Q. \( \mathrm{C_2H_6} \) shape.

Goal: Determine the molecular “shape” (geometry) of the substance with formula \( \mathrm{C_2H_6} \), which is ethane.

Step 1: Identify the central atom(s).

In ethane, \( \mathrm{C_2H_6} \), the carbon atoms are connected to each other. Each carbon can be treated as a central atom when considering its local geometry.

Step 2: Count valence electron domains around one carbon.

Focus on one carbon (say the first carbon). Its bonding pattern is:

• It is bonded to the other carbon by a single bond.
• It is bonded to three hydrogen atoms by single bonds.

So, around the carbon, there are four regions of electron density (electron domains):

• 1 bond to carbon
• 3 bonds to hydrogen

Therefore, the arrangement is \( 4 \) electron domains and there are no lone pairs on carbon.

Step 3: Use VSEPR theory to predict the molecular geometry.

With \( 4 \) electron domains and \( 0 \) lone pairs, the electron-domain geometry is tetrahedral, and the molecular shape is also tetrahedral.

For a carbon center:

\[
\mathrm{C}\text{ has }4\text{ sigma bonds} \Rightarrow \text{tetrahedral geometry}
\]

Step 4: Describe the shape for the whole molecule.

Ethane has two tetrahedral carbon centers connected by a single bond. Each carbon is locally tetrahedral, meaning each carbon has bond angles close to the tetrahedral angle.

Step 5: State the key bond angle.

For tetrahedral geometry, the ideal bond angle is:

\[
\theta \approx 109.5^\circ
\]

Final Answer (molecular shape of \( \mathrm{C_2H_6} \)).

\( \mathrm{C_2H_6} \) (ethane) consists of two \( sp^3 \) (tetrahedral) carbon centers; each carbon has a tetrahedral geometry with approximately \( 109.5^\circ \) bond angles.