{"id":241,"date":"2023-07-25T01:35:15","date_gmt":"2023-07-25T01:35:15","guid":{"rendered":"https:\/\/chemuza.org\/tr\/lewis-c2br4-yapisi\/"},"modified":"2023-07-25T01:35:15","modified_gmt":"2023-07-25T01:35:15","slug":"lewis-c2br4-yapisi","status":"publish","type":"post","link":"https:\/\/chemuza.org\/tr\/lewis-c2br4-yapisi\/","title":{"rendered":"6 ad\u0131mda c2br4 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-c2br4-lewis.jpg\" alt=\"Lewis yap\u0131s\u0131 C2Br4\" class=\"wp-image-3067\" 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\">C2Br4 Lewis yap\u0131s\u0131, iki karbon (C) atomu aras\u0131nda bir \u00e7ift ba\u011fa ve karbon (C) atomu ile bromin (Br) atomlar\u0131 aras\u0131nda tek bir ba\u011fa sahiptir. T\u00fcm bromin (Br) atomlar\u0131nda 3 yaln\u0131z \u00e7ift vard\u0131r.<\/mark><\/em><\/strong><\/p>\n<p> C2Br4&#8217;\u00fcn Lewis yap\u0131s\u0131n\u0131n yukar\u0131daki g\u00f6r\u00fcnt\u00fcs\u00fcnden hi\u00e7bir \u015fey anlamad\u0131ysan\u0131z, benimle kal\u0131n ve <a href=\"http:\/\/www.chemspider.com\/Chemical-Structure.59617.html\" target=\"_blank\" rel=\"noopener\">C2Br4&#8217;\u00fcn<\/a> Lewis yap\u0131s\u0131n\u0131n nas\u0131l \u00e7izilece\u011fine dair ayr\u0131nt\u0131l\u0131 ad\u0131m ad\u0131m a\u00e7\u0131klamay\u0131 alacaks\u0131n\u0131z.<\/p>\n<p> O halde C2Br4&#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>C2Br4 Lewis Yap\u0131s\u0131n\u0131 \u00c7izim Ad\u0131mlar\u0131<\/strong><\/h2>\n<h3 class=\"wp-block-heading\"> <strong>Ad\u0131m 1: C2Br4 molek\u00fcl\u00fcndeki toplam de\u011ferlik elektronu say\u0131s\u0131n\u0131 bulun<\/strong><\/h3>\n<p> Bir C2Br4 <a href=\"https:\/\/chemuza.org\/tr\/kimyanin-temel-tanimlari\/\">molek\u00fcl\u00fcndeki<\/a> <a href=\"https:\/\/energyeducation.ca\/encyclopedia\/Valence_and_core_electrons\" target=\"_blank\" rel=\"noopener\">de\u011ferlik elektronlar\u0131n\u0131n<\/a> toplam say\u0131s\u0131n\u0131 bulmak i\u00e7in \u00f6ncelikle brom atomunun yan\u0131 s\u0131ra karbon 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 hem karbonun hem de bromun de\u011ferlik elektronlar\u0131n\u0131 nas\u0131l kolayca bulaca\u011f\u0131n\u0131z\u0131 anlataca\u011f\u0131m.<\/p>\n<p> <strong>C2Br4 molek\u00fcl\u00fcndeki toplam de\u011ferlik elektronlar\u0131<\/strong><\/p>\n<p> <strong>\u2192 Karbon 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\/1.jpg\" alt=\"\" class=\"wp-image-25\" srcset=\"\" sizes=\"\"><\/figure>\n<p> Karbon periyodik tablonun 14. grubunda yer alan bir elementtir. <a href=\"https:\/\/pubchem.ncbi.nlm.nih.gov\/element\/6\" target=\"_blank\" rel=\"noopener\"><sup>[1]<\/sup><\/a> Bu nedenle karbonda bulunan de\u011ferlik elektronlar\u0131 <strong>4&#8217;t\u00fcr<\/strong> . <\/p>\n<figure class=\"wp-block-image aligncenter size-full\"><img decoding=\"async\" loading=\"lazy\" width=\"600\" height=\"230\" src=\"https:\/\/chemuza.org\/wp-content\/uploads\/2023\/08\/2.jpg\" alt=\"\" class=\"wp-image-26\" srcset=\"\" sizes=\"\"><\/figure>\n<p> Yukar\u0131daki resimde g\u00f6sterildi\u011fi gibi karbon atomunda bulunan 4 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:\/\/www.rsc.org\/periodic-table\/element\/35\/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>C2Br4 molek\u00fcl\u00fcndeki toplam de\u011ferlik elektronlar\u0131<\/strong> = 2 karbon atomu taraf\u0131ndan ba\u011f\u0131\u015flanan de\u011ferlik elektronlar\u0131 + 4 bromin atomu taraf\u0131ndan ba\u011f\u0131\u015flanan de\u011ferlik elektronlar\u0131 = <strong>4(2) + 7(4) = 36<\/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 C2Br4&#8217;t\u00fcr ve karbon (C) atomlar\u0131 ve bromin (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 karbon atomunun (C) ve bromin atomunun (Br) elektronegatiflik de\u011ferlerini g\u00f6rebilirsiniz.<\/p>\n<p> Karbon (C) ve bromun (Br) elektronegatiflik de\u011ferlerini kar\u015f\u0131la\u015ft\u0131r\u0131rsak, <a href=\"https:\/\/chemuza.org\/tr\/periyodik-tablonun-elektronegatifligi\/\">karbon atomu daha az elektronegatiftir<\/a> .<\/p>\n<p> Burada karbon (C) atomlar\u0131 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=\"99\" src=\"https:\/\/chemuza.org\/wp-content\/uploads\/2023\/08\/c2br4-etape-1.webp\" alt=\"C2Br4 ad\u0131m 1\" class=\"wp-image-3068\" 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 C2Br4 molek\u00fcl\u00fcnde <a href=\"https:\/\/doi.org\/10.1038\/s41570-018-0052-4\" target=\"_blank\" rel=\"noopener\">elektron \u00e7iftlerini<\/a> karbon-karbon atomlar\u0131 aras\u0131na ve karbon-brom atomlar\u0131 aras\u0131na yerle\u015ftirmeniz gerekiyor. <\/p>\n<figure class=\"wp-block-image aligncenter size-full\"><img decoding=\"async\" loading=\"lazy\" width=\"600\" height=\"103\" src=\"https:\/\/chemuza.org\/wp-content\/uploads\/2023\/08\/c2br4-etape-2.webp\" alt=\"C2Br4 ad\u0131m 2\" class=\"wp-image-3069\" srcset=\"\" sizes=\"\"><\/figure>\n<p> Bu, bu atomlar\u0131n bir C2Br4 molek\u00fcl\u00fcnde birbirine <a href=\"https:\/\/www.lamar.edu\/arts-sciences\/_files\/documents\/chemistry-biochemistry\/dorris\/chapter8.pdf\" target=\"_blank\" rel=\"noopener\">kimyasal olarak 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. Kalan de\u011ferlik elektron \u00e7iftini merkez atomun \u00fczerine yerle\u015ftirin.<\/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 C2Br4 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.wikipedia.org\/wiki\/Octet_rule\" 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=\"262\" src=\"https:\/\/chemuza.org\/wp-content\/uploads\/2023\/08\/c2br4-etape-3.webp\" alt=\"C2Br4 ad\u0131m 3\" class=\"wp-image-3070\" srcset=\"\" sizes=\"\"><\/figure>\n<p> Ek olarak, 1. ad\u0131mda C2Br4 molek\u00fcl\u00fcnde bulunan toplam de\u011ferlik elektronu say\u0131s\u0131n\u0131 hesaplad\u0131k.<\/p>\n<p> C2Br4 molek\u00fcl\u00fcnde toplam <strong>36 de\u011ferlik elektronu<\/strong> vard\u0131r ve yukar\u0131daki diyagramda bunlardan sadece <strong>34 de\u011ferlik elektronu<\/strong> kullan\u0131lm\u0131\u015ft\u0131r.<\/p>\n<p> Yani kalan elektron say\u0131s\u0131 = <strong>36 \u2013 34 = 2<\/strong> .<\/p>\n<p> Bu <strong>2<\/strong> elektronu yukar\u0131daki diyagramda C2Br4 molek\u00fcl\u00fcn\u00fcn iki merkezi karbon atomuna yerle\u015ftirmeniz gerekir. <\/p>\n<figure class=\"wp-block-image aligncenter size-full\"><img decoding=\"async\" loading=\"lazy\" width=\"600\" height=\"249\" src=\"https:\/\/chemuza.org\/wp-content\/uploads\/2023\/08\/c2br4-etape-4.webp\" alt=\"C2Br4 ad\u0131m 4\" class=\"wp-image-3071\" srcset=\"\" sizes=\"\"><\/figure>\n<p> \u015eimdi bir sonraki ad\u0131ma ge\u00e7elim.<\/p>\n<h3 class=\"wp-block-heading\"> <strong>Ad\u0131m 5: Merkezi atomdaki sekizliyi kontrol edin. Bayt\u0131 yoksa, yaln\u0131z \u00e7ifti ikili ba\u011fa veya \u00fc\u00e7l\u00fc ba\u011fa d\u00f6n\u00fc\u015ft\u00fcr\u00fcn.<\/strong><\/h3>\n<p> Bu ad\u0131mda merkezi karbon atomlar\u0131n\u0131n (C) kararl\u0131 olup olmad\u0131\u011f\u0131n\u0131 kontrol etmeniz gerekir.<\/p>\n<p> Merkezi karbon (C) atomlar\u0131n\u0131n kararl\u0131l\u0131\u011f\u0131n\u0131 kontrol etmek i\u00e7in bunlar\u0131n bir oktet olu\u015fturup olu\u015fturmad\u0131\u011f\u0131n\u0131 kontrol etmemiz gerekir.<\/p>\n<p> Maalesef karbon atomlar\u0131ndan biri burada bir oktet olu\u015fturmuyor. <\/p>\n<figure class=\"wp-block-image aligncenter size-full\"><img decoding=\"async\" loading=\"lazy\" width=\"600\" height=\"249\" src=\"https:\/\/chemuza.org\/wp-content\/uploads\/2023\/08\/c2br4-etape-5.webp\" alt=\"C2Br4 ad\u0131m 5\" class=\"wp-image-3072\" srcset=\"\" sizes=\"\"><\/figure>\n<p> \u015eimdi, bu karbon atomunu kararl\u0131 hale getirmek i\u00e7in, yaln\u0131z \u00e7ifti bir \u00e7ift ba\u011fa d\u00f6n\u00fc\u015ft\u00fcrmeniz gerekir, b\u00f6ylece karbon atomu 8 elektrona (yani bir oktete) sahip olabilir. <\/p>\n<figure class=\"wp-block-image aligncenter size-full\"><img decoding=\"async\" loading=\"lazy\" width=\"600\" height=\"138\" src=\"https:\/\/chemuza.org\/wp-content\/uploads\/2023\/08\/c2br4-etape-6.webp\" alt=\"C2Br4 ad\u0131m 6\" class=\"wp-image-3073\" srcset=\"\" sizes=\"\"><\/figure>\n<p> Bu elektron \u00e7ifti ikili ba\u011fa d\u00f6n\u00fc\u015ft\u00fcr\u00fcld\u00fckten sonra merkezi karbon atomu 2 elektron daha alacak ve b\u00f6ylece toplam elektron say\u0131s\u0131 8 olacakt\u0131r. <\/p>\n<figure class=\"wp-block-image aligncenter size-full\"><img decoding=\"async\" loading=\"lazy\" width=\"600\" height=\"251\" src=\"https:\/\/chemuza.org\/wp-content\/uploads\/2023\/08\/c2br4-etape-7.webp\" alt=\"C2Br4 ad\u0131m 7\" class=\"wp-image-3074\" srcset=\"\" sizes=\"\"><\/figure>\n<p> Yukar\u0131daki resimde iki karbon atomunun bir oktet olu\u015fturdu\u011funu g\u00f6rebilirsiniz.<\/p>\n<p> Ve b\u00f6ylece bu karbon atomlar\u0131 kararl\u0131d\u0131r.<\/p>\n<p> \u015eimdi C2Br4&#8217;\u00fcn Lewis yap\u0131s\u0131n\u0131n kararl\u0131 olup olmad\u0131\u011f\u0131n\u0131 kontrol etmek i\u00e7in son ad\u0131ma ge\u00e7elim.<\/p>\n<h3 class=\"wp-block-heading\"> <strong>Ad\u0131m 6: Lewis yap\u0131s\u0131n\u0131n kararl\u0131l\u0131\u011f\u0131n\u0131 kontrol edin<\/strong><\/h3>\n<p> Art\u0131k C2Br4&#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:\/\/en.wikipedia.org\/wiki\/Formal_charge\" target=\"_blank\" rel=\"noopener\">formal y\u00fck<\/a> kavram\u0131 kullan\u0131larak do\u011frulanabilir.<\/p>\n<p> K\u0131sacas\u0131, \u015fimdi C2Br4 molek\u00fcl\u00fcnde bulunan bromin atomlar\u0131n\u0131n (Br) yan\u0131 s\u0131ra karbon atomlar\u0131n\u0131n (C) 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 C2Br4 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=\"227\" src=\"https:\/\/chemuza.org\/wp-content\/uploads\/2023\/08\/c2br4-etape-8.webp\" alt=\"C2Br4 ad\u0131m 8\" class=\"wp-image-3075\" srcset=\"\" sizes=\"\"><\/figure>\n<p> <strong>Karbon atomu (C) i\u00e7in:<\/strong><strong><br \/><\/strong> De\u011ferlik elektronlar\u0131 = 4 (\u00e7\u00fcnk\u00fc karbon 14. gruptad\u0131r)<br \/> Ba\u011f elektronlar\u0131 = 8<br \/> Ba\u011flanmayan elektronlar = 0<\/p>\n<p> <strong>Brom atomu (Br) i\u00e7in:<\/strong><br \/> De\u011ferlik elektronlar\u0131 = 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\"> VS<\/td>\n<td> =<\/td>\n<td class=\"has-text-align-center\" data-align=\"center\"> 4<\/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\"> 0<\/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, karbon (C) atomlar\u0131n\u0131n yan\u0131 s\u0131ra bromin (Br) atomlar\u0131n\u0131n da <strong>&#8220;s\u0131f\u0131r&#8221;<\/strong> formal y\u00fcke sahip oldu\u011funu g\u00f6rebilirsiniz.<\/p>\n<p> Bu, C2Br4&#8217;\u00fcn yukar\u0131daki Lewis yap\u0131s\u0131n\u0131n stabil oldu\u011funu ve C2Br4&#8217;\u00fcn yukar\u0131daki yap\u0131s\u0131nda ba\u015fka bir de\u011fi\u015fiklik olmad\u0131\u011f\u0131n\u0131 g\u00f6sterir.<\/p>\n<p> C2Br4&#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 C2Br4&#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=\"282\" src=\"https:\/\/chemuza.org\/wp-content\/uploads\/2023\/08\/structure-de-Lewis-de-c2br4.jpg\" alt=\"C2Br4'\u00fcn Lewis yap\u0131s\u0131\" class=\"wp-image-3076\" 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-asf5-yapisi\/\">Lewis yap\u0131s\u0131 AsF5<\/a><\/td>\n<td> <a href=\"https:\/\/chemuza.org\/tr\/merhaba-lewis-yapisi\/\">Merhaba Lewis yap\u0131s\u0131<\/a><\/td>\n<\/tr>\n<tr>\n<td> <a href=\"https:\/\/chemuza.org\/tr\/yapi-po3-lewis\/\">PO3-Lewis yap\u0131s\u0131<\/a><\/td>\n<td> <a href=\"https:\/\/chemuza.org\/tr\/yapi-bbr3-lewis\/\">BBr3 Lewis yap\u0131s\u0131<\/a><\/td>\n<\/tr>\n<tr>\n<td> <a href=\"https:\/\/chemuza.org\/tr\/lewis-yapisi-if2\/\">IF2-Lewis yap\u0131s\u0131<\/a><\/td>\n<td> <a href=\"https:\/\/chemuza.org\/tr\/lewis-yapisi-brf2\/\">BrF2 \u2013 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. C2Br4 Lewis yap\u0131s\u0131, iki karbon (C) atomu aras\u0131nda bir \u00e7ift ba\u011fa ve karbon (C) atomu ile bromin (Br) atomlar\u0131 aras\u0131nda tek bir ba\u011fa sahiptir. T\u00fcm bromin (Br) atomlar\u0131nda 3 yaln\u0131z \u00e7ift vard\u0131r. C2Br4&#8217;\u00fcn Lewis yap\u0131s\u0131n\u0131n yukar\u0131daki g\u00f6r\u00fcnt\u00fcs\u00fcnden hi\u00e7bir \u015fey anlamad\u0131ysan\u0131z, benimle kal\u0131n ve C2Br4&#8217;\u00fcn Lewis &#8230; <a title=\"6 ad\u0131mda c2br4 lewis yap\u0131s\u0131 (resimlerle)\" class=\"read-more\" href=\"https:\/\/chemuza.org\/tr\/lewis-c2br4-yapisi\/\" aria-label=\"More on 6 ad\u0131mda c2br4 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>6 Ad\u0131mda C2Br4 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-c2br4-yapisi\/\" \/>\n<meta property=\"og:locale\" content=\"tr_TR\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"6 Ad\u0131mda C2Br4 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. C2Br4 Lewis yap\u0131s\u0131, iki karbon (C) atomu aras\u0131nda bir \u00e7ift ba\u011fa ve karbon (C) atomu ile bromin (Br) atomlar\u0131 aras\u0131nda tek bir ba\u011fa sahiptir. T\u00fcm bromin (Br) atomlar\u0131nda 3 yaln\u0131z \u00e7ift vard\u0131r. C2Br4&#8217;\u00fcn Lewis yap\u0131s\u0131n\u0131n yukar\u0131daki g\u00f6r\u00fcnt\u00fcs\u00fcnden hi\u00e7bir \u015fey anlamad\u0131ysan\u0131z, benimle kal\u0131n ve C2Br4&#8217;\u00fcn Lewis ... 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