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To draw Lewis structures for complicated molecules and molecular ions, it is helpful to follow a step-by-step procedure as outlined:

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.

For instance, consider SiH4, CHO2, NO+, and OF2 as examples for which this general guideline can be applied to determine their Lewis structures.

1. Determine the total number of valence (outer shell) electrons in the molecule or ion.

For a molecule like SiH4, the number of valence electrons on each atom in the molecule is added:

= [4 valence e/Si atom × 1 Si atom] + [ 1 valence e/H atom × 4 H atoms] = 8 valence e

#### CHO2–

For a negative ion, such as  CHO2, the numbers of valence electrons on the atoms are added to the number of negative charges on the ion (one electron is gained for each single negative charge):

= [4 valence e/C atom × 1 C atom] + [1 valence e/H atom × 1 H atom] + [6 valence e/O atom × 2 O atoms] + [1 additional e] = 18 valence e

#### NO+

For a positive ion, such as NO+, the numbers of valence electrons on the atoms in the ion are added, followed by subtraction of the number of positive charges on the ion (one electron is lost for every single positive charge) from the total number of valence electrons:

= [5 valence e/N atom × 1 N atom] + [6 valence e/O atom × 1 O atom] + [−1 e] = 10 valence e

#### OF2

OF2 being a neutral molecule, the number of valence electrons are simply added:

= [6 valence e/O atom × 1 O atom] + [7 valence e/F atom × 2 F atoms] = 20 valence e

1. 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 ions are denoted with brackets around the structure, and the ionic charge outside the brackets:)

In cases where several arrangements of atoms are possible, as for CHO2, experimental evidence is used to choose the correct one. In general, the less electronegative elements are more likely to be central atoms. In  CHO2,  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  ClO4.  An exception is that hydrogen is almost never a central atom. As the most electronegative element, fluorine also cannot be a central atom.
2. Distribute the remaining electrons as lone pairs on the terminal atoms (except hydrogen) to complete their valence shells with an octet of electrons. (With no remaining electrons on SiH4, its structure is unchanged.)
3. Place all remaining electrons on the central atom.
• For SiH4,  CHO2,  and NO+, there are no remaining electrons. For OF2, of the 16 electrons remaining, 12 are placed, thereby leaving 4 electrons to be placed on the central atom:
4. Rearrange the electrons of the outer atoms to make multiple bonds with the central atom, to obtain octets wherever possible.
• SiH4: Si already has an octet, so nothing needs to be done.
• CHO2:  The valence electrons are distributed as lone pairs on the oxygen atoms, but the carbon atom lacks an octet.
• Hence, one lone pair of electrons is donated from one of the oxygen to the carbon atom forming a double bond. Depending on which oxygen atom donated the electrons, there can be two possible structures, otherwise called resonance structures.
• NO+: For this ion, eight valence electrons are added, but neither atom has an octet. Additional electrons cannot be added since the total electrons are already used up. In this scenario, electrons must be moved to form multiple bonds. The nitrogen atom has two lone pairs of electrons and the oxygen atom has one.
• This still does not produce an octet, so another pair must be moved to form a triple bond.

• In OF2, nothing changes as each atom already has an octet.

This text is adapted from Openstax, Chemistry 2e, Section 7.3: Lewis Symbols and Structures.

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