{"id":294,"date":"2023-07-24T13:31:30","date_gmt":"2023-07-24T13:31:30","guid":{"rendered":"https:\/\/chemuza.org\/tr\/hbro3-lewis-yapisi\/"},"modified":"2023-07-24T13:31:30","modified_gmt":"2023-07-24T13:31:30","slug":"hbro3-lewis-yapisi","status":"publish","type":"post","link":"https:\/\/chemuza.org\/tr\/hbro3-lewis-yapisi\/","title":{"rendered":"6 ad\u0131mda hbro3 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-hbro3-lewis.jpg\" alt=\"Lewis yap\u0131s\u0131 HBrO3\" class=\"wp-image-3681\" 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\">HBrO3 Lewis yap\u0131s\u0131n\u0131n merkezinde iki oksijen (O) atomu ve bir OH grubu ile \u00e7evrelenmi\u015f bir brom (Br) atomu bulunur. Brom atomu (Br) ile oksijen atomu (O) aras\u0131nda 2 adet \u00e7ift ba\u011f vard\u0131r ve di\u011fer atomlar\u0131n geri kalan\u0131nda tek ba\u011f bulunur.<\/mark><\/em><\/strong><\/p>\n<p> HBrO3&#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=\"http:\/\/www.chemspider.com\/Chemical-Structure.22853.html\" target=\"_blank\" rel=\"noopener\">HBrO3&#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 HBrO3&#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>HBrO3 Lewis yap\u0131s\u0131n\u0131 \u00e7izme ad\u0131mlar\u0131<\/strong><\/h2>\n<h3 class=\"wp-block-heading\"> <strong>Ad\u0131m 1: HBrO3 molek\u00fcl\u00fcndeki toplam de\u011ferlik elektronu say\u0131s\u0131n\u0131 bulun<\/strong><\/h3>\n<p> Bir HBrO3 <a href=\"https:\/\/chemuza.org\/tr\/kimyanin-temel-tanimlari\/\">molek\u00fcl\u00fcndeki<\/a> toplam <a href=\"https:\/\/flexbooks.ck12.org\/cbook\/ck-12-middle-school-physical-science-flexbook-2.0\/section\/4.6\/primary\/lesson\/valence-electrons-ms-ps\/\" target=\"_blank\" rel=\"noopener\">de\u011ferlik elektron<\/a> say\u0131s\u0131n\u0131 bulmak i\u00e7in \u00f6ncelikle hidrojen atomunda, bromin 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 hidrojen, brom ve oksijenin de\u011ferlik elektronlar\u0131n\u0131 nas\u0131l kolayca bulaca\u011f\u0131n\u0131z\u0131 anlataca\u011f\u0131m.<\/p>\n<p> <strong>HBrO3 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 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:\/\/periodic.lanl.gov\/35.shtml\" 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:\/\/education.jlab.org\/itselemental\/ele008.html\" 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>HBrO3 molek\u00fcl\u00fcndeki toplam de\u011ferlik elektronlar\u0131<\/strong> = 1 hidrojen atomu taraf\u0131ndan ba\u011f\u0131\u015flanan de\u011ferlik elektronlar\u0131 + 1 brom 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 + 7 + 6(3) = 26<\/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 HBrO3&#8217;t\u00fcr ve hidrojen atomu (H), brom atomu (Br) ve oksijen atomlar\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 brom atomunun (Br) ve oksijen atomunun (O) elektronegatiflik de\u011ferlerini g\u00f6rebilirsiniz.<\/p>\n<p> Bromun (Br) ve oksijenin (O) elektronegatiflik de\u011ferlerini kar\u015f\u0131la\u015ft\u0131r\u0131rsak, <a href=\"https:\/\/chemuza.org\/tr\/periyodik-tablonun-elektronegatifligi\/\">bromin atomu daha az elektronegatiftir<\/a> .<\/p>\n<p> Burada brom (Br) 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=\"104\" src=\"https:\/\/chemuza.org\/wp-content\/uploads\/2023\/08\/hbro3-etape-1.webp\" alt=\"HBrO3 ad\u0131m 1\" class=\"wp-image-3682\" 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 HBrO3 molek\u00fcl\u00fcnde <a href=\"https:\/\/chemuza.org\/tr\/kimyanin-temel-tanimlari\/\">elektron \u00e7iftlerini<\/a> oksijen (O) ve hidrojen (H) atomlar\u0131 aras\u0131na ve oksijen (O) ve brom (Br) 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\/hbro3-etape-2.webp\" alt=\"HBrO3 ad\u0131m 2\" class=\"wp-image-3683\" srcset=\"\" sizes=\"\"><\/figure>\n<p> Bu, bu atomlar\u0131n bir HBrO3 molek\u00fcl\u00fcnde <a href=\"https:\/\/whs.rocklinusd.org\/documents\/Science\/2011_Reading_the_Different_Types_of_Bonds.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 HBrO3 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:\/\/openpress.usask.ca\/intro-organic-chemistry\/chapter\/1-2\/\" 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=\"266\" src=\"https:\/\/chemuza.org\/wp-content\/uploads\/2023\/08\/hbro3-etape-3.webp\" alt=\"HBrO3 ad\u0131m 3\" class=\"wp-image-3684\" srcset=\"\" sizes=\"\"><\/figure>\n<p> Ek olarak 1. ad\u0131mda HBrO3 molek\u00fcl\u00fcnde bulunan toplam de\u011ferlik elektronu say\u0131s\u0131n\u0131 hesaplad\u0131k.<\/p>\n<p> HBrO3 molek\u00fcl\u00fcnde toplam <strong>26 de\u011ferlik elektronu<\/strong> vard\u0131r ve yukar\u0131daki diyagramda bunlardan sadece <strong>24 de\u011ferlik elektronu<\/strong> kullan\u0131lm\u0131\u015ft\u0131r.<\/p>\n<p> Yani kalan elektron say\u0131s\u0131 = <strong>26 \u2013 24 = 2<\/strong> .<\/p>\n<p> Bu <strong>2<\/strong> elektronu yukar\u0131daki HBrO3 molek\u00fcl\u00fcn\u00fcn diyagram\u0131ndaki brom atomlar\u0131n\u0131n \u00fczerine yerle\u015ftirmeniz gerekiyor. <\/p>\n<figure class=\"wp-block-image aligncenter size-full\"><img decoding=\"async\" loading=\"lazy\" width=\"600\" height=\"234\" src=\"https:\/\/chemuza.org\/wp-content\/uploads\/2023\/08\/hbro3-etape-4.webp\" alt=\"HBrO3 ad\u0131m 4\" class=\"wp-image-3685\" 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 brom (Br) atomunun kararl\u0131 olup olmad\u0131\u011f\u0131n\u0131 kontrol etmeniz gerekir.<\/p>\n<p> Merkezi brom (Br) 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=\"297\" src=\"https:\/\/chemuza.org\/wp-content\/uploads\/2023\/08\/hbro3-etape-5.webp\" alt=\"HBrO3 ad\u0131m 5\" class=\"wp-image-3686\" srcset=\"\" sizes=\"\"><\/figure>\n<p> Yukar\u0131daki resimde brom atomunun bir oktet olu\u015fturdu\u011funu g\u00f6rebilirsiniz. Bu, 8 elektrona sahip oldu\u011fu anlam\u0131na gelir.<\/p>\n<p> Ve b\u00f6ylece merkezi brom atomu kararl\u0131d\u0131r.<\/p>\n<p> \u015eimdi HBrO3&#8217;\u00fcn Lewis yap\u0131s\u0131n\u0131n stabil 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 HBrO3&#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=\"http:\/\/home.miracosta.edu\/dlr\/info\/formal_charge.htm\" target=\"_blank\" rel=\"noopener\">formal y\u00fck<\/a> kavram\u0131 kullan\u0131larak do\u011frulanabilir.<\/p>\n<p> K\u0131sacas\u0131 art\u0131k HBrO3 molek\u00fcl\u00fcnde bulunan hidrojen (H), brom (Br) 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 HBrO3 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=\"286\" src=\"https:\/\/chemuza.org\/wp-content\/uploads\/2023\/08\/hbro3-etape-6.webp\" alt=\"HBrO3 ad\u0131m 6\" class=\"wp-image-3687\" 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>Brom atomu (Br) i\u00e7in:<\/strong><strong><br \/><\/strong> De\u011ferlik elektronlar\u0131 = 7 (\u00e7\u00fcnk\u00fc brom 17. gruptad\u0131r)<strong><br \/><\/strong> Ba\u011f elektronlar\u0131 = 6<br \/> Ba\u011flanmayan elektronlar = 2<\/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<p> <strong>Oksijen (O) atomu i\u00e7in (OH grubundan):<\/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\"> 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\"> 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>+2<\/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<tr>\n<td class=\"has-text-align-center\" data-align=\"center\"> O (OH grubundan)<\/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 bromin (Br) atomunun <strong>+2<\/strong> , iki oksijen (O) atomunun ise <strong>-1<\/strong> y\u00fck\u00fcne sahip oldu\u011funu g\u00f6rebilirsiniz.<\/p>\n<p> Bu nedenle yukar\u0131da elde edilen HBrO3&#8217;\u00fcn Lewis yap\u0131s\u0131 stabil de\u011fildir.<\/p>\n<p> Bu nedenle elektron \u00e7iftlerinin brom atomuna do\u011fru hareket ettirilmesiyle bu y\u00fcklerin en aza indirilmesi gerekir. <\/p>\n<figure class=\"wp-block-image aligncenter size-full\"><img decoding=\"async\" loading=\"lazy\" width=\"600\" height=\"152\" src=\"https:\/\/chemuza.org\/wp-content\/uploads\/2023\/08\/hbro3-etape-7.webp\" alt=\"HBrO3 ad\u0131m 7\" class=\"wp-image-3688\" srcset=\"\" sizes=\"\"><\/figure>\n<p> Elektron \u00e7iftlerinin oksijen atomlar\u0131ndan brom atomuna ta\u015f\u0131nmas\u0131ndan sonra HBrO3&#8217;\u00fcn Lewis yap\u0131s\u0131 daha kararl\u0131 hale gelir. <\/p>\n<figure class=\"wp-block-image aligncenter size-full\"><img decoding=\"async\" loading=\"lazy\" width=\"600\" height=\"140\" src=\"https:\/\/chemuza.org\/wp-content\/uploads\/2023\/08\/hbro3-etape-8.webp\" alt=\"HBrO3 ad\u0131m 8\" class=\"wp-image-3689\" srcset=\"\" sizes=\"\"><\/figure>\n<p> HBrO3&#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 HBrO3&#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=\"279\" src=\"https:\/\/chemuza.org\/wp-content\/uploads\/2023\/08\/structure-de-Lewis-de-hbro3.jpg\" alt=\"HBrO3'\u00fcn Lewis yap\u0131s\u0131\" class=\"wp-image-3690\" 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\/hbro4-lewis-yapisi\/\">Lewis yap\u0131s\u0131 HBrO4<\/a><\/td>\n<td> <a href=\"https:\/\/chemuza.org\/tr\/po2-lewis-yapisi\/\">PO2-Lewis yap\u0131s\u0131<\/a><\/td>\n<\/tr>\n<tr>\n<td> <a href=\"https:\/\/chemuza.org\/tr\/tef5-lewis-yapisi\/\">TeF5-Lewis yap\u0131s\u0131<\/a><\/td>\n<td> <a href=\"https:\/\/chemuza.org\/tr\/lewis-secl6nin-yapisi\/\">Lewis Yap\u0131s\u0131 SeCl6<\/a><\/td>\n<\/tr>\n<tr>\n<td> <a href=\"https:\/\/chemuza.org\/tr\/sebr2-yapisi-lewis\/\">Lewis yap\u0131s\u0131 SeBr2<\/a><\/td>\n<td> <a href=\"https:\/\/chemuza.org\/tr\/hcp-lewis-yapisi\/\">HCP Lewis&#8217;in 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. HBrO3 Lewis yap\u0131s\u0131n\u0131n merkezinde iki oksijen (O) atomu ve bir OH grubu ile \u00e7evrelenmi\u015f bir brom (Br) atomu bulunur. Brom atomu (Br) ile oksijen atomu (O) aras\u0131nda 2 adet \u00e7ift ba\u011f vard\u0131r ve di\u011fer atomlar\u0131n geri kalan\u0131nda tek ba\u011f bulunur. HBrO3&#8217;\u00fcn Lewis yap\u0131s\u0131n\u0131n yukar\u0131daki g\u00f6r\u00fcnt\u00fcs\u00fcnden &#8230; <a title=\"6 ad\u0131mda hbro3 lewis yap\u0131s\u0131 (resimlerle)\" class=\"read-more\" href=\"https:\/\/chemuza.org\/tr\/hbro3-lewis-yapisi\/\" aria-label=\"More on 6 ad\u0131mda hbro3 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 HBrO3 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\/hbro3-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 HBrO3 Lewis Yap\u0131s\u0131 (Resimlerle) - Chemuza\" \/>\n<meta property=\"og:description\" content=\"Yukar\u0131daki g\u00f6rseli zaten g\u00f6rd\u00fcn\u00fcz de\u011fil mi? 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