{"id":131,"date":"2023-07-26T00:00:49","date_gmt":"2023-07-26T00:00:49","guid":{"rendered":"https:\/\/chemuza.org\/tr\/h2co3-lewis-yapisi\/"},"modified":"2023-07-26T00:00:49","modified_gmt":"2023-07-26T00:00:49","slug":"h2co3-lewis-yapisi","status":"publish","type":"post","link":"https:\/\/chemuza.org\/tr\/h2co3-lewis-yapisi\/","title":{"rendered":"6 ad\u0131mda h2co3 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-h2co3-lewis.jpg\" alt=\"Lewis yap\u0131s\u0131 H2CO3\" class=\"wp-image-1807\" 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\">H2CO3 Lewis yap\u0131s\u0131n\u0131n merkezinde bir oksijen (O) atomu ve iki OH grubu ile \u00e7evrelenmi\u015f bir karbon (C) atomu bulunur. Karbon (C) 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. \u00dc\u00e7 oksijen (O) atomunda 2 yaln\u0131z \u00e7ift vard\u0131r.<\/mark><\/em><\/strong><\/p>\n<p> H2CO3&#8217;\u00fcn 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\/767\" target=\"_blank\" rel=\"noopener\">H2CO3&#8217;\u00fcn<\/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> O halde H2CO3 molek\u00fcl\u00fcn\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>H2CO3 Lewis Yap\u0131s\u0131n\u0131 \u00c7izim Ad\u0131mlar\u0131<\/strong><\/h2>\n<h3 class=\"wp-block-heading\"> <strong>Ad\u0131m 1: H2CO3 molek\u00fcl\u00fcndeki toplam de\u011ferlik elektronu say\u0131s\u0131n\u0131 bulun<\/strong><\/h3>\n<p> Bir H2CO3 <a href=\"https:\/\/chemuza.org\/tr\/kimyanin-temel-tanimlari\/\">molek\u00fcl\u00fcndeki<\/a> toplam <a href=\"https:\/\/energyeducation.ca\/encyclopedia\/Valence_and_core_electrons\" target=\"_blank\" rel=\"noopener\">de\u011ferlik elektronu<\/a> say\u0131s\u0131n\u0131 bulmak i\u00e7in \u00f6ncelikle <a href=\"https:\/\/chemuza.org\/tr\/kimyanin-temel-tanimlari\/\">hidrojen atomunda<\/a> , karbon 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 periyodik tabloyu kullanarak hidrojen, karbon ve oksijenin de\u011ferlik elektronlar\u0131n\u0131 nas\u0131l kolayca bulaca\u011f\u0131n\u0131z\u0131 anlataca\u011f\u0131m.<\/p>\n<p> <strong>H2CO3 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> Hidrojen periyodik tablonun 1. grup 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 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>[2]<\/sup><\/a> Bu nedenle karbonda bulunan de\u011ferlik elektronlar\u0131 <strong>4&#8217;t\u00fcr<\/strong> . <\/p>\n<figure class=\"wp-block-image aligncenter size-full\"><img decoding=\"async\" loading=\"lazy\" width=\"600\" height=\"230\" src=\"https:\/\/chemuza.org\/wp-content\/uploads\/2023\/08\/2.jpg\" alt=\"\" class=\"wp-image-26\" srcset=\"\" sizes=\"\"><\/figure>\n<p> Yukar\u0131daki resimde g\u00f6sterildi\u011fi gibi karbon atomunda bulunan 4 de\u011ferlik elektronunu g\u00f6rebilirsiniz.<\/p>\n<p> <strong>\u2192 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:\/\/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>H2CO3 molek\u00fcl\u00fcndeki toplam de\u011ferlik elektronlar\u0131<\/strong> = 2 hidrojen atomu taraf\u0131ndan ba\u011f\u0131\u015flanan de\u011ferlik elektronlar\u0131 + 1 karbon 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(2) + 4 + 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 hidrojen varsa, daima hidrojeni d\u0131\u015far\u0131ya koyun.)<\/p>\n<p> \u015eimdi burada verilen molek\u00fcl H2CO3&#8217;t\u00fcr ve hidrojen atomu (H), karbon atomu (C) 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 karbon atomunun (C) ve oksijen atomunun (O) elektronegatiflik de\u011ferlerini g\u00f6rebilirsiniz.<\/p>\n<p> Karbon (C) ve oksijenin (O) elektronegatiflik de\u011ferlerini kar\u015f\u0131la\u015ft\u0131r\u0131rsak, <a href=\"https:\/\/chemuza.org\/tr\/periyodik-tablonun-elektronegatifligi\/\">karbon atomu daha az elektronegatiftir<\/a> .<\/p>\n<p> Burada karbon (C) atomu merkez atom, 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=\"83\" src=\"https:\/\/chemuza.org\/wp-content\/uploads\/2023\/08\/h2co3-etape-1.webp\" alt=\"H2CO3 ad\u0131m 1\" class=\"wp-image-1808\" 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 H2CO3 molek\u00fcl\u00fcnde <a href=\"https:\/\/doi.org\/10.1038\/s41570-018-0052-4\" target=\"_blank\" rel=\"noopener\">elektron \u00e7iftlerini<\/a> oksijen (O) ve hidrojen (H) atomlar\u0131 aras\u0131na ve oksijen (O) ve karbon (C) 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=\"80\" src=\"https:\/\/chemuza.org\/wp-content\/uploads\/2023\/08\/h2co3-etape-2.webp\" alt=\"H2CO3 ad\u0131m 2\" class=\"wp-image-1809\" srcset=\"\" sizes=\"\"><\/figure>\n<p> Bu, bu atomlar\u0131n bir H2CO3 molek\u00fcl\u00fcnde <a href=\"https:\/\/www.lamar.edu\/arts-sciences\/_files\/documents\/chemistry-biochemistry\/dorris\/chapter8.pdf\" 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<\/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 H2CO3 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:\/\/en.wikipedia.org\/wiki\/Octet_rule\" 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=\"231\" src=\"https:\/\/chemuza.org\/wp-content\/uploads\/2023\/08\/h2co3-etape-3.webp\" alt=\"H2CO3 ad\u0131m 3\" class=\"wp-image-1810\" srcset=\"\" sizes=\"\"><\/figure>\n<p> Ek olarak, 1. ad\u0131mda H2CO3 molek\u00fcl\u00fcnde bulunan toplam de\u011ferlik elektronu say\u0131s\u0131n\u0131 hesaplad\u0131k.<\/p>\n<p> H2CO3 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 diyagramda kullan\u0131lm\u0131\u015ft\u0131r.<\/p>\n<p> Bu nedenle merkezi karbon atomunda tutulacak daha fazla elektron \u00e7ifti yoktur.<\/p>\n<p> \u015eimdi bir sonraki ad\u0131ma ge\u00e7elim.<\/p>\n<h3 class=\"wp-block-heading\"> <strong>Ad\u0131m 5: Merkezi atomdaki sekizliyi kontrol edin. Okteti yoksa yaln\u0131z \u00e7ifti \u00e7ift ba\u011f veya \u00fc\u00e7l\u00fc ba\u011f olu\u015fturacak \u015fekilde hareket ettirin.<\/strong><\/h3>\n<p> Bu ad\u0131mda merkezi karbon atomunun (C) kararl\u0131 olup olmad\u0131\u011f\u0131n\u0131 kontrol etmeniz gerekir.<\/p>\n<p> Merkezi karbon (C) atomunun stabilitesini kontrol etmek i\u00e7in oktet olu\u015fturup olu\u015fturmad\u0131\u011f\u0131n\u0131 kontrol etmemiz gerekir.<\/p>\n<p> Maalesef karbon atomu burada bir oktet olu\u015fturmuyor. Karbonun yaln\u0131zca 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=\"221\" src=\"https:\/\/chemuza.org\/wp-content\/uploads\/2023\/08\/h2co3-etape-4.webp\" alt=\"H2CO3 ad\u0131m 4\" class=\"wp-image-1811\" srcset=\"\" sizes=\"\"><\/figure>\n<p> \u015eimdi, bu karbon atomunu kararl\u0131 hale getirmek i\u00e7in, d\u0131\u015ftaki oksijen atomunun elektron \u00e7iftini, karbon 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=\"119\" src=\"https:\/\/chemuza.org\/wp-content\/uploads\/2023\/08\/h2co3-etape-5.webp\" alt=\"H2CO3 ad\u0131m 5\" class=\"wp-image-1812\" srcset=\"\" sizes=\"\"><\/figure>\n<p> Bu elektron \u00e7iftini hareket ettirdikten sonra merkezi karbon atomu 2 elektron daha alacak ve b\u00f6ylece toplam elektron say\u0131s\u0131 8 olacakt\u0131r. <\/p>\n<figure class=\"wp-block-image aligncenter size-full\"><img decoding=\"async\" loading=\"lazy\" width=\"600\" height=\"240\" src=\"https:\/\/chemuza.org\/wp-content\/uploads\/2023\/08\/h2co3-etape-6.webp\" alt=\"H2CO3 ad\u0131m 6\" class=\"wp-image-1813\" srcset=\"\" sizes=\"\"><\/figure>\n<p> Yukar\u0131daki resimde karbon 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 H2CO3&#8217;\u00fcn Lewis yap\u0131s\u0131n\u0131n stabilitesini kontrol etmeniz gereken son ad\u0131ma geldiniz.<\/p>\n<p> Lewis yap\u0131s\u0131n\u0131n kararl\u0131l\u0131\u011f\u0131 <a href=\"https:\/\/www.chem.ucalgary.ca\/courses\/353\/Carey5th\/Ch01\/ch1-3-2.html\" target=\"_blank\" rel=\"noopener\">formal y\u00fck<\/a> kavram\u0131 kullan\u0131larak do\u011frulanabilir.<\/p>\n<p> K\u0131saca art\u0131k H2CO3 molek\u00fcl\u00fcnde bulunan hidrojen (H), karbon (C) 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 H2CO3 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=\"288\" src=\"https:\/\/chemuza.org\/wp-content\/uploads\/2023\/08\/h2co3-etape-7.webp\" alt=\"H2CO3 ad\u0131m 7\" class=\"wp-image-1815\" 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>Karbon atomu (C) i\u00e7in:<\/strong><strong><br \/><\/strong> De\u011ferlik elektronlar\u0131 = 4 (\u00e7\u00fcnk\u00fc karbon 14. gruptad\u0131r)<strong><br \/><\/strong> Ba\u011f elektronlar\u0131 = 8<br \/> Ba\u011flanmayan elektronlar = 0<\/p>\n<p> <strong>\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>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<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\"> VS<\/td>\n<td> =<\/td>\n<td class=\"has-text-align-center\" data-align=\"center\"> 4<\/td>\n<td> \u2013<\/td>\n<td class=\"has-text-align-center\" data-align=\"center\"> 8\/2<\/td>\n<td> \u2013<\/td>\n<td class=\"has-text-align-center\" data-align=\"center\"> 0<\/td>\n<td> =<\/td>\n<td class=\"has-text-align-center\" data-align=\"center\"> <strong>0<\/strong><\/td>\n<\/tr>\n<tr>\n<td class=\"has-text-align-center\" data-align=\"center\"> 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)<\/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 hidrojen (H), karbon (C) 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, H2CO3&#8217;\u00fcn yukar\u0131daki Lewis yap\u0131s\u0131n\u0131n stabil oldu\u011funu ve H2CO3&#8217;\u00fcn yukar\u0131daki yap\u0131s\u0131nda ba\u015fka bir de\u011fi\u015fiklik olmad\u0131\u011f\u0131n\u0131 g\u00f6sterir.<\/p>\n<p> H2CO3&#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 size H2CO3&#8217;\u00fcn a\u015fa\u011f\u0131daki Lewis yap\u0131s\u0131n\u0131 verecektir. <\/p>\n<figure class=\"wp-block-image aligncenter size-full\"><img decoding=\"async\" loading=\"lazy\" width=\"600\" height=\"241\" src=\"https:\/\/chemuza.org\/wp-content\/uploads\/2023\/08\/structure-de-Lewis-de-h2co3.jpg\" alt=\"H2CO3'\u00fcn Lewis yap\u0131s\u0131\" class=\"wp-image-1814\" 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\/hocl-lewis-yapisi\/\">Lewis yap\u0131s\u0131 HOCl<\/a><\/td>\n<td> <a href=\"https:\/\/chemuza.org\/tr\/c6h6-benzen-lewis-yapisi\/\">Lewis yap\u0131s\u0131 C6H6 (Benzen)<\/a><\/td>\n<\/tr>\n<tr>\n<td> <a href=\"https:\/\/chemuza.org\/tr\/lewis-nbr3-yapisi\/\">Lewis yap\u0131s\u0131 NBr3<\/a><\/td>\n<td> <a href=\"https:\/\/chemuza.org\/tr\/lewis-sef4-yapisi\/\">Lewis yap\u0131s\u0131 SeF4<\/a><\/td>\n<\/tr>\n<tr>\n<td> <a href=\"https:\/\/chemuza.org\/tr\/lewis-h3po4un-yapisi\/\">Lewis yap\u0131s\u0131 H3PO4<\/a><\/td>\n<td> <a href=\"https:\/\/chemuza.org\/tr\/h2se-lewis-yapisi\/\">H2Se Lewis yap\u0131s\u0131<\/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. H2CO3 Lewis yap\u0131s\u0131n\u0131n merkezinde bir oksijen (O) atomu ve iki OH grubu ile \u00e7evrelenmi\u015f bir karbon (C) atomu bulunur. Karbon (C) 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. \u00dc\u00e7 oksijen (O) atomunda 2 &#8230; <a title=\"6 ad\u0131mda h2co3 lewis yap\u0131s\u0131 (resimlerle)\" class=\"read-more\" href=\"https:\/\/chemuza.org\/tr\/h2co3-lewis-yapisi\/\" aria-label=\"More on 6 ad\u0131mda h2co3 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 H2CO3 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\/h2co3-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 H2CO3 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. H2CO3 Lewis yap\u0131s\u0131n\u0131n merkezinde bir oksijen (O) atomu ve iki OH grubu ile \u00e7evrelenmi\u015f bir karbon (C) atomu bulunur. Karbon (C) 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. \u00dc\u00e7 oksijen (O) atomunda 2 ... 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Yukar\u0131daki g\u00f6rseli k\u0131saca a\u00e7\u0131klayay\u0131m. H2CO3 Lewis yap\u0131s\u0131n\u0131n merkezinde bir oksijen (O) atomu ve iki OH grubu ile \u00e7evrelenmi\u015f bir karbon (C) atomu bulunur. Karbon (C) 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. \u00dc\u00e7 oksijen (O) atomunda 2 ... 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