10th Grade: Chemical Bonding ⚛️🔗

Interactive Lesson • $\text{Ionic}$ $\text{vs.}$ $\text{Covalent}$ • $\text{VSEPR}$ $\text{Shapes}$

The $\text{Octet}$ $\text{Rule}$ 🤝

  $\text{Chemical}$ $\text{Bonds}$ form because atoms seek to achieve a stable electron configuration, typically by having **eight** $\text{valence}$ $\text{electrons}$ (the **$\text{Octet}$ $\text{Rule}$**). They do this by either $\text{transferring}$ ($\text{Ionic}$) or $\text{sharing}$ ($\text{Covalent}$) electrons.

 
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Two $\text{Main}$ $\text{Types}$ of $\text{Bonds}$ ⚖️

 
   

$\text{Ionic}$ $\text{Bond}$

   

Action: $\text{Transfer}$ of $\text{electrons}$ from one atom to another.

   

Elements: $\text{Metal}$ ($\text{loses } \text{e}^-$ to become $\text{cation}$) and $\text{Nonmetal}$ ($\text{gains } \text{e}^-$ to become $\text{anion}$).

   

Result: $\text{Electrostatic}$ $\text{attraction}$ between opposite charges.

 
 
   

$\text{Covalent}$ $\text{Bond}$

   

Action: $\text{Sharing}$ of $\text{valence}$ $\text{electrons}$ between atoms.

   

Elements: Typically two $\text{Nonmetals}$.

   

Result: Forms a discrete $\text{molecule}$.

 
 
   

$\text{Electronegativity}$ $\text{Difference}$

   

Bonds are $\text{Ionic}$ if $\Delta\text{EN} > 1.7$.

   

Bonds are $\text{Polar}$ $\text{Covalent}$ if $0.4 < \Delta\text{EN} \le 1.7$.

   

Bonds are $\text{Nonpolar}$ $\text{Covalent}$ if $\Delta\text{EN} \le 0.4$.

 

VSEPR $\text{Theory}$ ($\text{Molecular}$ $\text{Shapes}$) 📐

  **$\text{VSEPR}$** ($\text{Valence}$-$\text{Shell}$ $\text{Electron}$-$\text{Pair}$ $\text{Repulsion}$) $\text{Theory}$ states that electron pairs around a central atom repel each other, arranging themselves as far apart as possible to minimize repulsion. This dictates the shape of the molecule.

 
   

$\text{Linear}$

   

2 bonding pairs, 0 lone pairs. $\text{Example}$: $\text{CO}_2$

   

Angle: $180^\circ$

 
 
   

$\text{Trigonal}$ $\text{Planar}$

   

3 bonding pairs, 0 lone pairs. $\text{Example}$: $\text{BF}_3$

   

Angle: $120^\circ$

 
 
   

$\text{Tetrahedral}$

   

4 bonding pairs, 0 lone pairs. $\text{Example}$: $\text{CH}_4$

   

Angle: $109.5^\circ$

 
 
   

$\text{Bent}$ (or $\text{V}$-$\text{shaped}$)

   

2 bonding pairs, 2 lone pairs. $\text{Example}$: $\text{H}_2\text{O}$

   

Angle: $< 109.5^\circ$ (due to lone pair repulsion)

 

Practice Problems ($\text{Bond}$ $\text{Type}$ $\text{Check}$)

Based on the elements involved, classify the bond as **$\text{Ionic}$** ($\text{Metal}$ + $\text{Nonmetal}$) or **$\text{Covalent}$** ($\text{Nonmetal}$ + $\text{Nonmetal}$).

 
   
1) $\text{NaCl}$ ($\text{Sodium}$ $\text{Chloride}$)
   
               
   
 
 
   
2) $\text{CO}_2$ ($\text{Carbon}$ $\text{Dioxide}$)
   
               
   
 
 
   
3) $\text{MgF}_2$ ($\text{Magnesium}$ $\text{Fluoride}$)
   
               
   
 
 
   
4) $\text{NH}_3$ ($\text{Ammonia}$)
   
               
   
 

⚡ $\text{VSEPR}$ $\text{Shape}$ $\text{Challenge}$!

Match the description to the correct $\text{VSEPR}$ $\text{molecular}$ $\text{shape}$ (click the correct tile).

 
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Molecule with 4 bonding groups and 0 lone pairs.
 
     
 
 

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