Lewis Structures

We also use Lewis symbols to indicate the formation of covalent bonds, which are shown in Lewis structures, drawings that describe the bonding in molecules and polyatomic ions. For example, when two chlorine atoms form a chlorine molecule, they share one pair of electrons:

*

The Lewis structure indicates that each Cl atom has three pairs of electrons that are not used in bonding (called lone pairs) and one shared pair of electrons (written between the atoms). A dash (or line) is usually used to indicate a shared pair of electrons:

*

In the Lewis model, a single shared pair of electrons is a single bond. Each Cl atom interacts with eight valence electrons total: the six in the lone pairs and the two in the single bond.

You are watching: What is the central atom in a lewis structure



The Octet Rule

The other halogen molecules (F2, Br2, I2, and At2) form bonds like those in the chlorine molecule: one single bond between atoms and three lone pairs of electrons per atom. This allows each halogen atom to have a noble gas electron configuration. The tendency of main group atoms to form enough bonds to obtain eight valence electrons is known as the octet rule.

The number of bonds that an atom can form can often be predicted from the number of electrons needed to reach an octet (eight valence electrons); this is especially true of the nonmetals of the second period of the periodic table (C, N, O, and F). For example, each atom of a group 14 element has four electrons in its outermost shell and therefore requires four more electrons to reach an octet. These four electrons can be gained by forming four covalent bonds, as illustrated here for carbon in CCl4 (carbon tetrachloride) and silicon in SiH4 (silane). Because hydrogen only needs two electrons to fill its valence shell, it is an exception to the octet rule and only needs to form one bond. The transition elements and inner transition elements also do not follow the octet rule since they have d and f electrons involved in their valence shells.

*

Group 15 elements such as nitrogen have five valence electrons in the atomic Lewis symbol: one lone pair and three unpaired electrons. To obtain an octet, these atoms form three covalent bonds, as in NH3 (ammonia). Oxygen and other atoms in group 16 obtain an octet by forming two covalent bonds:

*
api/deki/files/66735/CNX_les-grizzlys-catalans.org_07_03_DoubleBond_img.jpg?revision=1" />

A triple bond forms when three electron pairs are shared by a pair of atoms, as in carbon monoxide (CO) and the cyanide ion (CN–):

*



Writing Lewis Structures with the Octet Rule

For very simple molecules and molecular ions, we can write the Lewis structures by merely pairing up the unpaired electrons on the constituent atoms. See these examples:

*

For more complicated molecules and molecular ions, it is helpful to follow the step-by-step procedure outlined here:

Determine the total number of valence (outer shell) electrons among all the atoms. For cations, subtract one electron for each positive charge. For anions, add one electron for each negative charge. Draw a skeleton structure of the molecule or ion, arranging the atoms around a central atom. (Generally, the least electronegative element should be placed in the center.) Connect each atom to the central atom with a single bond (one electron pair). Distribute the remaining electrons as lone pairs on the terminal atoms (except hydrogen), completing an octet around each atom. Place all remaining electrons on the central atom. Rearrange the electrons of the outer atoms to make multiple bonds with the central atom in order to obtain octets wherever possible.

Let us determine the Lewis structures of SiH4, (ceCHO2-), NO+, and OF2 as examples in following this procedure:

Determine the total number of valence (outer shell) electrons in the molecule or ion. For a molecule, we add the number of valence electrons on each atom in the molecule:

(eginalign &phantom+ceSiH4\ &phantom+ extrmSi: 4 valence electrons/atom × 1 atom = 4\ &underline extrm+H: 1 valence electron/atom × 4 atoms = 4\ &hspace271px extrm= 8 valence electrons endalign)

(ceCHO2-\ extrmC: 4 valence electrons/atom × 1 atom hspace6px= phantom14\ extrmH: 1 valence electron/atom × 1 atom hspace12px= phantom11\ extrmO: 6 valence electrons/atom × 2 atoms = 12\ underline+hspace100px extrm1 additional electron hspace9px= phantom11\ hspace264px extrm= 18 valence electrons)
(ceNO+\ extrmN: 5 valence electrons/atom × 1 atom = phantom−5\ extrmO: 6 valence electron/atom × 1 atomhspace5px = phantom−6\ underline extrm+ −1 electron (positive charge) hspace44px= −1\ hspace260px extrm= 10 valence electrons)
Since OF2 is a neutral molecule, we simply add the number of valence electrons:
Draw a skeleton structure of the molecule or ion, arranging the atoms around a central atom and connecting each atom to the central atom with a single (one electron pair) bond. (Note that we denote ions with brackets around the structure, indicating the charge outside the brackets:)
*
*

Rearrange the electrons of the outer atoms to make multiple bonds with the central atom in order to obtain octets wherever possible. SiH4: Si already has an octet, so nothing needs to be done. (ceCHO2-): We have distributed the valence electrons as lone pairs on the oxygen atoms, but the carbon atom lacks an octet:

*

NO+: For this ion, we added eight outer electrons, but neither atom has an octet. We cannot add any more electrons since we have already used the total that we found in Step 1, so we must move electrons to form a multiple bond:

*

In OF2, each atom has an octet as drawn, so nothing changes.
api/deki/files/142168/imageedit_3_8055985152.png?revision=1" />

Where needed, distribute electrons to the terminal atoms:

*

HCN: no electrons remain H3CCH3: no electrons remain HCCH: four electrons placed on carbon NH3: two electrons placed on nitrogen

Where needed, rearrange electrons to form multiple bonds in order to obtain an octet on each atom:

HCN: form two more C–N bonds H3CCH3: all atoms have the correct number of electrons HCCH: form a triple bond between the two carbon atoms NH3: all atoms have the correct number of electrons

*



api/deki/files/142169/imageedit_33_8502017829.png?revision=1" />


Fullerene les-grizzlys-catalans.orgistry

Carbon soot has been known to man since prehistoric times, but it was not until fairly recently that the molecular structure of the main component of soot was discovered. In 1996, the Nobel Prize in les-grizzlys-catalans.orgistry was awarded to Richard Smalley, Robert Curl, and Harold Kroto for their work in discovering a new form of carbon, the C60 buckminsterfullerene molecule. An entire class of compounds, including spheres and tubes of various shapes, were discovered based on C60. This type of molecule, called a fullerene, consists of a complex network of single- and double-bonded carbon atoms arranged in such a way that each carbon atom obtains a full octet of electrons. Because of their size and shape, fullerenes can encapsulate other molecules, so they have shown potential in various applications from hydrogen storage to targeted drug delivery systems. They also possess unique electronic and optical properties that have been put to good use in solar powered devices and les-grizzlys-catalans.orgical sensors.

See more: Lab: How Is The Surface Tension Of Water Affected By Soap ? How Does Detergent Break Surface Tension


api/deki/files/66748/CNX_les-grizzlys-catalans.org_07_03_NOsingle_img.jpg?revision=1&size=bestfit&width=114&height=61" />
Place all remaining electrons on the central atom. Since there are no remaining electrons, this step does not apply. Rearrange the electrons to make multiple bonds with the central atom in order to obtain octets wherever possible. We know that an odd-electron molecule cannot have an octet for every atom, but we want to get each atom as close to an octet as possible. In this case, nitrogen has only five electrons around it. To move closer to an octet for nitrogen, we take one of the lone pairs from oxygen and use it to form a NO double bond. (We cannot take another lone pair of electrons on oxygen and form a triple bond because nitrogen would then have nine electrons:)

*

An atom like the boron atom in BF3, which does not have eight electrons, is very reactive. It readily combines with a molecule containing an atom with a lone pair of electrons. For example, NH3 reacts with BF3 because the lone pair on nitrogen can be shared with the boron atom: