{"id":189,"date":"2023-07-25T12:02:28","date_gmt":"2023-07-25T12:02:28","guid":{"rendered":"https:\/\/chemuza.org\/tr\/yani-lewis-yapisi\/"},"modified":"2023-07-25T12:02:28","modified_gmt":"2023-07-25T12:02:28","slug":"yani-lewis-yapisi","status":"publish","type":"post","link":"https:\/\/chemuza.org\/tr\/yani-lewis-yapisi\/","title":{"rendered":"5 ad\u0131mda so 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\/donc-la-structure-de-Lewis.jpg\" alt=\"S.O. Lewis'in Yap\u0131s\u0131\" class=\"wp-image-2495\" 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\">SO Lewis yap\u0131s\u0131nda, aralar\u0131nda \u00e7ift ba\u011f i\u00e7eren bir k\u00fck\u00fcrt (S) atomu ve bir oksijen (O) atomu bulunur. K\u00fck\u00fcrt (S) atomunda ve oksijen (O) atomunda 2 yaln\u0131z \u00e7ift vard\u0131r.<\/mark><\/em><\/strong><\/p>\n<p> SO 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:\/\/en.wikipedia.org\/wiki\/Sulfur_monoxide\" target=\"_blank\" rel=\"noopener\">SO&#8217;nun<\/a> Lewis yap\u0131s\u0131n\u0131n nas\u0131l \u00e7izilece\u011fine dair ayr\u0131nt\u0131l\u0131 ad\u0131m ad\u0131m a\u00e7\u0131klamay\u0131 alacaks\u0131n\u0131z.<\/p>\n<p> \u015eimdi SO&#8217;nun <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>SO Lewis Yap\u0131s\u0131n\u0131 \u00c7izim Ad\u0131mlar\u0131<\/strong><\/h2>\n<h3 class=\"wp-block-heading\"> <strong>Ad\u0131m 1: SO 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\/\">SO molek\u00fcl\u00fcndeki<\/a> toplam <a href=\"https:\/\/chemed.chem.purdue.edu\/genchem\/topicreview\/bp\/ch8\/#valence\" target=\"_blank\" rel=\"noopener\">de\u011ferlik elektron<\/a> say\u0131s\u0131n\u0131 bulmak i\u00e7in \u00f6ncelikle bir k\u00fck\u00fcrt atomunda ve bir oksijen 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 oksijenin 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>SO 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:\/\/periodic.lanl.gov\/16.shtml\" 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 Oksijen 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\/2-2.jpg\" alt=\"\" class=\"wp-image-49\" srcset=\"\" sizes=\"\"><\/figure>\n<p> Oksijen periyodik tablonun 16. grubunda yer alan bir elementtir. <a href=\"https:\/\/education.jlab.org\/itselemental\/ele008.html\" target=\"_blank\" rel=\"noopener\"><sup>[2]<\/sup><\/a> Bu nedenle oksijende bulunan 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=\"238\" src=\"https:\/\/chemuza.org\/wp-content\/uploads\/2023\/08\/3-2.jpg\" alt=\"\" class=\"wp-image-50\" srcset=\"\" sizes=\"\"><\/figure>\n<p> Yukar\u0131daki resimde g\u00f6sterildi\u011fi gibi oksijen atomunda bulunan 6 de\u011ferlik elektronunu g\u00f6rebilirsiniz.<\/p>\n<p> Bu y\u00fczden,<\/p>\n<p> <strong>SO molek\u00fcl\u00fcndeki toplam de\u011ferlik elektronlar\u0131<\/strong> = 1 k\u00fck\u00fcrt atomu taraf\u0131ndan ba\u011f\u0131\u015flanan de\u011ferlik elektronlar\u0131 + 1 oksijen atomu taraf\u0131ndan ba\u011f\u0131\u015flanan de\u011ferlik elektronlar\u0131 = <strong>6 + 6 = 12<\/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 SO&#8217;dur. Yaln\u0131zca iki atomu vard\u0131r, dolay\u0131s\u0131yla bunlardan herhangi birini merkez atom olarak se\u00e7ebilirsiniz. <\/p>\n<figure class=\"wp-block-image aligncenter size-full\"><img decoding=\"async\" loading=\"lazy\" width=\"600\" height=\"58\" src=\"https:\/\/chemuza.org\/wp-content\/uploads\/2023\/08\/donc-etape-1.webp\" alt=\"Yani 1. ad\u0131m\" class=\"wp-image-2496\" srcset=\"\" sizes=\"\"><\/figure>\n<p> Oksijen atomunun merkezi bir atom oldu\u011funu varsayal\u0131m.<\/p>\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 SO molek\u00fcl\u00fcnde <a href=\"https:\/\/www.britannica.com\/science\/electron-pair\" target=\"_blank\" rel=\"noopener\">elektron \u00e7iftlerini<\/a> k\u00fck\u00fcrt atomu (S) ile oksijen atomu (O) aras\u0131na yerle\u015ftirmeniz gerekiyor. <\/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\/donc-etape-2.webp\" alt=\"Yani 2. ad\u0131m\" class=\"wp-image-2497\" srcset=\"\" sizes=\"\"><\/figure>\n<p> Bu, bir SO molek\u00fcl\u00fcnde k\u00fck\u00fcrt (S) atomunun ve oksijen (O) atomunun <a href=\"https:\/\/goldbook.iupac.org\/terms\/view\/CT07009\" 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 harici atomun stabilitesini kontrol etmeniz gerekir.<\/p>\n<p> Burada SO molek\u00fcl\u00fcn\u00fcn diyagram\u0131nda oksijen atomunun merkez atom oldu\u011funu varsayd\u0131k. Bu nedenle k\u00fck\u00fcrt d\u0131\u015f atomdur.<\/p>\n<p> Bu nedenle k\u00fck\u00fcrt atomunu kararl\u0131 hale getirmeliyiz.<\/p>\n<p> A\u015fa\u011f\u0131daki resimde k\u00fck\u00fcrt atomunun bir <a href=\"https:\/\/www.ck12.org\/c\/chemistry\/octet-rule\/lesson\/Octet-Rule-CHEM\/\" target=\"_blank\" rel=\"noopener\">oktet<\/a> olu\u015fturdu\u011funu ve bu nedenle kararl\u0131 oldu\u011funu g\u00f6rebilirsiniz. <\/p>\n<figure class=\"wp-block-image aligncenter size-full\"><img decoding=\"async\" loading=\"lazy\" width=\"600\" height=\"244\" src=\"https:\/\/chemuza.org\/wp-content\/uploads\/2023\/08\/donc-etape-3.webp\" alt=\"Yani 3. ad\u0131m\" class=\"wp-image-2498\" srcset=\"\" sizes=\"\"><\/figure>\n<p> Ek olarak, 1. ad\u0131mda SO molek\u00fcl\u00fcnde bulunan toplam de\u011ferlik elektronu say\u0131s\u0131n\u0131 hesaplad\u0131k.<\/p>\n<p> SO molek\u00fcl\u00fcnde toplam <strong>12 de\u011ferlik elektronu<\/strong> vard\u0131r ve yukar\u0131daki diyagramda bunlardan sadece <strong>8 de\u011ferlik elektronu<\/strong> kullan\u0131lm\u0131\u015ft\u0131r.<\/p>\n<p> Yani kalan elektron say\u0131s\u0131 = <strong>12 \u2013 8 = 4<\/strong> .<\/p>\n<p> Bu <strong>4<\/strong> elektronu yukar\u0131daki SO molek\u00fcl\u00fcn\u00fcn diyagram\u0131nda oksijen atomunun \u00fczerine yerle\u015ftirmeniz gerekiyor. <\/p>\n<figure class=\"wp-block-image aligncenter size-full\"><img decoding=\"async\" loading=\"lazy\" width=\"600\" height=\"217\" src=\"https:\/\/chemuza.org\/wp-content\/uploads\/2023\/08\/donc-etape-4.webp\" alt=\"Yani 4. ad\u0131m\" class=\"wp-image-2499\" 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. 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 oksijen (O) atomunun kararl\u0131 olup olmad\u0131\u011f\u0131n\u0131 kontrol etmeniz gerekir.<\/p>\n<p> Bu oksijen atomunun (O) stabilitesini kontrol etmek i\u00e7in oktet olu\u015fturup olu\u015fturmad\u0131\u011f\u0131n\u0131 kontrol etmemiz gerekir.<\/p>\n<p> Maalesef bu oksijen atomu burada bir oktet olu\u015fturmuyor. Oksijenin yaln\u0131zca 6 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\/donc-etape-5.webp\" alt=\"Yani 5. ad\u0131m\" class=\"wp-image-2500\" srcset=\"\" sizes=\"\"><\/figure>\n<p> \u015eimdi bu oksijen atomunu kararl\u0131 hale getirmek i\u00e7in elektron \u00e7iftini k\u00fck\u00fcrt atomundan uzakla\u015ft\u0131rman\u0131z gerekir. <\/p>\n<figure class=\"wp-block-image aligncenter size-full\"><img decoding=\"async\" loading=\"lazy\" width=\"600\" height=\"120\" src=\"https:\/\/chemuza.org\/wp-content\/uploads\/2023\/08\/donc-etape-6.webp\" alt=\"Yani 6. ad\u0131m\" class=\"wp-image-2501\" srcset=\"\" sizes=\"\"><\/figure>\n<p> Bu elektron \u00e7iftini hareket ettirdikten sonra oksijen 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=\"268\" src=\"https:\/\/chemuza.org\/wp-content\/uploads\/2023\/08\/donc-etape-7.webp\" alt=\"Yani 7. ad\u0131m\" class=\"wp-image-2502\" srcset=\"\" sizes=\"\"><\/figure>\n<p> Yukar\u0131daki resimde oksijen atomunun bir oktet olu\u015fturdu\u011funu g\u00f6rebilirsiniz.<\/p>\n<p> Ve b\u00f6ylece bu oksijen atomu kararl\u0131d\u0131r.<\/p>\n<p> \u015eimdi SO&#8217;nun 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 SO&#8217;nun 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:\/\/chem.libretexts.org\/Bookshelves\/Physical_and_Theoretical_Chemistry_Textbook_Maps\/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)\/Physical_Properties_of_Matter\/Atomic_and_Molecular_Properties\/Formal_Charges\/Formal_Charge\" target=\"_blank\" rel=\"noopener\">formal y\u00fck<\/a> kavram\u0131 kullan\u0131larak do\u011frulanabilir.<\/p>\n<p> K\u0131saca \u015fimdi SO molek\u00fcl\u00fcnde bulunan k\u00fck\u00fcrt atomunun (S) ve oksijen atomunun (O) 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 <a href=\"https:\/\/chemuza.org\/tr\/kimyanin-temel-tanimlari\/\">ba\u011f elektronlar\u0131n\u0131n<\/a> ve <a href=\"https:\/\/chemuza.org\/tr\/kimyanin-temel-tanimlari\/\">ba\u011f yapmayan 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=\"255\" src=\"https:\/\/chemuza.org\/wp-content\/uploads\/2023\/08\/donc-etape-8.webp\" alt=\"Yani 8. ad\u0131m\" class=\"wp-image-2503\" 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>Oksijen atomu (O) i\u00e7in:<\/strong><strong><br \/><\/strong> De\u011ferlik elektronlar\u0131 = 6 (\u00e7\u00fcnk\u00fc oksijen grup 16&#8217;dad\u0131r)<strong><br \/><\/strong> Ba\u011f elektronlar\u0131 = 4<br \/> Ba\u011flanmayan elektronlar = 4 <\/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\"> Ah<\/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<\/tbody>\n<\/table>\n<\/figure>\n<p> Yukar\u0131daki formal y\u00fck hesaplamalar\u0131ndan k\u00fck\u00fcrt (S) ve oksijen (O) atomlar\u0131n\u0131n <strong>&#8220;s\u0131f\u0131r&#8221;<\/strong> formal y\u00fcke sahip oldu\u011funu g\u00f6rebilirsiniz.<\/p>\n<p> Bu, SO&#8217;nun yukar\u0131daki Lewis yap\u0131s\u0131n\u0131n kararl\u0131 oldu\u011funu ve SO&#8217;nun yukar\u0131daki yap\u0131s\u0131nda ba\u015fka bir de\u011fi\u015fiklik olmad\u0131\u011f\u0131n\u0131 g\u00f6sterir.<\/p>\n<p> SO&#8217;nun yukar\u0131daki Lewis nokta yap\u0131s\u0131nda, her bir ba\u011f elektronu \u00e7iftini (:) tek bir ba\u011f (|) olarak da temsil edebilirsiniz. Bunu yapmak SO&#8217;nun 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-de-Lewis-de-donc.jpg\" alt=\"SO'nun Lewis yap\u0131s\u0131\" class=\"wp-image-2504\" 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-sef2-yapisi\/\">SeF2 Lewis yap\u0131s\u0131<\/a><\/td>\n<td> <a href=\"https:\/\/chemuza.org\/tr\/yapi-c2f2-lewis\/\">Lewis yap\u0131s\u0131 C2F2<\/a><\/td>\n<\/tr>\n<tr>\n<td> <a href=\"https:\/\/chemuza.org\/tr\/lewis-xecl4-yapisi\/\">Lewis yap\u0131s\u0131 XeCl4<\/a><\/td>\n<td> <a href=\"https:\/\/chemuza.org\/tr\/lewis-albr3-yapisi\/\">Lewis yap\u0131s\u0131 AlBr3<\/a><\/td>\n<\/tr>\n<tr>\n<td> <a href=\"https:\/\/chemuza.org\/tr\/lewis-alf3-yapisi\/\">Lewis yap\u0131s\u0131 AlF3<\/a><\/td>\n<td> <a href=\"https:\/\/chemuza.org\/tr\/ibr-lewis-yapisi\/\">Lewis yap\u0131s\u0131 IBr<\/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. SO Lewis yap\u0131s\u0131nda, aralar\u0131nda \u00e7ift ba\u011f i\u00e7eren bir k\u00fck\u00fcrt (S) atomu ve bir oksijen (O) atomu bulunur. K\u00fck\u00fcrt (S) atomunda ve oksijen (O) atomunda 2 yaln\u0131z \u00e7ift vard\u0131r. SO Lewis yap\u0131s\u0131n\u0131n yukar\u0131daki g\u00f6r\u00fcnt\u00fcs\u00fcnden hi\u00e7bir \u015fey anlamad\u0131ysan\u0131z, benimle kal\u0131n ve SO&#8217;nun Lewis yap\u0131s\u0131n\u0131n nas\u0131l \u00e7izilece\u011fine dair &#8230; <a title=\"5 ad\u0131mda so lewis yap\u0131s\u0131 (resimlerle)\" class=\"read-more\" href=\"https:\/\/chemuza.org\/tr\/yani-lewis-yapisi\/\" aria-label=\"More on 5 ad\u0131mda so 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 SO 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\/yani-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 SO 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. SO Lewis yap\u0131s\u0131nda, aralar\u0131nda \u00e7ift ba\u011f i\u00e7eren bir k\u00fck\u00fcrt (S) atomu ve bir oksijen (O) atomu bulunur. K\u00fck\u00fcrt (S) atomunda ve oksijen (O) atomunda 2 yaln\u0131z \u00e7ift vard\u0131r. SO Lewis yap\u0131s\u0131n\u0131n yukar\u0131daki g\u00f6r\u00fcnt\u00fcs\u00fcnden hi\u00e7bir \u015fey anlamad\u0131ysan\u0131z, benimle kal\u0131n ve SO&#8217;nun Lewis yap\u0131s\u0131n\u0131n nas\u0131l \u00e7izilece\u011fine dair ... 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Yukar\u0131daki g\u00f6rseli k\u0131saca a\u00e7\u0131klayay\u0131m. SO Lewis yap\u0131s\u0131nda, aralar\u0131nda \u00e7ift ba\u011f i\u00e7eren bir k\u00fck\u00fcrt (S) atomu ve bir oksijen (O) atomu bulunur. K\u00fck\u00fcrt (S) atomunda ve oksijen (O) atomunda 2 yaln\u0131z \u00e7ift vard\u0131r. SO Lewis yap\u0131s\u0131n\u0131n yukar\u0131daki g\u00f6r\u00fcnt\u00fcs\u00fcnden hi\u00e7bir \u015fey anlamad\u0131ysan\u0131z, benimle kal\u0131n ve SO&#8217;nun Lewis yap\u0131s\u0131n\u0131n nas\u0131l \u00e7izilece\u011fine dair ... 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