Question

Write reasonable Lewis structures for the following species, none of which follow the octet rule. (a) \(\mathrm{BeCl}_{2}\) (b) \(\mathrm{SeO}_{2}^{-}\) (c) \(\mathrm{ClO}_{3}\) (d) \(\mathrm{CH}_{3}\)

Short answer

Question: Draw the Lewis structures for the following species that do not follow the octet rule: (a) BeCl2, (b) SeO2-, (c) ClO3, (d) CH3. Answer: (a) BeCl2: Be in the center with single bonds to two Cl atoms. No lone pairs. Be - Cl | Cl (b) SeO2-: Se in the center with single bonds to two O atoms and three lone pairs. Each O has three lone pairs. O - Se - O : : : : (c) ClO3: Cl in the center with single bonds to three O atoms, one O has a double bond, and one lone pair on Cl. Each O atom has three lone pairs. O - Cl - O : || : O (d) CH3: C in the center with single bonds to three H atoms and one lone pair. H - C - H : H

Step-by-step solution

Determine valence electrons and central atom

First, count the number of valence electrons present in the species and identify the central atom. The central atom is usually the least electronegative (excluding hydrogen), and it is the one to which the other atoms will bond. (a) BeCl2: Be has 2 valence electrons and each Cl has 7, total valence electrons: 2 + 7*2 = 16. Central atom: Be (b) SeO2-: Se has 6 valence electrons and each O has 6, total valence electrons: 6 + 6*2 + 1(extra electron due to the negative charge) = 19. Central atom: Se (c) ClO3: Cl has 7 valence electrons and each O has 6, total valence electrons: 7 + 6*3 = 25. Central atom: Cl (d) CH3: C has 4 valence electrons and each H has 1, total valence electrons: 4 + 1*3 = 7. Central atom: C

Arrange atoms and draw single bonds

Arrange the atoms with the central atom in the center and the other atoms around it. Draw single bonds between the central atom and the surrounding atoms. (a) Be-Cl (single bond) and Be-Cl (single bond) (b) Se-O (single bond) and Se-O (single bond) (c) Cl-O (single bond), Cl-O (single bond), and Cl-O (single bond) (d) C-H (single bond), C-H (single bond), and C-H (single bond)

Distribute remaining valence electrons

Distribute the remaining valence electrons as lone pairs around the atoms to satisfy their valence electron requirements. Remember that we are dealing with species that do not follow the octet rule. (a) Be has 4 electrons already (due to 2 single bonds) and Cl atoms have 8 electrons each (2 from the bonds and 6 as lone pairs). So, no more electrons are needed. (b) Se has 4 electrons (due to 2 single bonds); add 3 lone pairs (6 electrons) to Se. Each O atom has 2 electrons (single bond), add 3 lone pairs (6 electrons) to each O. (c) Cl has 6 electrons (due to 3 single bonds); add 1 lone pair (2 electrons) to Cl. Each O atom has 2 electrons (single bond), add 3 lone pairs (6 electrons) to each O. (d) C has 6 electrons (due to 3 single bonds); add 1 lone pair (1 electron) to C. Each H atom has 2 electrons (single bond), so no more electrons are needed.

Check if Lewis structures follow the given condition

Check if the Lewis structures do not follow the octet rule. (a) Be has only 4 electrons, so it doesn't follow the octet rule. (b) Se now has 12 electrons, so it doesn't follow the octet rule. (c) Cl has 8 electrons, all three O atoms have 8 electrons, so it follows the octet rule, but we were given that ClO3 doesn't follow it. To not follow the octet rule, we can add another bond between one of the O atoms and Cl making it a double bond. (d) C has only 7 electrons, so it doesn't follow the octet rule. Now we have drawn the reasonable Lewis structures for the given species that don't follow the octet rule.