{"id":146,"date":"2023-07-25T20:57:56","date_gmt":"2023-07-25T20:57:56","guid":{"rendered":"https:\/\/chemuza.org\/pt\/estrutura-noc-lewis\/"},"modified":"2023-07-25T20:57:56","modified_gmt":"2023-07-25T20:57:56","slug":"estrutura-noc-lewis","status":"publish","type":"post","link":"https:\/\/chemuza.org\/pt\/estrutura-noc-lewis\/","title":{"rendered":"Estrutura cno-lewis em 6 etapas (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-cno-lewis.jpg\" alt=\"Estrutura CNO-Lewis\" class=\"wp-image-1974\" 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 CNO- (\u00edon fulminado) tem um \u00e1tomo de nitrog\u00eanio (N) no centro que \u00e9 cercado por um \u00e1tomo de carbono (C) e um \u00e1tomo de oxig\u00eanio (O). Existe uma liga\u00e7\u00e3o simples entre o \u00e1tomo de nitrog\u00eanio (N) e oxig\u00eanio (O) e uma liga\u00e7\u00e3o tripla entre carbono (C) e nitrog\u00eanio (N).<\/mark><\/em><\/strong><\/p>\n<p> Se voc\u00ea n\u00e3o entendeu nada da imagem acima da estrutura de Lewis do \u00edon CNO ( <a href=\"https:\/\/en.wikipedia.org\/wiki\/Fulminate\" target=\"_blank\" rel=\"noopener\">\u00edon fulminado<\/a> ), 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 \u00edon CNO.<\/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 CNO.<\/p>\n<h2 class=\"wp-block-heading\"> <strong>Passos para desenhar a estrutura CNO-Lewis<\/strong><\/h2>\n<h3 class=\"wp-block-heading\"> <strong>Etapa 1: Encontre o n\u00famero total de el\u00e9trons de val\u00eancia na mol\u00e9cula CNO<\/strong><\/h3>\n<p> Para encontrar o n\u00famero total <a href=\"http:\/\/www.chem.ucla.edu\/~harding\/IGOC\/V\/valence_electron.html\" target=\"_blank\" rel=\"noopener\">de el\u00e9trons de val\u00eancia<\/a> em um \u00edon CNO (\u00edon fulminado), primeiro voc\u00ea precisa saber os el\u00e9trons de val\u00eancia presentes no \u00e1tomo de carbono, no \u00e1tomo de nitrog\u00eanio e tamb\u00e9m no \u00e1tomo de oxig\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 carbono, nitrog\u00eanio e tamb\u00e9m do oxig\u00eanio usando uma tabela peri\u00f3dica.<\/p>\n<p> <strong>El\u00e9trons totais de val\u00eancia no \u00edon CNO<\/strong><\/p>\n<p> <strong>\u2192 El\u00e9trons de val\u00eancia dados pelo \u00e1tomo de carbono:<\/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=\"auto, \"><\/figure>\n<p> O carbono \u00e9 um elemento do grupo 14 da tabela peri\u00f3dica. <a href=\"https:\/\/periodic.lanl.gov\/6.shtml\" target=\"_blank\" rel=\"noopener\"><sup>[1]<\/sup><\/a> Portanto, os el\u00e9trons de val\u00eancia presentes no carbono s\u00e3o <strong>4<\/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=\"auto, \"><\/figure>\n<p> Voc\u00ea pode ver os 4 el\u00e9trons de val\u00eancia presentes no \u00e1tomo de carbono, conforme mostrado na imagem acima.<\/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> O nitrog\u00eanio \u00e9 um elemento do grupo 15 da tabela peri\u00f3dica. <a href=\"https:\/\/education.jlab.org\/itselemental\/ele007.html\" target=\"_blank\" rel=\"noopener\"><sup>[2]<\/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> <strong>\u2192 El\u00e9trons de val\u00eancia dados pelo \u00e1tomo de oxig\u00eanio:<\/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=\"auto, \"><\/figure>\n<p> O oxig\u00eanio \u00e9 um elemento do grupo 16 da tabela peri\u00f3dica. <a href=\"https:\/\/pubchem.ncbi.nlm.nih.gov\/element\/8\" target=\"_blank\" rel=\"noopener\"><sup>[3]<\/sup><\/a> Portanto, os el\u00e9trons de val\u00eancia presentes no oxig\u00eanio s\u00e3o <strong>6<\/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=\"auto, \"><\/figure>\n<p> Voc\u00ea pode ver os 6 el\u00e9trons de val\u00eancia presentes no \u00e1tomo de oxig\u00eanio, conforme mostrado na imagem acima.<\/p>\n<p> Ent\u00e3o,<\/p>\n<p> <strong>Total de el\u00e9trons de val\u00eancia no \u00edon CNO<\/strong> = el\u00e9trons de val\u00eancia doados por 1 \u00e1tomo de carbono + el\u00e9trons de val\u00eancia doados por 1 \u00e1tomo de nitrog\u00eanio + el\u00e9trons de val\u00eancia doados por 1 \u00e1tomo de oxig\u00eanio + 1 el\u00e9tron extra \u00e9 adicionado devido a 1 carga negativa = <strong>4 + 5 + 6 + 1 = 16<\/strong> .<\/p>\n<h3 class=\"wp-block-heading\"> <strong>Passo 2: Prepare um esbo\u00e7o<\/strong><\/h3>\n<p> Para fazer um esbo\u00e7o do NOC, basta observar sua f\u00f3rmula qu\u00edmica. Voc\u00ea pode ver que h\u00e1 um \u00e1tomo de nitrog\u00eanio (N) no centro e ele est\u00e1 cercado por um \u00e1tomo de carbono e tamb\u00e9m por um \u00e1tomo de oxig\u00eanio em ambos os lados.<\/p>\n<p> Ent\u00e3o, vamos fazer um esbo\u00e7o do mesmo. <\/p>\n<figure class=\"wp-block-image aligncenter size-full\"><img decoding=\"async\" loading=\"lazy\" width=\"600\" height=\"60\" src=\"https:\/\/chemuza.org\/wp-content\/uploads\/2023\/08\/pas-detape-1.webp\" alt=\"NOC - est\u00e1gio 1\" class=\"wp-image-1975\" 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 CNO voc\u00ea precisa colocar os <a href=\"https:\/\/www2.chem.wisc.edu\/deptfiles\/genchem\/netorial\/rottosen\/tutorial\/modules\/intermolecular_forces\/01review\/review3.htm\" target=\"_blank\" rel=\"noopener\">pares de el\u00e9trons<\/a> entre o \u00e1tomo de carbono (C), o \u00e1tomo de nitrog\u00eanio (N) e o \u00e1tomo de oxig\u00eanio (O). <\/p>\n<figure class=\"wp-block-image aligncenter size-full\"><img decoding=\"async\" loading=\"lazy\" width=\"600\" height=\"63\" src=\"https:\/\/chemuza.org\/wp-content\/uploads\/2023\/08\/pas-detape-2.webp\" alt=\"NOC - est\u00e1gio 2\" class=\"wp-image-1976\" srcset=\"\" sizes=\"auto, \"><\/figure>\n<p> Isso indica que o \u00e1tomo de carbono (C), o \u00e1tomo de nitrog\u00eanio (N) e o \u00e1tomo de oxig\u00eanio (O) est\u00e3o <a href=\"https:\/\/www.sydney.edu.au\/science\/chemistry\/~george\/1611\/ChemicalBonding.pdf\" target=\"_blank\" rel=\"noopener\">quimicamente ligados<\/a> entre si em uma mol\u00e9cula de CNO.<\/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 CNO, voc\u00ea pode ver que os \u00e1tomos externos s\u00e3o o \u00e1tomo de carbono e o \u00e1tomo de oxig\u00eanio.<\/p>\n<p> Esses \u00e1tomos externos de carbono e oxig\u00eanio formam um <a href=\"https:\/\/en.wikibooks.org\/wiki\/General_Chemistry\/Octet_Rule_and_Exceptions\" target=\"_blank\" rel=\"noopener\">octeto<\/a> e, portanto, s\u00e3o est\u00e1veis. <\/p>\n<figure class=\"wp-block-image aligncenter size-full\"><img decoding=\"async\" loading=\"lazy\" width=\"600\" height=\"269\" src=\"https:\/\/chemuza.org\/wp-content\/uploads\/2023\/08\/pas-detape-3.webp\" alt=\"NOC - est\u00e1gio 3\" class=\"wp-image-1977\" 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 CNO.<\/p>\n<p> O \u00edon CNO 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 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=\"257\" src=\"https:\/\/chemuza.org\/wp-content\/uploads\/2023\/08\/pas-detape-4.webp\" alt=\"NOC - est\u00e1gio 4\" class=\"wp-image-1978\" 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 carbono externo para que o \u00e1tomo de nitrog\u00eanio possa ter 8 el\u00e9trons (ou seja, um octeto).<\/p>\n<p> <strong>(Nota:<\/strong> Lembre-se de que voc\u00ea precisa mover o par de el\u00e9trons do \u00e1tomo que \u00e9 menos eletronegativo.<br \/> Na verdade, o \u00e1tomo menos eletronegativo tem maior tend\u00eancia a doar el\u00e9trons.<br \/> Aqui, se compararmos o \u00e1tomo de carbono e o \u00e1tomo de oxig\u00eanio, ent\u00e3o o \u00e1tomo de carbono \u00e9 menos eletronegativo.<br \/> Ent\u00e3o voc\u00ea precisa mover o par de el\u00e9trons do \u00e1tomo de carbono.) <\/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\/pas-detape-5.webp\" alt=\"NOC - etapa 5\" class=\"wp-image-1979\" srcset=\"\" sizes=\"auto, \"><\/figure>\n<p> Mas depois de mover um par de el\u00e9trons, o \u00e1tomo 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=\"240\" src=\"https:\/\/chemuza.org\/wp-content\/uploads\/2023\/08\/pas-detape-6.webp\" alt=\"NOC - est\u00e1gio 6\" class=\"wp-image-1980\" srcset=\"\" sizes=\"auto, \"><\/figure>\n<p> Novamente, precisamos mover um par extra de el\u00e9trons apenas do \u00e1tomo de carbono. (Porque o carbono \u00e9 menos eletronegativo que o oxig\u00eanio.) <\/p>\n<figure class=\"wp-block-image aligncenter size-full\"><img decoding=\"async\" loading=\"lazy\" width=\"600\" height=\"100\" src=\"https:\/\/chemuza.org\/wp-content\/uploads\/2023\/08\/pas-detape-7.webp\" alt=\"NOC - est\u00e1gio 7\" class=\"wp-image-1981\" 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=\"260\" src=\"https:\/\/chemuza.org\/wp-content\/uploads\/2023\/08\/pas-detape-8.webp\" alt=\"NOC - est\u00e1gio 8\" class=\"wp-image-1982\" srcset=\"\" sizes=\"auto, \"><\/figure>\n<p> Voc\u00ea pode ver na imagem acima que o \u00e1tomo de nitrog\u00eanio forma um octeto.<\/p>\n<p> E, portanto, o \u00e1tomo de nitrog\u00eanio \u00e9 est\u00e1vel.<\/p>\n<p> Agora vamos passar para a \u00faltima etapa para verificar se a estrutura de Lewis do CNO \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 da mol\u00e9cula CNO.<\/p>\n<p> A estabilidade da estrutura de Lewis pode ser verificada usando um conceito <a href=\"http:\/\/guweb2.gonzaga.edu\/faculty\/cronk\/CHEM101pub\/formal_charge.html\" target=\"_blank\" rel=\"noopener\">formal de carga<\/a> .<\/p>\n<p> Resumindo, devemos agora encontrar a carga formal dos \u00e1tomos de carbono (C), nitrog\u00eanio (N) e oxig\u00eanio (O) presentes na mol\u00e9cula de CNO.<\/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 CNO na imagem abaixo. <\/p>\n<figure class=\"wp-block-image aligncenter size-full\"><img decoding=\"async\" loading=\"lazy\" width=\"600\" height=\"226\" src=\"https:\/\/chemuza.org\/wp-content\/uploads\/2023\/08\/pas-detape-9.webp\" alt=\"NOC - est\u00e1gio 9\" class=\"wp-image-1983\" srcset=\"\" sizes=\"auto, \"><\/figure>\n<p> <strong>Para o \u00e1tomo de carbono (C):<\/strong><strong><br \/><\/strong> El\u00e9trons de val\u00eancia = 4 (porque o carbono est\u00e1 no grupo 14)<strong><br \/><\/strong> El\u00e9trons de liga\u00e7\u00e3o = 6<br \/> El\u00e9trons n\u00e3o ligantes = 2<\/p>\n<p> <strong>Para o \u00e1tomo 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 o \u00e1tomo de oxig\u00eanio (O):<\/strong><strong><br \/><\/strong> El\u00e9trons de val\u00eancia = 6 (porque o oxig\u00eanio est\u00e1 no grupo 16)<strong><br \/><\/strong> El\u00e9trons de liga\u00e7\u00e3o = 2<br \/> El\u00e9trons n\u00e3o ligantes = 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>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\"> 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\"> 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>+1<\/strong><\/td>\n<\/tr>\n<tr>\n<td class=\"has-text-align-center\" data-align=\"center\"> N\u00c3O<\/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>0<\/strong><\/td>\n<\/tr>\n<tr>\n<td class=\"has-text-align-center\" data-align=\"center\"> Oh<\/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> A partir dos c\u00e1lculos formais de carga acima, voc\u00ea pode ver que o \u00e1tomo de carbono (C) tem carga de <strong>-1<\/strong> e o \u00e1tomo de oxig\u00eanio (O) tem carga <strong>de +1<\/strong> .<\/p>\n<p> Ent\u00e3o vamos manter essas cargas nos respectivos \u00e1tomos da mol\u00e9cula CNO. <\/p>\n<figure class=\"wp-block-image aligncenter size-full\"><img decoding=\"async\" loading=\"lazy\" width=\"600\" height=\"84\" src=\"https:\/\/chemuza.org\/wp-content\/uploads\/2023\/08\/pas-detape-10.webp\" alt=\"NOC - est\u00e1gio 10\" class=\"wp-image-1984\" srcset=\"\" sizes=\"auto, \"><\/figure>\n<p> As cargas <strong>+1<\/strong> e <strong>-1<\/strong> s\u00e3o canceladas e a carga geral <strong>-1<\/strong> na mol\u00e9cula CNO \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\/pas-detape-11.webp\" alt=\"NOC - est\u00e1gio 11\" class=\"wp-image-1985\" srcset=\"\" sizes=\"auto, \"><\/figure>\n<p> Na estrutura de pontos de Lewis do \u00edon CNO acima, voc\u00ea tamb\u00e9m pode representar cada par de el\u00e9trons de liga\u00e7\u00e3o (:) como uma liga\u00e7\u00e3o simples (|). Ao fazer isso, voc\u00ea obter\u00e1 a seguinte estrutura de Lewis do \u00edon CNO. <\/p>\n<figure class=\"wp-block-image aligncenter size-full\"><img decoding=\"async\" loading=\"lazy\" width=\"600\" height=\"227\" src=\"https:\/\/chemuza.org\/wp-content\/uploads\/2023\/08\/structure-de-Lewis-de-cno.jpg\" alt=\"Estrutura de Lewis do NOC-\" class=\"wp-image-1986\" 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-bro2-lewis\/\">Estrutura BrO2-Lewis<\/a><\/td>\n<td> <a href=\"https:\/\/chemuza.org\/pt\/estrutura-de-lewis-n2o4\/\">Estrutura de Lewis N2O4<\/a><\/td>\n<\/tr>\n<tr>\n<td> <a href=\"https:\/\/chemuza.org\/pt\/estrutura-cof2-lewis\/\">Estrutura de Lewis COF2<\/a><\/td>\n<td> <a href=\"https:\/\/chemuza.org\/pt\/estrutura-lewis-scl4\/\">Estrutura de Lewis SCl4<\/a><\/td>\n<\/tr>\n<tr>\n<td> <a href=\"https:\/\/chemuza.org\/pt\/estrutura-lewis-pbr5\/\">Estrutura de Lewis PBr5<\/a><\/td>\n<td> <a href=\"https:\/\/chemuza.org\/pt\/estrutura-sis2-lewis\/\">Estrutura de Lewis SiS2<\/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 CNO- (\u00edon fulminado) tem um \u00e1tomo de nitrog\u00eanio (N) no centro que \u00e9 cercado por um \u00e1tomo de carbono (C) e um \u00e1tomo de oxig\u00eanio (O). Existe uma liga\u00e7\u00e3o simples entre o \u00e1tomo de nitrog\u00eanio (N) e &#8230; <a title=\"Estrutura cno-lewis em 6 etapas (com imagens)\" class=\"read-more\" href=\"https:\/\/chemuza.org\/pt\/estrutura-noc-lewis\/\" aria-label=\"Mais em Estrutura cno-lewis em 6 etapas (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-146","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>Estrutura CNO-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-noc-lewis\/\" \/>\n<meta property=\"og:locale\" content=\"pt_PT\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Estrutura CNO-Lewis em 6 passos (com imagens) - Chemuza\" \/>\n<meta property=\"og:description\" content=\"Ent\u00e3o voc\u00ea j\u00e1 viu a imagem acima, certo? 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Deixe-me explicar brevemente a imagem acima. A estrutura de Lewis CNO- (\u00edon fulminado) tem um \u00e1tomo de nitrog\u00eanio (N) no centro que \u00e9 cercado por um \u00e1tomo de carbono (C) e um \u00e1tomo de oxig\u00eanio (O). Existe uma liga\u00e7\u00e3o simples entre o \u00e1tomo de nitrog\u00eanio (N) e ... 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