{"id":124,"date":"2023-07-26T01:33:07","date_gmt":"2023-07-26T01:33:07","guid":{"rendered":"https:\/\/chemuza.org\/tr\/yapi-ch3br-lewis\/"},"modified":"2023-07-26T01:33:07","modified_gmt":"2023-07-26T01:33:07","slug":"yapi-ch3br-lewis","status":"publish","type":"post","link":"https:\/\/chemuza.org\/tr\/yapi-ch3br-lewis\/","title":{"rendered":"6 ad\u0131mda ch3br 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-ch3br-lewis.jpg\" alt=\"Lewis yap\u0131s\u0131 CH3Br\" class=\"wp-image-1730\" 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\">CH3Br Lewis yap\u0131s\u0131n\u0131n merkezinde \u00fc\u00e7 hidrojen (H) atomu ve bir bromin (Br) atomu ile \u00e7evrelenmi\u015f bir karbon (C) atomu bulunur. Karbon (C) ve bromin (Br) atomlar\u0131 aras\u0131nda ve ayr\u0131ca karbon (C) ve hidrojen (H) atomlar\u0131 aras\u0131nda tek bir ba\u011f vard\u0131r. Bromin (Br) atomunda 3 yaln\u0131z \u00e7ift vard\u0131r.<\/mark><\/em><\/strong><\/p>\n<p> CH3Br&#8217;nin Lewis yap\u0131s\u0131n\u0131n yukar\u0131daki g\u00f6r\u00fcnt\u00fcs\u00fcnden hi\u00e7bir \u015fey anlamad\u0131ysan\u0131z, benimle kal\u0131n ve <a href=\"https:\/\/pubchem.ncbi.nlm.nih.gov\/compound\/Bromomethane\" target=\"_blank\" rel=\"noopener\">CH3Br&#8217;nin<\/a> 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 CH3Br&#8217;nin <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>CH3Br Lewis yap\u0131s\u0131n\u0131 \u00e7izme ad\u0131mlar\u0131<\/strong><\/h2>\n<h3 class=\"wp-block-heading\"> <strong>Ad\u0131m 1: CH3Br molek\u00fcl\u00fcndeki toplam de\u011ferlik elektronu say\u0131s\u0131n\u0131 bulun<\/strong><\/h3>\n<p> Bir CH3Br <a href=\"https:\/\/chemuza.org\/tr\/kimyanin-temel-tanimlari\/\">molek\u00fcl\u00fcndeki<\/a> toplam <a href=\"https:\/\/chemuza.org\/tr\/kimyanin-temel-tanimlari\/\">de\u011ferlik elektronu<\/a> say\u0131s\u0131n\u0131 bulmak i\u00e7in \u00f6ncelikle <a href=\"https:\/\/chemuza.org\/tr\/kimyanin-temel-tanimlari\/\">karbon atomunda<\/a> , hidrojen atomunda ve ayr\u0131ca brom 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 karbon, hidrojen ve bromun de\u011ferlik elektronlar\u0131n\u0131 nas\u0131l kolayca bulaca\u011f\u0131n\u0131z\u0131 anlataca\u011f\u0131m.<\/p>\n<p> <strong>CH3Br 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:\/\/www.rsc.org\/periodic-table\/element\/6\/carbon\" 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 Hidrojen 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\/3.jpg\" alt=\"\" class=\"wp-image-27\" srcset=\"\" sizes=\"\"><\/figure>\n<p> Hidrojen periyodik tablonun 1. grup elementidir. <a href=\"https:\/\/education.jlab.org\/itselemental\/ele001.html\" target=\"_blank\" rel=\"noopener\"><sup>[2]<\/sup><\/a> Bu nedenle hidrojende bulunan de\u011ferlik elektronu <strong>1&#8217;dir<\/strong> . <\/p>\n<figure class=\"wp-block-image aligncenter size-full\"><img decoding=\"async\" loading=\"lazy\" width=\"600\" height=\"177\" src=\"https:\/\/chemuza.org\/wp-content\/uploads\/2023\/08\/4.jpg\" alt=\"\" class=\"wp-image-28\" srcset=\"\" sizes=\"\"><\/figure>\n<p> Yukar\u0131daki resimde g\u00f6sterildi\u011fi gibi hidrojen atomunda yaln\u0131zca bir de\u011ferlik elektronunun bulundu\u011funu 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:\/\/periodic.lanl.gov\/35.shtml\" target=\"_blank\" rel=\"noopener\"><sup>[3]<\/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>CH3Br molek\u00fcl\u00fcndeki toplam de\u011ferlik elektronlar\u0131<\/strong> = 1 karbon atomu taraf\u0131ndan ba\u011f\u0131\u015flanan de\u011ferlik elektronlar\u0131 + 3 hidrojen atomu taraf\u0131ndan ba\u011f\u0131\u015flanan de\u011ferlik elektronlar\u0131 + 1 bromin atomu taraf\u0131ndan ba\u011f\u0131\u015flanan de\u011ferlik elektronlar\u0131 = <strong>4 + 1(3) + 7 = 14<\/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> <strong>(Unutmay\u0131n:<\/strong> e\u011fer verilen molek\u00fclde hidrojen varsa, daima hidrojeni d\u0131\u015far\u0131ya koyun.)<\/p>\n<p> \u015eimdi burada verilen molek\u00fcl CH3Br&#8217;dir ve karbon atomu (C), hidrojen atomlar\u0131 (H) ve brom atomu (Br) i\u00e7erir.<\/p>\n<p> Yani kurala g\u00f6re hidrojeni d\u0131\u015far\u0131da tutmal\u0131y\u0131z. <\/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> Art\u0131k 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) atomu merkez atom, brom (Br) atomu ise d\u0131\u015f atomdur. <\/p>\n<figure class=\"wp-block-image aligncenter size-full\"><img decoding=\"async\" loading=\"lazy\" width=\"600\" height=\"231\" src=\"https:\/\/chemuza.org\/wp-content\/uploads\/2023\/08\/ch3br-etape-1.webp\" alt=\"CH3Br ad\u0131m 1\" class=\"wp-image-1731\" 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 CH3Br molek\u00fcl\u00fcnde <a href=\"https:\/\/chemuza.org\/tr\/kimyanin-temel-tanimlari\/\">elektron \u00e7iftlerini<\/a> karbon (C) ve brom (Br) atomlar\u0131 aras\u0131na ve karbon (C) ve hidrojen (H) 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=\"225\" src=\"https:\/\/chemuza.org\/wp-content\/uploads\/2023\/08\/ch3br-etape-2.webp\" alt=\"CH3Br ad\u0131m 2\" class=\"wp-image-1732\" srcset=\"\" sizes=\"\"><\/figure>\n<p> Bu, bu atomlar\u0131n bir CH3Br molek\u00fcl\u00fcnde birbirine <a href=\"https:\/\/www.britannica.com\/science\/chemical-bonding\" 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<\/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 CH3Br molek\u00fcl\u00fcn\u00fcn \u00e7iziminde d\u0131\u015ftaki atomlar\u0131n hidrojen atomlar\u0131 ve brom atomlar\u0131 oldu\u011funu g\u00f6rebilirsiniz.<\/p>\n<p> Bu hidrojen ve brom atomlar\u0131 s\u0131ras\u0131yla bir <a href=\"https:\/\/chemuza.org\/tr\/kimyanin-temel-tanimlari\/\">ikili<\/a> ve bir oktet olu\u015fturur ve bu nedenle stabildir. <\/p>\n<figure class=\"wp-block-image aligncenter size-full\"><img decoding=\"async\" loading=\"lazy\" width=\"600\" height=\"355\" src=\"https:\/\/chemuza.org\/wp-content\/uploads\/2023\/08\/ch3br-etape-3.webp\" alt=\"CH3Br ad\u0131m 3\" class=\"wp-image-1733\" srcset=\"\" sizes=\"\"><\/figure>\n<p> Ek olarak 1. ad\u0131mda CH3Br molek\u00fcl\u00fcnde bulunan toplam de\u011ferlik elektron say\u0131s\u0131n\u0131 hesaplad\u0131k.<\/p>\n<p> CH3Br molek\u00fcl\u00fcn\u00fcn toplam <strong>14 de\u011ferlik elektronu<\/strong> vard\u0131r ve bu de\u011ferlik elektronlar\u0131n\u0131n t\u00fcm\u00fc yukar\u0131daki CH3Br diyagram\u0131nda kullan\u0131lm\u0131\u015ft\u0131r.<\/p>\n<p> Bu nedenle merkez atomda tutulacak daha fazla elektron \u00e7ifti yoktur.<\/p>\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<\/strong><\/h3>\n<p> Bu ad\u0131mda merkezi karbon atomunun (C) kararl\u0131 olup olmad\u0131\u011f\u0131n\u0131 kontrol etmeniz gerekir.<\/p>\n<p> Merkezi karbon (C) atomunun stabilitesini kontrol etmek i\u00e7in oktet olu\u015fturup olu\u015fturmad\u0131\u011f\u0131n\u0131 kontrol etmemiz gerekir. <\/p>\n<figure class=\"wp-block-image aligncenter size-full\"><img decoding=\"async\" loading=\"lazy\" width=\"600\" height=\"242\" src=\"https:\/\/chemuza.org\/wp-content\/uploads\/2023\/08\/ch3br-etape-4.webp\" alt=\"CH3Br ad\u0131m 4\" class=\"wp-image-1734\" srcset=\"\" sizes=\"\"><\/figure>\n<p> Yukar\u0131daki resimde karbon atomunun bir oktet olu\u015fturdu\u011funu g\u00f6rebilirsiniz. Bu, 8 elektrona sahip oldu\u011fu anlam\u0131na gelir.<\/p>\n<p> Ve b\u00f6ylece merkezi karbon atomu kararl\u0131d\u0131r.<\/p>\n<p> \u015eimdi CH3Br&#8217;nin 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 CH3Br&#8217;nin 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:\/\/bluebox.creighton.edu\/demo\/modules\/en-boundless-old\/www.boundless.com\/chemistry\/definition\/formal-charge\/index.html\" target=\"_blank\" rel=\"noopener\">formal y\u00fck<\/a> kavram\u0131 kullan\u0131larak do\u011frulanabilir.<\/p>\n<p> K\u0131sacas\u0131 art\u0131k CH3Br molek\u00fcl\u00fcnde bulunan karbon (C), hidrojen (H) ve bromin (Br) atomlar\u0131n\u0131n 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 CH3Br molek\u00fcl\u00fcn\u00fcn her bir atomu i\u00e7in ba\u011flanan elektronlar\u0131n ve ba\u011flanmayan elektronlar\u0131n say\u0131s\u0131n\u0131 g\u00f6rebilirsiniz. <\/p>\n<figure class=\"wp-block-image aligncenter size-full\"><img decoding=\"async\" loading=\"lazy\" width=\"600\" height=\"330\" src=\"https:\/\/chemuza.org\/wp-content\/uploads\/2023\/08\/ch3br-etape-5.webp\" alt=\"CH3Br ad\u0131m 5\" class=\"wp-image-1735\" 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)<strong><br \/><\/strong> Ba\u011f elektronlar\u0131 = 8<br \/> Ba\u011flanmayan elektronlar = 0<\/p>\n<p> <strong>Hidrojen atomu (H) i\u00e7in:<\/strong><br \/> De\u011ferlik elektronu = 1 (\u00e7\u00fcnk\u00fc hidrojen grup 1&#8217;dedir)<br \/> Ba\u011f elektronlar\u0131 = 2<br \/> Ba\u011flanmayan elektronlar = 0<\/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\"> 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\"> H<\/td>\n<td> =<\/td>\n<td class=\"has-text-align-center\" data-align=\"center\"> 1<\/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\"> 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 atomunun (C), hidrojen atomunun (H) ve bromin atomunun (Br) <strong>&#8220;s\u0131f\u0131r&#8221;<\/strong> formal y\u00fcke sahip oldu\u011funu g\u00f6rebilirsiniz.<\/p>\n<p> Bu, CH3Br&#8217;nin yukar\u0131daki Lewis yap\u0131s\u0131n\u0131n stabil oldu\u011funu ve CH3Br&#8217;nin yukar\u0131daki yap\u0131s\u0131nda ba\u015fka bir de\u011fi\u015fiklik olmad\u0131\u011f\u0131n\u0131 g\u00f6sterir.<\/p>\n<p> CH3Br&#8217;nin yukar\u0131daki Lewis nokta yap\u0131s\u0131nda, her bir ba\u011f elektronu \u00e7iftini (:) <a href=\"https:\/\/www.britannica.com\/science\/single-bond\" target=\"_blank\" rel=\"noopener\">tek bir ba\u011f<\/a> (|) olarak da temsil edebilirsiniz. Bunu yapmak CH3Br&#8217;nin 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=\"305\" src=\"https:\/\/chemuza.org\/wp-content\/uploads\/2023\/08\/structure-de-Lewis-de-ch3br.jpg\" alt=\"CH3Br'nin Lewis yap\u0131s\u0131\" class=\"wp-image-1736\" 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\/ch3och3-lewisin-yapisi\/\">Lewis yap\u0131s\u0131 CH3OCH3<\/a><\/td>\n<td> <a href=\"https:\/\/chemuza.org\/tr\/formik-asit-hcoohun-lewis-yapisi\/\">HCOOH (formik asit) Lewis yap\u0131s\u0131<\/a><\/td>\n<\/tr>\n<tr>\n<td> <a href=\"https:\/\/chemuza.org\/tr\/if3-lewis-yapisi\/\">Lewis yap\u0131s\u0131 IF3<\/a><\/td>\n<td> <a href=\"https:\/\/chemuza.org\/tr\/xeo4-lewis-yapisi\/\">Lewis yap\u0131s\u0131 XeO4<\/a><\/td>\n<\/tr>\n<tr>\n<td> <a href=\"https:\/\/chemuza.org\/tr\/sf3-lewis-yapisi\/\">SF3+ Lewis yap\u0131s\u0131<\/a><\/td>\n<td> <a href=\"https:\/\/chemuza.org\/tr\/xeo3-lewis-yapisi\/\">Lewis yap\u0131s\u0131 XeO3<\/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. CH3Br Lewis yap\u0131s\u0131n\u0131n merkezinde \u00fc\u00e7 hidrojen (H) atomu ve bir bromin (Br) atomu ile \u00e7evrelenmi\u015f bir karbon (C) atomu bulunur. Karbon (C) ve bromin (Br) atomlar\u0131 aras\u0131nda ve ayr\u0131ca karbon (C) ve hidrojen (H) atomlar\u0131 aras\u0131nda tek bir ba\u011f vard\u0131r. Bromin (Br) atomunda 3 yaln\u0131z &#8230; <a title=\"6 ad\u0131mda ch3br lewis yap\u0131s\u0131 (resimlerle)\" class=\"read-more\" href=\"https:\/\/chemuza.org\/tr\/yapi-ch3br-lewis\/\" aria-label=\"More on 6 ad\u0131mda ch3br 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 CH3Br 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\/yapi-ch3br-lewis\/\" \/>\n<meta property=\"og:locale\" content=\"tr_TR\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"6 Ad\u0131mda CH3Br 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. CH3Br Lewis yap\u0131s\u0131n\u0131n merkezinde \u00fc\u00e7 hidrojen (H) atomu ve bir bromin (Br) atomu ile \u00e7evrelenmi\u015f bir karbon (C) atomu bulunur. Karbon (C) ve bromin (Br) atomlar\u0131 aras\u0131nda ve ayr\u0131ca karbon (C) ve hidrojen (H) atomlar\u0131 aras\u0131nda tek bir ba\u011f vard\u0131r. Bromin (Br) atomunda 3 yaln\u0131z ... 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Yukar\u0131daki g\u00f6rseli k\u0131saca a\u00e7\u0131klayay\u0131m. CH3Br Lewis yap\u0131s\u0131n\u0131n merkezinde \u00fc\u00e7 hidrojen (H) atomu ve bir bromin (Br) atomu ile \u00e7evrelenmi\u015f bir karbon (C) atomu bulunur. Karbon (C) ve bromin (Br) atomlar\u0131 aras\u0131nda ve ayr\u0131ca karbon (C) ve hidrojen (H) atomlar\u0131 aras\u0131nda tek bir ba\u011f vard\u0131r. Bromin (Br) atomunda 3 yaln\u0131z ... 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