Chemical Bonding: Question 7

Worked solution

Part (a): How each C=O double bond forms

Carbon has $4$ outer-shell electrons, and it uses all $4$ of them to form two double bonds, one to each oxygen atom. Each double bond consists of two shared electron pairs between carbon and that oxygen atom:

• One $\sigma$ bond, formed by direct, end-on overlap of orbitals along the C–O axis.
• One $\pi$ bond, formed by sideways overlap of unhybridised $p$ orbitals above and below the axis.

So $\boxed{2}$ electron pairs are shared between carbon and each oxygen atom (one $\sigma$ pair and one $\pi$ pair), giving $4$ shared pairs in total across the whole molecule and leaving carbon with no lone pairs.

Part (b): Shape and bond angle of CO₂

For VSEPR purposes, a double bond (however many electron pairs it actually contains) counts as one region of electron density, because all the electrons of that double bond point in the same direction, towards the same oxygen atom.

Carbon therefore has: $\text{Number of regions} = \underbrace{1}_{\text{C=O bond 1}} + \underbrace{1}_{\text{C=O bond 2}} + \underbrace{0}_{\text{lone pairs}} = 2$

By VSEPR theory, $2$ regions of electron density repel each other as far apart as possible, i.e. to opposite sides of the carbon atom. This gives a linear molecule, with an O–C–O bond angle of $\boxed{180^\circ}$.

Part (c): Why CO₂ has no overall dipole moment

Oxygen is more electronegative than carbon, so each C=O bond is polar: the shared electrons are pulled towards the oxygen atom, giving each oxygen a small $\delta^-$ charge and the carbon atom a $\delta^+$ character along each bond direction. Each individual bond therefore has a bond dipole pointing from C towards that O atom.

However, from part (b), the molecule is linear and symmetric: the two oxygen atoms are on exactly opposite sides of the carbon atom, $180^\circ$ apart. The two bond dipoles are:

• Equal in magnitude (both C=O bonds are identical).
• Opposite in direction (one points left, the other points right, along the same line).

Two equal and opposite dipoles cancel exactly, so the vector sum of the two bond dipoles is zero. This means $\text{CO}_2$ has no net (overall) dipole moment, even though the individual bonds within it are polar. The molecule is non-polar overall purely because of its symmetric linear shape.

Final answers

• (a) Each C=O bond shares $\boxed{2}$ electron pairs between C and O (one $\sigma$, one $\pi$).
• (b) $2$ electron-pair regions around C; shape $=$ linear; bond angle $= \boxed{180^\circ}$.
• (c) The two equal, oppositely-directed C=O bond dipoles cancel exactly in the symmetric linear shape, so $\text{CO}_2$ has no overall dipole moment and is non-polar, despite having polar bonds.