Lewis Dot Structure For K

Affiliate 9. Chemical Bonds

Lewis Electron Dot Diagrams

Depict a Lewis electron dot diagram for an atom or a monatomic ion.

In almost all cases, chemical bonds are formed by interactions of valence electrons in atoms. To facilitate our understanding of how valence electrons collaborate, a simple manner of representing those valence electrons would be useful.

A (or electron dot diagram or a Lewis diagram or a Lewis structure) is a representation of the valence electrons of an atom that uses dots around the symbol of the chemical element. The number of dots equals the number of valence electrons in the atom. These dots are arranged to the right and left and to a higher place and below the symbol, with no more than two dots on a side. (It does non matter what society the positions are used.) For example, the Lewis electron dot diagram for hydrogen is simply:

$\LARGE \ce{\Lewis{0.,H}}$

Because the side is non important, the Lewis electron dot diagram could also exist fatigued equally follows:

$\Lewis{2.,H}\hspace{10 mm}\text{or}\hspace{10 mm}\Lewis{4.,H}\hspace{10 mm}\text{or}\hspace{10 mm}\Lewis{6.,H}$

The electron dot diagram for helium, with two valence electrons, is equally follows:

$\LARGE \ce{\Lewis{0:,He}}$

By putting the two electrons together on the same side, nosotros emphasize the fact that these two electrons are both in the 1s subshell; this is the common convention we will adopt, although there will exist exceptions later. The next atom, lithium, has an electron configuration of 1due south ^twotwos ^one, so information technology has only 1 electron in its valence shell. Its electron dot diagram resembles that of hydrogen, except the symbol for lithium is used:

$\LARGE \ce{\Lewis{0.,Li}}$

Beryllium has two valence electrons in its 2s shell, so its electron dot diagram is like that of helium:

$\LARGE \Lewis{0:,Be}$

The side by side atom is boron. Its valence electron shell is iis ^two2p ¹, then it has iii valence electrons. The third electron will keep another side of the symbol:

$\LARGE \Lewis{0:2.,B}$

Again, it does not thing on which sides of the symbol the electron dots are positioned.

For carbon, in that location are four valence electrons, two in the twosouthward subshell and two in the 2p subshell. As usual, we will draw two dots together on one side, to represent the 2south electrons. Nonetheless, conventionally, we draw the dots for the two p electrons on dissimilar sides. Equally such, the electron dot diagram for carbon is as follows:

$\huge \Lewis{0:2.4.,C}$

With nitrogen, which has three p electrons, we put a unmarried dot on each of the three remaining sides:

$\huge \Lewis{0:2.4.6.,N}$

For oxygen, which has four p electrons, we at present have to outset doubling upwards on the dots on one other side of the symbol. When doubling up electrons, brand certain that a side has no more than two electrons.

$\huge \Lewis{0:2:4.6.,O}$

Fluorine and neon have 7 and eight dots, respectively:

$\LARGE \Lewis{0:2:4:6.,F}\hspace{10mm}\Lewis{0:2:4:6:,Ne}$

With the adjacent element, sodium, the process starts over with a unmarried electron because sodium has a unmarried electron in its highest-numbered shell, the n = 3 shell. By going through the periodic table, we encounter that the Lewis electron dot diagrams of atoms will never have more eight dots around the atomic symbol.

Problem

What is the Lewis electron dot diagram for each element?

aluminum
selenium

Solution

The valence electron configuration for aluminum is 3due south ⁱⁱ3p ¹. So it would take three dots effectually the symbol for aluminum, two of them paired to stand for the 3s electrons:
$\LARGE \Lewis{0:2.,Al}$
The valence electron configuration for selenium is 4s ²4p ⁴. In the highest-numbered shell, the n = 4 shell, there are 6 electrons. Its electron dot diagram is equally follows:
$\huge \Lewis{0:2.4.6:,Se}$

Test Yourself

What is the Lewis electron dot diagram for each element?

phosphorus
argon

Reply

$\LARGE \Lewis{0:2.4.6.,P}\hspace{10mm}\Lewis{0:2:4:6:,Ar}$

For atoms with partially filled d or f subshells, these electrons are typically omitted from Lewis electron dot diagrams. For example, the electron dot diagram for iron (valence beat out configuration ivs ^twothreed ^half-dozen) is as follows:

$\LARGE \Lewis{0:,Fe}$

Elements in the same column of the periodic table accept like Lewis electron dot diagrams considering they have the same valence shell electron configuration. Thus the electron dot diagrams for the first column of elements are every bit follows:

$\large \Lewis{0.,H}\hspace{10mm}\Lewis{0.,Li}\hspace{10mm}\Lewis{0.,Na}\hspace{10mm}\Lewis{0.,K}\hspace{10mm}\Lewis{0.,Rb}\hspace{10mm}\Lewis{0.,Cs}\hspace{10mm}$

Monatomic ions are atoms that take either lost (for cations) or gained (for anions) electrons. Electron dot diagrams for ions are the same every bit for atoms, except that some electrons take been removed for cations, while some electrons accept been added for anions. Thus in comparing the electron configurations and electron dot diagrams for the Na cantlet and the Na⁺ ion, we note that the Na atom has a single valence electron in its Lewis diagram, while the Na⁺ ion has lost that one valence electron:

$\begin{array}{lll} \text{Lewis dot diagram:}&\Lewis{0.,Na}&\ce{Na^+} \\ \text{Electron configuration:}&\ce{[Ne]}3s^1&\ce{[Ne]} \end{array}$

Technically, the valence shell of the Na⁺ ion is now the n = 2 beat, which has eight electrons in it. Then why exercise we not put 8 dots effectually Na⁺? Conventionally, when nosotros show electron dot diagrams for ions, we testify the original valence trounce of the atom, which in this case is the n = iii shell and empty in the Na⁺ ion.

In making cations, electrons are showtime lost from the highest numbered trounce, non necessarily the concluding subshell filled. For case, in going from the neutral Fe atom to the Iron²⁺ ion, the Fe cantlet loses its ii 4s electrons first, not its 3d electrons, despite the fact that the threed subshell is the concluding subshell existence filled. Thus we have:

$\begin{array}{lll} \text{Lewis dot diagram:}&\Lewis{0:,Fe}&\ce{Fe^{2+}} \\ \text{Electron configuration:}&\ce{[Ar]}4s^23d^6&\ce{[Ar]}3d^6 \end{array}$

Anions have extra electrons when compared to the original atom. Here is a comparison of the Cl atom with the Cl⁻ ion:

$\begin{array}{lll} \text{Lewis dot diagram:}&\Lewis{0.2:4:6:,Cl}&\ce{\Lewis{0:2:4:6:,Cl}^-} \\ \text{Electron configuration:}&\ce{[Ne]}3s^23p^5&\ce{[Ne]}3s^23p^6 \end{array}$

Problem

What is the Lewis electron dot diagram for each ion?

Ca²⁺
O^two−

Solution

Having lost its two original valence electrons, the Lewis electron dot diagram is just Ca^two+.
The O²⁻ ion has gained two electrons in its valence shell, and then its Lewis electron dot diagram is as follows:
$\large \ce{\Lewis{0:2:4:6:,O}}^{2-}$

Exam Yourself

The valence electron configuration of thallium, whose symbol is Tl, is 6south ²5d ¹⁰half dozenp ^ane. What is the Lewis electron dot diagram for the Tl⁺ ion?

Answer

$\ce{\Lewis{0:,Tl}^+}$

Lewis electron dot diagrams use dots to represent valence electrons effectually an diminutive symbol.
Lewis electron dot diagrams for ions have fewer (for cations) or more (for anions) dots than the corresponding atom.