{"id":84,"date":"2023-07-26T09:16:40","date_gmt":"2023-07-26T09:16:40","guid":{"rendered":"https:\/\/chemuza.org\/pt\/estrutura-de-lewis-n3\/"},"modified":"2023-07-26T09:16:40","modified_gmt":"2023-07-26T09:16:40","slug":"estrutura-de-lewis-n3","status":"publish","type":"post","link":"https:\/\/chemuza.org\/pt\/estrutura-de-lewis-n3\/","title":{"rendered":"N3- estrutura de lewis em 6 passos (com imagens)"},"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-de-Lewis-n3.jpg\" alt=\"Estrutura N3-Lewis\" class=\"wp-image-601\" srcset=\"\" sizes=\"auto, \"><\/figure>\n<p> Ent\u00e3o voc\u00ea j\u00e1 viu a imagem acima, certo?<\/p>\n<p> Deixe-me explicar brevemente a imagem acima.<\/p>\n<p> <strong><em><mark style=\"background-color:rgba(0, 0, 0, 0);color:#ff0000\" class=\"has-inline-color\">A estrutura de Lewis N3- (\u00edon azida) possui tr\u00eas \u00e1tomos de nitrog\u00eanio (N). Existem 2 liga\u00e7\u00f5es duplas entre cada \u00e1tomo de nitrog\u00eanio (N). Existem 2 pares solit\u00e1rios nos dois \u00e1tomos de nitrog\u00eanio externos. Existe uma carga formal -1 em um dos \u00e1tomos externos de nitrog\u00eanio (N).<\/mark><\/em><\/strong><\/p>\n<p> Se voc\u00ea n\u00e3o entendeu nada da imagem acima da estrutura de Lewis N3- (\u00edon azida), ent\u00e3o fique comigo e voc\u00ea obter\u00e1 uma explica\u00e7\u00e3o detalhada passo a passo sobre como desenhar uma estrutura de Lewis do <a href=\"https:\/\/pubchem.ncbi.nlm.nih.gov\/compound\/Azide-ion\" target=\"_blank\" rel=\"noopener\">\u00edon N3-<\/a> .<\/p>\n<p> Ent\u00e3o, vamos prosseguir para as etapas de desenho da <a href=\"https:\/\/chemuza.org\/pt\/definicoes-basicas-de-quimica\/\">estrutura de Lewis<\/a> do \u00edon N3.<\/p>\n<h2 class=\"wp-block-heading\"> <strong>Etapas para desenhar a estrutura N3 Lewis<\/strong><\/h2>\n<h3 class=\"wp-block-heading\"> <strong>Etapa 1: Encontre o n\u00famero total de el\u00e9trons de val\u00eancia no \u00edon N3<\/strong><\/h3>\n<p> Para encontrar o n\u00famero total <a href=\"https:\/\/www.chemeurope.com\/en\/encyclopedia\/Valence_electron.html\" target=\"_blank\" rel=\"noopener\">de el\u00e9trons de val\u00eancia<\/a> no \u00edon N3 (\u00edon azida), primeiro voc\u00ea precisa saber os el\u00e9trons de val\u00eancia presentes em um \u00fanico <a href=\"https:\/\/chemuza.org\/pt\/definicoes-basicas-de-quimica\/\">\u00e1tomo<\/a> de nitrog\u00eanio.<br \/> (El\u00e9trons de val\u00eancia s\u00e3o os el\u00e9trons presentes na <a href=\"https:\/\/chemuza.org\/pt\/definicoes-basicas-de-quimica\/\">\u00f3rbita<\/a> mais externa de qualquer \u00e1tomo.)<\/p>\n<p> Aqui vou lhe dizer como encontrar facilmente os el\u00e9trons de val\u00eancia do nitrog\u00eanio usando uma <a href=\"https:\/\/chemuza.org\/pt\/definicoes-basicas-de-quimica\/\">tabela peri\u00f3dica<\/a> .<\/p>\n<p class=\"has-medium-font-size\"> <strong>El\u00e9trons totais de val\u00eancia no \u00edon N3<\/strong><\/p>\n<p> <strong>\u2192 El\u00e9trons de val\u00eancia dados pelo \u00e1tomo de nitrog\u00eanio:<\/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=\"auto, \"><\/figure>\n<p> <a href=\"https:\/\/chemuza.org\/pt\/definicoes-basicas-de-quimica\/\">O nitrog\u00eanio<\/a> \u00e9 um elemento <a href=\"https:\/\/chemuza.org\/pt\/definicoes-basicas-de-quimica\/\">do grupo 15<\/a> da tabela peri\u00f3dica. <a href=\"https:\/\/pubchem.ncbi.nlm.nih.gov\/element\/7\" target=\"_blank\" rel=\"noopener\"><sup>[1]<\/sup><\/a> Portanto, os el\u00e9trons de val\u00eancia presentes no nitrog\u00eanio s\u00e3o <strong>5<\/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=\"auto, \"><\/figure>\n<p> Voc\u00ea pode ver os 5 el\u00e9trons de val\u00eancia presentes no \u00e1tomo de nitrog\u00eanio, conforme mostrado na imagem acima.<\/p>\n<p> Ent\u00e3o,<\/p>\n<p> <strong>Total de el\u00e9trons de val\u00eancia no \u00edon N3<\/strong> = el\u00e9trons de val\u00eancia doados por 3 \u00e1tomos de nitrog\u00eanio + 1 el\u00e9tron extra \u00e9 adicionado devido a 1 carga negativa = <strong>5(3) + 1 = 16<\/strong> .<\/p>\n<h3 class=\"wp-block-heading\"> <strong>Passo 2: Selecione o \u00e1tomo central<\/strong><\/h3>\n<p> Para selecionar o \u00e1tomo central, devemos lembrar que o \u00e1tomo menos <a href=\"https:\/\/chemuza.org\/pt\/definicoes-basicas-de-quimica\/\">eletronegativo<\/a> permanece no centro.<\/p>\n<p> Agora aqui a mol\u00e9cula dada \u00e9 N3. Todos os tr\u00eas \u00e1tomos s\u00e3o id\u00eanticos, ent\u00e3o voc\u00ea pode selecionar qualquer um dos \u00e1tomos como \u00e1tomo central. <\/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\/n3-etape-1.webp\" alt=\"N3-passo 1\" class=\"wp-image-602\" srcset=\"\" sizes=\"auto, \"><\/figure>\n<h3 class=\"wp-block-heading\"> <strong>Etapa 3: Conecte cada \u00e1tomo colocando um par de el\u00e9trons entre eles<\/strong><\/h3>\n<p> Agora, na mol\u00e9cula de N3, voc\u00ea precisa colocar os <a href=\"https:\/\/en.wikipedia.org\/wiki\/Electron_pair\" target=\"_blank\" rel=\"noopener\">pares de el\u00e9trons<\/a> entre os tr\u00eas \u00e1tomos de nitrog\u00eanio (N). <\/p>\n<figure class=\"wp-block-image aligncenter size-full\"><img decoding=\"async\" loading=\"lazy\" width=\"600\" height=\"59\" src=\"https:\/\/chemuza.org\/wp-content\/uploads\/2023\/08\/n3-etape-2.webp\" alt=\"N3-passo 2\" class=\"wp-image-603\" srcset=\"\" sizes=\"auto, \"><\/figure>\n<p> Isso indica que os tr\u00eas \u00e1tomos de nitrog\u00eanio (N) est\u00e3o <a href=\"https:\/\/chem.fsu.edu\/chemlab\/chm1045\/bonding.html\" target=\"_blank\" rel=\"noopener\">quimicamente ligados<\/a> entre si na mol\u00e9cula de N3.<\/p>\n<h3 class=\"wp-block-heading\"> <strong>Etapa 4: tornar os \u00e1tomos externos est\u00e1veis<\/strong><\/h3>\n<p> Nesta etapa voc\u00ea precisa verificar a estabilidade dos \u00e1tomos externos.<\/p>\n<p> Aqui no esbo\u00e7o da mol\u00e9cula de N3 voc\u00ea pode ver que os \u00e1tomos externos s\u00e3o apenas \u00e1tomos de nitrog\u00eanio.<\/p>\n<p> Esses \u00e1tomos externos de nitrog\u00eanio formam um <a href=\"https:\/\/en.wikipedia.org\/wiki\/Octet_rule\" target=\"_blank\" rel=\"noopener\">octeto<\/a> e s\u00e3o, portanto, est\u00e1veis. <\/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\/n3-etape-3.webp\" alt=\"N3-etapa 3\" class=\"wp-image-604\" srcset=\"\" sizes=\"auto, \"><\/figure>\n<p> Al\u00e9m disso, na etapa 1, calculamos o n\u00famero total de el\u00e9trons de val\u00eancia presentes no \u00edon N3-.<\/p>\n<p> O \u00edon N3- tem um total <strong>de 16 el\u00e9trons de val\u00eancia<\/strong> e todos esses el\u00e9trons de val\u00eancia s\u00e3o usados no diagrama acima.<\/p>\n<p> Portanto, n\u00e3o h\u00e1 mais pares de el\u00e9trons para manter no \u00e1tomo central.<\/p>\n<p> Ent\u00e3o agora vamos para a pr\u00f3xima etapa.<\/p>\n<h3 class=\"wp-block-heading\"> <strong>Etapa 5: verifique o octeto no \u00e1tomo central. Se n\u00e3o tiver octeto, mova o par solit\u00e1rio para formar uma liga\u00e7\u00e3o dupla ou tripla.<\/strong><\/h3>\n<p> Nesta etapa, voc\u00ea precisa verificar se o \u00e1tomo central de nitrog\u00eanio (N) \u00e9 est\u00e1vel ou n\u00e3o.<\/p>\n<p> Para verificar a estabilidade do \u00e1tomo central de nitrog\u00eanio (N), precisamos verificar se ele forma um octeto ou n\u00e3o.<\/p>\n<p> Infelizmente, o \u00e1tomo central de nitrog\u00eanio n\u00e3o forma um octeto aqui. O nitrog\u00eanio tem apenas 4 el\u00e9trons e \u00e9 inst\u00e1vel. <\/p>\n<figure class=\"wp-block-image aligncenter size-full\"><img decoding=\"async\" loading=\"lazy\" width=\"600\" height=\"254\" src=\"https:\/\/chemuza.org\/wp-content\/uploads\/2023\/08\/n3-etape-4.webp\" alt=\"N3-passo 4\" class=\"wp-image-605\" srcset=\"\" sizes=\"auto, \"><\/figure>\n<p> Agora, para tornar este \u00e1tomo de nitrog\u00eanio est\u00e1vel, voc\u00ea precisa deslocar o par de el\u00e9trons do \u00e1tomo de nitrog\u00eanio externo para que o \u00e1tomo de nitrog\u00eanio central possa ter 8 el\u00e9trons (ou seja, um octeto). <\/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\/n3-etape-5.webp\" alt=\"N3-passo 5\" class=\"wp-image-606\" srcset=\"\" sizes=\"auto, \"><\/figure>\n<p> Mas depois de mover um par de el\u00e9trons, o \u00e1tomo central de nitrog\u00eanio ainda n\u00e3o forma um octeto, pois possui apenas 6 el\u00e9trons. <\/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\/n3-etape-6.webp\" alt=\"N3-passo 6\" class=\"wp-image-607\" srcset=\"\" sizes=\"auto, \"><\/figure>\n<p> Novamente, precisamos mover um par extra de el\u00e9trons do outro \u00e1tomo de nitrog\u00eanio. <\/p>\n<figure class=\"wp-block-image aligncenter size-full\"><img decoding=\"async\" loading=\"lazy\" width=\"600\" height=\"105\" src=\"https:\/\/chemuza.org\/wp-content\/uploads\/2023\/08\/n3-etape-7.webp\" alt=\"N3-passo 7\" class=\"wp-image-608\" srcset=\"\" sizes=\"auto, \"><\/figure>\n<p> Ap\u00f3s movimentar esse par de el\u00e9trons, o \u00e1tomo central de nitrog\u00eanio receber\u00e1 mais 2 el\u00e9trons e seu total de el\u00e9trons passar\u00e1 a ser 8. <\/p>\n<figure class=\"wp-block-image aligncenter size-full\"><img decoding=\"async\" loading=\"lazy\" width=\"600\" height=\"235\" src=\"https:\/\/chemuza.org\/wp-content\/uploads\/2023\/08\/n3-etape-8.webp\" alt=\"N3-passo 8\" class=\"wp-image-609\" srcset=\"\" sizes=\"auto, \"><\/figure>\n<p> Voc\u00ea pode ver na imagem acima que o \u00e1tomo central de nitrog\u00eanio forma um octeto.<\/p>\n<p> E ent\u00e3o este \u00e1tomo de nitrog\u00eanio \u00e9 est\u00e1vel.<\/p>\n<p> Agora vamos passar para a \u00faltima etapa para verificar se a estrutura de Lewis de N3 \u00e9 est\u00e1vel ou n\u00e3o.<\/p>\n<h3 class=\"wp-block-heading\"> <strong>Passo 6: Verifique a estabilidade da estrutura de Lewis<\/strong><\/h3>\n<p> Agora voc\u00ea chegou \u00e0 \u00faltima etapa em que precisa verificar a estabilidade da estrutura de Lewis de N3.<\/p>\n<p> A estabilidade da estrutura de Lewis pode ser verificada usando um conceito <a href=\"https:\/\/www.chem.ucalgary.ca\/courses\/353\/Carey5th\/Ch01\/ch1-3-2.html\" target=\"_blank\" rel=\"noopener\">formal de carga<\/a> .<\/p>\n<p> Resumindo, precisamos agora de determinar a carga formal de todos os \u00e1tomos de azoto (N) presentes na mol\u00e9cula de N3.<\/p>\n<p> Para calcular o imposto formal, deve-se utilizar a seguinte f\u00f3rmula:<\/p>\n<p class=\"has-background\" style=\"background-color:#ffe9cf\"> <strong>Carga formal = El\u00e9trons de val\u00eancia \u2013 (El\u00e9trons ligantes)\/2 \u2013 El\u00e9trons n\u00e3o ligantes<\/strong><\/p>\n<p> Voc\u00ea pode ver o n\u00famero de <a href=\"https:\/\/chemuza.org\/pt\/definicoes-basicas-de-quimica\/\">el\u00e9trons ligantes<\/a> e <a href=\"https:\/\/chemuza.org\/pt\/definicoes-basicas-de-quimica\/\">el\u00e9trons n\u00e3o ligantes<\/a> para cada \u00e1tomo da mol\u00e9cula de N3 na imagem abaixo. <\/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\/n3-etape-9.webp\" alt=\"N3-passo 9\" class=\"wp-image-610\" srcset=\"\" sizes=\"auto, \"><\/figure>\n<p> <strong>Para o \u00e1tomo central de nitrog\u00eanio (N):<\/strong><strong><br \/><\/strong> El\u00e9trons de val\u00eancia = 5 (porque o nitrog\u00eanio est\u00e1 no grupo 15)<strong><br \/><\/strong> El\u00e9trons de liga\u00e7\u00e3o = 8<br \/> El\u00e9trons n\u00e3o ligantes = 0<\/p>\n<p> <strong>Para \u00e1tomos externos de nitrog\u00eanio (N):<\/strong><strong><br \/><\/strong> El\u00e9trons de val\u00eancia = 5 (porque o nitrog\u00eanio est\u00e1 no grupo 15)<strong><br \/><\/strong> El\u00e9trons de liga\u00e7\u00e3o = 4<br \/> El\u00e9trons n\u00e3o ligantes = 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>Acusa\u00e7\u00e3o formal<\/strong><\/td>\n<td> <strong>=<\/strong><\/td>\n<td class=\"has-text-align-center\" data-align=\"center\"> <strong>el\u00e9trons de val\u00eancia<\/strong><\/td>\n<td> <strong>\u2013<\/strong><\/td>\n<td class=\"has-text-align-center\" data-align=\"center\"> <strong>(El\u00e9trons de liga\u00e7\u00e3o)\/2<\/strong><\/td>\n<td> <strong>\u2013<\/strong><\/td>\n<td class=\"has-text-align-center\" data-align=\"center\"> <strong>El\u00e9trons n\u00e3o ligantes<\/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\"> N (central)<\/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\"> 02\/08<\/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\"> N (exterior)<\/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\"> 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>-1<\/strong><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/figure>\n<p> A partir dos c\u00e1lculos formais de carga acima, voc\u00ea pode ver que o \u00e1tomo central de nitrog\u00eanio (N) tem uma carga de <strong>-1<\/strong> e os \u00e1tomos externos de nitrog\u00eanio t\u00eam uma carga de <strong>-1<\/strong> .<\/p>\n<p> Ent\u00e3o vamos manter essas cargas nos respectivos \u00e1tomos da mol\u00e9cula de N3. <\/p>\n<figure class=\"wp-block-image aligncenter size-full\"><img decoding=\"async\" loading=\"lazy\" width=\"600\" height=\"89\" src=\"https:\/\/chemuza.org\/wp-content\/uploads\/2023\/08\/n3-etape-10.webp\" alt=\"N3-passo 10\" class=\"wp-image-611\" srcset=\"\" sizes=\"auto, \"><\/figure>\n<p> As cargas <strong>+1<\/strong> e <strong>-1<\/strong> no diagrama acima s\u00e3o canceladas e a \u00fanica carga <strong>-1<\/strong> permanece no diagrama acima, resultando em uma carga formal <strong>-1<\/strong> na mol\u00e9cula de N3.<\/p>\n<p> Esta carga geral <strong>-1<\/strong> na mol\u00e9cula de N3 \u00e9 mostrada na imagem abaixo. <\/p>\n<figure class=\"wp-block-image aligncenter size-full\"><img decoding=\"async\" loading=\"lazy\" width=\"600\" height=\"99\" src=\"https:\/\/chemuza.org\/wp-content\/uploads\/2023\/08\/n3-etape-11.webp\" alt=\"N3-passo 11\" class=\"wp-image-612\" srcset=\"\" sizes=\"auto, \"><\/figure>\n<p> Na estrutura de pontos de Lewis do \u00edon N3 acima, voc\u00ea tamb\u00e9m pode representar cada par de el\u00e9trons de liga\u00e7\u00e3o (:) como uma liga\u00e7\u00e3o simples (|). Isso lhe dar\u00e1 a seguinte estrutura de Lewis do \u00edon N3. <\/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\/structure-de-Lewis-de-n3.jpg\" alt=\"Estrutura de Lewis de N3-\" class=\"wp-image-613\" srcset=\"\" sizes=\"auto, \"><\/figure>\n<p> Espero que voc\u00ea tenha entendido completamente todas as etapas acima.<\/p>\n<p> Para mais pr\u00e1tica e melhor compreens\u00e3o, voc\u00ea pode tentar outras estruturas de Lewis listadas abaixo.<\/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>Experimente (ou pelo menos veja) estas estruturas de Lewis para uma melhor compreens\u00e3o:<\/strong><\/p>\n<figure class=\"wp-block-table\">\n<table>\n<tbody>\n<tr>\n<td> <a href=\"https:\/\/chemuza.org\/pt\/estrutura-bh3-lewis\/\">Estrutura de Lewis BH3<\/a><\/td>\n<td> <a href=\"https:\/\/chemuza.org\/pt\/estrutura-c2h6-lewis\/\">Estrutura de Lewis C2H6<\/a><\/td>\n<\/tr>\n<tr>\n<td> <a href=\"https:\/\/chemuza.org\/pt\/estrutura-cocl2-lewis\/\">Estrutura de Lewis COCl2<\/a><\/td>\n<td> <a href=\"https:\/\/chemuza.org\/pt\/estrutura-de-lewis-brf5\/\">Estrutura BrF5 Lewis<\/a><\/td>\n<\/tr>\n<tr>\n<td> <a href=\"https:\/\/chemuza.org\/pt\/estrutura-de-lewis-ncl3\/\">Estrutura de Lewis NCl3<\/a><\/td>\n<td> <a href=\"https:\/\/chemuza.org\/pt\/estrutura-chcl3-lewis\/\">Estrutura de Lewis CHCl3<\/a><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/figure>\n","protected":false},"excerpt":{"rendered":"<p>Ent\u00e3o voc\u00ea j\u00e1 viu a imagem acima, certo? Deixe-me explicar brevemente a imagem acima. A estrutura de Lewis N3- (\u00edon azida) possui tr\u00eas \u00e1tomos de nitrog\u00eanio (N). Existem 2 liga\u00e7\u00f5es duplas entre cada \u00e1tomo de nitrog\u00eanio (N). Existem 2 pares solit\u00e1rios nos dois \u00e1tomos de nitrog\u00eanio externos. Existe uma carga formal -1 em um dos &#8230; <a title=\"N3- estrutura de lewis em 6 passos (com imagens)\" class=\"read-more\" href=\"https:\/\/chemuza.org\/pt\/estrutura-de-lewis-n3\/\" aria-label=\"Mais em N3- estrutura de lewis em 6 passos (com imagens)\">Ler mais<\/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":[],"class_list":["post-84","post","type-post","status-publish","format-standard","hentry","category-estrutura-de-lewis"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v21.4 - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>N3- Estrutura de Lewis em 6 passos (com imagens) - 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\/pt\/estrutura-de-lewis-n3\/\" \/>\n<meta property=\"og:locale\" content=\"pt_PT\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"N3- Estrutura de Lewis em 6 passos (com imagens) - Chemuza\" \/>\n<meta property=\"og:description\" content=\"Ent\u00e3o voc\u00ea j\u00e1 viu a imagem acima, certo? Deixe-me explicar brevemente a imagem acima. A estrutura de Lewis N3- (\u00edon azida) possui tr\u00eas \u00e1tomos de nitrog\u00eanio (N). Existem 2 liga\u00e7\u00f5es duplas entre cada \u00e1tomo de nitrog\u00eanio (N). Existem 2 pares solit\u00e1rios nos dois \u00e1tomos de nitrog\u00eanio externos. Existe uma carga formal -1 em um dos ... 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