Lewis structures – Chemistry for the Health Sciences (2024)

The Octet Rule

The other halogen molecules (F₂, Br₂, I_2,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 theoctet 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. The transition elements and inner transition elements also do not follow the octet rule:

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:

Double and Triple Bonds

As previously mentioned, when a pair of atoms share 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 CH₂O (formaldehyde) and between the two carbon atoms in C₂H₄(ethylene):

Atriple bondforms when three electron pairs are shared by a pair of atoms, as in carbon monoxide (CO) and the cyanide ion (CN^–):

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 (VE) on each atom in the molecule:

   SiH₄(Si: 4 VE/atom × 1 atom=4 VE) + (H: 1 VE/atom × 4 atoms=4 VE) = 8 VE

   For anegative ion, such asCHO₂⁻,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):

   CHO₂⁻ (C: 4 VE/atom × 1 atom=4 VE) + (H: 1 VE/atom × 1 atom=1 VE) +
   (O: 6 VE/atom × 2 atoms=12 VE) + (1 additional electron=1 VE) = 18 VE

   For apositive 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 VE/atom × 1 atom=5 VE) + (O: 6 VE/atom × 1 atom=6 VE) −
   1 electron (positive charge)=10 VE

   See Also

   Lewis Electron Dot Structures - Detailed Explanation with Examples & VideosICl2- Lewis Structure in 6 Steps (With Images)Lewis Structure | Brilliant Math & Science Wiki10.4: Lewis Structures of Molecular Compounds

   Since OF₂is a neutral molecule, we simply add the number of valence electrons:

   OF₂ (O: 6 VE/atom × 1 atom=6 VE) + (F: 7 VE/atom × 2 atoms=14 VE)=20 VE

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 forCHO₂⁻,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 inClO₄⁻.

   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,CHO₂⁻,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.CHO₂⁻: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.

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) = 10H₃CCH₃: (1×3) + (2×4) + (1×3) = 14HCCH: (1×1) + (2×4) + (1×1) = 10NH₃: (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 remainH₃CCH₃: no electrons remainHCCH: four electrons placed on carbonNH₃: 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 bondsH₃CCH₃: all atoms have the correct number of electronsHCCH: form a triple bond between the two carbon atomsNH₃: all atoms have the correct number of electrons
   
   

Check Your Learning: Both carbon monoxide, CO, and carbon dioxide, CO₂, are products of the combustion of fossil fuels. Both of these gases also cause problems: CO is toxic and CO₂has been implicated in global climate change. What are the Lewis structures of these two molecules?

ANSWER: