{"id":132,"date":"2023-07-25T23:51:19","date_gmt":"2023-07-25T23:51:19","guid":{"rendered":"https:\/\/chemuza.org\/tr\/lewis-sbr2-yapisi\/"},"modified":"2023-07-25T23:51:19","modified_gmt":"2023-07-25T23:51:19","slug":"lewis-sbr2-yapisi","status":"publish","type":"post","link":"https:\/\/chemuza.org\/tr\/lewis-sbr2-yapisi\/","title":{"rendered":"6 ad\u0131mda sbr2 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-sbr2-lewis.jpg\" alt=\"Lewis yap\u0131s\u0131 SBr2\" class=\"wp-image-1819\" 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\">SBr2 Lewis yap\u0131s\u0131n\u0131n merkezinde iki 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 2 tekli ba\u011f vard\u0131r. K\u00fck\u00fcrt (S) atomunda 2 yaln\u0131z \u00e7ift ve iki bromin (Br) atomunda 3 yaln\u0131z \u00e7ift bulunur.<\/mark><\/em><\/strong><\/p>\n<p> SBr2&#8217;nin Lewis yap\u0131s\u0131n\u0131n (s\u00fclf\u00fcr dibrom\u00fcr) yukar\u0131daki g\u00f6r\u00fcnt\u00fcs\u00fcnden hi\u00e7bir \u015fey anlamad\u0131ysan\u0131z, o zaman benimle kal\u0131n ve <a href=\"https:\/\/en.wikipedia.org\/wiki\/Sulfur_dibromide\" target=\"_blank\" rel=\"noopener\">SBr2&#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 SBr2&#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>SBr2 Lewis yap\u0131s\u0131n\u0131 \u00e7izme ad\u0131mlar\u0131<\/strong><\/h2>\n<h3 class=\"wp-block-heading\"> <strong>Ad\u0131m 1: SBr2 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\/\">SBr2 (k\u00fck\u00fcrt dibromid) molek\u00fcl\u00fcndeki<\/a> toplam <a href=\"https:\/\/www.chemeurope.com\/en\/encyclopedia\/Valence_electron.html\" target=\"_blank\" rel=\"noopener\">de\u011ferlik elektron<\/a> say\u0131s\u0131n\u0131 bulmak i\u00e7in \u00f6ncelikle <a href=\"https:\/\/chemuza.org\/tr\/kimyanin-temel-tanimlari\/\">brom<\/a> 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>SBr2 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:\/\/education.jlab.org\/itselemental\/ele016.html\" 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:\/\/periodic.lanl.gov\/35.shtml\" 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>SBr2 molek\u00fcl\u00fcndeki toplam de\u011ferlik elektronlar\u0131<\/strong> = 1 k\u00fck\u00fcrt atomu taraf\u0131ndan ba\u011f\u0131\u015flanan de\u011ferlik elektronlar\u0131 + 2 brom atomu taraf\u0131ndan ba\u011f\u0131\u015flanan de\u011ferlik elektronlar\u0131 = <strong>6 + 7(2) = 20<\/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 SBr2&#8217;dir (k\u00fck\u00fcrt dibromit) ve k\u00fck\u00fcrt (S) 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 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=\"61\" src=\"https:\/\/chemuza.org\/wp-content\/uploads\/2023\/08\/sbr2-etape-1.webp\" alt=\"SBr2 ad\u0131m 1\" class=\"wp-image-1820\" 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 SBr2 molek\u00fcl\u00fcnde <a href=\"https:\/\/en.wikipedia.org\/wiki\/Electron_pair\" target=\"_blank\" rel=\"noopener\">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=\"54\" src=\"https:\/\/chemuza.org\/wp-content\/uploads\/2023\/08\/sbr2-etape-2.webp\" alt=\"SBr2 a\u015fama 2\" class=\"wp-image-1821\" srcset=\"\" sizes=\"\"><\/figure>\n<p> Bu, k\u00fck\u00fcrt (S) ve bromin (Br)&#8217;nin bir SBr2 molek\u00fcl\u00fcnde <a href=\"https:\/\/chem.fsu.edu\/chemlab\/chm1045\/bonding.html\" 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. 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 SBr2 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:\/\/chem.libretexts.org\/Bookshelves\/Physical_and_Theoretical_Chemistry_Textbook_Maps\/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)\/Electronic_Structure_of_Atoms_and_Molecules\/Electronic_Configurations\/The_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\/sbr2-etape-3.webp\" alt=\"SBr2 ad\u0131m 3\" class=\"wp-image-1822\" srcset=\"\" sizes=\"\"><\/figure>\n<p> Ek olarak, 1. ad\u0131mda SBr2 molek\u00fcl\u00fcnde bulunan toplam de\u011ferlik elektronu say\u0131s\u0131n\u0131 hesaplad\u0131k.<\/p>\n<p> SBr2 molek\u00fcl\u00fcnde toplam <strong>20 de\u011ferlik elektronu<\/strong> vard\u0131r ve yukar\u0131daki diyagramda bunlardan sadece <strong>16 de\u011ferlik elektronu<\/strong> kullan\u0131lm\u0131\u015ft\u0131r.<\/p>\n<p> Yani kalan elektron say\u0131s\u0131 = <strong>20 \u2013 16 = 4<\/strong> .<\/p>\n<p> Bu <strong>4<\/strong> elektronu yukar\u0131daki \u015femada SBr2 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=\"239\" src=\"https:\/\/chemuza.org\/wp-content\/uploads\/2023\/08\/sbr2-etape-4.webp\" alt=\"SBr2 ad\u0131m 4\" class=\"wp-image-1823\" 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<\/strong><\/h3>\n<p> Bu ad\u0131mda merkezi k\u00fck\u00fcrt atomunun (S) kararl\u0131 olup olmad\u0131\u011f\u0131n\u0131 kontrol etmeniz gerekir.<\/p>\n<p> Merkezi brom (Br) 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=\"271\" src=\"https:\/\/chemuza.org\/wp-content\/uploads\/2023\/08\/sbr2-etape-5.webp\" alt=\"SBr2 ad\u0131m 5\" class=\"wp-image-1825\" srcset=\"\" sizes=\"\"><\/figure>\n<p> Yukar\u0131daki resimde k\u00fck\u00fcrt atomunun bir oktet olu\u015fturdu\u011funu g\u00f6rebilirsiniz. Bu, 8 elektrona sahip oldu\u011fu anlam\u0131na gelir.<\/p>\n<p> Ve b\u00f6ylece merkezi k\u00fck\u00fcrt atomu kararl\u0131d\u0131r.<\/p>\n<p> \u015eimdi SBr2&#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 SBr2&#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:\/\/openstax.org\/books\/chemistry-2e\/pages\/7-4-formal-charges-and-resonance\" target=\"_blank\" rel=\"noopener\">formal y\u00fck<\/a> kavram\u0131 kullan\u0131larak do\u011frulanabilir.<\/p>\n<p> K\u0131sacas\u0131, \u015fimdi SBr2 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 SBr2 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=\"354\" src=\"https:\/\/chemuza.org\/wp-content\/uploads\/2023\/08\/sbr2-etape-6.webp\" alt=\"SBr2 ad\u0131m 6\" class=\"wp-image-1826\" 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 = 4<br \/> Ba\u011flanmayan elektronlar = 4<\/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\"> 4\/2<\/td>\n<td> \u2013<\/td>\n<td class=\"has-text-align-center\" data-align=\"center\"> 4<\/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, SBr2&#8217;nin yukar\u0131daki Lewis yap\u0131s\u0131n\u0131n stabil oldu\u011funu ve SBr2&#8217;nin yukar\u0131daki yap\u0131s\u0131nda ba\u015fka bir de\u011fi\u015fiklik olmad\u0131\u011f\u0131n\u0131 g\u00f6sterir.<\/p>\n<p> SBr2&#8217;nin yukar\u0131daki Lewis nokta yap\u0131s\u0131nda, her bir ba\u011f elektronu \u00e7iftini (:) tek bir ba\u011f (|) olarak da temsil edebilirsiniz. Bunu yapmak, SBr2&#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=\"230\" src=\"https:\/\/chemuza.org\/wp-content\/uploads\/2023\/08\/structure-de-Lewis-de-sbr2.jpg\" alt=\"SBr2'nin Lewis yap\u0131s\u0131\" class=\"wp-image-1827\" 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\/hocl-lewis-yapisi\/\">Lewis yap\u0131s\u0131 HOCl<\/a><\/td>\n<td> <a href=\"https:\/\/chemuza.org\/tr\/c6h6-benzen-lewis-yapisi\/\">Lewis yap\u0131s\u0131 C6H6 (Benzen)<\/a><\/td>\n<\/tr>\n<tr>\n<td> <a href=\"https:\/\/chemuza.org\/tr\/lewis-nbr3-yapisi\/\">Lewis yap\u0131s\u0131 NBr3<\/a><\/td>\n<td> <a href=\"https:\/\/chemuza.org\/tr\/lewis-sef4-yapisi\/\">Lewis yap\u0131s\u0131 SeF4<\/a><\/td>\n<\/tr>\n<tr>\n<td> <a href=\"https:\/\/chemuza.org\/tr\/lewis-h3po4un-yapisi\/\">Lewis yap\u0131s\u0131 H3PO4<\/a><\/td>\n<td> <a href=\"https:\/\/chemuza.org\/tr\/h2se-lewis-yapisi\/\">H2Se 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. SBr2 Lewis yap\u0131s\u0131n\u0131n merkezinde iki 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 2 tekli ba\u011f vard\u0131r. K\u00fck\u00fcrt (S) atomunda 2 yaln\u0131z \u00e7ift ve iki bromin (Br) atomunda 3 yaln\u0131z \u00e7ift bulunur. SBr2&#8217;nin Lewis &#8230; <a title=\"6 ad\u0131mda sbr2 lewis yap\u0131s\u0131 (resimlerle)\" class=\"read-more\" href=\"https:\/\/chemuza.org\/tr\/lewis-sbr2-yapisi\/\" aria-label=\"More on 6 ad\u0131mda sbr2 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 SBr2 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-sbr2-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 SBr2 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. SBr2 Lewis yap\u0131s\u0131n\u0131n merkezinde iki 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 2 tekli ba\u011f vard\u0131r. K\u00fck\u00fcrt (S) atomunda 2 yaln\u0131z \u00e7ift ve iki bromin (Br) atomunda 3 yaln\u0131z \u00e7ift bulunur. SBr2&#8217;nin Lewis ... 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Yukar\u0131daki g\u00f6rseli k\u0131saca a\u00e7\u0131klayay\u0131m. SBr2 Lewis yap\u0131s\u0131n\u0131n merkezinde iki 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 2 tekli ba\u011f vard\u0131r. K\u00fck\u00fcrt (S) atomunda 2 yaln\u0131z \u00e7ift ve iki bromin (Br) atomunda 3 yaln\u0131z \u00e7ift bulunur. SBr2&#8217;nin Lewis ... 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