Name: drawing lewis structures with valence electron counts
Uploaded: 2020-09-03T13:38:28.000Z
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Chemistry

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Lewis Structures of Molecular Compounds and Polyatomic Ions

ルイス構造は 原子間の化学結合を 簡略化したものです エテンやアンモニアのような 中性化合物や ヒドロニウムやリン酸塩のような 多原子イオンの ルイス構造を書くには次のような 一連の手順を踏む必要があります まず 分子内の価電子の総数を 計算します 例えばエテンを考えてみましょう 炭素原子には 4個の原子価電子があり 水素原子には1個の 原子価電子があるので 炭素原子が2個 水素原子が4個の場合 エテンは合計12個の 原子価電子を持っています 次に 分子の骨格構造を 描きましょう 一般的には 化学式の 最初の原子 または 電子数の少ない原子が 中心に配置されています この原子は中心原子とも呼ばれ 少なくとも2つの結合を 形成することが できる必要があります 水素原子または それ以上の電子陰性度の原子は 中心原子の周りに 末端的に配置されています ここでは 炭素原子は 水素原子を 取り囲んで 中心位置を占めています 次のステップは オクテット補完である まず 2つの原子の間に 電子の対を配置して 結合対を表します 残りの電子は デュエットまたは オクテット構成を満たすために 末端原子と中央原子に ローンペアとして 分配されます この場合 水素は デュエットに達しているので 残りの電子は炭素原子に 転送されます 次にオクテットを欠いた原子間に 複数の結合を形成するために ローンペアを使用します エテンでは 炭素原子は まだオクテットに達していません そのためローンペアを結合領域に 移動して二重結合を形成し すべての原子が 安定した電子配置に 到達することを可能になります ヒドロニウムやリン酸塩のような 電荷を帯びた多原子イオンの ルイス構造を書くには イオン電荷に対応するために 修正を加えて 同じアプローチに従います ヒドロニウムの ルイス構造を書くには まず イオンの価電子の数を 計算します ヒドロニウムイオンは 正の電荷を持っているので 全原子価電子から 電子を1個取り除き 8個に減らします 次に 分子の骨格構造を 図示してみましょう ここで すべての原子の間に 電子の対を置き 次に中心の原子の上に 電子の対を置きます このようにして 水素原子と 酸素原子は それぞれデュエットと オクテットを満たしています 最後に ルイス構造を 右上に正の電荷を 括弧で書きます

分子化合物及び多原子イオンのルイス構造

複雑な分子や分子イオンのルイス構造を描くには、以下の手順に従うといいでしょう。 
 
 <ol>
 <li>価（外殻）電子の総数を求めます。陽イオンの場合、正電荷1個につき電子1個を引きます。陰イオンの場合は、負の電荷1個につき電子1個を加えます。 </li>
 <li>分子やイオンの骨格構造を、中心となる原子を中心に原子を配置して描いてみます。(一般的には、電気陰性度の低い元素を中心に配置します。) 各原子と中心原子を単結合（1つの電子対）で結びます。 </li>
 <li>残りの電子を孤立電子対として末端原子（水素を除く）に分配し、各原子の周りにオクテットを完成させます。 </li>
 <li>残りのすべての電子を中心原子に配置します。 </li>
 <li>可能な限りオクテットになるように、外側の原子の電子を並べ替えて中心原子と多重結合させます。 </li>
 </ol>
 
 例えば、この一般的なガイドラインを適用してルイス構造を決定する例として、SiH4、CHO2–、NO+、OF2を検討します。 
 
 <ol>
 <li>分子またはイオンの原価 ( 外殻 ) 電子の総数を求めます。  </li>
 </ol>
 
 SiH4 のような分子では、分子の各原子の価電子の数が加算されます。
 
 = [4 valence e − / Si Atom × 1 Si atom] + [1 valence e − /H atom × 4 H atoms] = 8 valence e − 
 
 <h4>CHO2 - </h4>
 
 CHO2-のような陰イオンの場合、原子の価電子の数とイオンの負電荷の数が加算されます（1つの負電荷に対して1つの電子が得られる）。 
 
 = [4 valence e − /C atom × 1 C atom ] + [1 valence e − /H atom × 1 H atom ] + [6 valence e − /O atom × 2 O atoms] + [1 additional e − ] = 18 valence e − 
 
 <h4>NO + </h4>
 
 NO+のような正イオンの場合、イオンを構成する原子の価電子の数を加算した後、価電子の総数からイオンの正電荷の数（正電荷1個につき電子1個が失われる）を減算します。 
 
 = [5 valence e − /N atom × 1 N atom ] + [6 valence e − /O atom × 1 O atom ] + [-1 e − ] = 10 valence e − 
 
 <h4>OF 2</h4>
 
 OF2は中性分子なので、価電子の数は単純に加算されます。 
 
 = [6 valence e−/O atom × 1 O atom] + [7 valence e−/F atom × 2 F atoms] = 20 valence e−
 
 <ol start="2">
 <li>分子またはイオンの骨格構造を描く。中心原子の周りに原子を配置し、各原子と中心原子を単結合（1電子対）で結びます。(なお、イオンは構造体を括弧で囲み、イオン電荷は括弧の外に記載しています：）。 
 <img alt="Figure1" src="https://www.jovecore.com/CDNSource/lessons/1063/1063Image_1.svg" style="height: 80px;" /> 
 CHO2−のように複数の原子配置が可能な場合は、実験的証拠をもとに正しいものを選択します。一般的には、電気陰性度の低い元素が中心原子になりやすいと言われています。CHO2−では、電気陰性度の低い炭素原子が中心に位置し、その周りを酸素原子と水素原子が取り囲んでいます。他の例としては、POCl3のP、SO2のS、ClO4−のClなどがあります。例外として、水素が中心原子になることはほとんどありません。最も電気陰性な元素であるフッ素も中心原子にはなれません。 </li>
 <li>残りの電子を孤立電子対として末端原子（水素を除く）に分配し、原子価殻に8個の電子を完成させます。(SiH4には電子が残っていないので、構造は変わらない) 
 <img alt="Figure2" src="https://www.jovecore.com/CDNSource/lessons/1063/1063image_2.svg" style="height: 80px;" /></li>
 <li>残りのすべての電子を中心原子に配置します
 <ul>
 <li>SiH4、CHO2−、NO+では、残りの電子はありません。OFでは、16個の電子のうち12個が配置され、4個の電子が中心原子に配置されます。 
 <img alt="Figure3" src="https://www.jovecore.com/CDNSource/lessons/1063/1063Image_3.svg" style="height: 40px;" /></li>
 </ul>
 </li>
 <li>外側の原子の電子を並べ替えて中心の原子と多重結合させ、可能な限りオクテットにします。
 <ul>
 <li>SiH4 ： Si にはすでにオクテットがあるので、何もする必要はありません。   
 <img alt="Figure4" src="https://www.jovecore.com/CDNSource/lessons/1063/1063Image_4.svg" style="height: 70px;" /></li>
 <li>CHO2 −：価電子は酸素原子上で孤立電子対として分配されますが、炭素原子にはオクテットがありません。   
 <img alt="Figure5" src="https://www.jovecore.com/CDNSource/lessons/1063/1063Image_5.svg" style="height: 80px;" /></li>
 <li>そのため、一方の酸素から炭素原子に孤立電子が1つ提供され、二重結合が形成されます。どちらの酸素原子から電子が提供されたかによって、2つの構造が考えられます。  </li>
 <li>NO+ ：このイオンでは、8個の価電子が追加されていますが、どちらの原子もオクテットを持っていません。電子の総量はすでに使い切っているので、さらに電子を追加することはできません。このシナリオでは、多重結合を形成するために電子を移動させる必要があります。窒素原子には2つの孤立電子対があり、酸素原子には1つあります。   
 <img alt="Figure6" src="https://www.jovecore.com/CDNSource/lessons/1063/1063Image_6.svg" style="height: 55px;" /></li>
 <li>これでもオクテットは生成されないため、別のペアを移動して三重結合を形成する必要があります。 
  <img alt="Figure7" src="https://www.jovecore.com/CDNSource/lessons/1063/1063Image_7.svg" style="height: 50px;" /></li>
 <li>OF2では、各原子がすでにオクテットを持っているため、何も変化しません。 
 <img alt="Figure8" src="https://www.jovecore.com/CDNSource/lessons/1063/1063Image_8.svg" style="height: 40px;" /></li>
 </ul>
 </li>
 </ol>
 
 このテキストは <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 から引用しています。</a>

複雑な分子や分子イオンのルイス構造を描くには、次のように段階的な手順に従うと便利です。

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Chemical Bonding: Basic Concepts

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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.

Drawing Lewis Structures with Valence Electron Counts | JoVE K12

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Physical Science

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Matter and Its Interactions

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Drawing Lewis Structures with Valence Electron Counts

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