Name: drawing lewis structures with valence electron counts
Uploaded: 2020-09-03T13:38:28.000Z
Duration: 2 min 54 s
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K12 Standards

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North Carolina K-12 Science Standard Course of Study

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Chemistry

Matter and its Interactions

PS.Chm.3.1

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Lewis Structures of Molecular Compounds and Polyatomic Ions

Struktury Lewisa są uproszczonymi reprezentacjami wiązań chemicznych między atomami. Pisanie struktur Lewisa dla związków obojętnych, takich jak eten lub amoniak, lub dla jonów wieloatomowych, takich jak hydroniów lub fosforan, obejmuje następującą sekwencję kroków.

Najpierw oblicz całkowitą liczbę elektronów walencyjnych w cząsteczce. Weźmy na przykład eten. Każdy atom węgla ma 4 elektrony walencyjne, a każdy atom wodoru ma 1 elektron walencyjny; Z 2 atomami węgla i 4 atomami wodoru eten ma w sumie 12 elektronów walencyjnych.

Następnie narysuj strukturę szkieletową cząsteczki. Ogólnie rzecz biorąc, pierwszy atom we wzorze chemicznym lub mniej elektroujemny atom jest umieszczony centralnie. Atom ten nazywany jest również atomem centralnym i musi być w stanie utworzyć co najmniej dwa wiązania. 

Atomy wodoru lub więcej atomów elektroujemnych są umieszczone na końcu wokół atomu centralnego. Tutaj atomy węgla zajmują centralne pozycje, a otaczające je atomy wodoru są otoczone.

Kolejnym krokiem jest kompletacja oktetu. Najpierw umieść parę elektronów między każdymi dwoma atomami, aby reprezentować pary wiązań. Pozostałe elektrony są rozprowadzane jako samotne pary do atomów końcowych i centralnych, aby spełnić konfiguracje duetu lub oktetu. W tym przypadku, ponieważ wodór osiągnął duet, pozostałe elektrony są przenoszone do atomów węgla. 

Teraz użyj samotnych par, aby utworzyć wiele wiązań między atomami, które nie mają oktetu. W etenie atomy węgla nie osiągnęły jeszcze oktetu. Dlatego samotna para jest przenoszona do obszaru wiązania, aby utworzyć podwójne wiązanie, dzięki czemu wszystkie atomy osiągają stabilną konfigurację elektronową. 

Aby zapisać strukturę Lewisa dla naładowanych jonów wieloatomowych, takich jak hydronim lub fosforan, stosuje się to samo podejście, z modyfikacją w celu uwzględnienia ładunku jonowego. 

Aby narysować strukturę Lewisa hydronii, najpierw oblicz liczbę elektronów walencyjnych na jonie. Ponieważ jon hydroniowy ma ładunek dodatni, usuń elektron z całkowitej liczby elektronów walencyjnych, zmniejszając liczbę do 8. Następnie zilustruj cząsteczkę na podstawie jej struktury szkieletowej. 

Teraz umieść parę elektronów między każdym atomem, a następnie samotną parę elektronów na centralnym atomie. W ten sposób atomy wodoru i tlenu zaspokajają odpowiednio duet i oktet. Na koniec napisz strukturę Lewisa w nawiasach kwadratowych z ładunkiem dodatnim w prawym górnym rogu.

struktury lewisa związków molekularnych i jonów wieloatomowych

Aby narysować struktury Lewisa dla skomplikowanych cząsteczek i jonów molekularnych, warto postępować zgodnie z następującą procedurą:

<ol style="display: flex; flex-direction: column; gap: 10px;">
	<li>Określ całkowitą liczbę elektronów walencyjnych (powłoki zewnętrznej). W przypadku kationów odejmij jeden elektron na każdy ładunek dodatni. W przypadku anionów dodaj jeden elektron na każdy ładunek ujemny.</li>
	<li>Narysuj szkieletową strukturę cząsteczki lub jonu, układając atomy wokół centralnego atomu (zasadniczo element najmniej elektroujemny powinien być umieszczony w środku). Połącz każdy atom z atomem centralnym wiązaniem pojedynczym jedną parą elektronów.</li>
	<li>Rozłóż pozostałe elektrony jako wolne pary na atomach końcowych (z wyjątkiem wodoru), tworząc oktet wokół każdego atomu.</li>
	<li>Umieść wszystkie pozostałe elektrony na atomie centralnym.</li>
	<li>Zmień układ elektronów atomów zewnętrznych, aby utworzyć wiązania wielokrotne z atomem centralnym, aby uzyskać oktety, jeśli to możliwe.</li>
</ol>

Na przykład rozważ SiH4, CHO2–, NO+ i OF2 jako przykłady, do których można zastosować niniejsze ogólne wytyczne w celu określenia ich struktur Lewisa.

Określ całkowitą liczbę elektronów walencyjnych (powłoki zewnętrznej) w cząsteczce lub jonie.

W przypadku cząsteczki takiej jak SiH4 dodaje się liczbę elektronów walencyjnych na każdym atomie w cząsteczce:

= [4 wartościowość e−/atom Si × 1 atom Si] + [ 1 wartościowość e−/atom H × 4 atomy H] = 8 wartościowość e−

<h4>CHO2–</h4>

W przypadku jonu ujemnego, takiego jak CHO2–, liczbę elektronów walencyjnych na atomach dodaje się do liczby ładunków ujemnych na jonie (na każdy pojedynczy ładunek ujemny zyskuje się jeden elektron):

= [4 wartościowość e-/atom C × 1 atom C] + [1 wartościowość e-/atom H × 1 atom H] + [6 wartościowość e-/atom O × 2 atomy O] + [1 dodatkowy e-] = 18 wartościowość e−

<h4>NIE+</h4>

W przypadku jonu dodatniego, takiego jak NO+, dodaje się liczbę elektronów walencyjnych na atomach jonu, a następnie odejmuje liczbę ładunków dodatnich na jonie (na każdy pojedynczy ładunek dodatni traci się jeden elektron) od całkowitej liczby elektronów walencyjnych:

= [5 wartościowość e−/N atom × 1 N atom] + [6 wartościowość e−/O atom × 1 atom O] + [−1 e−] = 10 wartościowość e−

<h4>OF2</h4>

OF2 jest cząsteczką obojętną, po prostu dodaje się liczbę elektronów walencyjnych:

= [6 wartościowości atom e−/O × 1 atom O] + [7 wartościowość e−/atom F × 2 atomy F] = 20 wartościowość e−

Narysuj szkieletową strukturę cząsteczki lub jonu, układając atomy wokół centralnego atomu i łącząc każdy atom z centralnym atomem pojedynczym wiązaniem (jedna para elektronów). Zauważ, że jony są oznaczone nawiasami wokół struktury, a ładunek jonowy znajduje się poza nawiasami.

<img alt="Figure1" src="https://www.jove.com/files/core/CDNSource/lessons/1063/1063Image_1.svg" style="height: 80px;">

W przypadkach, gdy możliwych jest kilka układów atomów, jak w przypadku CHO2-, do wybrania właściwego układu wykorzystuje się dowody eksperymentalne. Ogólnie rzecz biorąc, pierwiastki mniej elektroujemne są raczej atomami centralnymi. W CHO2− mniej elektroujemny atom węgla zajmuje centralną pozycję wraz z otaczającymi go atomami tlenu i wodoru. Inne przykłady obejmują P w POCl3, S w SO2 i Cl w ClO4−. Wyjątkiem jest to, że wodór prawie nigdy nie jest atomem centralnym. Jako pierwiastek najbardziej elektroujemny, fluor również nie może być atomem centralnym.

Rozłóż pozostałe elektrony jako wolne pary na atomach końcowych (z wyjątkiem wodoru), aby uzupełnić ich powłoki walencyjne oktetem elektronów. Bez pozostałych elektronów na SiH4 jego struktura pozostaje niezmieniona.

<img alt="Figure2" src="https://www.jove.com/files/core/CDNSource/lessons/1063/1063image_2.svg" style="height: 80px;">

Umieść wszystkie pozostałe elektrony na atomie centralnym.

W przypadku SiH4, CHO2− i NO+ nie ma pozostałych elektronów. W przypadku OF2 z pozostałych 16 elektronów umieszcza się 12, pozostawiając w ten sposób 4 elektrony do umieszczenia na atomie centralnym:

<img alt="Figure3" src="https://www.jove.com/files/core/CDNSource/lessons/1063/1063Image_3.svg" style="height: 40px;">

Zmień rozmieszczenie elektronów atomów zewnętrznych, aby utworzyć wiązania wielokrotne z atomem centralnym, aby uzyskać oktety, jeśli to możliwe.

W SiH4, Si ma już oktet, więc nic nie trzeba zmieniać.

<img alt="Figure4" src="https://www.jove.com/files/core/CDNSource/lessons/1063/1063Image_4.svg" style="height: 70px;">

CHO2-: Elektrony walencyjne są rozmieszczone jako samotne pary na atomach tlenu, ale atom węgla nie ma oktetu.

<img alt="Figure5" src="https://www.jove.com/files/core/CDNSource/lessons/1063/1063Image_5.svg" style="height: 80px;">

Jedna wolna para elektronów jest przekazywana z jednego tlenu do atomu węgla, tworząc wiązanie podwójne. W zależności od tego, który atom tlenu oddał elektrony, mogą istnieć dwie możliwe struktury, zwane inaczej strukturami rezonansowymi.

NIE+: Do tego jonu dodaje się osiem elektronów walencyjnych, ale żaden atom nie ma oktetu. Nie można dodać dodatkowych elektronów, ponieważ cała ilość elektronów została już wykorzystana. W tym scenariuszu elektrony muszą zostać przesunięte, aby utworzyć wiązania wielokrotne. Atom azotu ma dwie wolne pary elektronów, a atom tlenu jedną.

<img alt="Figure6" src="https://www.jove.com/files/core/CDNSource/lessons/1063/1063Image_6.svg" style="height: 55px;">

To nadal nie tworzy oktetu, więc należy przesunąć kolejną parę, aby utworzyć wiązanie potrójne.

<img alt="Figure7" src="https://www.jove.com/files/core/CDNSource/lessons/1063/1063Image_7.svg" style="height: 50px;">

W OF2 nic się nie zmienia, ponieważ każdy atom ma już oktet.

<img alt="Figure8" src="https://www.jove.com/files/core/CDNSource/lessons/1063/1063Image_8.svg" style="height: 40px;">

Struktury Lewisa związków molekularnych i jonów wieloatomowych

Aby narysować struktury Lewisa dla skomplikowanych cząsteczek i jonów molekularnych, pomocne jest postępowanie zgodnie z opisaną procedurą krok po kroku:

Określ ...

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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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Compared to ionic bonds, which results from the transfer of electrons between metallic and nonmetallic atoms, covalent bonds result from the mutual ...

Electronegativity

Whether a bond is nonpolar or polar covalent is determined by a property of the bonding atoms called electronegativity.&#xA0;

Electronegativity values of ...

Bond Polarity, Dipole Moment, and Percent Ionic Character

Bond Polarity

The absolute value of the difference in electronegativity (&Delta;EN) of two bonded atoms provides a rough measure of the polarity to be ...

Analyze and interpret data to explain the mechanisms and properties of the two main types of intramolecular (ionic and covalent) bonds.

Analyze the structure of atoms and isotopes.

Drawing Lewis Structures with Valence Electron Counts

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