{"id":146,"date":"2023-07-25T20:57:56","date_gmt":"2023-07-25T20:57:56","guid":{"rendered":"https:\/\/chemuza.org\/tr\/noc-lewis-yapisi\/"},"modified":"2023-07-25T20:57:56","modified_gmt":"2023-07-25T20:57:56","slug":"noc-lewis-yapisi","status":"publish","type":"post","link":"https:\/\/chemuza.org\/tr\/noc-lewis-yapisi\/","title":{"rendered":"6 ad\u0131mda cno-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-cno-lewis.jpg\" alt=\"CNO-Lewis yap\u0131s\u0131\" class=\"wp-image-1974\" 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\">CNO- (fulminat iyonu) Lewis yap\u0131s\u0131n\u0131n merkezinde bir karbon (C) atomu ve bir oksijen (O) atomu ile \u00e7evrelenmi\u015f bir nitrojen (N) atomu bulunur. Azot (N) ile oksijen (O) atomu aras\u0131nda tekli ba\u011f, karbon (C) ile azot (N) aras\u0131nda ise \u00fc\u00e7l\u00fc ba\u011f bulunur.<\/mark><\/em><\/strong><\/p>\n<p> CNO-iyonunun Lewis yap\u0131s\u0131n\u0131n ( <a href=\"https:\/\/en.wikipedia.org\/wiki\/Fulminate\" target=\"_blank\" rel=\"noopener\">fulminat iyonu<\/a> ) yukar\u0131daki g\u00f6r\u00fcnt\u00fcs\u00fcnden hi\u00e7bir \u015fey anlamad\u0131ysan\u0131z, o zaman benimle kal\u0131n ve CNO iyonunun 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 CNO iyonunun <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>CNO-Lewis yap\u0131s\u0131n\u0131 \u00e7izme ad\u0131mlar\u0131<\/strong><\/h2>\n<h3 class=\"wp-block-heading\"> <strong>Ad\u0131m 1: CNO molek\u00fcl\u00fcndeki toplam de\u011ferlik elektronu say\u0131s\u0131n\u0131 bulun<\/strong><\/h3>\n<p> Bir CNO iyonundaki (fulminat iyonu) <a href=\"http:\/\/www.chem.ucla.edu\/~harding\/IGOC\/V\/valence_electron.html\" target=\"_blank\" rel=\"noopener\">de\u011ferlik elektronlar\u0131n\u0131n<\/a> toplam say\u0131s\u0131n\u0131 bulmak i\u00e7in \u00f6ncelikle karbon atomunda, nitrojen atomunda ve ayr\u0131ca 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 karbon, nitrojen ve oksijenin de\u011ferlik elektronlar\u0131n\u0131 nas\u0131l kolayca bulaca\u011f\u0131n\u0131z\u0131 anlataca\u011f\u0131m.<\/p>\n<p> <strong>CNO iyonundaki 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:\/\/periodic.lanl.gov\/6.shtml\" 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 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:\/\/education.jlab.org\/itselemental\/ele007.html\" target=\"_blank\" rel=\"noopener\"><sup>[2]<\/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> <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:\/\/pubchem.ncbi.nlm.nih.gov\/element\/8\" target=\"_blank\" rel=\"noopener\"><sup>[3]<\/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>CNO iyonundaki toplam de\u011ferlik elektronlar\u0131<\/strong> = 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 + 1 oksijen atomu taraf\u0131ndan ba\u011f\u0131\u015flanan de\u011ferlik elektronlar\u0131 + 1 negatif y\u00fck nedeniyle eklenen 1 ekstra elektron = <strong>4 + 5 + 6 + 1 = 16<\/strong> .<\/p>\n<h3 class=\"wp-block-heading\"> <strong>Ad\u0131m 2: Bir eskiz haz\u0131rlay\u0131n<\/strong><\/h3>\n<p> NOC&#8217;nin bir tasla\u011f\u0131n\u0131 \u00e7izmek i\u00e7in kimyasal form\u00fcl\u00fcne bakman\u0131z yeterli. Merkezde bir nitrojen (N) atomunun bulundu\u011funu ve her iki taraf\u0131nda da bir karbon atomu ve bir oksijen atomu ile \u00e7evrelendi\u011fini g\u00f6rebilirsiniz.<\/p>\n<p> \u015eimdi bunun kaba bir tasla\u011f\u0131n\u0131 yapal\u0131m. <\/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\/pas-detape-1.webp\" alt=\"NOC - a\u015fama 1\" class=\"wp-image-1975\" 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 CNO 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> karbon atomu (C), nitrojen atomu (N) ve oksijen atomu (O) aras\u0131na yerle\u015ftirmeniz gerekir. <\/p>\n<figure class=\"wp-block-image aligncenter size-full\"><img decoding=\"async\" loading=\"lazy\" width=\"600\" height=\"63\" src=\"https:\/\/chemuza.org\/wp-content\/uploads\/2023\/08\/pas-detape-2.webp\" alt=\"NOC - a\u015fama 2\" class=\"wp-image-1976\" srcset=\"\" sizes=\"\"><\/figure>\n<p> Bu, bir CNO molek\u00fcl\u00fcnde karbon atomunun (C), nitrojen atomunun (N) ve oksijen atomunun (O) 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<\/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 CNO molek\u00fcl\u00fcn\u00fcn \u00e7iziminde d\u0131\u015ftaki atomlar\u0131n karbon atomu ve oksijen atomu oldu\u011funu g\u00f6rebilirsiniz.<\/p>\n<p> Bu d\u0131\u015f karbon ve oksijen atomlar\u0131 bir <a href=\"https:\/\/en.wikibooks.org\/wiki\/General_Chemistry\/Octet_Rule_and_Exceptions\" 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=\"269\" src=\"https:\/\/chemuza.org\/wp-content\/uploads\/2023\/08\/pas-detape-3.webp\" alt=\"NOC - a\u015fama 3\" class=\"wp-image-1977\" srcset=\"\" sizes=\"\"><\/figure>\n<p> Ek olarak, 1. ad\u0131mda CNO iyonunda bulunan toplam de\u011ferlik elektronu say\u0131s\u0131n\u0131 hesaplad\u0131k.<\/p>\n<p> CNO-iyonunun 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 nitrojen (N) atomunun kararl\u0131 olup olmad\u0131\u011f\u0131n\u0131 kontrol etmeniz gerekir.<\/p>\n<p> Merkezi nitrojen (N) atomunun stabilitesini kontrol etmek i\u00e7in onun bir oktet olu\u015fturup olu\u015fturmad\u0131\u011f\u0131n\u0131 kontrol etmemiz gerekir.<\/p>\n<p> Ne yaz\u0131k ki nitrojen atomu burada bir oktet olu\u015fturmuyor. Azotun sadece 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=\"257\" src=\"https:\/\/chemuza.org\/wp-content\/uploads\/2023\/08\/pas-detape-4.webp\" alt=\"NOC - a\u015fama 4\" class=\"wp-image-1978\" srcset=\"\" sizes=\"\"><\/figure>\n<p> \u015eimdi, bu nitrojen atomunu kararl\u0131 hale getirmek i\u00e7in, d\u0131\u015ftaki karbon atomunun elektron \u00e7iftini, nitrojen atomunun 8 elektrona (yani bir oktete) sahip olaca\u011f\u0131 \u015fekilde kayd\u0131rman\u0131z gerekir.<\/p>\n<p> <strong>(Not:<\/strong> Daha az elektronegatif olan atomun elektron \u00e7iftini hareket ettirmeniz gerekti\u011fini unutmay\u0131n.<br \/> Ger\u00e7ekten de, elektronegatifli\u011fi daha az olan atomun elektron verme e\u011filimi daha y\u00fcksektir.<br \/> Burada karbon atomu ile oksijen atomunu kar\u015f\u0131la\u015ft\u0131r\u0131rsak, karbon atomu daha az elektronegatiftir.<br \/> Yani karbon atomunun elektron \u00e7iftini hareket ettirmeniz 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\/pas-detape-5.webp\" alt=\"NOC - ad\u0131m 5\" class=\"wp-image-1979\" srcset=\"\" sizes=\"\"><\/figure>\n<p> Ancak bir \u00e7ift elektronu hareket ettirdikten sonra nitrojen 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=\"240\" src=\"https:\/\/chemuza.org\/wp-content\/uploads\/2023\/08\/pas-detape-6.webp\" alt=\"NOC - a\u015fama 6\" class=\"wp-image-1980\" srcset=\"\" sizes=\"\"><\/figure>\n<p> Yine, yaln\u0131zca karbon atomundan fazladan bir \u00e7ift elektronu hareket ettirmemiz gerekiyor. (\u00c7\u00fcnk\u00fc karbon oksijenden daha az elektronegatiftir.) <\/p>\n<figure class=\"wp-block-image aligncenter size-full\"><img decoding=\"async\" loading=\"lazy\" width=\"600\" height=\"100\" src=\"https:\/\/chemuza.org\/wp-content\/uploads\/2023\/08\/pas-detape-7.webp\" alt=\"NOC - a\u015fama 7\" class=\"wp-image-1981\" srcset=\"\" sizes=\"\"><\/figure>\n<p> Bu elektron \u00e7iftini hareket ettirdikten sonra merkezi nitrojen 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=\"260\" src=\"https:\/\/chemuza.org\/wp-content\/uploads\/2023\/08\/pas-detape-8.webp\" alt=\"NOC - a\u015fama 8\" class=\"wp-image-1982\" srcset=\"\" sizes=\"\"><\/figure>\n<p> Yukar\u0131daki resimde nitrojen atomunun bir oktet olu\u015fturdu\u011funu g\u00f6rebilirsiniz.<\/p>\n<p> Ve bu nedenle nitrojen atomu kararl\u0131d\u0131r.<\/p>\n<p> \u015eimdi CNO&#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 CNO 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=\"http:\/\/guweb2.gonzaga.edu\/faculty\/cronk\/CHEM101pub\/formal_charge.html\" target=\"_blank\" rel=\"noopener\">formal y\u00fck<\/a> kavram\u0131 kullan\u0131larak do\u011frulanabilir.<\/p>\n<p> K\u0131sacas\u0131 art\u0131k CNO molek\u00fcl\u00fcnde bulunan karbon (C), nitrojen (N) ve oksijen (O) atomlar\u0131n\u0131n formal y\u00fck\u00fcn\u00fc bulmam\u0131z gerekiyor.<\/p>\n<p> Resmi vergiyi hesaplamak i\u00e7in a\u015fa\u011f\u0131daki form\u00fcl\u00fc kullanman\u0131z gerekir:<\/p>\n<p class=\"has-background\" style=\"background-color:#ffe9cf\"> <strong>Resmi y\u00fck = De\u011ferlik elektronlar\u0131 \u2013 (Ba\u011f elektronlar\u0131)\/2 \u2013 Ba\u011f yapmayan elektronlar<\/strong><\/p>\n<p> A\u015fa\u011f\u0131daki resimde CNO 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=\"226\" src=\"https:\/\/chemuza.org\/wp-content\/uploads\/2023\/08\/pas-detape-9.webp\" alt=\"NOC - a\u015fama 9\" class=\"wp-image-1983\" srcset=\"\" sizes=\"\"><\/figure>\n<p> <strong>Karbon atomu (C) i\u00e7in:<\/strong><strong><br \/><\/strong> De\u011ferlik elektronlar\u0131 = 4 (\u00e7\u00fcnk\u00fc karbon 14. gruptad\u0131r)<strong><br \/><\/strong> Ba\u011f elektronlar\u0131 = 6<br \/> Ba\u011flanmayan elektronlar = 2<\/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 = 8<br \/> Ba\u011flanmayan elektronlar = 0<\/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 = 2<br \/> Ba\u011flanmayan elektronlar = 6 <\/p>\n<figure class=\"wp-block-table\">\n<table>\n<tbody>\n<tr>\n<td class=\"has-text-align-center\" data-align=\"center\"> <strong>Resmi su\u00e7lama<\/strong><\/td>\n<td> <strong>=<\/strong><\/td>\n<td class=\"has-text-align-center\" data-align=\"center\"> <strong>de\u011ferlik elektronlar\u0131<\/strong><\/td>\n<td> <strong>\u2013<\/strong><\/td>\n<td class=\"has-text-align-center\" data-align=\"center\"> <strong>(Elektronlar\u0131n ba\u011flanmas\u0131)\/2<\/strong><\/td>\n<td> <strong>\u2013<\/strong><\/td>\n<td class=\"has-text-align-center\" data-align=\"center\"> <strong>Ba\u011flanmayan elektronlar<\/strong> <\/td>\n<td><\/td>\n<td class=\"has-text-align-center\" data-align=\"center\"><\/td>\n<\/tr>\n<tr>\n<td class=\"has-text-align-center\" data-align=\"center\"> VS<\/td>\n<td> =<\/td>\n<td class=\"has-text-align-center\" data-align=\"center\"> 4<\/td>\n<td> \u2013<\/td>\n<td class=\"has-text-align-center\" data-align=\"center\"> 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>+1<\/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\"> 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\"> 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\"> 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>-1<\/strong><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/figure>\n<p> Yukar\u0131daki formal y\u00fck hesaplamalar\u0131ndan karbon (C) atomunun <strong>-1<\/strong> , oksijen (O) atomunun ise <strong>+1<\/strong> y\u00fck\u00fcne sahip oldu\u011funu g\u00f6rebilirsiniz.<\/p>\n<p> \u00d6yleyse bu y\u00fckleri CNO molek\u00fcl\u00fcn\u00fcn ilgili atomlar\u0131nda tutal\u0131m. <\/p>\n<figure class=\"wp-block-image aligncenter size-full\"><img decoding=\"async\" loading=\"lazy\" width=\"600\" height=\"84\" src=\"https:\/\/chemuza.org\/wp-content\/uploads\/2023\/08\/pas-detape-10.webp\" alt=\"NOC - a\u015fama 10\" class=\"wp-image-1984\" srcset=\"\" sizes=\"\"><\/figure>\n<p> <strong>+1<\/strong> ve <strong>-1<\/strong> y\u00fckleri iptal edilir ve CNO molek\u00fcl\u00fcndeki toplam <strong>-1<\/strong> y\u00fck\u00fc a\u015fa\u011f\u0131daki g\u00f6r\u00fcnt\u00fcde g\u00f6sterilmektedir. <\/p>\n<figure class=\"wp-block-image aligncenter size-full\"><img decoding=\"async\" loading=\"lazy\" width=\"600\" height=\"99\" src=\"https:\/\/chemuza.org\/wp-content\/uploads\/2023\/08\/pas-detape-11.webp\" alt=\"NOC - a\u015fama 11\" class=\"wp-image-1985\" srcset=\"\" sizes=\"\"><\/figure>\n<p> CNO iyonunun yukar\u0131daki Lewis nokta yap\u0131s\u0131nda, her bir ba\u011f elektronu \u00e7iftini (:) tek bir ba\u011f (|) olarak da temsil edebilirsiniz. Bunu yaparak CNO iyonunun a\u015fa\u011f\u0131daki Lewis yap\u0131s\u0131n\u0131 elde edeceksiniz. <\/p>\n<figure class=\"wp-block-image aligncenter size-full\"><img decoding=\"async\" loading=\"lazy\" width=\"600\" height=\"227\" src=\"https:\/\/chemuza.org\/wp-content\/uploads\/2023\/08\/structure-de-Lewis-de-cno.jpg\" alt=\"NOC'nin Lewis yap\u0131s\u0131-\" class=\"wp-image-1986\" 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\/bro2-lewis-yapisi\/\">BrO2-Lewis yap\u0131s\u0131<\/a><\/td>\n<td> <a href=\"https:\/\/chemuza.org\/tr\/lewis-n2o4un-yapisi\/\">Lewis yap\u0131s\u0131 N2O4<\/a><\/td>\n<\/tr>\n<tr>\n<td> <a href=\"https:\/\/chemuza.org\/tr\/cof2-lewis-yapisi\/\">Lewis yap\u0131s\u0131 COF2<\/a><\/td>\n<td> <a href=\"https:\/\/chemuza.org\/tr\/lewis-scl4-yapisi\/\">Lewis yap\u0131s\u0131 SCl4<\/a><\/td>\n<\/tr>\n<tr>\n<td> <a href=\"https:\/\/chemuza.org\/tr\/lewis-pbr5-yapisi\/\">Lewis yap\u0131s\u0131 PBr5<\/a><\/td>\n<td> <a href=\"https:\/\/chemuza.org\/tr\/yapi-sis2-lewis\/\">Lewis yap\u0131s\u0131 SiS2<\/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. CNO- (fulminat iyonu) Lewis yap\u0131s\u0131n\u0131n merkezinde bir karbon (C) atomu ve bir oksijen (O) atomu ile \u00e7evrelenmi\u015f bir nitrojen (N) atomu bulunur. Azot (N) ile oksijen (O) atomu aras\u0131nda tekli ba\u011f, karbon (C) ile azot (N) aras\u0131nda ise \u00fc\u00e7l\u00fc ba\u011f bulunur. CNO-iyonunun Lewis yap\u0131s\u0131n\u0131n ( &#8230; <a title=\"6 ad\u0131mda cno-lewis yap\u0131s\u0131 (resimlerle)\" class=\"read-more\" href=\"https:\/\/chemuza.org\/tr\/noc-lewis-yapisi\/\" aria-label=\"More on 6 ad\u0131mda cno-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 CNO-Lewis yap\u0131s\u0131 (resimlerle birlikte) - 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\/noc-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 CNO-Lewis yap\u0131s\u0131 (resimlerle birlikte) - 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. CNO- (fulminat iyonu) Lewis yap\u0131s\u0131n\u0131n merkezinde bir karbon (C) atomu ve bir oksijen (O) atomu ile \u00e7evrelenmi\u015f bir nitrojen (N) atomu bulunur. Azot (N) ile oksijen (O) atomu aras\u0131nda tekli ba\u011f, karbon (C) ile azot (N) aras\u0131nda ise \u00fc\u00e7l\u00fc ba\u011f bulunur. CNO-iyonunun Lewis yap\u0131s\u0131n\u0131n ( ... 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Yukar\u0131daki g\u00f6rseli k\u0131saca a\u00e7\u0131klayay\u0131m. CNO- (fulminat iyonu) Lewis yap\u0131s\u0131n\u0131n merkezinde bir karbon (C) atomu ve bir oksijen (O) atomu ile \u00e7evrelenmi\u015f bir nitrojen (N) atomu bulunur. Azot (N) ile oksijen (O) atomu aras\u0131nda tekli ba\u011f, karbon (C) ile azot (N) aras\u0131nda ise \u00fc\u00e7l\u00fc ba\u011f bulunur. CNO-iyonunun Lewis yap\u0131s\u0131n\u0131n ( ... 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