{"id":220,"date":"2023-07-25T05:52:31","date_gmt":"2023-07-25T05:52:31","guid":{"rendered":"https:\/\/chemuza.org\/tr\/brcn-lewis-yapisi\/"},"modified":"2023-07-25T05:52:31","modified_gmt":"2023-07-25T05:52:31","slug":"brcn-lewis-yapisi","status":"publish","type":"post","link":"https:\/\/chemuza.org\/tr\/brcn-lewis-yapisi\/","title":{"rendered":"6 ad\u0131mda brcn 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-de-brcn-lewis.jpg\" alt=\"Lewis yap\u0131s\u0131 BrCN\" class=\"wp-image-2823\" 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\">BrCN Lewis yap\u0131s\u0131n\u0131n merkezinde bir brom (Br) atomu ve bir nitrojen (N) atomu ile \u00e7evrelenmi\u015f bir karbon (C) atomu bulunur. Karbon (C) ve bromin (Br) atomlar\u0131 aras\u0131nda 1 adet tekli ba\u011f, karbon (C) ve nitrojen (N) aras\u0131nda ise 1 adet \u00fc\u00e7l\u00fc ba\u011f bulunmaktad\u0131r.<\/mark><\/em><\/strong><\/p>\n<p> BrCN&#8217;nin Lewis yap\u0131s\u0131na ili\u015fkin yukar\u0131daki g\u00f6r\u00fcnt\u00fcden hi\u00e7bir \u015fey anlamad\u0131ysan\u0131z, benimle kal\u0131n ve<a href=\"https:\/\/pubchem.ncbi.nlm.nih.gov\/compound\/Cyanogen-bromide\" target=\"_blank\" rel=\"noopener\">BrCN<\/a> molek\u00fcl\u00fcn\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> O halde BrCN&#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>BrCN Lewis Yap\u0131s\u0131n\u0131 \u00c7izim Ad\u0131mlar\u0131<\/strong><\/h2>\n<h3 class=\"wp-block-heading\"> <strong>Ad\u0131m 1: BrCN molek\u00fcl\u00fcndeki toplam de\u011ferlik elektronu say\u0131s\u0131n\u0131 bulun<\/strong><\/h3>\n<p> Bir BrCN <a href=\"https:\/\/chemuza.org\/tr\/kimyanin-temel-tanimlari\/\">molek\u00fcl\u00fcndeki<\/a> toplam <a href=\"http:\/\/www.chem.ucla.edu\/~harding\/IGOC\/V\/valence_electron.html\" target=\"_blank\" rel=\"noopener\">de\u011ferlik elektron<\/a> say\u0131s\u0131n\u0131 bulmak i\u00e7in \u00f6ncelikle bromin atomunda, karbon atomunda ve ayr\u0131ca nitrojen 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, karbonun ve nitrojenin de\u011ferlik elektronlar\u0131n\u0131 nas\u0131l kolayca bulaca\u011f\u0131n\u0131z\u0131 anlataca\u011f\u0131m.<\/p>\n<p> <strong>BrCN molek\u00fcl\u00fcndeki toplam de\u011ferlik elektronlar\u0131<\/strong><\/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>[1]<\/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> <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:\/\/education.jlab.org\/itselemental\/ele006.html\" target=\"_blank\" rel=\"noopener\"><sup>[2]<\/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 Azot 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=\"302\" src=\"https:\/\/chemuza.org\/wp-content\/uploads\/2023\/08\/2-4.jpg\" alt=\"\" class=\"wp-image-84\" srcset=\"\" sizes=\"\"><\/figure>\n<p> Azot periyodik tablonun 15. grubunda yer alan bir elementtir. <a href=\"https:\/\/pubchem.ncbi.nlm.nih.gov\/element\/7\" target=\"_blank\" rel=\"noopener\"><sup>[3]<\/sup><\/a> Bu nedenle nitrojende bulunan de\u011ferlik elektronlar\u0131 <strong>5&#8217;tir<\/strong> . <\/p>\n<figure class=\"wp-block-image aligncenter size-full\"><img decoding=\"async\" loading=\"lazy\" width=\"600\" height=\"222\" src=\"https:\/\/chemuza.org\/wp-content\/uploads\/2023\/08\/3-4.jpg\" alt=\"\" class=\"wp-image-85\" srcset=\"\" sizes=\"\"><\/figure>\n<p> Yukar\u0131daki resimde g\u00f6sterildi\u011fi gibi nitrojen atomunda bulunan 5 de\u011ferlik elektronunu g\u00f6rebilirsiniz.<\/p>\n<p> Bu y\u00fczden,<\/p>\n<p> <strong>BrCN molek\u00fcl\u00fcndeki toplam de\u011ferlik elektronlar\u0131<\/strong> = 1 bromin atomu taraf\u0131ndan ba\u011f\u0131\u015flanan de\u011ferlik elektronlar\u0131 + 1 karbon atomu taraf\u0131ndan ba\u011f\u0131\u015flanan de\u011ferlik elektronlar\u0131 + 1 nitrojen atomu taraf\u0131ndan ba\u011f\u0131\u015flanan de\u011ferlik elektronlar\u0131 = <strong>7 + 4 + 5 = 16<\/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 BrCN&#8217;dir ve bir bromin atomu (Br), bir karbon atomu (C) ve bir nitrojen atomu (N) 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 brom atomunun (Br), karbon atomunun (C) ve nitrojen atomunun (N) elektronegatiflik de\u011ferlerini g\u00f6rebilirsiniz.<\/p>\n<p> Brom atomu (Br), karbon atomu (C) ve nitrojen atomunun (N) 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 ve nitrojen (N) atomu ise d\u0131\u015f atomlard\u0131r. <\/p>\n<figure class=\"wp-block-image aligncenter size-full\"><img decoding=\"async\" loading=\"lazy\" width=\"600\" height=\"59\" src=\"https:\/\/chemuza.org\/wp-content\/uploads\/2023\/08\/Brcn-etape-1.webp\" alt=\"CNbr ad\u0131m 1\" class=\"wp-image-2824\" 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 BrCN molek\u00fcl\u00fcnde <a href=\"https:\/\/www2.chem.wisc.edu\/deptfiles\/genchem\/netorial\/rottosen\/tutorial\/modules\/intermolecular_forces\/01review\/review3.htm\" target=\"_blank\" rel=\"noopener\">elektron \u00e7iftlerini<\/a> bromin atomu (Br), karbon atomu (C) ve nitrojen atomu (N) aras\u0131na yerle\u015ftirmeniz gerekir. <\/p>\n<figure class=\"wp-block-image aligncenter size-full\"><img decoding=\"async\" loading=\"lazy\" width=\"600\" height=\"55\" src=\"https:\/\/chemuza.org\/wp-content\/uploads\/2023\/08\/Brcn-etape-2.webp\" alt=\"BrCN a\u015fama 2\" class=\"wp-image-2825\" srcset=\"\" sizes=\"\"><\/figure>\n<p> Bu, bir BrCN molek\u00fcl\u00fcnde bromin atomunun (Br), karbon atomunun (C) ve nitrojen atomunun (N) birbirine <a href=\"https:\/\/uen.pressbooks.pub\/introductorychemistry\/chapter\/the-covalent-bond\/\" 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 BrCN molek\u00fcl\u00fcn\u00fcn \u00e7iziminde d\u0131\u015ftaki atomlar\u0131n brom atomu ve nitrojen atomu oldu\u011funu g\u00f6rebilirsiniz.<\/p>\n<p> Bu harici brom ve nitrojen atomlar\u0131 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=\"264\" src=\"https:\/\/chemuza.org\/wp-content\/uploads\/2023\/08\/Brcn-etape-3.webp\" alt=\"BrCN ad\u0131m 3\" class=\"wp-image-2826\" srcset=\"\" sizes=\"\"><\/figure>\n<p> Ek olarak, 1. ad\u0131mda BrCN molek\u00fcl\u00fcnde bulunan toplam de\u011ferlik elektronu say\u0131s\u0131n\u0131 hesaplad\u0131k.<\/p>\n<p> BrCN molek\u00fcl\u00fcn\u00fcn toplam <strong>16 de\u011ferlik elektronu<\/strong> vard\u0131r ve bu de\u011ferlik elektronlar\u0131n\u0131n t\u00fcm\u00fc yukar\u0131daki diyagramda 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. Okteti yoksa yaln\u0131z \u00e7ifti \u00e7ift ba\u011f veya \u00fc\u00e7l\u00fc ba\u011f olu\u015fturacak \u015fekilde hareket ettirin.<\/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 <a href=\"https:\/\/en.wikibooks.org\/wiki\/General_Chemistry\/Octet_Rule_and_Exceptions\" target=\"_blank\" rel=\"noopener\">oktet<\/a> olu\u015fturup olu\u015fturmad\u0131\u011f\u0131n\u0131 kontrol etmemiz gerekir.<\/p>\n<p> Maalesef karbon atomu burada bir oktet olu\u015fturmuyor. Karbonun yaln\u0131zca 4 elektronu vard\u0131r ve karars\u0131zd\u0131r. <\/p>\n<figure class=\"wp-block-image aligncenter size-full\"><img decoding=\"async\" loading=\"lazy\" width=\"600\" height=\"243\" src=\"https:\/\/chemuza.org\/wp-content\/uploads\/2023\/08\/Brcn-etape-4.webp\" alt=\"BrCN ad\u0131m 4\" class=\"wp-image-2827\" srcset=\"\" sizes=\"\"><\/figure>\n<p> \u015eimdi, bu karbon atomunu kararl\u0131 hale getirmek i\u00e7in, d\u0131\u015f nitrojen atomunun elektron \u00e7iftini, karbon atomunun 8 elektrona (yani bir oktete) sahip olaca\u011f\u0131 \u015fekilde kayd\u0131rman\u0131z gerekir.<\/p>\n<p> <strong>(Not:<\/strong> Burada 2 se\u00e7ene\u011finiz vard\u0131r. Bromun veya nitrojenin elektron \u00e7iftini hareket ettirebilirsiniz. Ancak halojenler genellikle tek bir ba\u011f olu\u015fturur. Yani burada nitrojenin elektron \u00e7iftini hareket ettirmeniz gerekir. ) <\/p>\n<figure class=\"wp-block-image aligncenter size-full\"><img decoding=\"async\" loading=\"lazy\" width=\"600\" height=\"105\" src=\"https:\/\/chemuza.org\/wp-content\/uploads\/2023\/08\/Brcn-etape-5.webp\" alt=\"BrCN ad\u0131m 5\" class=\"wp-image-2828\" srcset=\"\" sizes=\"\"><\/figure>\n<p> Ancak bir \u00e7ift elektronu hareket ettirdikten sonra karbon atomu yaln\u0131zca 6 elektrona sahip oldu\u011fundan hala bir oktet olu\u015fturmuyor. <\/p>\n<figure class=\"wp-block-image aligncenter size-full\"><img decoding=\"async\" loading=\"lazy\" width=\"600\" height=\"232\" src=\"https:\/\/chemuza.org\/wp-content\/uploads\/2023\/08\/Brcn-etape-6.webp\" alt=\"BrCN ad\u0131m 6\" class=\"wp-image-2829\" srcset=\"\" sizes=\"\"><\/figure>\n<p> Yine, yaln\u0131zca nitrojen atomundan fazladan bir \u00e7ift elektronu hareket ettirmemiz gerekiyor. <\/p>\n<figure class=\"wp-block-image aligncenter size-full\"><img decoding=\"async\" loading=\"lazy\" width=\"600\" height=\"105\" src=\"https:\/\/chemuza.org\/wp-content\/uploads\/2023\/08\/Brcn-etape-7.webp\" alt=\"BrCN ad\u0131m 7\" class=\"wp-image-2830\" srcset=\"\" sizes=\"\"><\/figure>\n<p> Bu elektron \u00e7iftini hareket ettirdikten 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=\"233\" src=\"https:\/\/chemuza.org\/wp-content\/uploads\/2023\/08\/Brcn-etape-8.webp\" alt=\"BrCN ad\u0131m 8\" class=\"wp-image-2831\" srcset=\"\" sizes=\"\"><\/figure>\n<p> Yukar\u0131daki resimde karbon atomunun bir oktet olu\u015fturdu\u011funu g\u00f6rebilirsiniz.<\/p>\n<p> Ve bu nedenle karbon atomu kararl\u0131d\u0131r.<\/p>\n<p> \u015eimdi BrCN&#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 BrCN molek\u00fcl\u00fcn\u00fcn Lewis yap\u0131s\u0131n\u0131n stabilitesini kontrol etmeniz gereken son ad\u0131ma geldiniz.<\/p>\n<p> Lewis yap\u0131s\u0131n\u0131n kararl\u0131l\u0131\u011f\u0131 <a href=\"https:\/\/pressbooks-dev.oer.hawaii.edu\/chemistry\/chapter\/formal-charges-and-resonance\/\" target=\"_blank\" rel=\"noopener\">formal y\u00fck<\/a> kavram\u0131 kullan\u0131larak do\u011frulanabilir.<\/p>\n<p> K\u0131sacas\u0131 art\u0131k BrCN molek\u00fcl\u00fcnde bulunan bromin (Br), karbon (C) ve nitrojen (N) 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 g\u00f6r\u00fcnt\u00fcde BrCN 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=\"235\" src=\"https:\/\/chemuza.org\/wp-content\/uploads\/2023\/08\/Brcn-etape-9.webp\" alt=\"BrCN ad\u0131m 9\" class=\"wp-image-2832\" srcset=\"\" sizes=\"\"><\/figure>\n<p> <strong>Brom atomu (Br) i\u00e7in:<\/strong><strong><br \/><\/strong> De\u011ferlik elektronlar\u0131 = 7 (\u00e7\u00fcnk\u00fc brom 17. gruptad\u0131r)<strong><br \/><\/strong> Ba\u011f elektronlar\u0131 = 2<br \/> Ba\u011flanmayan elektronlar = 6<\/p>\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>Azot atomu (N) i\u00e7in:<\/strong><strong><br \/><\/strong> De\u011ferlik elektronlar\u0131 = 5 (\u00e7\u00fcnk\u00fc nitrojen 15. gruptad\u0131r)<strong><br \/><\/strong> Ba\u011f elektronlar\u0131 = 6<br \/> Ba\u011flanmayan elektronlar = 2 <\/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\"> 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<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\"> OLUMSUZ<\/td>\n<td> =<\/td>\n<td class=\"has-text-align-center\" data-align=\"center\"> 5<\/td>\n<td> \u2013<\/td>\n<td class=\"has-text-align-center\" data-align=\"center\"> 6\/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<\/tbody>\n<\/table>\n<\/figure>\n<p> Yukar\u0131daki formal y\u00fck hesaplamalar\u0131ndan, brom atomunun (Br), karbon atomunun (C) ve nitrojen atomunun (N) <strong>&#8220;s\u0131f\u0131r&#8221;<\/strong> formal y\u00fcke sahip oldu\u011funu g\u00f6rebilirsiniz.<\/p>\n<p> Bu, BrCN&#8217;nin yukar\u0131daki Lewis yap\u0131s\u0131n\u0131n stabil oldu\u011funu ve BrCN&#8217;nin yukar\u0131daki yap\u0131s\u0131nda ba\u015fka bir de\u011fi\u015fiklik olmad\u0131\u011f\u0131n\u0131 g\u00f6sterir.<\/p>\n<p> BrCN&#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 BrCN&#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=\"213\" src=\"https:\/\/chemuza.org\/wp-content\/uploads\/2023\/08\/Structure-Lewis-de-Brcn.jpg\" alt=\"BrCN'nin Lewis yap\u0131s\u0131\" class=\"wp-image-2833\" 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\/bei2-lewis-yapisi\/\">Lewis yap\u0131s\u0131 BeI2<\/a><\/td>\n<td> <a href=\"https:\/\/chemuza.org\/tr\/chbr3-yapisi-lewis\/\">Lewis yap\u0131s\u0131 CHBr3<\/a><\/td>\n<\/tr>\n<tr>\n<td> <a href=\"https:\/\/chemuza.org\/tr\/sicl2br2-lewis-yapisi\/\">Lewis yap\u0131s\u0131 SiCl2Br2<\/a><\/td>\n<td> <a href=\"https:\/\/chemuza.org\/tr\/lewis-sbf5in-yapisi\/\">Lewis yap\u0131s\u0131 SbF5<\/a><\/td>\n<\/tr>\n<tr>\n<td> <a href=\"https:\/\/chemuza.org\/tr\/yapi-clbr3-lewis\/\">Lewis yap\u0131s\u0131 ClBr3<\/a><\/td>\n<td> <a href=\"https:\/\/chemuza.org\/tr\/geh4-lewis-yapisi\/\">Lewis yap\u0131s\u0131 GeH4<\/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. BrCN Lewis yap\u0131s\u0131n\u0131n merkezinde bir brom (Br) atomu ve bir nitrojen (N) atomu ile \u00e7evrelenmi\u015f bir karbon (C) atomu bulunur. Karbon (C) ve bromin (Br) atomlar\u0131 aras\u0131nda 1 adet tekli ba\u011f, karbon (C) ve nitrojen (N) aras\u0131nda ise 1 adet \u00fc\u00e7l\u00fc ba\u011f bulunmaktad\u0131r. BrCN&#8217;nin Lewis &#8230; <a title=\"6 ad\u0131mda brcn lewis yap\u0131s\u0131 (resimlerle)\" class=\"read-more\" href=\"https:\/\/chemuza.org\/tr\/brcn-lewis-yapisi\/\" aria-label=\"More on 6 ad\u0131mda brcn 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 BrCN Lewis Yap\u0131s\u0131 (Resimlerle) \u2013 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\/brcn-lewis-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 BrCN Lewis Yap\u0131s\u0131 (Resimlerle) \u2013 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. BrCN Lewis yap\u0131s\u0131n\u0131n merkezinde bir brom (Br) atomu ve bir nitrojen (N) atomu ile \u00e7evrelenmi\u015f bir karbon (C) atomu bulunur. Karbon (C) ve bromin (Br) atomlar\u0131 aras\u0131nda 1 adet tekli ba\u011f, karbon (C) ve nitrojen (N) aras\u0131nda ise 1 adet \u00fc\u00e7l\u00fc ba\u011f bulunmaktad\u0131r. BrCN&#8217;nin Lewis ... 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Yukar\u0131daki g\u00f6rseli k\u0131saca a\u00e7\u0131klayay\u0131m. BrCN Lewis yap\u0131s\u0131n\u0131n merkezinde bir brom (Br) atomu ve bir nitrojen (N) atomu ile \u00e7evrelenmi\u015f bir karbon (C) atomu bulunur. Karbon (C) ve bromin (Br) atomlar\u0131 aras\u0131nda 1 adet tekli ba\u011f, karbon (C) ve nitrojen (N) aras\u0131nda ise 1 adet \u00fc\u00e7l\u00fc ba\u011f bulunmaktad\u0131r. BrCN&#8217;nin Lewis ... 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