ÌâÄ¿ÄÚÈÝ

¡¾ÌâÄ¿¡¿·ú¼°Æ仯ºÏÎïÓÃ;·Ç³£¹ã·º¡£»Ø´ðÏÂÁÐÎÊÌ⣺

(1)»ù̬FÔ­×ӵļ۲ãµç×ӵĹìµÀ±í´ïʽΪ____¡£

(2)[H2F]+[SbF6]-(·úÌàËá)ÊÇÒ»ÖÖ³¬Ç¿Ëᣬ´æÔÚ[H2F]+£¬¸ÃÀë×ӵĿռ乹ÐÍΪ_____£¬Óë[H2F]+¾ßÓÐÏàͬ¿Õ¼ä¹¹ÐͺͼüºÏÐÎʽµÄ·Ö×ÓºÍÒõÀë×Ó·Ö±ðÊÇ_____ºÍ_____(¸÷¾ÙÒ»Àý)¡£

(3)NH4F(·ú»¯ï§)¿ÉÓÃÓÚ²£Á§µÄÊ´¿Ì·À¸¯¼Á¡¢Ïû¶¾¼Á¡£ÖÐÖÐÐÄÔ­×ÓµÄÔÓ»¯ÀàÐÍÊÇ_____£»·ú»¯ï§ÖдæÔڵĻ¯Ñ§¼üÊÇ_____(Ìî×Öĸ)¡£

A.Àë×Ó¼ü B.¦Ò¼ü C.¦Ð¼ü D.Çâ¼ü

(4)SF6±»¹ã·ºÓÃ×÷¸ßѹµçÆøÉ豸µÄ¾øÔµ½éÖÊ¡£SF6ÊÇÒ»ÖÖ¹²¼Û»¯ºÏÎ¿Éͨ¹ýÀàËÆÓÚBorn£­HaberÑ­»·ÄÜÁ¿¹¹½¨ÄÜÁ¿Í¼¼ÆËãÏà¹Ø¼üÄÜ¡£ÔòF£­F¼üµÄ¼üÄÜΪ____kJ¡¤mol-1£¬S£­F¼üµÄ¼üÄÜΪ____kJ¡¤mol-1¡£

¡¾´ð°¸¡¿ VÐÎ H2O sp3 AB 155 327

¡¾½âÎö¡¿

ͨ¹ý¼Û²ãµç×Ó¶ÔÊýÀ´ÅжÏÖÐÐÄÔ­×ÓµÄÔÓ»¯·½Ê½¡¢Á£×ӵļ¸ºÎ¹¹ÐÍ£»ÓÉͼ¿ÉµÃ£¬3F2(g)=6F(g)¡÷H=+465 kJ¡¤mol-1£¬ÔòF-F¼üµÄ¼üÄÜΪ465 kJ¡¤mol-1¡Â3=155 kJ¡¤mol-1£¬Í¬ÀíÇóS-FµÄ¼üÄÜ¡£

(1)FµÄÔ­×ÓÐòÊýΪ9£¬ºËÍâµç×ÓÅŲ¼Ê½Îª1s22s22p5£¬¼Ûµç×ÓΪ2s22p5£¬Òò´Ë»ù̬FÔ­×ӵļ۵ç×ÓÅŲ¼Í¼Îª¡£¹Ê´ð°¸Îª£º£»

(2) [H2F]+Àë×ÓÖÐÖÐÐÄÔ­×ÓFµÄ¹Âµç×Ó¶ÔÊý=£¬¦Ò¼üµç×Ó¶ÔÊýΪ2£¬¼Û²ãµç×Ó¶ÔÊýΪ4£¬VSEPRÄ£ÐÍΪËÄÃæÌåÐÍ£¬¿Õ¼ä¹¹ÐÍΪVÐÍ£»Óë[H2F]+¾ßÓÐÏàͬ¿Õ¼ä¹¹ÐͺͼüºÏÐÎʽµÄ·Ö×ÓºÍÒõÀë×Ó·Ö±ðÊÇH2OºÍ¡£¹Ê´ð°¸Îª£ºVÐÍ£»H2O£»£»

(3)ÖÐNÔ­×ÓÔÓ»¯¹ìµÀÊýΪ4+£¬N²ÉÈ¡sp3ÔÓ»¯£»·ú»¯ï§ÖÐ笠ùÀë×ÓÓë·úÀë×ÓÖ®¼ä´æÔÚÀë×Ó¼ü£¬ï§¸ùÀë×ÓÖеªÔ­×ÓÓëÇâÔ­×ÓÐγɵĵªÇâ¼üΪ¦Ò¼ü£¬²»´æÔÚË«¼üÔòûÓЦмü£¬Í¬Ê±Ò²²»´æÔÚÇâ¼ü£¬¹Ê´ð°¸Ñ¡AB¡£¹Ê´ð°¸Îª£ºsp3£»AB£»

(4)ÓÉͼ¿ÉµÃ£¬3F2(g)=6F(g)¡÷H=+465 kJ¡¤mol-1£¬ÔòF-F¼üµÄ¼üÄÜΪ465 kJ¡¤mol-1¡Â3=155 kJ¡¤mol-1£¬6F(g)+S(g)=SF6(g)£¬¡÷H=-1962 kJ¡¤mol-1£¬ÔòS-F¼üµÄ¼üÄÜΪ1962kJ¡¤mol-1¡Â6=327 kJ¡¤mol-1¡££¬¹Ê´ð°¸Îª£º155£»327¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

Î¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨µç»°£º027-86699610 ¾Ù±¨ÓÊÏ䣺58377363@163.com

¾«Ó¢¼Ò½ÌÍø