The shape of the molecule of Carbon(IV) oxide is

1. Analyze the question

The question asks for the geometric shape of the Carbon(IV) oxide ($CO_2$) molecule. To determine this, we need to look at the bonding arrangement around the central carbon atom.

2. Explain the concept

The shape of a molecule is determined by the Valence Shell Electron Pair Repulsion (VSEPR) theory. This theory states that electron pairs around a central atom will arrange themselves as far apart as possible to minimize repulsion.

• Step 1: Draw the Lewis structure. Carbon is in Group 4, so it has 4 valence electrons. Oxygen is in Group 6, so it has 6 valence electrons. In $CO_2$, the carbon atom forms a double bond with each of the two oxygen atoms ($O=C=O$).
• Step 2: Count the electron domains. Around the central carbon atom, there are two double bonds. In VSEPR theory, double bonds are treated as a single "domain" or region of electron density.
• Step 3: Determine the geometry. With two electron domains around the central atom and no lone pairs on the carbon, the electron pairs push each other to opposite sides to achieve a bond angle of 180°. This results in a straight-line geometry.

3. Show the solution

• Central Atom: Carbon (C)
• Bonding: Two double bonds to Oxygen ($O=C=O$)
• Lone Pairs on Carbon: 0
• Arrangement: The two double bonds repel each other equally, forcing the molecule into a straight line.
• Bond Angle: 180°

Since there are no lone pairs on the central carbon to distort the shape, the molecule remains perfectly straight.

4. Identify the answer

Comparing our findings to the options: A. Tetrahedral (e.g., $CH_4$) B. Pyramidal (e.g., $NH_3$) C. Planar (e.g., $BF_3$) D. Linear (e.g., $CO_2$)

The correct shape is linear.

5. Tip

Memory Trick: Think of "Linear" as a straight line. Any molecule with a central atom bonded to only two other atoms and no lone pairs (like $CO_2$, $BeCl_2$, or $CS_2$) will always be linear with a 180° bond angle.