÇ»îÆõķǽðÊô£¬Ã¾ÊÇ»îÆõĽðÊô£¬¶þÕßÐγɵĻ¯Ñ§¼üÊÇÀë×Ó¼ü¡£
£¨3£©A¡¢CµÄµ¥ÖÊÖ±½Ó»¯ºÏÐγɵĻ¯ºÏÎïÊÇË®£¬ËùÒÔºÍÂÈÆø·´Ó¦µÄÀë×Ó·½³ÌʽÊÇCl2£«H2O===H£«£«Cl££«HClO¡£
£¨4£©¸ù¾Ý¼Û²ãµç×Ó¶Ô»¥³âÀíÂÛ¿ÉÖª£¬ÔÚ°±Æø·Ö×ÓÖеªÔªËغ¬ÓеŶԵç×Ó¶ÔÊýÊÇ£¨5£1¡Á3£©¡Â2£½1£¬ËùÒÔ°±ÆøÊÇÈý½Ç׶Ðνṹ£¬µªÔ×ÓÊÇsp3ÔÓ»¯£¬ÓÉÓÚÕý¸ºµçºÉ²»ÄÜÖغϣ¬ËùÒÔ°±ÆøÊôÓÚ¼«ÐÔ·Ö×Ó¡£
£¨5£©ÁòËáÍÖеçÀë³öµÄÌúÀë×ÓÄܺÍË®·Ö×ÓÐγÉÅäλ¼ü£¬ÆäÖÐË®ÊÇÅäÌ壬ÅäλÊýÊÇ4£¬ËùÒÔ·´Ó¦µÄÀë×Ó·½³ÌʽÊÇCu2++4H2O=[Cu(H2O)4]2+ ¡£
£¨6£©·Ç½ðÊôÐÔԽǿ£¬µç¸ºÐÔÔ½´ó¡£ÓÉÓÚÑõÔªËصķǽðÊôÐÔ×îÇ¿£¬ËùÒÔÑõÔªËصĵ縺ÐÔ×î´ó¡£
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿