{"id":46,"date":"2023-07-26T16:39:04","date_gmt":"2023-07-26T16:39:04","guid":{"rendered":"https:\/\/chemuza.org\/tr\/hno3-lewis-yapisi\/"},"modified":"2023-07-26T16:39:04","modified_gmt":"2023-07-26T16:39:04","slug":"hno3-lewis-yapisi","status":"publish","type":"post","link":"https:\/\/chemuza.org\/tr\/hno3-lewis-yapisi\/","title":{"rendered":"6 ad\u0131mda hno3 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-hno3-lewis.jpg\" alt=\"HNO3 Lewis yap\u0131s\u0131\" class=\"wp-image-717\" 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\">HNO3&#8217;\u00fcn (nitrik asit) Lewis yap\u0131s\u0131nda, merkezde iki oksijen (O) atomu ve bir OH grubu ile \u00e7evrelenmi\u015f bir nitrojen (N) atomu bulunur. Azot (N) atomu ile oksijen (O) atomu aras\u0131nda 1 adet \u00e7ift ba\u011f bulunur, di\u011fer atomlar\u0131n geri kalan\u0131nda ise tek ba\u011f bulunur.<\/mark><\/em><\/strong><\/p>\n<p> HNO3&#8217;\u00fcn (nitrik asit) 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:\/\/pubchem.ncbi.nlm.nih.gov\/compound\/Nitric-Acid\" target=\"_blank\" rel=\"noopener\">HNO3&#8217;\u00fcn<\/a> Lewis yap\u0131s\u0131n\u0131n \u00e7izilmesine ili\u015fkin ayr\u0131nt\u0131l\u0131 ad\u0131m ad\u0131m a\u00e7\u0131klamay\u0131 alacaks\u0131n\u0131z.<\/p>\n<p> O halde HNO3&#8217;\u00fcn <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>HNO3 Lewis Yap\u0131s\u0131n\u0131 \u00c7izim Ad\u0131mlar\u0131<\/strong><\/h2>\n<h3 class=\"wp-block-heading\"> <strong>Ad\u0131m 1: HNO3 molek\u00fcl\u00fcndeki toplam de\u011ferlik elektronu say\u0131s\u0131n\u0131 bulun<\/strong><\/h3>\n<p> Bir HNO3 <a href=\"https:\/\/chemuza.org\/tr\/kimyanin-temel-tanimlari\/\">molek\u00fcl\u00fcndeki<\/a> toplam <a href=\"https:\/\/chem.libretexts.org\/Courses\/Purdue\/Purdue%3A_Chem_26505%3A_Organic_Chemistry_I_(Lipton)\/Chapter_1._Electronic_Structure_and_Chemical_Bonding\/1.03_Valence_electrons_and_open_valences\" target=\"_blank\" rel=\"noopener\">de\u011ferlik elektron<\/a> say\u0131s\u0131n\u0131 bulmak i\u00e7in \u00f6ncelikle <a href=\"https:\/\/chemuza.org\/tr\/kimyanin-temel-tanimlari\/\">hidrojen atomunda<\/a> , nitrojen atomunda ve 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 <a href=\"https:\/\/chemuza.org\/tr\/kimyanin-temel-tanimlari\/\">periyodik tabloyu<\/a> kullanarak hidrojen, nitrojen ve oksijenin de\u011ferlik elektronlar\u0131n\u0131 nas\u0131l kolayca bulaca\u011f\u0131n\u0131z\u0131 anlataca\u011f\u0131m.<\/p>\n<p class=\"has-medium-font-size\"> <strong>HNO3 molek\u00fcl\u00fcndeki toplam de\u011ferlik elektronlar\u0131<\/strong><\/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> <a href=\"https:\/\/chemuza.org\/tr\/kimyanin-temel-tanimlari\/\">Hidrojen<\/a> periyodik tablonun <a href=\"https:\/\/chemuza.org\/tr\/kimyanin-temel-tanimlari\/\">1. grup<\/a> elementidir. <a href=\"https:\/\/en.wikipedia.org\/wiki\/Hydrogen\" target=\"_blank\" rel=\"noopener\"><sup>[1]<\/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> <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> <a href=\"https:\/\/chemuza.org\/tr\/kimyanin-temel-tanimlari\/\">Azot<\/a> periyodik tablonun <a href=\"https:\/\/chemuza.org\/tr\/kimyanin-temel-tanimlari\/\">15. grubunda yer alan<\/a> bir elementtir. <a href=\"https:\/\/pubchem.ncbi.nlm.nih.gov\/element\/7\" 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> <a href=\"https:\/\/chemuza.org\/tr\/kimyanin-temel-tanimlari\/\">Oksijen<\/a> periyodik tablonun <a href=\"https:\/\/chemuza.org\/tr\/kimyanin-temel-tanimlari\/\">16. grubunda yer alan<\/a> bir elementtir. <a href=\"https:\/\/www.rsc.org\/periodic-table\/element\/8\/oxygen\" 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>HNO3 molek\u00fcl\u00fcndeki toplam de\u011ferlik elektronlar\u0131<\/strong> = 1 hidrojen atomu taraf\u0131ndan ba\u011f\u0131\u015flanan de\u011ferlik elektronlar\u0131 + 1 nitrojen atomu taraf\u0131ndan ba\u011f\u0131\u015flanan de\u011ferlik elektronlar\u0131 + 3 oksijen atomu taraf\u0131ndan ba\u011f\u0131\u015flanan de\u011ferlik elektronlar\u0131 = <strong>1 + 5 + 6(3) = 24<\/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 <a href=\"https:\/\/chemuza.org\/tr\/kimyanin-temel-tanimlari\/\">hidrojen<\/a> varsa, daima hidrojeni d\u0131\u015far\u0131ya koyun.)<\/p>\n<p> \u015eimdi burada verilen molek\u00fcl HNO3&#8217;t\u00fcr (nitrik asit) ve hidrojen atomu (H), nitrojen atomu (N) ve oksijen atomlar\u0131n\u0131 (O) i\u00e7erir.<\/p>\n<p> Yani kurala g\u00f6re hidrojeni d\u0131\u015far\u0131da tutmal\u0131y\u0131z. <\/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> Art\u0131k yukar\u0131daki periyodik tabloda nitrojen (N) atomu ve oksijen (O) atomunun elektronegatiflik de\u011ferlerini g\u00f6rebilirsiniz.<\/p>\n<p> Azot (N) ve oksijenin (O) elektronegatiflik de\u011ferlerini kar\u015f\u0131la\u015ft\u0131r\u0131rsak <a href=\"https:\/\/chemuza.org\/tr\/periyodik-tablonun-elektronegatifligi\/\">azot atomu daha az elektronegatiftir<\/a> .<\/p>\n<p> Burada nitrojen (N) atomu merkez atom, oksijen (O) atomlar\u0131 ise d\u0131\u015f atomdur. <\/p>\n<figure class=\"wp-block-image aligncenter size-full\"><img decoding=\"async\" loading=\"lazy\" width=\"600\" height=\"109\" src=\"https:\/\/chemuza.org\/wp-content\/uploads\/2023\/08\/hno3-etape-1.webp\" alt=\"HNO3 ad\u0131m 1\" class=\"wp-image-718\" 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 HNO3 molek\u00fcl\u00fcnde elektron \u00e7iftlerini oksijen (O) ve hidrojen (H) atomlar\u0131 aras\u0131na ve oksijen (O) ve nitrojen (N) atomlar\u0131 aras\u0131na yerle\u015ftirmeniz gerekiyor. <\/p>\n<figure class=\"wp-block-image aligncenter size-full\"><img decoding=\"async\" loading=\"lazy\" width=\"600\" height=\"108\" src=\"https:\/\/chemuza.org\/wp-content\/uploads\/2023\/08\/hno3-etape-2.webp\" alt=\"HNO3 ad\u0131m 2\" class=\"wp-image-719\" srcset=\"\" sizes=\"\"><\/figure>\n<p> Bu, bir HNO3 molek\u00fcl\u00fcnde bu atomlar\u0131n birbirine <a href=\"http:\/\/hyperphysics.phy-astr.gsu.edu\/hbase\/Chemical\/bond.html\" 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 HNO3 molek\u00fcl\u00fcn\u00fcn \u00e7iziminde d\u0131\u015ftaki atomlar\u0131n hidrojen ve oksijen atomlar\u0131 oldu\u011funu g\u00f6rebilirsiniz.<\/p>\n<p> Bu hidrojen ve oksijen atomlar\u0131 s\u0131ras\u0131yla bir <a href=\"https:\/\/chemuza.org\/tr\/kimyanin-temel-tanimlari\/\">ikili<\/a> ve <a href=\"https:\/\/doi.org\/10.1016\/S0010-8545(02)00102-9\" target=\"_blank\" rel=\"noopener\">bir 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=\"268\" src=\"https:\/\/chemuza.org\/wp-content\/uploads\/2023\/08\/hno3-etape-3.webp\" alt=\"HNO3 ad\u0131m 3\" class=\"wp-image-720\" srcset=\"\" sizes=\"\"><\/figure>\n<p> Ek olarak, 1. ad\u0131mda HNO3 molek\u00fcl\u00fcnde bulunan toplam de\u011ferlik elektronu say\u0131s\u0131n\u0131 hesaplad\u0131k.<\/p>\n<p> HNO3 molek\u00fcl\u00fcn\u00fcn toplam <strong>24 de\u011ferlik elektronu<\/strong> vard\u0131r ve bu de\u011ferlik elektronlar\u0131n\u0131n t\u00fcm\u00fc yukar\u0131daki HNO3 diyagram\u0131nda kullan\u0131lm\u0131\u015ft\u0131r.<\/p>\n<p> Bu nedenle merkezi nitrojen atomunda tutulacak elektron \u00e7ifti kalmad\u0131.<\/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 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=\"245\" src=\"https:\/\/chemuza.org\/wp-content\/uploads\/2023\/08\/hno3-etape-4.webp\" alt=\"HNO3 ad\u0131m 4\" class=\"wp-image-721\" srcset=\"\" sizes=\"\"><\/figure>\n<p> \u015eimdi bu nitrojen atomunu stabil hale getirmek i\u00e7in, d\u0131\u015ftaki oksijen atomunun elektron \u00e7iftini, nitrojen atomunun 8 elektrona (yani bir oktete) sahip olaca\u011f\u0131 \u015fekilde kayd\u0131rman\u0131z gerekir. <\/p>\n<figure class=\"wp-block-image aligncenter size-full\"><img decoding=\"async\" loading=\"lazy\" width=\"600\" height=\"142\" src=\"https:\/\/chemuza.org\/wp-content\/uploads\/2023\/08\/hno3-etape-5.webp\" alt=\"HNO3 ad\u0131m 5\" class=\"wp-image-722\" 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=\"252\" src=\"https:\/\/chemuza.org\/wp-content\/uploads\/2023\/08\/hno3-etape-6.webp\" alt=\"HNO3 ad\u0131m 6\" class=\"wp-image-723\" srcset=\"\" sizes=\"\"><\/figure>\n<p> Yukar\u0131daki resimde nitrojen atomunun 8 elektrona sahip olmas\u0131 nedeniyle bir oktet olu\u015fturdu\u011funu g\u00f6rebilirsiniz.<\/p>\n<p> \u015eimdi yukar\u0131daki 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 HNO3&#8217;\u00fcn 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:\/\/en.wikipedia.org\/wiki\/Formal_charge\" target=\"_blank\" rel=\"noopener\">formal y\u00fck<\/a> kavram\u0131 kullan\u0131larak do\u011frulanabilir.<\/p>\n<p> K\u0131saca art\u0131k HNO3 molek\u00fcl\u00fcnde bulunan hidrojen (H), 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 HNO3 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=\"301\" src=\"https:\/\/chemuza.org\/wp-content\/uploads\/2023\/08\/hno3-etape-7.webp\" alt=\"HNO3 ad\u0131m 7\" class=\"wp-image-724\" srcset=\"\" sizes=\"\"><\/figure>\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<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>\u00c7ift ba\u011fl\u0131 oksijen (O) atomu 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<p> <strong>Soldaki tek ba\u011fl\u0131 oksijen (O) atomu 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<p> <strong>D\u00fcz, tek ba\u011fl\u0131 oksijen (O) atomu 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\"> 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<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>+1<\/strong><\/td>\n<\/tr>\n<tr>\n<td class=\"has-text-align-center\" data-align=\"center\"> O (\u00e7ift atlama)<\/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\"> O (tek ba\u011f, sol)<\/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\"> O (tek ba\u011f, sa\u011f)<\/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 resmi y\u00fck hesaplamalar\u0131ndan nitrojen (N) atomunun <strong>+1<\/strong> , tek ba\u011fl\u0131 oksijen atomunun (sa\u011f tarafta) <strong>-1<\/strong> y\u00fck\u00fcne sahip oldu\u011funu g\u00f6rebilirsiniz.<\/p>\n<p> O halde bu y\u00fckleri HNO3 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=\"158\" src=\"https:\/\/chemuza.org\/wp-content\/uploads\/2023\/08\/hno3-etape-8.webp\" alt=\"HNO3 ad\u0131m 8\" class=\"wp-image-725\" srcset=\"\" sizes=\"\"><\/figure>\n<p> Yukar\u0131daki \u00e7izimde <strong>+1<\/strong> ve <strong>-1<\/strong> y\u00fckleri iptal edilmi\u015ftir ve HNO3&#8217;\u00fcn yukar\u0131daki Lewis nokta yap\u0131s\u0131, kararl\u0131 Lewis yap\u0131s\u0131d\u0131r.<\/p>\n<p> HNO3&#8217;\u00fcn yukar\u0131daki Lewis nokta yap\u0131s\u0131nda, her bir ba\u011f elektronu \u00e7iftini (:) tek bir ba\u011f (|) olarak da temsil edebilirsiniz. Bunu yapmak HNO3&#8217;\u00fcn 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=\"270\" src=\"https:\/\/chemuza.org\/wp-content\/uploads\/2023\/08\/structure-de-Lewis-de-hno3.jpg\" alt=\"HNO3'\u00fcn Lewis yap\u0131s\u0131\" class=\"wp-image-726\" 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\/scn-yapisi-lewis\/\">SCN-Lewis yap\u0131s\u0131<\/a><\/td>\n<td> <a href=\"https:\/\/chemuza.org\/tr\/yapi-clf3-lewis\/\">Lewis yap\u0131s\u0131 ClF3<\/a><\/td>\n<\/tr>\n<tr>\n<td> <a href=\"https:\/\/chemuza.org\/tr\/cl2-lewis-yapisi\/\">Lewis yap\u0131s\u0131 Cl2<\/a><\/td>\n<td> <a href=\"https:\/\/chemuza.org\/tr\/lewis-hf-yapisi\/\">HF Lewis yap\u0131s\u0131<\/a><\/td>\n<\/tr>\n<tr>\n<td> <a href=\"https:\/\/chemuza.org\/tr\/lewis-scl2-yapisi\/\">Lewis yap\u0131s\u0131 SCl2<\/a><\/td>\n<td> <a href=\"https:\/\/chemuza.org\/tr\/lewis-pf5-yapisi\/\">Lewis yap\u0131s\u0131 PF5<\/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. HNO3&#8217;\u00fcn (nitrik asit) Lewis yap\u0131s\u0131nda, merkezde iki oksijen (O) atomu ve bir OH grubu ile \u00e7evrelenmi\u015f bir nitrojen (N) atomu bulunur. Azot (N) atomu ile oksijen (O) atomu aras\u0131nda 1 adet \u00e7ift ba\u011f bulunur, di\u011fer atomlar\u0131n geri kalan\u0131nda ise tek ba\u011f bulunur. HNO3&#8217;\u00fcn (nitrik asit) &#8230; <a title=\"6 ad\u0131mda hno3 lewis yap\u0131s\u0131 (resimlerle)\" class=\"read-more\" href=\"https:\/\/chemuza.org\/tr\/hno3-lewis-yapisi\/\" aria-label=\"More on 6 ad\u0131mda hno3 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 HNO3 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\/hno3-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 HNO3 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. HNO3&#8217;\u00fcn (nitrik asit) Lewis yap\u0131s\u0131nda, merkezde iki oksijen (O) atomu ve bir OH grubu ile \u00e7evrelenmi\u015f bir nitrojen (N) atomu bulunur. Azot (N) atomu ile oksijen (O) atomu aras\u0131nda 1 adet \u00e7ift ba\u011f bulunur, di\u011fer atomlar\u0131n geri kalan\u0131nda ise tek ba\u011f bulunur. HNO3&#8217;\u00fcn (nitrik asit) ... Devam\u0131n\u0131 oku\" \/>\n<meta property=\"og:url\" content=\"https:\/\/chemuza.org\/tr\/hno3-lewis-yapisi\/\" \/>\n<meta property=\"og:site_name\" content=\"Chemuza\" \/>\n<meta property=\"article:published_time\" content=\"2023-07-26T16:39:04+00:00\" \/>\n<meta property=\"og:image\" content=\"https:\/\/chemuza.org\/wp-content\/uploads\/2023\/08\/structure-hno3-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=\"6 dakika\" \/>\n<script type=\"application\/ld+json\" class=\"yoast-schema-graph\">{\"@context\":\"https:\/\/schema.org\",\"@graph\":[{\"@type\":\"Article\",\"@id\":\"https:\/\/chemuza.org\/tr\/hno3-lewis-yapisi\/#article\",\"isPartOf\":{\"@id\":\"https:\/\/chemuza.org\/tr\/hno3-lewis-yapisi\/\"},\"author\":{\"name\":\"Edit\u00f6r ekibi\",\"@id\":\"https:\/\/chemuza.org\/tr\/#\/schema\/person\/5d92b9b82b9b43578b70f5a7cd51fa3e\"},\"headline\":\"6 ad\u0131mda hno3 lewis yap\u0131s\u0131 (resimlerle)\",\"datePublished\":\"2023-07-26T16:39:04+00:00\",\"dateModified\":\"2023-07-26T16:39:04+00:00\",\"mainEntityOfPage\":{\"@id\":\"https:\/\/chemuza.org\/tr\/hno3-lewis-yapisi\/\"},\"wordCount\":1277,\"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\/hno3-lewis-yapisi\/#respond\"]}]},{\"@type\":\"WebPage\",\"@id\":\"https:\/\/chemuza.org\/tr\/hno3-lewis-yapisi\/\",\"url\":\"https:\/\/chemuza.org\/tr\/hno3-lewis-yapisi\/\",\"name\":\"6 Ad\u0131mda HNO3 Lewis Yap\u0131s\u0131 (Resimlerle) - Chemuza\",\"isPartOf\":{\"@id\":\"https:\/\/chemuza.org\/tr\/#website\"},\"datePublished\":\"2023-07-26T16:39:04+00:00\",\"dateModified\":\"2023-07-26T16:39:04+00:00\",\"breadcrumb\":{\"@id\":\"https:\/\/chemuza.org\/tr\/hno3-lewis-yapisi\/#breadcrumb\"},\"inLanguage\":\"tr\",\"potentialAction\":[{\"@type\":\"ReadAction\",\"target\":[\"https:\/\/chemuza.org\/tr\/hno3-lewis-yapisi\/\"]}]},{\"@type\":\"BreadcrumbList\",\"@id\":\"https:\/\/chemuza.org\/tr\/hno3-lewis-yapisi\/#breadcrumb\",\"itemListElement\":[{\"@type\":\"ListItem\",\"position\":1,\"name\":\"Ev\",\"item\":\"https:\/\/chemuza.org\/tr\/\"},{\"@type\":\"ListItem\",\"position\":2,\"name\":\"6 ad\u0131mda hno3 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":"6 Ad\u0131mda HNO3 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\/hno3-lewis-yapisi\/","og_locale":"tr_TR","og_type":"article","og_title":"6 Ad\u0131mda HNO3 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. HNO3&#8217;\u00fcn (nitrik asit) Lewis yap\u0131s\u0131nda, merkezde iki oksijen (O) atomu ve bir OH grubu ile \u00e7evrelenmi\u015f bir nitrojen (N) atomu bulunur. Azot (N) atomu ile oksijen (O) atomu aras\u0131nda 1 adet \u00e7ift ba\u011f bulunur, di\u011fer atomlar\u0131n geri kalan\u0131nda ise tek ba\u011f bulunur. HNO3&#8217;\u00fcn (nitrik asit) ... Devam\u0131n\u0131 oku","og_url":"https:\/\/chemuza.org\/tr\/hno3-lewis-yapisi\/","og_site_name":"Chemuza","article_published_time":"2023-07-26T16:39:04+00:00","og_image":[{"url":"https:\/\/chemuza.org\/wp-content\/uploads\/2023\/08\/structure-hno3-lewis.jpg"}],"author":"Edit\u00f6r ekibi","twitter_card":"summary_large_image","twitter_misc":{"Yazan:":"Edit\u00f6r ekibi","Tahmini okuma s\u00fcresi":"6 dakika"},"schema":{"@context":"https:\/\/schema.org","@graph":[{"@type":"Article","@id":"https:\/\/chemuza.org\/tr\/hno3-lewis-yapisi\/#article","isPartOf":{"@id":"https:\/\/chemuza.org\/tr\/hno3-lewis-yapisi\/"},"author":{"name":"Edit\u00f6r ekibi","@id":"https:\/\/chemuza.org\/tr\/#\/schema\/person\/5d92b9b82b9b43578b70f5a7cd51fa3e"},"headline":"6 ad\u0131mda hno3 lewis yap\u0131s\u0131 (resimlerle)","datePublished":"2023-07-26T16:39:04+00:00","dateModified":"2023-07-26T16:39:04+00:00","mainEntityOfPage":{"@id":"https:\/\/chemuza.org\/tr\/hno3-lewis-yapisi\/"},"wordCount":1277,"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\/hno3-lewis-yapisi\/#respond"]}]},{"@type":"WebPage","@id":"https:\/\/chemuza.org\/tr\/hno3-lewis-yapisi\/","url":"https:\/\/chemuza.org\/tr\/hno3-lewis-yapisi\/","name":"6 Ad\u0131mda HNO3 Lewis Yap\u0131s\u0131 (Resimlerle) - Chemuza","isPartOf":{"@id":"https:\/\/chemuza.org\/tr\/#website"},"datePublished":"2023-07-26T16:39:04+00:00","dateModified":"2023-07-26T16:39:04+00:00","breadcrumb":{"@id":"https:\/\/chemuza.org\/tr\/hno3-lewis-yapisi\/#breadcrumb"},"inLanguage":"tr","potentialAction":[{"@type":"ReadAction","target":["https:\/\/chemuza.org\/tr\/hno3-lewis-yapisi\/"]}]},{"@type":"BreadcrumbList","@id":"https:\/\/chemuza.org\/tr\/hno3-lewis-yapisi\/#breadcrumb","itemListElement":[{"@type":"ListItem","position":1,"name":"Ev","item":"https:\/\/chemuza.org\/tr\/"},{"@type":"ListItem","position":2,"name":"6 ad\u0131mda hno3 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\/46"}],"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=46"}],"version-history":[{"count":0,"href":"https:\/\/chemuza.org\/tr\/wp-json\/wp\/v2\/posts\/46\/revisions"}],"wp:attachment":[{"href":"https:\/\/chemuza.org\/tr\/wp-json\/wp\/v2\/media?parent=46"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/chemuza.org\/tr\/wp-json\/wp\/v2\/categories?post=46"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/chemuza.org\/tr\/wp-json\/wp\/v2\/tags?post=46"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}