Name: drawing lewis structures with valence electron counts
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Lewis Structures of Molecular Compounds and Polyatomic Ions

As estruturas Lewis são representações simplificadas de ligações químicas entre átomos. Escrever estruturas de Lewis para compostos neutros como o eteno ou amoníaco, ou para íons poliatômicos como o hidrónio ou fosfato envolve a seguinte sequência de passos. Primeiro, calcular o número total de elétrons de valência na molécula.Considere o eteno, por exemplo. Cada átomo de carbono tem 4 valências de elétrons, e cada átomo de hidrogênio tem 1 elétron de valência;com 2 átomos de carbono e 4 átomos de hidrogênio, O eteno tem um total de 12 valências de elétrons. Em seguida, desenhar a estrutura esquelética da molécula.Geralmente, o primeiro átomo da fórmula química ou o átomo menos eletronegativo é colocado de forma centralizada. Este átomo também é chamado de átomo central e deve ser capaz de formar pelo menos duas ligações. Átomos de hidrogênio ou átomos mais eletronegativos são posicionados terminantemente em torno do átomo central.Aqui, os átomos de carbono ocupam as posições centrais com átomos de hidrogênio à sua volta. O passo seguinte é a conclusão do octeto. Primeiro, colocar um par de elétrons entre cada dois átomos para representar pares de ligação.Os restantes elétrons são distribuídos em pares solitários a átomos terminais e centrais para satisfazer as configurações de o dueto ou o octeto. Neste caso, desde que o hidrogênio atingiu um dueto, os restantes elétrons são transferidos aos átomos de carbono. Agora, use os pares solitários para formar múltiplas ligações entre os átomos que carecem de um octeto.No eteno, os átomos de carbono ainda não chegaram a um octeto. Portanto, o par solitário é movido para a regíon de colagem para formar uma dupla ligação permitindo a todos os átomos alcançar uma configuração eletrônica estável. Para escrever a estrutura de Lewis para íons poliatômicos carregados hidrônio ou fosfato, a mesma abordagem é seguido, com uma modificação para acomodar a carga iónica.Desenhar a estrutura Lewis do hidrônio, primeiro, calcular o número de elétrons de valência sobre o íon. Uma vez que o íon hidrônico tem uma carga positiva, remover um elétron da valência total dos elétrons, reduzindo a contagem para 8. A seguir, ilustrar a molécula pela sua estrutura esquelética.Agora coloque um par de elétrons entre cada átomo, seguido pelo par solitário de elétrons no átomo central. Desta forma, átomos de hidrogênio e oxigênio satisfazem o dueto e o octeto, respectivamente. Finalmente, escreva a estrutura Lewis entre parênteses com uma carga positiva no seu canto superior direito.

estruturas de lewis de compostos moleculares e iões poliatómicos

Para desenhar estruturas Lewis para moléculas complicadas e iões moleculares, é útil seguir um procedimento passo a passo, conforme descrito:

<ol>
<li>Determine o número total de eletrões de valência (camada exterior). Para catiões, subtraia um eletrão por cada carga positiva. Para aniões, adicione um eletrão por cada carga negativa.</li>
<li>Desenhe uma estrutura de esqueleto da molécula ou do ião, organizando os átomos em torno de um átomo central. (Geralmente, o elemento menos eletronegativo deve ser colocado no centro.) Ligue cada átomo ao átomo central com uma única ligação (um par de eletrões).</li>
<li>Distribua os eletrões restantes como pares solitários nos átomos terminais (excepto hidrogénio), completando um octeto em torno de cada átomo.</li>
<li>Coloque todos os eletrões restantes no átomo central.</li>
<li>Reorganize os eletrões dos átomos exteriores para fazer múltiplas ligações com o átomo central, de forma a obter octetos sempre que possível.</li>
</ol>

Por exemplo, considere SiH4, CHO2–, NO+, e OF2 como exemplos para os quais esta diretriz geral pode ser aplicada para determinar as suas estruturas de Lewis.

<ol>
<li>Determine o número total de eletrões de valência (camada exterior) na molécula ou no ião. </li>
</ol>

Para uma molécula como SiH4, é adicionado o número de eletrões de valência em cada átomo da molécula: 

= [4 e− valência/átomo Si × 1 átomo Si] + [ 1 e− valência/átomo H × 4 átomos H] = 8 e− valência

<h4>CHO2–</h4>

Para um ião negativo, como  CHO2–, os números de eletrões de valência nos átomos são adicionados ao número de cargas negativas no ião (é obtido um eletrão por cada carga negativa única):

= [4 e− valência/átomo C × 1 átomo C] + [1 e− valência/átomo H × 1 átomo H] + [6 e− valência/átomo O × 2 átomos O] + [1 e− adicional] = 18 e− valência

<h4>NO+</h4>

Para um ião positivo, como NO+, são adicionados os números de eletrões de valência nos átomos no ião, seguido da subtração do número de cargas positivas no ião (um eletrão é perdido por cada carga positiva única) do número total de eletrões de valência:

= [5 e− valência/átomo N × 1 átomo N] + [6 e− valência/átomo O × 1 átomo O] + [−1 e−] = 10 e− valência

<h4>OF2 </h4>

Sendo OF2 uma molécula neutra, o número de eletrões de valência é simplesmente adicionado:

= [6 e− valência/átomo O × 1 átomo O] + [7 e− valência/átomo F × 2 átomos F] = 20 e− valência

<ol start="2">
<li>Desenhe uma estrutura de esqueleto da molécula ou do ião, organizando os átomos em torno de um átomo central e ligando cada átomo ao átomo central com uma única ligação (um par de eletrões). (Note que os iões são indicados com parênteses à volta da estrutura, e a carga iónica fora dos parênteses:) 
<img alt="Figure1" src="https://www.jove.com/files/core/CDNSource/lessons/1063/1063Image_1.svg" style="height: 80px;" /> 
Nos casos em que vários arranjos de átomos são possíveis, como para CHO2−, são usadas evidências experimentais para escolher o certo. No geral, os elementos menos eletronegativos têm maior probabilidade de serem átomos centrais. Em CHO2−,  o átomo de carbono menos eletronegativo ocupa a posição central com os átomos de oxigénio e hidrogénio em torno dele. Outros exemplos incluem o P em POCl3, S em SO2, e Cl em  ClO4−.  Uma excepção é que o hidrogénio quase nunca é um átomo central. Como o elemento mais electronegativo, o flúor também não pode ser um átomo central.</li>
<li>Distribua os eletrões restantes como pares solitários nos átomos terminais (excepto hidrogénio) para completar as suas camadas de valência com um octeto de eletrões. (Sem eletrões restantes em SiH4, a sua estrutura permanece inalterada.) 
<img alt="Figure2" src="https://www.jove.com/files/core/CDNSource/lessons/1063/1063image_2.svg" style="height: 80px;" /></li>
<li>Coloque todos os eletrões restantes no átomo central. 
<ul>
<li>Para SiH4,  CHO2−,  e NO+, não há eletrões restantes. Para OF2, dos 16 eletrões restantes, 12 estão colocados, deixando assim 4 eletrões para serem colocados no átomo central: 
<img alt="Figure3" src="https://www.jove.com/files/core/CDNSource/lessons/1063/1063Image_3.svg" style="height: 40px;" /></li>
</ul>
</li>
<li>Reorganize os eletrões dos átomos exteriores para fazer múltiplas ligações com o átomo central, para obter octetos sempre que possível. 
<ul>
<li>SiH4: Si já tem um octeto, então nada precisa ser feito.  
<img alt="Figure4" src="https://www.jove.com/files/core/CDNSource/lessons/1063/1063Image_4.svg" style="height: 70px;" /></li>
<li>CHO2−:  Os eletrões de valência estão distribuídos como pares solitários nos átomos de oxigénio, mas falta um octeto ao átomo de carbono.  
<img alt="Figure5" src="https://www.jove.com/files/core/CDNSource/lessons/1063/1063Image_5.svg" style="height: 80px;" /></li>
<li>Assim, um par solitário de eletrões é doado de um do oxigénio para o átomo de carbono, formando uma ligação dupla. Dependendo do átomo de oxigénio que doou os eletrões, podem existir duas estruturas possíveis, também chamadas estruturas de ressonância. </li>
<li>NO+: Para este ião, estão adicionados oito eletrões de valência, mas nenhum átomo tem um octeto. Não é possível adicionar eletrões adicionais, uma vez que os eletrões totais já estão a ser utilizados. Neste cenário, os eletrões devem ser movidos para formar múltiplas ligações. O átomo de azoto tem dois pares isolados de eletrões e o átomo de oxigénio tem um.  
<img alt="Figure6" src="https://www.jove.com/files/core/CDNSource/lessons/1063/1063Image_6.svg" style="height: 55px;" /></li>
<li>Isto ainda não produz um octeto, pelo que outro par deve ser movido para formar uma ligação tripla. 
 <img alt="Figure7" src="https://www.jove.com/files/core/CDNSource/lessons/1063/1063Image_7.svg" style="height: 50px;" /></li>
<li>Em OF2, nada muda, pois cada átomo já tem um octeto. 
<img alt="Figure8" src="https://www.jove.com/files/core/CDNSource/lessons/1063/1063Image_8.svg" style="height: 40px;" /></li>
</ul>
</li>
</ol>

Este texto é adaptado de <a href="https://openstax.org/books/chemistry-2e/pages/7-3-lewis-symbols-and-structures">Openstax, Chemistry 2e, Section 7.3: Lewis Symbols and Structures.</a>

Estruturas de Lewis de Compostos Moleculares e Iões Poliatómicos

Para desenhar estruturas de Lewis para moléculas complicadas e íons moleculares, é útil seguir um procedimento passo a passo, conforme descrito:

Determine o ...

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drawing lewis structures with valence electron counts

Drawing Lewis structures starts with counting valence electrons and building the atom arrangement step by step. Valence electrons are the outer-shell electrons that help show how atoms bond. For ions, a positive charge means one electron is removed for each plus sign, and a negative charge means one electron is added for each minus sign.

The next step is to sketch the skeleton structure. The least electronegative atom usually goes in the center, while hydrogen is almost never central. Fluorine also cannot be a central atom because it is the most electronegative element. Ions are written in brackets, with the charge shown outside the brackets.

After the skeleton is drawn, connect the atoms to the central atom with single bonds. Each single bond represents one electron pair. Then place the remaining electrons as lone pairs on the terminal atoms, except hydrogen, until those atoms have octets when possible. Any electrons still left go on the central atom.

If the central atom still does not have an octet, move lone pairs to make double or triple bonds. This can create resonance structures, which are different valid arrangements of electrons. For the formate ion, CHO2−, one oxygen can donate a lone pair to make a double bond with carbon. That gives two possible resonance forms.

The examples in the model show how the procedure works in real cases. SiH4 has 8 valence electrons total, so its structure does not need more changes once the single bonds are drawn. CHO2− has 18 valence electrons. NO+ has 10 valence electrons, and its atoms must be rearranged into a triple bond to satisfy octets as much as possible. OF2 has 20 valence electrons, and after the lone pairs are placed, its structure stays unchanged because each atom already has an octet.

Drawing Lewis Structures with Valence Electron Counts

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Drawing Lewis Structures with Valence Electron Counts

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