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Two sets of Lewis dot structures are shown. The left structures depict five C l symbols in a cross shape with eight dots around each, the word “or” and the same five C l symbols, connected by four single bonds in a cross shape. The name “Carbon tetrachloride” is written below the structure. The right hand structures show a S i symbol, surrounded by eight dots and four H symbols in a cross shape. The word “or” separates this from an S i symbol with four single bonds connecting the four H symbols in a cross shape. The name “Silane” is written below these diagrams.

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 NH 3 (ammonia). Oxygen and other atoms in group 16 obtain an octet by forming two covalent bonds:

Three Lewis structures labeled, “Ammonia,” “Water,” and “Hydrogen fluoride” are shown. The left structure shows a nitrogen atom with a lone pair of electrons and single bonded to three hydrogen atoms. The middle structure shows an oxygen atom with two lone pairs of electrons and two singly-bonded hydrogen atoms. The right structure shows a hydrogen atom single bonded to a fluorine atom that has three lone pairs of electrons.

Double and triple bonds

As previously mentioned, when a pair of atoms shares one pair of electrons, we call this a single bond. However, a pair of atoms may need to share more than one pair of electrons in order to achieve the requisite octet. A double bond    forms when two pairs of electrons are shared between a pair of atoms, as between the carbon and oxygen atoms in CH 2 O (formaldehyde) and between the two carbon atoms in C 2 H 4 (ethylene):

Two pairs of Lewis structures are shown. The left pair of structures shows a carbon atom forming single bonds to two hydrogen atoms. There are four electrons between the C atom and an O atom. The O atom also has two pairs of dots. The word “or” separates this structure from the same diagram, except this time there is a double bond between the C atom and O atom. The name, “Formaldehyde” is written below these structures. A right-facing arrow leads to two more structures. The left shows two C atoms with four dots in between them and each forming single bonds to two H atoms. The word “or” lies to the left of the second structure, which is the same except that the C atoms form double bonds with one another. The name, “Ethylene” is written below these structures.

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 ):

Two pairs of Lewis structures are shown and connected by a right-facing arrow. The left pair of structures show a C atom and an O atom with six dots in between them and a lone pair on each. The word “or” and the same structure with a triple bond in between the C atom and O atom also are shown. The name “Carbon monoxide” is written below this structure. The right pair of structures show a C atom and an N atom with six dots in between them and a lone pair on each. The word “or” and the same structure with a triple bond in between the C atom and N atom also are shown. The name “Cyanide ion” is written below this structure.

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:

Three reactions are shown with Lewis dot diagrams. The first shows a hydrogen with one red dot, a plus sign and a bromine with seven dots, one of which is red, connected by a right-facing arrow to a hydrogen and bromine with a pair of red dots in between them. There are also three lone pairs on the bromine. The second reaction shows a hydrogen with a coefficient of two and one red dot, a plus sign, and a sulfur atom with six dots, two of which are red, connected by a right facing arrow to two hydrogen atoms and one sulfur atom. There are two red dots in between the two hydrogen atoms and the sulfur atom. Both pairs of these dots are red. The sulfur atom also has two lone pairs of dots. The third reaction shows two nitrogen atoms each with five dots, three of which are red, separated by a plus sign, and connected by a right-facing arrow to two nitrogen atoms with six red electron dots in between one another. Each nitrogen atom also has one lone pair of electrons.

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

  1. Determine the total number of valence (outer shell) electrons. For cations, subtract one electron for each positive charge. For anions, add one electron for each negative charge.
  2. 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).
  3. Distribute the remaining electrons as lone pairs on the terminal atoms (except hydrogen), completing an octet around each atom.
  4. Place all remaining electrons on the central atom.
  5. 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 SiH 4 , CHO 2 , NO + , and OF 2 as examples in following this procedure:

  1. 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:
      SiH 4 Si: 4 valence electrons/atom × 1 atom = 4 + H: 1 valence electron/atom × 4 atoms = 4 ¯ = 8 valence electrons
    • For a negative ion , such as CHO 2 , we add the number of valence electrons on the atoms to the number of negative charges on the ion (one electron is gained for each single negative charge):
      CHO 2 C: 4 valence electrons/atom × 1 atom = 4 H: 1 valence electron/atom × 1 atom = 1 O: 6 valence electrons/atom × 2 atoms = 12 + 1 additional electron = 1 ¯ = 18 valence electrons
    • For a positive ion , such as NO + , we add the number of valence electrons on the atoms in the ion and then subtract the number of positive charges on the ion (one electron is lost for each single positive charge) from the total number of valence electrons:
      NO + N: 5 valence electrons/atom × 1 atom = 5 O: 6 valence electron/atom × 1 atom = 6 + −1 electron (positive charge) = −1 ¯ = 10 valence electrons
    • Since OF 2 is a neutral molecule, we simply add the number of valence electrons:
      OF 2 O: 6 valence electrons/atom × 1 atom = 6 + F: 7 valence electrons/atom × 2 atoms = 14 ¯ = 20 valence electrons
  2. 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:)
    Four Lewis diagrams are shown. The first shows one silicon single boned to four hydrogen atoms. The second shows a carbon which forms a single bond with an oxygen and a hydrogen and a double bond with a second oxygen. This structure is surrounded by brackets and has a superscripted negative sign near the upper right corner. The third structure shows a nitrogen single bonded to an oxygen and surrounded by brackets with a superscripted plus sign in the upper right corner. The last structure shows two fluorine atoms single bonded to a central oxygen.
    When several arrangements of atoms are possible, as for CHO 2 , we must use experimental evidence to choose the correct one. In general, the less electronegative elements are more likely to be central atoms. In CHO 2 , the less electronegative carbon atom occupies the central position with the oxygen and hydrogen atoms surrounding it. Other examples include P in POCl 3 , S in SO 2 , and Cl in ClO 4 . An exception is that hydrogen is almost never a central atom. As the most electronegative element, fluorine also cannot be a central atom.
  3. Distribute the remaining electrons as lone pairs on the terminal atoms (except hydrogen) to complete their valence shells with an octet of electrons.
    • There are no remaining electrons on SiH 4 , so it is unchanged:
      Four Lewis structures are shown. The first shows one silicon single boned to four hydrogen atoms. The second shows a carbon single bonded to two oxygen atoms that each have three lone pairs and single bonded to a hydrogen. This structure is surrounded by brackets and has a superscripted negative sign near the upper right corner. The third structure shows a nitrogen single bonded to an oxygen, each with three lone pairs of electrons. This structure is surrounded by brackets with a superscripted plus sign in the upper right corner. The last structure shows two fluorine atoms, each with three lone pairs of electrons, single bonded to a central oxygen.
  4. Place all remaining electrons on the central atom.
    • For SiH 4 , CHO 2 , and NO + , there are no remaining electrons; we already placed all of the electrons determined in Step 1.
    • For OF 2 , we had 16 electrons remaining in Step 3, and we placed 12, leaving 4 to be placed on the central atom:
      A Lewis structure shows two fluorine atoms, each with three lone pairs of electrons, single bonded to a central oxygen which has two lone pairs of electrons.
  5. Rearrange the electrons of the outer atoms to make multiple bonds with the central atom in order to obtain octets wherever possible.
    • SiH 4 : Si already has an octet, so nothing needs to be done.
    • CHO 2 : We have distributed the valence electrons as lone pairs on the oxygen atoms, but the carbon atom lacks an octet:
      Two Lewis diagrams are shown with the word “gives” in between them. The left diagram, surrounded by brackets and with a superscripted negative sign, shows a carbon atom single bonded to two oxygen atoms, each with three lone pairs of electrons. The carbon atom also forms a single bond with a hydrogen atom. A curved arrow points from a lone pair on one of the oxygen atoms to the carbon atom. The right diagram, surrounded by brackets and with a superscripted negative sign, shows a carbon atom single bonded to an oxygen atom with three lone pairs of electrons, double bonded to an oxygen atom with two lone pairs of electrons, and single bonded to a hydrogen atom.
    • NO + : For this ion, we added eight valence 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:
      Two Lewis diagrams are shown with the word “gives” in between them. The left diagram, surrounded by brackets and with a superscripted positive sign, shows a nitrogen atom single bonded to an oxygen atom, each with two lone pairs of electrons. The right diagram, surrounded by brackets and with a superscripted positive sign, shows a nitrogen atom double bonded to an oxygen atom. The nitrogen atom has two lone pairs of electrons and the oxygen atom has one.
      This still does not produce an octet, so we must move another pair, forming a triple bond:
      A Lewis structure shows a nitrogen atom with one lone pair of electrons triple bonded to an oxygen with a lone pair of electrons. The structure is surrounded by brackets and has a superscripted positive sign.
    • In OF 2 , each atom has an octet as drawn, so nothing changes.
Practice Key Terms 9

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Source:  OpenStax, Chemistry. OpenStax CNX. May 20, 2015 Download for free at http://legacy.cnx.org/content/col11760/1.9
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