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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:

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:)
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:
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:
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:
• 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:

This still does not produce an octet, so we must move another pair, forming a triple bond:

• 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:
3. Where needed, distribute electrons to the terminal 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:
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

### 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.

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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