Question

chemistry i – introduction to lewis structures
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.

directions: using the steps above (on a separate sheet of paper), name the molecule and draw the lewis structure (lines and dots) for each of the following compounds:

molecular formula	rough draft	electron dot diagram	lewis structure
f₂
h₂s
geh₄
asbr₃
sio₂

Answer:

To solve for the Lewis structures of each compound, we follow the 5 - step procedure for Lewis structures:

1. For \( \boldsymbol{HCl} \)

• Step 1: Determine total valence electrons
• Hydrogen (\( H \)) has 1 valence electron, Chlorine (\( Cl \)) has 7 valence electrons.
• Total valence electrons \( = 1 + 7=8 \).
• Step 2: Draw skeleton structure
• The less electronegative atom ( \( H \)) is bonded to \( Cl \) with a single bond: \( H - Cl \).
• Step 3: Distribute remaining electrons as lone pairs on terminal atoms
• After the single bond (2 electrons), we have \( 8 - 2 = 6 \) electrons left. These are placed as lone pairs on \( Cl \) to complete its octet: \( H-\ddot{\underset{\cdot\cdot}{Cl}} \).
• Step 4: Place remaining electrons on central atom
• There are no remaining electrons to place on the central atom ( \( Cl \) already has an octet from the lone pairs and the bond).
• Step 5: Rearrange electrons for multiple bonds (not needed here)
• \( HCl \) has a single bond and \( Cl \) has an octet, so no multiple bonds are required.

2. For \( \boldsymbol{F_2} \)

• Step 1: Determine total valence electrons
• Each Fluorine (\( F \)) atom has 7 valence electrons. For \( F_2 \), total valence electrons \( = 7\times2 = 14 \).
• Step 2: Draw skeleton structure
• The two \( F \) atoms are bonded with a single bond: \( F - F \).
• Step 3: Distribute remaining electrons as lone pairs on terminal atoms
• After the single bond (2 electrons), we have \( 14 - 2=12 \) electrons left. These are distributed as lone pairs on each \( F \) atom. Each \( F \) atom gets 3 lone pairs (6 electrons) to complete their octets: \( \ddot{\underset{\cdot\cdot}{F}}-\ddot{\underset{\cdot\cdot}{F}} \).
• Step 4: Place remaining electrons on central atom
• There are no remaining electrons to place on the central atom (both \( F \) atoms have octets).
• Step 5: Rearrange electrons for multiple bonds (not needed here)
• The octets are already satisfied with a single bond and lone pairs.

3. For \( \boldsymbol{H_2S} \)

• Step 1: Determine total valence electrons
• Hydrogen (\( H \)) has 1 valence electron (2 atoms, so \( 2\times1 = 2 \) valence electrons), Sulfur (\( S \)) has 6 valence electrons.
• Total valence electrons \( = 2 + 6=8 \).
• Step 2: Draw skeleton structure
• Sulfur (\( S \)) is the central atom (less electronegative than \( H \) in terms of bonding for this molecule) and is bonded to two \( H \) atoms: \( H - S - H \).
• Step 3: Distribute remaining electrons as lone pairs on terminal atoms
• After the two single bonds (4 electrons), we have \( 8 - 4 = 4 \) electrons left. These are placed as two lone pairs on \( S \) to complete its octet: \( H-\ddot{S}-H \).
• Step 4: Place remaining electrons on central atom
• The remaining electrons (the two lone pairs) are already placed on the central atom (\( S \)).
• Step 5: Rearrange electrons for multiple bonds (not needed here)
• The octet of \( S \) is satisfied with the two single bonds and two lone pairs.

4. For \( \boldsymbol{GeH_4} \)

• Step 1: Determine total valence electrons
• Germanium (\( Ge \)) has 4 valence electrons, Hydrogen (\( H \)) has 1 valence electron (4 atoms, so \( 4\times1 = 4 \) valence electrons).
• Total valence electrons \( = 4+4 = 8 \).
• Step 2: Draw skeleton structure
• Germanium (\( Ge \)) is the central atom and is bonded to four \( H \) atoms: \( H - Ge - H \) (with two more \( H \) atoms bonded to \( Ge \) in a tetrahedral arrangement, but for Lewis structure, we can represent it as \( H_3Ge - H \) or simply show the four single bonds).
• Step 3: Distribute remaining electrons as lone pairs on terminal atoms
• After the four single bonds (8 electrons), there are no remaining electrons to distribute as lone pairs on the terminal (\( H \)) atoms ( \( H \) only needs 2 electrons for a duet, which is satisfied by the single bond).
• Step 4: Place remaining electrons on central atom
• There are no remaining electrons to place on the central atom (\( Ge \) has a complete octet from the four single bonds).
• Step 5: Rearrange electrons for multiple bonds (not needed here)
• The octet of \( Ge \) is satisfied with the four single bonds.

5. For \( \boldsymbol{AsBr_3} \)

• Step 1: Determine total valence electrons
• Arsenic (\( As \)) has 5 valence electrons, Bromine (\( Br \)) has 7 valence electrons (3 atoms, so \( 3\times7 = 21 \) valence electrons).
• Total valence electrons \( = 5+21 = 26 \).
• Step 2: Draw skeleton structure
• Arsenic (\( As \)) is the central atom and is bonded to three \( Br \) atoms with single bonds: \( Br - As - Br \) (and one more \( Br \) bonded to \( As \)).
• Step 3: Distribute remaining electrons as lone pairs on terminal atoms
• After the three single bonds (6 electrons), we have \( 26 - 6=20 \) electrons left. Each \( Br \) atom needs 6 more electrons to complete its octet. For 3 \( Br \) atoms, we use \( 3\times6 = 18 \) electrons.
• After placing lone pairs on \( Br \) atoms: \( \ddot{\underset{\cdot\cdot}{Br}}-As-\ddot{\underset{\cdot\cdot}{Br}} \) (with the third \( Br \) also having 6 lone pair electrons: \( \ddot{\underset{\cdot\cdot}{Br}}-As-\ddot{\underset{\cdot\cdot}{Br}}-\ddot{\underset{\cdot\cdot}{Br}} \)).
• Step 4: Place remaining electrons on central atom
• We have used \( 6 + 18=24 \) electrons, so we have \( 26 - 24 = 2 \) electrons left. These are placed as a lone pair on \( As \): \( \ddot{\underset{\cdot\cdot}{Br}}-\underset{\cdot\cdot}{\ddot{As}}-\ddot{\underset{\cdot\cdot}{Br}}-\ddot{\underset{\cdot\cdot}{Br}} \).
• Step 5: Rearrange electrons for multiple bonds (not needed here)
• The octet of \( As \) (5 + 3 (bonds) + 2 (lone pair) = 10, which is an expanded octet, but this is acceptable for elements in the third period and below) and the octets of \( Br \) are satisfied.

6. For \( \boldsymbol{SiO_2} \)

• Step 1: Determine total valence electrons
• Silicon (\( Si \)) has 4 valence electrons, Oxygen (\( O \)) has 6 valence electrons (2 atoms, so \( 2\times6 = 12 \) valence electrons).
• Total valence electrons \( = 4 + 12=16 \).
• Step 2: Draw skeleton structure
• Silicon (\( Si \)) is the central atom and is bonded to two \( O \) atoms: \( O - Si - O \).
• Step 3: Distribute remaining electrons as lone pairs on terminal atoms
• After the two single bonds (4 electrons), we have \( 16 - 4 = 12 \) electrons left. We start by placing 6 electrons (3 lone pairs) on each \( O \) atom. But if we do this, \( Si \) will only have 4 electrons (from the two single bonds), which is less than an octet.
• Step 4: Place remaining electrons on central atom
• Since \( Si \) does not have an octet, we need to form multiple bonds.
• Step 5: Rearrange electrons for multiple bonds
• We convert one lone pair from each \( O \) atom into a double bond with \( Si \). The Lewis structure becomes \( O = Si = O \), with each \( O \) atom having two remaining lone pairs: \( \ddot{O}=Si=\ddot{O} \). In this structure, \( Si \) has an octet (4 electrons from the two double bonds) and each \( O \) has an octet (4 electrons from the double bond and 4 electrons from the two lone pairs).

Naming the Molecules

• \( HCl \): Hydrochloric acid (or Hydrogen chloride)
• \( F_2 \): Fluorine
• \( H_2S \): Hydrogen sulfide
• \( GeH_4 \): Germanium hydride (or Germane)
• \( AsBr_3 \): Arsenic tribromide
• \( SiO_2 \): Silicon dioxide

Lewis Structures Summary

Molecular Formula	Rough Draft (Skeleton + Lone Pairs)	Electron Dot Diagram (Lone Pairs as Dots, Bonds as Lines)	Lewis Structure (Formal)
\( F_2 \)	\( \ddot{\underset{\cdot\cdot}{F}}-\ddot{\underset{\cdot\cdot}{F}} \)	\( :\ddot{\underset{\cdot\cdot}{F}}-\ddot{\underset{\cdot\cdot}{F}}: \)	\( :\ddot{F}-\ddot{F}: \)
\( H_2S \)	\( H-\ddot{S}-H \)	\( H:\ddot{S}:H \)	\( H - \ddot{S}-H \)
\( GeH_4 \)	\( H_3Ge - H \) (with 4 single bonds)	\( H:H:Ge:H:H \) (simplified, showing bonds as electron pairs)	\( H_3Ge - H \) (tetrahedral with 4 single bonds)
\( AsBr_3 \)	\( \ddot{\underset{\cdot\cdot}{Br}}-\underset{\cdot\cdot}{\ddot{As}}-\ddot{\underset{\cdot\cdot}{Br}}-\ddot{\underset{\cdot\cdot}{Br}} \)	\( :\ddot{Br}:\underset{\cdot\cdot}{\ddot{As}}:\ddot{Br}: :\ddot{Br}: \) (bonds as electron pairs between atoms)	\( \ddot{Br}-As-\ddot{Br}-\ddot{Br} \) (with lone pairs on \( As \) and \( Br \))
\( SiO_2 \)	\( \ddot{O}=Si=\ddot{O} \)	\( :\ddot{O}=Si=\ddot{O}: \)	\( O = Si = O \) (with lone pairs on \( O \))

The key steps for each compound involve calculating valence electrons, drawing the skeleton, distributing lone pairs, and forming multiple bonds when necessary to satisfy the octet rule.