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Writing Lewis Structures With the Octet Rule

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 SiH4, \({\text{CHO}}_{2}{}^{\text{−}},\) NO+, and OF2 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:

      \(\begin{array}{r r l} \text{SiH}_4 & & \\[1em] & \text{Si: 4 valence electrons/atom} \times 1 \;\text{atom} & = 4 \\[1em] \rule[-0.5ex]{21em}{0.1ex}\hspace{-21em} + & \text{H: 1 valence electron/atom} \times 4 \;\text{atoms} & = 4 \\[1em] & & = 8 \;\text{valence electrons} \end{array}\)

    • For a negative ion, such as \({\text{CHO}}_{2}{}^{\text{−}},\) 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):

      \(\begin{array}{r r l} {\text{CHO}_2}^{-} & & \\[1em] & \text{C: 4 valence electrons/atom} \times 1 \;\text{atom} & = 4 \\[1em] & \text{H: 1 valence electron/atom} \times 1 \;\text{atom} & = 1 \\[1em] & \text{O: 6 valence electrons/atom} \times 2 \;\text{atoms} & = 12 \\[1em] \rule[-0.5ex]{21.5em}{0.1ex}\hspace{-21.5em} + & 1\;\text{additional electron} & = 1 \\[1em] & & = 18 \;\text{valence electrons} \end{array}\)

    • 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:

      \(\begin{array}{r r l} \text{NO}^{+} & & \\[1em] & \text{N: 5 valence electrons/atom} \times 1 \;\text{atom} & = 5 \\[1em] & \text{O: 6 valence electrons/atom} \times 1 \;\text{atom} & = 6 \\[1em] \rule[-0.5ex]{21em}{0.1ex}\hspace{-21em} + & -1 \;\text{electron (positive charge)} & = -1 \\[1em] & & = 10 \;\text{valence electrons} \end{array}\)

    • Since OF2 is a neutral molecule, we simply add the number of valence electrons:
      \(\begin{array}{r r l} \text{OF}_{2} & & \\[1em] & \text{O: 6 valence electrons/atom} \times 1 \;\text{atom} & = 6 \\[1em] \rule[-0.5ex]{21em}{0.1ex}\hspace{-21em} + & \text{F: 7 valence electrons/atom} \times 2 \;\text{atoms} & = 14 \\[1em] & & = 20 \;\text{valence electrons} \end{array}\)
  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 \({\text{CHO}}_{2}{}^{\text{−}},\) we must use experimental evidence to choose the correct one. In general, the less electronegative elements are more likely to be central atoms. In \({\text{CHO}}_{2}{}^{\text{−}},\) the less electronegative carbon atom occupies the central position with the oxygen and hydrogen atoms surrounding it. Other examples include P in POCl3, S in SO2, and Cl in \({\text{ClO}}_{4}{}^{\text{−}}.\) 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 SiH4, 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 SiH4, \({\text{CHO}}_{2}{}^{\text{−}},\) and NO+, there are no remaining electrons; we already placed all of the electrons determined in Step 1.
    • For OF2, 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.
    • SiH4: Si already has an octet, so nothing needs to be done.
    • \({\text{CHO}}_{2}{}^{\text{−}}:\) 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 OF2, each atom has an octet as drawn, so nothing changes.

Example: Writing Lewis Structures

NASA’s Cassini-Huygens mission detected a large cloud of toxic hydrogen cyanide (HCN) on Titan, one of Saturn’s moons. Titan also contains ethane (H3CCH3), acetylene (HCCH), and ammonia (NH3). What are the Lewis structures of these molecules?

Solution

  1. Calculate the number of valence electrons.
    HCN: (1 \(×\) 1) + (4 \(×\) 1) + (5 \(×\) 1) = 10
    H3CCH3: (1 \(×\) 3) + (2 \(×\) 4) + (1 \(×\) 3) = 14
    HCCH: (1 \(×\) 1) + (2 \(×\) 4) + (1 \(×\) 1) = 10
    NH3: (5 \(×\) 1) + (3 \(×\) 1) = 8
  2. Draw a skeleton and connect the atoms with single bonds. Remember that H is never a central atom:
    Four Lewis structures are shown. The first structure shows a carbon atom single bonded to a hydrogen atom and a nitrogen atom. The second structure shows two carbon atoms single bonded to one another. Each is single bonded to three hydrogen atoms. The third structure shows two carbon atoms single bonded to one another and each single bonded to one hydrogen atom. The fourth structure shows a nitrogen atom single bonded to three hydrogen atoms.
  3. Where needed, distribute electrons to the terminal atoms:
    Four Lewis structures are shown. The first structure shows a carbon atom single bonded to a hydrogen atom and a nitrogen atom, which has three lone pairs of electrons. The second structure shows two carbon atoms single bonded to one another. Each is single bonded to three hydrogen atoms. The third structure shows two carbon atoms single bonded to one another and each single bonded to one hydrogen atom. The fourth structure shows a nitrogen atom single bonded to three hydrogen atoms.
    HCN: six electrons placed on N
    H3CCH3: no electrons remain
    HCCH: no terminal atoms capable of accepting electrons
    NH3: no terminal atoms capable of accepting electrons
  4. Where needed, place remaining electrons on the central atom:
    Four Lewis structures are shown. The first structure shows a carbon atom single bonded to a hydrogen atom and a nitrogen atom, which has three lone pairs of electrons. The second structure shows two carbon atoms single bonded to one another. Each is single bonded to three hydrogen atoms. The third structure shows two carbon atoms, each with a lone pair of electrons, single bonded to one another and each single bonded to one hydrogen atom. The fourth structure shows a nitrogen atom with a lone pair of electrons single bonded to three hydrogen atoms.
    HCN: no electrons remain
    H3CCH3: no electrons remain
    HCCH: four electrons placed on carbon
    NH3: two electrons placed on nitrogen
  5. 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
    Four Lewis structures are shown. The first structure shows a carbon atom single bonded to a hydrogen atom and a nitrogen atom, which has three lone pairs of electrons. Two curved arrows point from the nitrogen to the carbon. Below this structure is the word “gives” and below that is the same structure, but this time there is a triple bond between the carbon and nitrogen. The second structure shows two carbons single bonded to one another and each single bonded to three hydrogen atoms. The third structure shows two carbon atoms, each with a lone pair of electrons, single bonded to one another and each single bonded to one hydrogen atom. Two curved arrows point from the carbon atoms to the space in between the two. Below this structure is the word “gives” and the same structure, but this time with a triple bond between the two carbons. The fourth structure shows a nitrogen atom with a lone pair of electrons single bonded to three hydrogen atoms.

Fullerene Chemistry

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 Chemistry was awarded to Richard Smalley (see the figure below), Robert Curl, and Harold Kroto for their work in discovering a new form of carbon, the C60 buckminsterfullerene molecule (see this lesson).

An entire class of compounds, including spheres and tubes of various shapes, were discovered based on C60. This type of molecule, called a fullerene, shows promise in a variety of applications. 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 chemical sensors.

A photo of Richard Smalley is shown.

Richard Smalley (1943–2005), a professor of physics, chemistry, and astronomy at Rice University, was one of the leading advocates for fullerene chemistry. Upon his death in 2005, the US Senate honored him as the “Father of Nanotechnology.” (credit: United States Department of Energy)

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