{"id":181,"date":"2023-07-25T13:35:05","date_gmt":"2023-07-25T13:35:05","guid":{"rendered":"https:\/\/chemuza.org\/tr\/ch2-lewis-yapisi\/"},"modified":"2023-07-25T13:35:05","modified_gmt":"2023-07-25T13:35:05","slug":"ch2-lewis-yapisi","status":"publish","type":"post","link":"https:\/\/chemuza.org\/tr\/ch2-lewis-yapisi\/","title":{"rendered":"5 ad\u0131mda ch2 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-ch2-lewis.jpg\" alt=\"Lewis yap\u0131s\u0131 CH2\" class=\"wp-image-2397\" 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\">CH2 Lewis yap\u0131s\u0131n\u0131n merkezinde iki hidrojen (H) atomu ile \u00e7evrelenmi\u015f bir karbon (C) atomu bulunur. Karbon atomu (C) ile her hidrojen atomu (H) aras\u0131nda 2 tekli ba\u011f vard\u0131r. Karbon atomunun (C) 1 ortak \u00e7ifti vard\u0131r.<\/mark><\/em><\/strong><\/p>\n<p> CH2&#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\/Methylene\" target=\"_blank\" rel=\"noopener\">CH2<\/a> molek\u00fcl\u00fcn\u00fcn Lewis yap\u0131s\u0131n\u0131n \u00e7izilmesiyle ilgili ayr\u0131nt\u0131l\u0131 ad\u0131m ad\u0131m a\u00e7\u0131klamay\u0131 alacaks\u0131n\u0131z.<\/p>\n<p> \u015eimdi CH2&#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>CH2 Lewis Yap\u0131s\u0131n\u0131 \u00c7izim Ad\u0131mlar\u0131<\/strong><\/h2>\n<h3 class=\"wp-block-heading\"> <strong>Ad\u0131m 1: CH2 molek\u00fcl\u00fcndeki toplam de\u011ferlik elektronu say\u0131s\u0131n\u0131 bulun<\/strong><\/h3>\n<p> Bir CH2 <a href=\"https:\/\/chemuza.org\/tr\/kimyanin-temel-tanimlari\/\">molek\u00fcl\u00fcndeki<\/a> <a href=\"https:\/\/www.nde-ed.org\/Physics\/AtomElements\/electrons.xhtml\" target=\"_blank\" rel=\"noopener\">de\u011ferlik elektronlar\u0131n\u0131n<\/a> toplam say\u0131s\u0131n\u0131 bulmak i\u00e7in \u00f6ncelikle hidrojen 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 hidrojenin de\u011ferlik elektronlar\u0131n\u0131 nas\u0131l kolayca bulaca\u011f\u0131n\u0131z\u0131 anlataca\u011f\u0131m.<\/p>\n<p> <strong>CH2 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 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:\/\/www.rsc.org\/periodic-table\/element\/1\/hydrogen\" 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> Bu y\u00fczden,<\/p>\n<p> <strong>CH2 molek\u00fcl\u00fcndeki toplam de\u011ferlik elektronlar\u0131<\/strong> = 1 karbon atomu taraf\u0131ndan ba\u011f\u0131\u015flanan de\u011ferlik elektronlar\u0131 + 2 hidrojen atomu taraf\u0131ndan ba\u011f\u0131\u015flanan de\u011ferlik elektronlar\u0131 = <strong>4 + 1(2) = 6<\/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 CH2&#8217;dir ve karbon (C) atomu ve hidrojen (H) atomlar\u0131n\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 hidrojen atomunun (H) elektronegatiflik de\u011ferlerini g\u00f6rebilirsiniz.<\/p>\n<p> Karbon (C) ve hidrojenin (H) elektronegatiflik de\u011ferlerini kar\u015f\u0131la\u015ft\u0131r\u0131rsak <a href=\"https:\/\/chemuza.org\/tr\/periyodik-tablonun-elektronegatifligi\/\">hidrojen atomu daha az elektronegatiftir<\/a> . Ama kural gere\u011fi hidrojeni d\u0131\u015far\u0131da tutmak zorunday\u0131z.<\/p>\n<p> Burada karbon (C) atomu merkez atom, hidrojen (H) 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=\"60\" src=\"https:\/\/chemuza.org\/wp-content\/uploads\/2023\/08\/ch2-etape-1.webp\" alt=\"CH2 ad\u0131m 1\" class=\"wp-image-2398\" 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 CH2 molek\u00fcl\u00fcnde elektron \u00e7iftlerini karbon atomu (C) ile hidrojen atomlar\u0131 (H) aras\u0131na yerle\u015ftirmeniz gerekir. <\/p>\n<figure class=\"wp-block-image aligncenter size-full\"><img decoding=\"async\" loading=\"lazy\" width=\"600\" height=\"60\" src=\"https:\/\/chemuza.org\/wp-content\/uploads\/2023\/08\/ch2-etape-2.webp\" alt=\"CH2 ad\u0131m 2\" class=\"wp-image-2399\" srcset=\"\" sizes=\"\"><\/figure>\n<p> Bu, karbon (C) ve hidrojenin (H) bir CH2 molek\u00fcl\u00fcnde birbirine <a href=\"https:\/\/www.sydney.edu.au\/science\/chemistry\/~george\/1611\/ChemicalBonding.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 CH2 molek\u00fcl\u00fcn\u00fcn \u00e7iziminde d\u0131\u015ftaki atomlar\u0131n hidrojen atomlar\u0131 oldu\u011funu g\u00f6rebilirsiniz.<\/p>\n<p> Bu harici hidrojen atomlar\u0131 bir <a href=\"https:\/\/chemuza.org\/tr\/kimyanin-temel-tanimlari\/\">ikili<\/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=\"185\" src=\"https:\/\/chemuza.org\/wp-content\/uploads\/2023\/08\/ch2-etape-3.webp\" alt=\"CH2 ad\u0131m 3\" class=\"wp-image-2400\" srcset=\"\" sizes=\"\"><\/figure>\n<p> Ek olarak, 1. ad\u0131mda CH2 molek\u00fcl\u00fcnde bulunan toplam de\u011ferlik elektronu say\u0131s\u0131n\u0131 hesaplad\u0131k.<\/p>\n<p> CH2 molek\u00fcl\u00fcn\u00fcn toplam <strong>6 de\u011ferlik elektronu<\/strong> vard\u0131r ve yukar\u0131daki diyagramda bunlardan sadece <strong>4 de\u011ferlik elektronu<\/strong> kullan\u0131lm\u0131\u015ft\u0131r.<\/p>\n<p> Yani kalan elektron say\u0131s\u0131 = <strong>6 \u2013 4 = 2<\/strong> .<\/p>\n<p> Bu <strong>2<\/strong> elektronu yukar\u0131daki \u015femada CH2 molek\u00fcl\u00fcn\u00fcn merkezi karbon atomuna yerle\u015ftirmeniz gerekiyor. <\/p>\n<figure class=\"wp-block-image aligncenter size-full\"><img decoding=\"async\" loading=\"lazy\" width=\"600\" height=\"219\" src=\"https:\/\/chemuza.org\/wp-content\/uploads\/2023\/08\/ch2-etape-4.webp\" alt=\"CH2 ad\u0131m 4\" class=\"wp-image-2401\" srcset=\"\" sizes=\"\"><\/figure>\n<p> \u015eimdi bir sonraki ad\u0131ma ge\u00e7elim.<\/p>\n<h3 class=\"wp-block-heading\"> <strong>Ad\u0131m 5: Resmi \u00dccreti Kontrol Edin<\/strong><\/h3>\n<p> Art\u0131k CH2&#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:\/\/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 CH2 molek\u00fcl\u00fcnde bulunan hidrojen atomlar\u0131n\u0131n (H) 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 CH2 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=\"348\" src=\"https:\/\/chemuza.org\/wp-content\/uploads\/2023\/08\/ch2-etape-5.webp\" alt=\"CH2 ad\u0131m 5\" class=\"wp-image-2402\" 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 = 4<br \/> Ba\u011flanmayan elektronlar = 2<\/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<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\"> 4\/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\"> 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<\/tbody>\n<\/table>\n<\/figure>\n<p> Yukar\u0131daki formal y\u00fck hesaplamalar\u0131ndan, hem karbon (C) atomlar\u0131n\u0131n hem de hidrojen (H) atomlar\u0131n\u0131n <strong>&#8220;s\u0131f\u0131r&#8221;<\/strong> formal y\u00fcke sahip oldu\u011funu g\u00f6rebilirsiniz.<\/p>\n<p> Bu, CH2&#8217;nin yukar\u0131daki Lewis yap\u0131s\u0131n\u0131n stabil oldu\u011funu ve CH2&#8217;nin yukar\u0131daki yap\u0131s\u0131nda art\u0131k bir de\u011fi\u015fiklik olmad\u0131\u011f\u0131n\u0131 g\u00f6sterir.<\/p>\n<p> CH2&#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 CH2&#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=\"206\" src=\"https:\/\/chemuza.org\/wp-content\/uploads\/2023\/08\/structure-de-Lewis-de-ch2.jpg\" alt=\"CH2'nin Lewis yap\u0131s\u0131\" class=\"wp-image-2403\" 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-c2hcl-yapisi\/\">Lewis yap\u0131s\u0131 C2HCl<\/a><\/td>\n<td> <a href=\"https:\/\/chemuza.org\/tr\/s2o-lewisin-yapisi\/\">S2O Lewis yap\u0131s\u0131<\/a><\/td>\n<\/tr>\n<tr>\n<td> <a href=\"https:\/\/chemuza.org\/tr\/lewis-yapisi-brcl3\/\">Lewis yap\u0131s\u0131 BrCl3<\/a><\/td>\n<td> <a href=\"https:\/\/chemuza.org\/tr\/yapi-no2cl-lewis\/\">Lewis Yap\u0131s\u0131 NO2Cl<\/a><\/td>\n<\/tr>\n<tr>\n<td> <a href=\"https:\/\/chemuza.org\/tr\/tef4-lewis-yapisi\/\">Lewis yap\u0131s\u0131 TeF4<\/a><\/td>\n<td> <a href=\"https:\/\/chemuza.org\/tr\/clf-lewis-yapisi\/\">Lewis yap\u0131s\u0131 ClF<\/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. CH2 Lewis yap\u0131s\u0131n\u0131n merkezinde iki hidrojen (H) atomu ile \u00e7evrelenmi\u015f bir karbon (C) atomu bulunur. Karbon atomu (C) ile her hidrojen atomu (H) aras\u0131nda 2 tekli ba\u011f vard\u0131r. Karbon atomunun (C) 1 ortak \u00e7ifti vard\u0131r. CH2&#8217;nin Lewis yap\u0131s\u0131na ili\u015fkin yukar\u0131daki g\u00f6r\u00fcnt\u00fcden hi\u00e7bir \u015fey anlamad\u0131ysan\u0131z, benimle &#8230; <a title=\"5 ad\u0131mda ch2 lewis yap\u0131s\u0131 (resimlerle)\" class=\"read-more\" href=\"https:\/\/chemuza.org\/tr\/ch2-lewis-yapisi\/\" aria-label=\"More on 5 ad\u0131mda ch2 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 CH2 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\/ch2-lewis-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 CH2 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. CH2 Lewis yap\u0131s\u0131n\u0131n merkezinde iki hidrojen (H) atomu ile \u00e7evrelenmi\u015f bir karbon (C) atomu bulunur. Karbon atomu (C) ile her hidrojen atomu (H) aras\u0131nda 2 tekli ba\u011f vard\u0131r. Karbon atomunun (C) 1 ortak \u00e7ifti vard\u0131r. CH2&#8217;nin Lewis yap\u0131s\u0131na ili\u015fkin yukar\u0131daki g\u00f6r\u00fcnt\u00fcden hi\u00e7bir \u015fey anlamad\u0131ysan\u0131z, benimle ... Devam\u0131n\u0131 oku\" \/>\n<meta property=\"og:url\" content=\"https:\/\/chemuza.org\/tr\/ch2-lewis-yapisi\/\" \/>\n<meta property=\"og:site_name\" content=\"Chemuza\" \/>\n<meta property=\"article:published_time\" content=\"2023-07-25T13:35:05+00:00\" \/>\n<meta property=\"og:image\" content=\"https:\/\/chemuza.org\/wp-content\/uploads\/2023\/08\/structure-ch2-lewis.jpg\" \/>\n<meta name=\"author\" content=\"Edit\u00f6r ekibi\" \/>\n<meta name=\"twitter:card\" content=\"summary_large_image\" \/>\n<meta name=\"twitter:label1\" content=\"Yazan:\" \/>\n\t<meta name=\"twitter:data1\" content=\"Edit\u00f6r ekibi\" \/>\n\t<meta name=\"twitter:label2\" content=\"Tahmini okuma s\u00fcresi\" \/>\n\t<meta name=\"twitter:data2\" content=\"5 dakika\" \/>\n<script type=\"application\/ld+json\" class=\"yoast-schema-graph\">{\"@context\":\"https:\/\/schema.org\",\"@graph\":[{\"@type\":\"Article\",\"@id\":\"https:\/\/chemuza.org\/tr\/ch2-lewis-yapisi\/#article\",\"isPartOf\":{\"@id\":\"https:\/\/chemuza.org\/tr\/ch2-lewis-yapisi\/\"},\"author\":{\"name\":\"Edit\u00f6r ekibi\",\"@id\":\"https:\/\/chemuza.org\/tr\/#\/schema\/person\/5d92b9b82b9b43578b70f5a7cd51fa3e\"},\"headline\":\"5 ad\u0131mda ch2 lewis yap\u0131s\u0131 (resimlerle)\",\"datePublished\":\"2023-07-25T13:35:05+00:00\",\"dateModified\":\"2023-07-25T13:35:05+00:00\",\"mainEntityOfPage\":{\"@id\":\"https:\/\/chemuza.org\/tr\/ch2-lewis-yapisi\/\"},\"wordCount\":974,\"commentCount\":0,\"publisher\":{\"@id\":\"https:\/\/chemuza.org\/tr\/#organization\"},\"articleSection\":[\"Lewis yap\u0131s\u0131\"],\"inLanguage\":\"tr\",\"potentialAction\":[{\"@type\":\"CommentAction\",\"name\":\"Comment\",\"target\":[\"https:\/\/chemuza.org\/tr\/ch2-lewis-yapisi\/#respond\"]}]},{\"@type\":\"WebPage\",\"@id\":\"https:\/\/chemuza.org\/tr\/ch2-lewis-yapisi\/\",\"url\":\"https:\/\/chemuza.org\/tr\/ch2-lewis-yapisi\/\",\"name\":\"5 Ad\u0131mda CH2 Lewis Yap\u0131s\u0131 (Resimlerle) - Chemuza\",\"isPartOf\":{\"@id\":\"https:\/\/chemuza.org\/tr\/#website\"},\"datePublished\":\"2023-07-25T13:35:05+00:00\",\"dateModified\":\"2023-07-25T13:35:05+00:00\",\"breadcrumb\":{\"@id\":\"https:\/\/chemuza.org\/tr\/ch2-lewis-yapisi\/#breadcrumb\"},\"inLanguage\":\"tr\",\"potentialAction\":[{\"@type\":\"ReadAction\",\"target\":[\"https:\/\/chemuza.org\/tr\/ch2-lewis-yapisi\/\"]}]},{\"@type\":\"BreadcrumbList\",\"@id\":\"https:\/\/chemuza.org\/tr\/ch2-lewis-yapisi\/#breadcrumb\",\"itemListElement\":[{\"@type\":\"ListItem\",\"position\":1,\"name\":\"Ev\",\"item\":\"https:\/\/chemuza.org\/tr\/\"},{\"@type\":\"ListItem\",\"position\":2,\"name\":\"5 ad\u0131mda ch2 lewis yap\u0131s\u0131 (resimlerle)\"}]},{\"@type\":\"WebSite\",\"@id\":\"https:\/\/chemuza.org\/tr\/#website\",\"url\":\"https:\/\/chemuza.org\/tr\/\",\"name\":\"Chemuza\",\"description\":\"Kimyasal ke\u015fiflere a\u00e7\u0131lan kap\u0131n\u0131z!\",\"publisher\":{\"@id\":\"https:\/\/chemuza.org\/tr\/#organization\"},\"potentialAction\":[{\"@type\":\"SearchAction\",\"target\":{\"@type\":\"EntryPoint\",\"urlTemplate\":\"https:\/\/chemuza.org\/tr\/?s={search_term_string}\"},\"query-input\":\"required name=search_term_string\"}],\"inLanguage\":\"tr\"},{\"@type\":\"Organization\",\"@id\":\"https:\/\/chemuza.org\/tr\/#organization\",\"name\":\"Chemuza\",\"url\":\"https:\/\/chemuza.org\/tr\/\",\"logo\":{\"@type\":\"ImageObject\",\"inLanguage\":\"tr\",\"@id\":\"https:\/\/chemuza.org\/tr\/#\/schema\/logo\/image\/\",\"url\":\"https:\/\/chemuza.org\/tr\/wp-content\/uploads\/2023\/10\/chemuza-logo.png\",\"contentUrl\":\"https:\/\/chemuza.org\/tr\/wp-content\/uploads\/2023\/10\/chemuza-logo.png\",\"width\":387,\"height\":70,\"caption\":\"Chemuza\"},\"image\":{\"@id\":\"https:\/\/chemuza.org\/tr\/#\/schema\/logo\/image\/\"}},{\"@type\":\"Person\",\"@id\":\"https:\/\/chemuza.org\/tr\/#\/schema\/person\/5d92b9b82b9b43578b70f5a7cd51fa3e\",\"name\":\"Edit\u00f6r ekibi\",\"image\":{\"@type\":\"ImageObject\",\"inLanguage\":\"tr\",\"@id\":\"https:\/\/chemuza.org\/tr\/#\/schema\/person\/image\/\",\"url\":\"https:\/\/secure.gravatar.com\/avatar\/0b4093cbb11daf811af20208005bf6b5?s=96&d=mm&r=g\",\"contentUrl\":\"https:\/\/secure.gravatar.com\/avatar\/0b4093cbb11daf811af20208005bf6b5?s=96&d=mm&r=g\",\"caption\":\"Edit\u00f6r ekibi\"},\"sameAs\":[\"http:\/\/chemuza.org\/tr\"]}]}<\/script>\n<!-- \/ Yoast SEO plugin. -->","yoast_head_json":{"title":"5 Ad\u0131mda CH2 Lewis Yap\u0131s\u0131 (Resimlerle) - Chemuza","robots":{"index":"index","follow":"follow","max-snippet":"max-snippet:-1","max-image-preview":"max-image-preview:large","max-video-preview":"max-video-preview:-1"},"canonical":"https:\/\/chemuza.org\/tr\/ch2-lewis-yapisi\/","og_locale":"tr_TR","og_type":"article","og_title":"5 Ad\u0131mda CH2 Lewis Yap\u0131s\u0131 (Resimlerle) - Chemuza","og_description":"Yukar\u0131daki g\u00f6rseli zaten g\u00f6rd\u00fcn\u00fcz de\u011fil mi? Yukar\u0131daki g\u00f6rseli k\u0131saca a\u00e7\u0131klayay\u0131m. CH2 Lewis yap\u0131s\u0131n\u0131n merkezinde iki hidrojen (H) atomu ile \u00e7evrelenmi\u015f bir karbon (C) atomu bulunur. Karbon atomu (C) ile her hidrojen atomu (H) aras\u0131nda 2 tekli ba\u011f vard\u0131r. Karbon atomunun (C) 1 ortak \u00e7ifti vard\u0131r. CH2&#8217;nin Lewis yap\u0131s\u0131na ili\u015fkin yukar\u0131daki g\u00f6r\u00fcnt\u00fcden hi\u00e7bir \u015fey anlamad\u0131ysan\u0131z, benimle ... Devam\u0131n\u0131 oku","og_url":"https:\/\/chemuza.org\/tr\/ch2-lewis-yapisi\/","og_site_name":"Chemuza","article_published_time":"2023-07-25T13:35:05+00:00","og_image":[{"url":"https:\/\/chemuza.org\/wp-content\/uploads\/2023\/08\/structure-ch2-lewis.jpg"}],"author":"Edit\u00f6r ekibi","twitter_card":"summary_large_image","twitter_misc":{"Yazan:":"Edit\u00f6r ekibi","Tahmini okuma s\u00fcresi":"5 dakika"},"schema":{"@context":"https:\/\/schema.org","@graph":[{"@type":"Article","@id":"https:\/\/chemuza.org\/tr\/ch2-lewis-yapisi\/#article","isPartOf":{"@id":"https:\/\/chemuza.org\/tr\/ch2-lewis-yapisi\/"},"author":{"name":"Edit\u00f6r ekibi","@id":"https:\/\/chemuza.org\/tr\/#\/schema\/person\/5d92b9b82b9b43578b70f5a7cd51fa3e"},"headline":"5 ad\u0131mda ch2 lewis yap\u0131s\u0131 (resimlerle)","datePublished":"2023-07-25T13:35:05+00:00","dateModified":"2023-07-25T13:35:05+00:00","mainEntityOfPage":{"@id":"https:\/\/chemuza.org\/tr\/ch2-lewis-yapisi\/"},"wordCount":974,"commentCount":0,"publisher":{"@id":"https:\/\/chemuza.org\/tr\/#organization"},"articleSection":["Lewis yap\u0131s\u0131"],"inLanguage":"tr","potentialAction":[{"@type":"CommentAction","name":"Comment","target":["https:\/\/chemuza.org\/tr\/ch2-lewis-yapisi\/#respond"]}]},{"@type":"WebPage","@id":"https:\/\/chemuza.org\/tr\/ch2-lewis-yapisi\/","url":"https:\/\/chemuza.org\/tr\/ch2-lewis-yapisi\/","name":"5 Ad\u0131mda CH2 Lewis Yap\u0131s\u0131 (Resimlerle) - Chemuza","isPartOf":{"@id":"https:\/\/chemuza.org\/tr\/#website"},"datePublished":"2023-07-25T13:35:05+00:00","dateModified":"2023-07-25T13:35:05+00:00","breadcrumb":{"@id":"https:\/\/chemuza.org\/tr\/ch2-lewis-yapisi\/#breadcrumb"},"inLanguage":"tr","potentialAction":[{"@type":"ReadAction","target":["https:\/\/chemuza.org\/tr\/ch2-lewis-yapisi\/"]}]},{"@type":"BreadcrumbList","@id":"https:\/\/chemuza.org\/tr\/ch2-lewis-yapisi\/#breadcrumb","itemListElement":[{"@type":"ListItem","position":1,"name":"Ev","item":"https:\/\/chemuza.org\/tr\/"},{"@type":"ListItem","position":2,"name":"5 ad\u0131mda ch2 lewis yap\u0131s\u0131 (resimlerle)"}]},{"@type":"WebSite","@id":"https:\/\/chemuza.org\/tr\/#website","url":"https:\/\/chemuza.org\/tr\/","name":"Chemuza","description":"Kimyasal ke\u015fiflere a\u00e7\u0131lan kap\u0131n\u0131z!","publisher":{"@id":"https:\/\/chemuza.org\/tr\/#organization"},"potentialAction":[{"@type":"SearchAction","target":{"@type":"EntryPoint","urlTemplate":"https:\/\/chemuza.org\/tr\/?s={search_term_string}"},"query-input":"required name=search_term_string"}],"inLanguage":"tr"},{"@type":"Organization","@id":"https:\/\/chemuza.org\/tr\/#organization","name":"Chemuza","url":"https:\/\/chemuza.org\/tr\/","logo":{"@type":"ImageObject","inLanguage":"tr","@id":"https:\/\/chemuza.org\/tr\/#\/schema\/logo\/image\/","url":"https:\/\/chemuza.org\/tr\/wp-content\/uploads\/2023\/10\/chemuza-logo.png","contentUrl":"https:\/\/chemuza.org\/tr\/wp-content\/uploads\/2023\/10\/chemuza-logo.png","width":387,"height":70,"caption":"Chemuza"},"image":{"@id":"https:\/\/chemuza.org\/tr\/#\/schema\/logo\/image\/"}},{"@type":"Person","@id":"https:\/\/chemuza.org\/tr\/#\/schema\/person\/5d92b9b82b9b43578b70f5a7cd51fa3e","name":"Edit\u00f6r ekibi","image":{"@type":"ImageObject","inLanguage":"tr","@id":"https:\/\/chemuza.org\/tr\/#\/schema\/person\/image\/","url":"https:\/\/secure.gravatar.com\/avatar\/0b4093cbb11daf811af20208005bf6b5?s=96&d=mm&r=g","contentUrl":"https:\/\/secure.gravatar.com\/avatar\/0b4093cbb11daf811af20208005bf6b5?s=96&d=mm&r=g","caption":"Edit\u00f6r ekibi"},"sameAs":["http:\/\/chemuza.org\/tr"]}]}},"yoast_meta":{"yoast_wpseo_title":"","yoast_wpseo_metadesc":"","yoast_wpseo_canonical":""},"_links":{"self":[{"href":"https:\/\/chemuza.org\/tr\/wp-json\/wp\/v2\/posts\/181"}],"collection":[{"href":"https:\/\/chemuza.org\/tr\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/chemuza.org\/tr\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/chemuza.org\/tr\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/chemuza.org\/tr\/wp-json\/wp\/v2\/comments?post=181"}],"version-history":[{"count":0,"href":"https:\/\/chemuza.org\/tr\/wp-json\/wp\/v2\/posts\/181\/revisions"}],"wp:attachment":[{"href":"https:\/\/chemuza.org\/tr\/wp-json\/wp\/v2\/media?parent=181"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/chemuza.org\/tr\/wp-json\/wp\/v2\/categories?post=181"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/chemuza.org\/tr\/wp-json\/wp\/v2\/tags?post=181"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}