{"id":264,"date":"2023-07-24T20:31:52","date_gmt":"2023-07-24T20:31:52","guid":{"rendered":"https:\/\/chemuza.org\/tr\/hbro-lewis-yapisi\/"},"modified":"2023-07-24T20:31:52","modified_gmt":"2023-07-24T20:31:52","slug":"hbro-lewis-yapisi","status":"publish","type":"post","link":"https:\/\/chemuza.org\/tr\/hbro-lewis-yapisi\/","title":{"rendered":"6 ad\u0131mda hbro lewis&#39;in 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-hbro-lewis.jpg\" alt=\"Lewis yap\u0131s\u0131 HBrO\" class=\"wp-image-3336\" 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\">HBrO (veya HOBr) Lewis yap\u0131s\u0131n\u0131n merkezinde bir hidrojen (H) atomu ve bir bromin (Br) atomu ile \u00e7evrelenmi\u015f bir oksijen (O) atomu vard\u0131r. Hidrojen (H) ve oksijen (O) atomlar\u0131 aras\u0131nda ve ayr\u0131ca oksijen (O) ve brom (Br) atomlar\u0131 aras\u0131nda tek bir ba\u011f vard\u0131r.<\/mark><\/em><\/strong><\/p>\n<p> HBrO Lewis yap\u0131s\u0131n\u0131n yukar\u0131daki g\u00f6r\u00fcnt\u00fcs\u00fcnden hi\u00e7bir \u015fey anlamad\u0131ysan\u0131z, benimle kal\u0131n ve <a href=\"http:\/\/www.chemspider.com\/Chemical-Structure.75379.html\" target=\"_blank\" rel=\"noopener\">HBrO<\/a> Lewis yap\u0131s\u0131n\u0131n \u00e7izimiyle ilgili ayr\u0131nt\u0131l\u0131 ad\u0131m ad\u0131m a\u00e7\u0131klamay\u0131 alacaks\u0131n\u0131z.<\/p>\n<p> O halde HBrO&#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>HBrO Lewis Yap\u0131s\u0131n\u0131 \u00c7izim Ad\u0131mlar\u0131<\/strong><\/h2>\n<h3 class=\"wp-block-heading\"> <strong>Ad\u0131m 1: HBrO molek\u00fcl\u00fcndeki toplam de\u011ferlik elektronu say\u0131s\u0131n\u0131 bulun<\/strong><\/h3>\n<p> Bir HBrO (veya HOBr) <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 hidrojen atomunda, oksijen atomunda ve ayr\u0131ca brom 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, oksijen ve bromun de\u011ferlik elektronlar\u0131n\u0131 nas\u0131l kolayca bulaca\u011f\u0131n\u0131z\u0131 anlataca\u011f\u0131m.<\/p>\n<p> <strong>HBrO 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:\/\/education.jlab.org\/itselemental\/ele001.html\" 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 Brom 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\/2-9.jpg\" alt=\"\" class=\"wp-image-653\" srcset=\"\" sizes=\"\"><\/figure>\n<p> Brom, periyodik tablonun 17. grubunda yer alan bir elementtir. <a href=\"https:\/\/pubchem.ncbi.nlm.nih.gov\/element\/35\" target=\"_blank\" rel=\"noopener\"><sup>[2]<\/sup><\/a> Bu nedenle bromda bulunan de\u011ferlik elektronlar\u0131 <strong>7&#8217;dir<\/strong> . <\/p>\n<figure class=\"wp-block-image aligncenter size-full\"><img decoding=\"async\" loading=\"lazy\" width=\"600\" height=\"266\" src=\"https:\/\/chemuza.org\/wp-content\/uploads\/2023\/08\/3-9.jpg\" alt=\"\" class=\"wp-image-654\" srcset=\"\" sizes=\"\"><\/figure>\n<p> Yukar\u0131daki resimde g\u00f6sterildi\u011fi gibi brom atomunda bulunan 7 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>HBrO molek\u00fcl\u00fcndeki toplam de\u011ferlik elektronlar\u0131<\/strong> = 1 hidrojen 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 bromin atomu taraf\u0131ndan ba\u011f\u0131\u015flanan de\u011ferlik elektronlar\u0131 = <strong>1 + 6 + 7 = 14<\/strong> .<\/p>\n<h3 class=\"wp-block-heading\"> <strong>Ad\u0131m 2: Tasla\u011f\u0131 haz\u0131rlay\u0131n<\/strong><\/h3>\n<p> HOBr (veya HBrO) molek\u00fcl\u00fcn\u00fcn bir tasla\u011f\u0131n\u0131 \u00e7izmek i\u00e7in kimyasal form\u00fcl\u00fcne bakman\u0131z yeterlidir. Merkezde bir oksijen (O) atomunun bulundu\u011funu ve her iki yan\u0131nda bir hidrojen (H) atomu ve bir bromin (Br) 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=\"56\" src=\"https:\/\/chemuza.org\/wp-content\/uploads\/2023\/08\/hbro-etape-1.webp\" alt=\"HBrO ad\u0131m 1\" class=\"wp-image-3337\" srcset=\"\" sizes=\"\"><\/figure>\n<p> <strong>(Not:<\/strong> Burada bromu de\u011fil oksijen atomunu merkezde tuttuk. Brom atomunu merkezde tutarsan\u0131z son Lewis yap\u0131s\u0131 kararl\u0131 olmayacakt\u0131r. Bu nedenle oksijen merkezde tutulur.)<\/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 HBrO molek\u00fcl\u00fcnde elektron \u00e7iftlerini hidrojen (H) ile oksijen (O) atomu aras\u0131na ve oksijen (O) ile bromin (Br) atomu aras\u0131na yerle\u015ftirmeniz gerekir. <\/p>\n<figure class=\"wp-block-image aligncenter size-full\"><img decoding=\"async\" loading=\"lazy\" width=\"600\" height=\"57\" src=\"https:\/\/chemuza.org\/wp-content\/uploads\/2023\/08\/hbro-etape-2.webp\" alt=\"HBrO ad\u0131m 2\" class=\"wp-image-3338\" srcset=\"\" sizes=\"\"><\/figure>\n<p> Bu, bu atomlar\u0131n bir HBrO molek\u00fcl\u00fcnde <a href=\"http:\/\/hyperphysics.phy-astr.gsu.edu\/hbase\/Chemical\/bond.html\" 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 d\u0131\u015f atomlar\u0131n kararl\u0131l\u0131\u011f\u0131n\u0131 kontrol etmeniz gerekir.<\/p>\n<p> Burada HBrO molek\u00fcl\u00fcn\u00fcn \u00e7iziminde d\u0131\u015ftaki atomlar\u0131n hidrojen atomu ve brom atomu oldu\u011funu g\u00f6rebilirsiniz.<\/p>\n<p> Bu hidrojen ve brom atomlar\u0131 s\u0131ras\u0131yla bir <a href=\"https:\/\/chemuza.org\/tr\/kimyanin-temel-tanimlari\/\">ikili<\/a> ve bir oktet olu\u015fturur ve bu nedenle stabildir. <\/p>\n<figure class=\"wp-block-image aligncenter size-full\"><img decoding=\"async\" loading=\"lazy\" width=\"600\" height=\"261\" src=\"https:\/\/chemuza.org\/wp-content\/uploads\/2023\/08\/hbro-etape-3.webp\" alt=\"HBrO ad\u0131m 3\" class=\"wp-image-3339\" srcset=\"\" sizes=\"\"><\/figure>\n<p> Ek olarak, 1. ad\u0131mda HBrO molek\u00fcl\u00fcnde bulunan toplam de\u011ferlik elektron say\u0131s\u0131n\u0131 hesaplad\u0131k.<\/p>\n<p> HBrO molek\u00fcl\u00fcnde toplam <strong>14 de\u011ferlik elektronu<\/strong> vard\u0131r ve yukar\u0131daki diyagramda bunlardan sadece <strong>10 de\u011ferlik elektronu<\/strong> kullan\u0131lm\u0131\u015ft\u0131r.<\/p>\n<p> Yani kalan elektron say\u0131s\u0131 = <strong>14 \u2013 10 = 4<\/strong> .<\/p>\n<p> Bu <strong>4<\/strong> elektronu yukar\u0131daki HBrO molek\u00fcl\u00fcn\u00fcn diyagram\u0131nda merkezi oksijen atomuna yerle\u015ftirmeniz gerekir. <\/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\/hbro-etape-4.webp\" alt=\"HBrO ad\u0131m 4\" class=\"wp-image-3340\" 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<\/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> Merkezi oksijen (O) atomunun stabilitesini kontrol etmek i\u00e7in oktet olu\u015fturup olu\u015fturmad\u0131\u011f\u0131n\u0131 kontrol etmemiz gerekir. <\/p>\n<figure class=\"wp-block-image aligncenter size-full\"><img decoding=\"async\" loading=\"lazy\" width=\"600\" height=\"263\" src=\"https:\/\/chemuza.org\/wp-content\/uploads\/2023\/08\/hbro-etape-5.webp\" alt=\"HBrO ad\u0131m 5\" class=\"wp-image-3341\" srcset=\"\" sizes=\"\"><\/figure>\n<p> Yukar\u0131daki resimde oksijen atomunun bir oktet olu\u015fturdu\u011funu g\u00f6rebilirsiniz. Bu, 8 elektrona sahip oldu\u011fu anlam\u0131na gelir.<\/p>\n<p> Ve b\u00f6ylece merkezi oksijen atomu stabildir.<\/p>\n<p> \u015eimdi HBrO&#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 HOBr&#8217;nin Lewis yap\u0131s\u0131n\u0131n kararl\u0131l\u0131\u011f\u0131n\u0131 kontrol etmeniz gereken son ad\u0131ma geldiniz.<\/p>\n<p> Lewis yap\u0131s\u0131n\u0131n kararl\u0131l\u0131\u011f\u0131 <a href=\"https:\/\/en.wikipedia.org\/wiki\/Formal_charge\" target=\"_blank\" rel=\"noopener\">formal y\u00fck<\/a> kavram\u0131 kullan\u0131larak do\u011frulanabilir.<\/p>\n<p> K\u0131sacas\u0131, \u015fimdi HBrO molek\u00fcl\u00fcnde bulunan hidrojen atomunun (H), oksijen atomunun (O) ve ayr\u0131ca bromin atomunun (Br) resmi 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 HBrO molek\u00fcl\u00fcn\u00fcn her 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=\"335\" src=\"https:\/\/chemuza.org\/wp-content\/uploads\/2023\/08\/hbro-etape-6.webp\" alt=\"HBrO ad\u0131m 6\" class=\"wp-image-3342\" 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>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<p> <strong>Brom atomu (Br) i\u00e7in:<\/strong><br \/> De\u011ferlik elektronu = 7 (\u00e7\u00fcnk\u00fc brom 17. gruptad\u0131r)<br \/> 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\"> 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<tr>\n<td class=\"has-text-align-center\" data-align=\"center\"> karde\u015fim<\/td>\n<td> =<\/td>\n<td class=\"has-text-align-center\" data-align=\"center\"> 7<\/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>0<\/strong><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/figure>\n<p> Yukar\u0131daki formal y\u00fck hesaplamalar\u0131ndan hidrojen atomunun (H), oksijen atomunun (O) ve bromin atomunun (Br) <strong>\u201cs\u0131f\u0131r\u201d<\/strong> formal <strong>y\u00fcke<\/strong> sahip oldu\u011funu g\u00f6rebilirsiniz.<\/p>\n<p> Bu, HBrO&#8217;nun yukar\u0131daki Lewis yap\u0131s\u0131n\u0131n stabil oldu\u011funu ve HBrO&#8217;nun yukar\u0131daki yap\u0131s\u0131nda ba\u015fka bir de\u011fi\u015fiklik olmad\u0131\u011f\u0131n\u0131 g\u00f6sterir.<\/p>\n<p> HBrO&#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 HBrO&#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=\"210\" src=\"https:\/\/chemuza.org\/wp-content\/uploads\/2023\/08\/Structure-Lewis-de-Hbro.jpg\" alt=\"HBrO'nun Lewis yap\u0131s\u0131\" class=\"wp-image-3343\" 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-io2nin-yapisi\/\">IO2-Lewis yap\u0131s\u0131<\/a><\/td>\n<td> <a href=\"https:\/\/chemuza.org\/tr\/ci4-lewis-yapisi\/\">Lewis yap\u0131s\u0131 CI4<\/a><\/td>\n<\/tr>\n<tr>\n<td> <a href=\"https:\/\/chemuza.org\/tr\/bi3-lewis-yapisi\/\">BI3 Lewis yap\u0131s\u0131<\/a><\/td>\n<td> <a href=\"https:\/\/chemuza.org\/tr\/ch3i-lewis-yapisi\/\">Lewis yap\u0131s\u0131 CH3I<\/a><\/td>\n<\/tr>\n<tr>\n<td> <a href=\"https:\/\/chemuza.org\/tr\/kardes-lewis-yapisi\/\">BrO-Lewis yap\u0131s\u0131<\/a><\/td>\n<td> <a href=\"https:\/\/chemuza.org\/tr\/lewis-seof2nin-yapisi\/\">Lewis yap\u0131s\u0131 SeOF2<\/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. HBrO (veya HOBr) Lewis yap\u0131s\u0131n\u0131n merkezinde bir hidrojen (H) atomu ve bir bromin (Br) atomu ile \u00e7evrelenmi\u015f bir oksijen (O) atomu vard\u0131r. Hidrojen (H) ve oksijen (O) atomlar\u0131 aras\u0131nda ve ayr\u0131ca oksijen (O) ve brom (Br) atomlar\u0131 aras\u0131nda tek bir ba\u011f vard\u0131r. HBrO Lewis yap\u0131s\u0131n\u0131n &#8230; <a title=\"6 ad\u0131mda hbro lewis&#039;in yap\u0131s\u0131 (resimlerle)\" class=\"read-more\" href=\"https:\/\/chemuza.org\/tr\/hbro-lewis-yapisi\/\" aria-label=\"More on 6 ad\u0131mda hbro lewis&#039;in 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 HBrO Lewis&#039;in 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\/hbro-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 HBrO Lewis&#039;in 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. HBrO (veya HOBr) Lewis yap\u0131s\u0131n\u0131n merkezinde bir hidrojen (H) atomu ve bir bromin (Br) atomu ile \u00e7evrelenmi\u015f bir oksijen (O) atomu vard\u0131r. Hidrojen (H) ve oksijen (O) atomlar\u0131 aras\u0131nda ve ayr\u0131ca oksijen (O) ve brom (Br) atomlar\u0131 aras\u0131nda tek bir ba\u011f vard\u0131r. HBrO Lewis yap\u0131s\u0131n\u0131n ... 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Yukar\u0131daki g\u00f6rseli k\u0131saca a\u00e7\u0131klayay\u0131m. HBrO (veya HOBr) Lewis yap\u0131s\u0131n\u0131n merkezinde bir hidrojen (H) atomu ve bir bromin (Br) atomu ile \u00e7evrelenmi\u015f bir oksijen (O) atomu vard\u0131r. Hidrojen (H) ve oksijen (O) atomlar\u0131 aras\u0131nda ve ayr\u0131ca oksijen (O) ve brom (Br) atomlar\u0131 aras\u0131nda tek bir ba\u011f vard\u0131r. HBrO Lewis yap\u0131s\u0131n\u0131n ... 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