{"id":281,"date":"2023-07-24T16:39:28","date_gmt":"2023-07-24T16:39:28","guid":{"rendered":"https:\/\/chemuza.org\/tr\/lewis-sbr4-yapisi\/"},"modified":"2023-07-24T16:39:28","modified_gmt":"2023-07-24T16:39:28","slug":"lewis-sbr4-yapisi","status":"publish","type":"post","link":"https:\/\/chemuza.org\/tr\/lewis-sbr4-yapisi\/","title":{"rendered":"5 ad\u0131mda sbr4 lewis yap\u0131s\u0131 (resimlerle)"},"content":{"rendered":"<figure class=\"wp-block-image aligncenter size-full\"><img decoding=\"async\" loading=\"lazy\" width=\"650\" height=\"447\" src=\"https:\/\/chemuza.org\/wp-content\/uploads\/2023\/08\/structure-sbr4-lewis.jpg\" alt=\"Lewis yap\u0131s\u0131 SBr4\" class=\"wp-image-3537\" srcset=\"\" sizes=\"\"><\/figure>\n<p> Yukar\u0131daki g\u00f6rseli zaten g\u00f6rd\u00fcn\u00fcz de\u011fil mi?<\/p>\n<p> Yukar\u0131daki g\u00f6rseli k\u0131saca a\u00e7\u0131klayay\u0131m.<\/p>\n<p> <strong><em><mark style=\"background-color:rgba(0, 0, 0, 0);color:#ff0000\" class=\"has-inline-color\">SBr4 Lewis yap\u0131s\u0131n\u0131n merkezinde d\u00f6rt bromin (Br) atomu ile \u00e7evrelenmi\u015f bir k\u00fck\u00fcrt (S) atomu bulunur. K\u00fck\u00fcrt (S) atomu ile her Brom (Br) atomu aras\u0131nda 4 adet tekli ba\u011f vard\u0131r. K\u00fck\u00fcrt (S) atomunda 1 yaln\u0131z \u00e7ift, d\u00f6rt bromin (Br) atomunda ise 3 yaln\u0131z \u00e7ift bulunur.<\/mark><\/em><\/strong><\/p>\n<p> SBr4&#8217;\u00fcn Lewis yap\u0131s\u0131n\u0131n yukar\u0131daki g\u00f6r\u00fcnt\u00fcs\u00fcnden hi\u00e7bir \u015fey anlamad\u0131ysan\u0131z, benimle kal\u0131n ve SBr4&#8217;\u00fcn Lewis yap\u0131s\u0131n\u0131n \u00e7izilmesine ili\u015fkin ayr\u0131nt\u0131l\u0131 ad\u0131m ad\u0131m a\u00e7\u0131klamay\u0131 alacaks\u0131n\u0131z.<\/p>\n<p> \u015eimdi SBr4&#8217;\u00fcn <a href=\"https:\/\/chemuza.org\/tr\/kimyanin-temel-tanimlari\/\">Lewis yap\u0131s\u0131n\u0131<\/a> \u00e7izme ad\u0131mlar\u0131na ge\u00e7elim.<\/p>\n<h2 class=\"wp-block-heading\"> <strong>SBr4 Lewis yap\u0131s\u0131n\u0131 \u00e7izme ad\u0131mlar\u0131<\/strong><\/h2>\n<h3 class=\"wp-block-heading\"> <strong>Ad\u0131m 1: SBr4 molek\u00fcl\u00fcndeki toplam de\u011ferlik elektronu say\u0131s\u0131n\u0131 bulun<\/strong><\/h3>\n<p> <a href=\"https:\/\/chemuza.org\/tr\/kimyanin-temel-tanimlari\/\">SBr4 molek\u00fcl\u00fcndeki<\/a> toplam <a href=\"https:\/\/www.lanl.gov\/orgs\/nmt\/nmtdo\/AQarchive\/04spring\/VO.html\" target=\"_blank\" rel=\"noopener\">de\u011ferlik elektron<\/a> say\u0131s\u0131n\u0131 bulmak i\u00e7in \u00f6ncelikle brom atomunun yan\u0131 s\u0131ra k\u00fck\u00fcrt atomunda bulunan de\u011ferlik elektronlar\u0131n\u0131 bilmeniz gerekir.<br \/> (De\u011ferlik elektronlar\u0131 herhangi bir atomun en d\u0131\u015f <a href=\"https:\/\/chemuza.org\/tr\/kimyanin-temel-tanimlari\/\">y\u00f6r\u00fcngesinde<\/a> bulunan elektronlard\u0131r.)<\/p>\n<p> Burada size periyodik tabloyu kullanarak bromun yan\u0131 s\u0131ra k\u00fck\u00fcrt\u00fcn de\u011ferlik elektronlar\u0131n\u0131 nas\u0131l kolayca bulaca\u011f\u0131n\u0131z\u0131 anlataca\u011f\u0131m.<\/p>\n<p> <strong>SBr4 molek\u00fcl\u00fcndeki toplam de\u011ferlik elektronlar\u0131<\/strong><\/p>\n<p> <strong>\u2192 K\u00fck\u00fcrt atomunun verdi\u011fi de\u011ferlik elektronlar\u0131:<\/strong> <\/p>\n<figure class=\"wp-block-image aligncenter size-full\"><img decoding=\"async\" loading=\"lazy\" width=\"600\" height=\"300\" src=\"https:\/\/chemuza.org\/wp-content\/uploads\/2023\/08\/2-3.jpg\" alt=\"\" class=\"wp-image-66\" srcset=\"\" sizes=\"\"><\/figure>\n<p> K\u00fck\u00fcrt periyodik tablonun 16. grubunda yer alan bir elementtir. <a href=\"https:\/\/www.rsc.org\/periodic-table\/element\/16\/sulfur\" target=\"_blank\" rel=\"noopener\"><sup>[1]<\/sup><\/a> Bu nedenle k\u00fck\u00fcrtteki de\u011ferlik elektronlar\u0131 <strong>6&#8217;d\u0131r<\/strong> . <\/p>\n<figure class=\"wp-block-image aligncenter size-full\"><img decoding=\"async\" loading=\"lazy\" width=\"600\" height=\"273\" src=\"https:\/\/chemuza.org\/wp-content\/uploads\/2023\/08\/3-3.jpg\" alt=\"\" class=\"wp-image-67\" srcset=\"\" sizes=\"\"><\/figure>\n<p> Yukar\u0131daki resimde g\u00f6sterildi\u011fi gibi k\u00fck\u00fcrt atomunda bulunan 6 de\u011ferlik elektronunu g\u00f6rebilirsiniz.<\/p>\n<p> <strong>\u2192 Brom atomunun verdi\u011fi de\u011ferlik elektronlar\u0131:<\/strong> <\/p>\n<figure class=\"wp-block-image aligncenter size-full\"><img decoding=\"async\" loading=\"lazy\" width=\"600\" height=\"300\" src=\"https:\/\/chemuza.org\/wp-content\/uploads\/2023\/08\/2-9.jpg\" alt=\"\" class=\"wp-image-653\" srcset=\"\" sizes=\"\"><\/figure>\n<p> Brom, periyodik tablonun 17. grubunda yer alan bir elementtir. <a href=\"https:\/\/en.wikipedia.org\/wiki\/Bromine\" target=\"_blank\" rel=\"noopener\"><sup>[2]<\/sup><\/a> Bu nedenle bromda bulunan de\u011ferlik elektronlar\u0131 <strong>7&#8217;dir<\/strong> . <\/p>\n<figure class=\"wp-block-image aligncenter size-full\"><img decoding=\"async\" loading=\"lazy\" width=\"600\" height=\"266\" src=\"https:\/\/chemuza.org\/wp-content\/uploads\/2023\/08\/3-9.jpg\" alt=\"\" class=\"wp-image-654\" srcset=\"\" sizes=\"\"><\/figure>\n<p> Yukar\u0131daki resimde g\u00f6sterildi\u011fi gibi brom atomunda bulunan 7 de\u011ferlik elektronunu g\u00f6rebilirsiniz.<\/p>\n<p> Bu y\u00fczden,<\/p>\n<p> <strong>SBr4 molek\u00fcl\u00fcndeki toplam de\u011ferlik elektronlar\u0131<\/strong> = 1 k\u00fck\u00fcrt atomu taraf\u0131ndan ba\u011f\u0131\u015flanan de\u011ferlik elektronlar\u0131 + 4 brom atomu taraf\u0131ndan ba\u011f\u0131\u015flanan de\u011ferlik elektronlar\u0131 = <strong>6 + 7(4) = 34<\/strong> .<\/p>\n<h3 class=\"wp-block-heading\"> <strong>Ad\u0131m 2: Merkez atomu se\u00e7in<\/strong><\/h3>\n<p> Merkez atomu se\u00e7mek i\u00e7in en az <a href=\"https:\/\/chemuza.org\/tr\/kimyanin-temel-tanimlari\/\">elektronegatif<\/a> atomun merkezde kald\u0131\u011f\u0131n\u0131 unutmamal\u0131y\u0131z.<\/p>\n<p> \u015eimdi burada verilen molek\u00fcl SBr4&#8217;t\u00fcr ve k\u00fck\u00fcrt (S) atomlar\u0131 ve brom (Br) atomlar\u0131 i\u00e7erir. <\/p>\n<figure class=\"wp-block-image aligncenter size-full\"><img decoding=\"async\" loading=\"lazy\" width=\"800\" height=\"478\" src=\"https:\/\/chemuza.org\/wp-content\/uploads\/2023\/08\/5.jpg\" alt=\"\" class=\"wp-image-29\" srcset=\"\" sizes=\"\"><\/figure>\n<p> Yukar\u0131daki periyodik tabloda k\u00fck\u00fcrt atomu (S) ve brom atomunun (Br) elektronegatiflik de\u011ferlerini g\u00f6rebilirsiniz.<\/p>\n<p> K\u00fck\u00fcrt (S) ve bromun (Br) elektronegatiflik de\u011ferlerini kar\u015f\u0131la\u015ft\u0131r\u0131rsak <a href=\"https:\/\/chemuza.org\/tr\/periyodik-tablonun-elektronegatifligi\/\">k\u00fck\u00fcrt atomu daha az elektronegatiftir<\/a> .<\/p>\n<p> Burada k\u00fck\u00fcrt (S) atomu merkez atom, brom (Br) atomlar\u0131 ise d\u0131\u015f atomlard\u0131r. <\/p>\n<figure class=\"wp-block-image aligncenter size-full\"><img decoding=\"async\" loading=\"lazy\" width=\"600\" height=\"227\" src=\"https:\/\/chemuza.org\/wp-content\/uploads\/2023\/08\/sbr4-etape-1.webp\" alt=\"SBr4 a\u015fama 1\" class=\"wp-image-3538\" srcset=\"\" sizes=\"\"><\/figure>\n<h3 class=\"wp-block-heading\"> <strong>Ad\u0131m 3: Her atomu aralar\u0131na bir \u00e7ift elektron yerle\u015ftirerek ba\u011flay\u0131n<\/strong><\/h3>\n<p> \u015eimdi SBr4 molek\u00fcl\u00fcnde <a href=\"https:\/\/chemuza.org\/tr\/kimyanin-temel-tanimlari\/\">elektron \u00e7iftlerini<\/a> k\u00fck\u00fcrt atomu (S) ile bromin atomlar\u0131 (Br) aras\u0131na yerle\u015ftirmeniz gerekir. <\/p>\n<figure class=\"wp-block-image aligncenter size-full\"><img decoding=\"async\" loading=\"lazy\" width=\"600\" height=\"225\" src=\"https:\/\/chemuza.org\/wp-content\/uploads\/2023\/08\/sbr4-etape-2.webp\" alt=\"SBr4 a\u015fama 2\" class=\"wp-image-3539\" srcset=\"\" sizes=\"\"><\/figure>\n<p> Bu, k\u00fck\u00fcrt (S) ve bromin (Br)&#8217;nin bir SBr4 molek\u00fcl\u00fcnde <a href=\"https:\/\/doi.org\/10.1016\/B978-0-12-804492-6.00002-2\" target=\"_blank\" rel=\"noopener\">kimyasal olarak birbirine ba\u011fland\u0131\u011f\u0131n\u0131<\/a> g\u00f6sterir.<\/p>\n<h3 class=\"wp-block-heading\"> <strong>Ad\u0131m 4: D\u0131\u015f Atomlar\u0131 Kararl\u0131 Hale Getirin<\/strong><\/h3>\n<p> Bu ad\u0131mda d\u0131\u015f atomlar\u0131n kararl\u0131l\u0131\u011f\u0131n\u0131 kontrol etmeniz gerekir.<\/p>\n<p> Burada SBr4 molek\u00fcl\u00fcn\u00fcn \u00e7iziminde d\u0131\u015ftaki atomlar\u0131n brom atomlar\u0131 oldu\u011funu g\u00f6rebilirsiniz.<\/p>\n<p> Bu harici brom atomlar\u0131 bir <a href=\"https:\/\/en.wikibooks.org\/wiki\/General_Chemistry\/Octet_Rule_and_Exceptions\" target=\"_blank\" rel=\"noopener\">oktet<\/a> olu\u015fturur ve bu nedenle stabildir. <\/p>\n<figure class=\"wp-block-image aligncenter size-full\"><img decoding=\"async\" loading=\"lazy\" width=\"600\" height=\"388\" src=\"https:\/\/chemuza.org\/wp-content\/uploads\/2023\/08\/sbr4-etape-3.webp\" alt=\"SBr4 ad\u0131m 3\" class=\"wp-image-3540\" srcset=\"\" sizes=\"\"><\/figure>\n<p> Ek olarak 1. ad\u0131mda SBr4 molek\u00fcl\u00fcnde bulunan toplam de\u011ferlik elektron say\u0131s\u0131n\u0131 hesaplad\u0131k.<\/p>\n<p> SBr4 molek\u00fcl\u00fcnde toplam <strong>34 de\u011ferlik elektronu<\/strong> vard\u0131r ve yukar\u0131daki diyagramda bunlardan sadece <strong>32 de\u011ferlik elektronu<\/strong> kullan\u0131lm\u0131\u015ft\u0131r.<\/p>\n<p> Yani kalan elektron say\u0131s\u0131 = <strong>34 \u2013 32 = 2<\/strong> .<\/p>\n<p> Bu <strong>2<\/strong> elektronu yukar\u0131daki \u015femada SBr4 molek\u00fcl\u00fcn\u00fcn merkezi k\u00fck\u00fcrt atomuna yerle\u015ftirmeniz gerekiyor. <\/p>\n<figure class=\"wp-block-image aligncenter size-full\"><img decoding=\"async\" loading=\"lazy\" width=\"600\" height=\"329\" src=\"https:\/\/chemuza.org\/wp-content\/uploads\/2023\/08\/sbr4-etape-4.webp\" alt=\"SBr4 ad\u0131m 4\" class=\"wp-image-3541\" srcset=\"\" sizes=\"\"><\/figure>\n<p> \u015eimdi bir sonraki ad\u0131ma ge\u00e7elim.<\/p>\n<h3 class=\"wp-block-heading\"> <strong>Ad\u0131m 5: Lewis yap\u0131s\u0131n\u0131n kararl\u0131l\u0131\u011f\u0131n\u0131 kontrol edin<\/strong><\/h3>\n<p> Art\u0131k SBr4&#8217;\u00fcn Lewis yap\u0131s\u0131n\u0131n kararl\u0131l\u0131\u011f\u0131n\u0131 kontrol etmeniz gereken son ad\u0131ma geldiniz.<\/p>\n<p> Lewis yap\u0131s\u0131n\u0131n kararl\u0131l\u0131\u011f\u0131 <a href=\"https:\/\/employees.csbsju.edu\/cschaller\/Principles%20Chem\/molecules\/1_5_formal_charge.htm\" target=\"_blank\" rel=\"noopener\">formal y\u00fck<\/a> kavram\u0131 kullan\u0131larak do\u011frulanabilir.<\/p>\n<p> K\u0131sacas\u0131, \u015fimdi SBr4 molek\u00fcl\u00fcnde bulunan bromin atomlar\u0131n\u0131n (Br) yan\u0131 s\u0131ra k\u00fck\u00fcrt atomlar\u0131n\u0131n (S) formal y\u00fck\u00fcn\u00fc bulmam\u0131z gerekiyor.<\/p>\n<p> Resmi vergiyi hesaplamak i\u00e7in a\u015fa\u011f\u0131daki form\u00fcl\u00fc kullanman\u0131z gerekir:<\/p>\n<p class=\"has-background\" style=\"background-color:#ffe9cf\"> <strong>Resmi y\u00fck = De\u011ferlik elektronlar\u0131 \u2013 (Ba\u011f elektronlar\u0131)\/2 \u2013 Ba\u011f yapmayan elektronlar<\/strong><\/p>\n<p> A\u015fa\u011f\u0131daki resimde SBr4 molek\u00fcl\u00fcn\u00fcn her bir atomu i\u00e7in <a href=\"https:\/\/chemuza.org\/tr\/kimyanin-temel-tanimlari\/\">ba\u011flanan elektronlar\u0131n<\/a> ve <a href=\"https:\/\/chemuza.org\/tr\/kimyanin-temel-tanimlari\/\">ba\u011flanmayan elektronlar\u0131n<\/a> say\u0131s\u0131n\u0131 g\u00f6rebilirsiniz. <\/p>\n<figure class=\"wp-block-image aligncenter size-full\"><img decoding=\"async\" loading=\"lazy\" width=\"600\" height=\"324\" src=\"https:\/\/chemuza.org\/wp-content\/uploads\/2023\/08\/sbr4-etape-5.webp\" alt=\"SBr4 ad\u0131m 5\" class=\"wp-image-3542\" srcset=\"\" sizes=\"\"><\/figure>\n<p> <strong>K\u00fck\u00fcrt (S) atomu i\u00e7in:<\/strong><strong><br \/><\/strong> De\u011ferlik elektronlar\u0131 = 6 (\u00e7\u00fcnk\u00fc k\u00fck\u00fcrt 16. gruptad\u0131r)<strong><br \/><\/strong> Ba\u011f elektronlar\u0131 = 8<br \/> Ba\u011flanmayan elektronlar = 2<\/p>\n<p> <strong>Brom atomu (Br) i\u00e7in:<\/strong><br \/> De\u011ferlik elektronu = 7 (\u00e7\u00fcnk\u00fc brom 17. gruptad\u0131r)<br \/> Ba\u011f elektronlar\u0131 = 2<br \/> Ba\u011flanmayan elektronlar = 6 <\/p>\n<figure class=\"wp-block-table\">\n<table>\n<tbody>\n<tr>\n<td class=\"has-text-align-center\" data-align=\"center\"> <strong>Resmi su\u00e7lama<\/strong><\/td>\n<td> <strong>=<\/strong><\/td>\n<td class=\"has-text-align-center\" data-align=\"center\"> <strong>de\u011ferlik elektronlar\u0131<\/strong><\/td>\n<td> <strong>\u2013<\/strong><\/td>\n<td class=\"has-text-align-center\" data-align=\"center\"> <strong>(Elektronlar\u0131n ba\u011flanmas\u0131)\/2<\/strong><\/td>\n<td> <strong>\u2013<\/strong><\/td>\n<td class=\"has-text-align-center\" data-align=\"center\"> <strong>Ba\u011flanmayan elektronlar<\/strong> <\/td>\n<td><\/td>\n<td class=\"has-text-align-center\" data-align=\"center\"><\/td>\n<\/tr>\n<tr>\n<td class=\"has-text-align-center\" data-align=\"center\"> S<\/td>\n<td> =<\/td>\n<td class=\"has-text-align-center\" data-align=\"center\"> 6<\/td>\n<td> \u2013<\/td>\n<td class=\"has-text-align-center\" data-align=\"center\"> 8\/2<\/td>\n<td> \u2013<\/td>\n<td class=\"has-text-align-center\" data-align=\"center\"> 2<\/td>\n<td> =<\/td>\n<td class=\"has-text-align-center\" data-align=\"center\"> <strong>0<\/strong><\/td>\n<\/tr>\n<tr>\n<td class=\"has-text-align-center\" data-align=\"center\"> karde\u015fim<\/td>\n<td> =<\/td>\n<td class=\"has-text-align-center\" data-align=\"center\"> 7<\/td>\n<td> \u2013<\/td>\n<td class=\"has-text-align-center\" data-align=\"center\"> 2\/2<\/td>\n<td> \u2013<\/td>\n<td class=\"has-text-align-center\" data-align=\"center\"> 6<\/td>\n<td> =<\/td>\n<td class=\"has-text-align-center\" data-align=\"center\"> <strong>0<\/strong><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/figure>\n<p> Yukar\u0131daki formal y\u00fck hesaplamalar\u0131ndan, hem k\u00fck\u00fcrt (S) atomunun hem de bromin (Br) atomunun <strong>&#8220;s\u0131f\u0131r&#8221;<\/strong> formal y\u00fcke sahip oldu\u011funu g\u00f6rebilirsiniz.<\/p>\n<p> Bu, SBr4&#8217;\u00fcn yukar\u0131daki Lewis yap\u0131s\u0131n\u0131n stabil oldu\u011funu ve SBr4&#8217;\u00fcn yukar\u0131daki yap\u0131s\u0131nda ba\u015fka bir de\u011fi\u015fiklik olmad\u0131\u011f\u0131n\u0131 g\u00f6sterir.<\/p>\n<p> SBr4&#8217;\u00fcn yukar\u0131daki Lewis nokta yap\u0131s\u0131nda, her bir ba\u011f elektronu \u00e7iftini (:) tek bir ba\u011f (|) olarak da temsil edebilirsiniz. Bunu yapmak, SBr4&#8217;\u00fcn a\u015fa\u011f\u0131daki Lewis yap\u0131s\u0131na yol a\u00e7acakt\u0131r. <\/p>\n<figure class=\"wp-block-image aligncenter size-full\"><img decoding=\"async\" loading=\"lazy\" width=\"600\" height=\"311\" src=\"https:\/\/chemuza.org\/wp-content\/uploads\/2023\/08\/structure-de-Lewis-de-sbr4.jpg\" alt=\"SBr4'\u00fcn Lewis yap\u0131s\u0131\" class=\"wp-image-3543\" srcset=\"\" sizes=\"\"><\/figure>\n<p> Umar\u0131m yukar\u0131daki t\u00fcm ad\u0131mlar\u0131 tamamen anlam\u0131\u015fs\u0131n\u0131zd\u0131r.<\/p>\n<p> Daha fazla pratik yapmak ve daha iyi anlamak i\u00e7in a\u015fa\u011f\u0131da listelenen di\u011fer Lewis yap\u0131lar\u0131n\u0131 deneyebilirsiniz.<\/p>\n<style>\n.wp-block-table table, .wp-block-table td, .wp-block-table th {\n    border: 0;\n}\n<\/style>\n<p><strong>Daha iyi anlamak i\u00e7in \u015fu Lewis yap\u0131lar\u0131n\u0131 deneyin (veya en az\u0131ndan g\u00f6r\u00fcn):<\/strong><\/p>\n<figure class=\"wp-block-table\">\n<table>\n<tbody>\n<tr>\n<td> <a href=\"https:\/\/chemuza.org\/tr\/lewis-cfcl3un-yapisi\/\">Lewis yap\u0131s\u0131 CFCl3<\/a><\/td>\n<td> <a href=\"https:\/\/chemuza.org\/tr\/lewis-ncl2nin-yapisi\/\">NCl2-Lewis yap\u0131s\u0131<\/a><\/td>\n<\/tr>\n<tr>\n<td> <a href=\"https:\/\/chemuza.org\/tr\/lewis-ascl5-yapisi\/\">Lewis yap\u0131s\u0131 AsCl5<\/a><\/td>\n<td> <a href=\"https:\/\/chemuza.org\/tr\/cobr2-lewis-yapisi\/\">Lewis yap\u0131s\u0131 COBr2<\/a><\/td>\n<\/tr>\n<tr>\n<td> <a href=\"https:\/\/chemuza.org\/tr\/gef4-lewis-yapisi\/\">GeF4 Lewis yap\u0131s\u0131<\/a><\/td>\n<td> <a href=\"https:\/\/chemuza.org\/tr\/lewis-yapisi-cl2o2\/\">Cl2O2&#8217;nin Lewis yap\u0131s\u0131<\/a><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/figure>\n","protected":false},"excerpt":{"rendered":"<p>Yukar\u0131daki g\u00f6rseli zaten g\u00f6rd\u00fcn\u00fcz de\u011fil mi? Yukar\u0131daki g\u00f6rseli k\u0131saca a\u00e7\u0131klayay\u0131m. SBr4 Lewis yap\u0131s\u0131n\u0131n merkezinde d\u00f6rt bromin (Br) atomu ile \u00e7evrelenmi\u015f bir k\u00fck\u00fcrt (S) atomu bulunur. K\u00fck\u00fcrt (S) atomu ile her Brom (Br) atomu aras\u0131nda 4 adet tekli ba\u011f vard\u0131r. K\u00fck\u00fcrt (S) atomunda 1 yaln\u0131z \u00e7ift, d\u00f6rt bromin (Br) atomunda ise 3 yaln\u0131z \u00e7ift bulunur. SBr4&#8217;\u00fcn &#8230; <a title=\"5 ad\u0131mda sbr4 lewis yap\u0131s\u0131 (resimlerle)\" class=\"read-more\" href=\"https:\/\/chemuza.org\/tr\/lewis-sbr4-yapisi\/\" aria-label=\"More on 5 ad\u0131mda sbr4 lewis yap\u0131s\u0131 (resimlerle)\">Devam\u0131n\u0131 oku<\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[12],"tags":[],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v21.4 - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>5 Ad\u0131mda SBr4 Lewis Yap\u0131s\u0131 (Resimlerle) - Chemuza<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/chemuza.org\/tr\/lewis-sbr4-yapisi\/\" \/>\n<meta property=\"og:locale\" content=\"tr_TR\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"5 Ad\u0131mda SBr4 Lewis Yap\u0131s\u0131 (Resimlerle) - Chemuza\" \/>\n<meta property=\"og:description\" content=\"Yukar\u0131daki g\u00f6rseli zaten g\u00f6rd\u00fcn\u00fcz de\u011fil mi? Yukar\u0131daki g\u00f6rseli k\u0131saca a\u00e7\u0131klayay\u0131m. SBr4 Lewis yap\u0131s\u0131n\u0131n merkezinde d\u00f6rt bromin (Br) atomu ile \u00e7evrelenmi\u015f bir k\u00fck\u00fcrt (S) atomu bulunur. K\u00fck\u00fcrt (S) atomu ile her Brom (Br) atomu aras\u0131nda 4 adet tekli ba\u011f vard\u0131r. K\u00fck\u00fcrt (S) atomunda 1 yaln\u0131z \u00e7ift, d\u00f6rt bromin (Br) atomunda ise 3 yaln\u0131z \u00e7ift bulunur. SBr4&#8217;\u00fcn ... 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