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

Las estructuras de Lewis son representaciones simplificadas de enlaces químicos entre átomos. Escribir estructuras de Lewis para compuestos neutros como eteno o amoniaco, o para iones poliatómicos como hidronio o fosfato implica la siguiente secuencia de pasos. Primero, calcule el número total de electrones de valencia en la molécula.Considere el eteno, por ejemplo. Cada átomo de carbono tiene 4 electrones de valencia y cada átomo de hidrógeno tiene 1 electrón de valencia;con 2 átomos de carbono y 4 átomos de hidrógeno, el eteno tiene un total de 12 electrones de valencia. A continuación, dibuja la estructura esquelética de la molécula.Generalmente, el primer átomo de la fórmula química o el átomo menos electronegativo se coloca en el centro. Este átomo también se llama átomo central y debe poder formar al menos dos enlaces. Los átomos de hidrógeno o átomos más electronegativos se colocan terminalmente alrededor del átomo central.Aquí, los átomos de carbono ocupan las posiciones centrales con los átomos de hidrógeno rodeándolos. El siguiente paso es completar el octeto. Primero, coloque un par de electrones entre cada dos átomos para representar pares de enlace.Los electrones restantes se distribuyen como pares solitarios a los átomos terminales y centrales para satisfacer las configuraciones de dúo u octeto. En este caso, dado que el hidrógeno ha alcanzado un dúo, los electrones restantes se transfieren a los átomos de carbono. Ahora, use los pares solitarios para formar múltiples enlaces entre átomos que carecen de un octeto.En el eteno, los átomos de carbono aún no han alcanzado un octeto. Por lo tanto, el par solitario se mueve a la región de enlace para formar un doble enlace que permite que todos los átomos alcancen una configuración electrónica estable. Para escribir la estructura de Lewis para iones poliatómicos cargados como hidronio o fosfato, se sigue el mismo enfoque, con una modificación para acomodar la carga iónica.Para dibujar la estructura de Lewis del hidronio, calcule el número de electrones de valencia del ion primero. Puesto que el ion hidronio tiene carga positiva, elimine un electrón del total de electrones de valencia, reduciendo el recuento a 8 A continuación, ilustre la molécula por su estructura esquelética. Ahora coloque un par de electrones entre cada átomo, seguido del par solitario de electrones en el átomo central.De esta forma, los átomos de hidrógeno y oxígeno satisfacen el dúo y el octeto, respectivamente. Finalmente, escriba la estructura de Lewis entre paréntesis con una carga positiva en su esquina superior derecha.

estructuras de lewis de compuestos moleculares e iones poliatómicos

Para dibujar estructuras de Lewis para moléculas complicadas e iones moleculares, es útil seguir un procedimiento paso a paso como se describe a continuación:
 
 <ol>
 <li>Determine el número total de electrones de valencia (capa externa). Para cationes, reste un electrón por cada carga positiva. Para aniones, agregue un electrón por cada carga negativa.</li>
 <li>Dibuje una estructura de esqueleto de la molécula o ión, colocando los átomos alrededor de un átomo central. (Generalmente, el elemento menos electronegativo debe colocarse en el centro.) Conecte cada átomo al átomo central con un solo enlace (un par de electrones).</li>
 <li>Distribuya los electrones restantes como pares solitarios en los átomos terminales (excepto el hidrógeno), completando un octeto alrededor de cada átomo.</li>
 <li>Coloque todos los electrones restantes en el átomo central.</li>
 <li>Reorganice los electrones de los átomos externos para hacer múltiples enlaces con el átomo central con el fin de obtener octetos siempre que sea posible.</li>
 </ol>
 
 Por ejemplo, considere SiH4, CHO2–, NO+ y OF2 como ejemplos para los cuales esta directriz general puede ser aplicada para determinar sus estructuras de Lewis.
 
 <ol>
 <li>Determine el número total de electrones de valencia (capa externa) en la molécula o el ion. </li>
 </ol>
 
 Para una molécula como SiH4, se agrega el número de electrones de valencia en cada átomo de la molécula: 
 
 = [4 e−de valencia/átomo de Si × 1Si átomo] + [ 1 e− de valencia/átomo de H × 4H átomos] = 8 e−de valencia
 
 <h4>CHO2–</h4>
 
 Para un ión negativo, como CHO2–, el número de electrones de valencia en los átomos se añaden al número de cargas negativas en el ión (se gana un electrón por cada carga negativa):
 
 = [4 e−de valencia/ C átomo × 1C átomo] + [1 e−de valencia/ H átomo × 1H átomo] + [6 e−de valencia/ O átomo × 2O átomos] + [1 e−adicional] = 18 e−
 
 <h4>NO+ </h4>
 
 Para un ión positivo, como el NO+, se añaden el número de electrones de valencia en los átomos del ion, seguidos de la resta del número de cargas positivas en el ion (se pierde un electrón por cada carga positiva) a partir del número total de electrones de valencia:
 
 = [5 e− de valencia/N átomo × 1N átomo] + [6 e−de valencia/O átomo × 1 O átomo] + [−1 e−] = 10 e−de valencia
 
 <h4>OF2 </h4>
 
 OF2 Siendo una molécula neutra, simplemente se añade el número de electrones de valencia:
 
 = [6 e−de valencia/O átomo × 1O átomo] + [7 e−de valencia/F átomo × 2 F átomos] = 20 e−de valencia
 
 <ol start="2">
 <li>Dibuje una estructura de esqueleto de la molécula o ion, organizando los átomos alrededor de un átomo central y conectando cada átomo al átomo central con un enlace único (un par de electrones). (Tenga en cuenta que los iones son denotados con brazos alrededor de la estructura y la carga iónica fuera de los brazos:) 
 <img alt="Figure1" src="https://www.jovecore.com/CDNSource/lessons/1063/1063Image_1.svg" style="height: 80px;" /> 
 En los casos en los que son posibles varios arreglos de átomos, como en el caso del CHO2−, se utiliza evidencia experimental para elegir la correcta. En general, los elementos menos electronegativos son más propensos a ser átomos centrales. En el CHO2−, el átomo de carbono menos electronegativo ocupa la posición central con los átomos de oxígeno e hidrógeno rodeándolo. Otros ejemplos incluyen el P en el POCl3, el S en SO2 y el Cl en ClO4−.  Una excepción es que el hidrógeno casi nunca es un átomo central. Como el elemento más electronegativo, el flúor tampoco puede ser un átomo central.</li>
 <li>Distribuya los electrones restantes como pares solitarios en los átomos terminales (excepto en el hidrógeno) para completar sus capas de valencia con un octeto de electrones. (Sin electrones restantes en SiH4, su estructura no cambia.) 
 <img alt="Figure2" src="https://www.jovecore.com/CDNSource/lessons/1063/1063image_2.svg" style="height: 80px;" /></li>
 <li>Coloque todos los electrones restantes en el átomo central. 
 <ul>
 <li>Para SiH4, CHO2− y NO+, no quedan electrones. Para OF2, de los 16 electrones restantes, se colocan 12, dejando así 4 electrones para ser colocados en el átomo central: 
 <img alt="Figure3" src="https://www.jovecore.com/CDNSource/lessons/1063/1063Image_3.svg" style="height: 40px;" /></li>
 </ul>
 </li>
 <li>Reorganice los electrones de los átomos externos para formar múltiples enlaces con el átomo central, para obtener octetos en donde sea posible. 
 <ul>
 <li>SiH4: El Si ya tiene un octeto, así que no hay que hacer nada.  
 <img alt="Figure4" src="https://www.jovecore.com/CDNSource/lessons/1063/1063Image_4.svg" style="height: 70px;" /></li>
 <li>CHO2−:   Los electrones de valencia se distribuyen como pares solitarios en los átomos de oxígeno, pero el átomo de carbono carece de un octeto.   
 <img alt="Figure5" src="https://www.jovecore.com/CDNSource/lessons/1063/1063Image_5.svg" style="height: 80px;" /></li>
 <li>Por lo tanto, un par de electrones solitario es donado desde uno de los oxígenos al átomo de carbono formando un enlace doble. Dependiendo del átomo de oxígeno que donó los electrones, puede haber dos estructuras posibles, también llamadas estructuras de resonancia. </li>
 <li>NO+: Para este ion, se agregan ocho electrones de valencia, pero ninguno de los dos átomos tiene un octeto. No se pueden añadir electrones adicionales ya que los electrones totales ya están usados. En este escenario, los electrones deben ser movidos para formar múltiples enlaces. El átomo de nitrógeno tiene dos pares de electrones solitarios y el átomo de oxígeno tiene uno.  
 <img alt="Figure6" src="https://www.jovecore.com/CDNSource/lessons/1063/1063Image_6.svg" style="height: 55px;" /></li>
 <li>Esto todavía no produce un octeto, así que otro par debe ser movido para formar un enlace triple. 
  <img alt="Figure7" src="https://www.jovecore.com/CDNSource/lessons/1063/1063Image_7.svg" style="height: 50px;" /></li>
 <li>En OF2, nada cambia debido a que cada átomo ya tiene un octeto. 
 <img alt="Figure8" src="https://www.jovecore.com/CDNSource/lessons/1063/1063Image_8.svg" style="height: 40px;" /></li>
 </ul>
 </li>
 </ol>
 
 Este texto es adaptado de<a href="https://openstax.org/books/chemistry-2e/pages/7-3-lewis-symbols-and-structures">Openstax, Química 2e, Sección 7.3: Símbolos y estructuras de Lewis.</a>

Estructuras de Lewis de Compuestos Moleculares e Iones Poliatómicos

Para dibujar estructuras de Lewis para moléculas complicadas e iones moleculares, es útil seguir un procedimiento paso a paso como se describe:

Determine el ...

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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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Learn the step-by-step method for drawing Lewis structures, counting valence electrons, placing lone pairs, and handling ions, resonance, and multiple bonds.

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Drawing Lewis Structures with Valence Electron Counts

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