Q. Draw the Lewis structure of \( \mathrm{CH_4} \).

We want the Lewis structure for methane, \( \mathrm{CH_4} \).

Step 1: Count the total valence electrons.

Carbon is in group 14, so it has \(4\) valence electrons.

Hydrogen is in group 1, so each hydrogen has \(1\) valence electron.

For \( \mathrm{CH_4} \):

\[ \text{Total valence electrons} = 4 + 4(1) = 8 \]

Step 2: Choose the central atom.

Carbon is the central atom because it can form bonds to multiple hydrogens.

So we place \( \mathrm{C} \) in the center and four \( \mathrm{H} \) atoms around it.

Step 3: Place bonds to use the electrons.

We want carbon to satisfy the octet rule. Carbon typically forms four covalent bonds.

Each single \( \mathrm{C-H} \) bond uses \(2\) electrons (one from carbon and one from hydrogen).

With four single bonds, we will use all \(8\) valence electrons:

\[ 4 \text{ bonds} \times 2 \text{ electrons per bond} = 8 \]

Step 4: Complete the Lewis structure and check octets/duets.

In the final structure:

• Carbon makes four single bonds, so it has \(8\) electrons around it (an octet).
• Each hydrogen forms one single bond and ends up with \(2\) electrons around it (a duet).

There are no leftover electrons, so there are no lone pairs.

Final Lewis structure of \( \mathrm{CH_4} \).

Draw carbon in the center with four single bonds to hydrogen:

\[ \mathrm{H{-}C{-}H} \]

Continue with two more bonds on the remaining sides to total four \( \mathrm{C-H} \) bonds, giving the tetrahedral arrangement.

In a simple diagram form:

\[ \begin{matrix}
\mathrm{\ \ H} \\
\ \ \ \ \ \ \ \ \vert \\
\mathrm{H – C – H} \\
\ \ \ \ \ \ \ \ \vert \\
\mathrm{\ \ H}
\end{matrix} \]

This structure shows four single \( \mathrm{C-H} \) bonds and no lone pairs on carbon or hydrogen.