Predict the hybridization and molecular geometry for carbon dioxide and methane.

• A. CO2: sp, linear; CH4: sp3, tetrahedral
• B. CO2: sp2, bent; CH4: sp3, trigonal pyramidal
• C. CO2: dsp3, see-saw; CH4: sp, linear
• D. CO2: sp, trigonal planar; CH4: sp2, bent

Answer: (A)

The concept being tested is how the number of electron domains around a central atom determines both hybridization and molecular geometry. Carbon dioxide has two regions of electron density around carbon (two C–O bonds) and no lone pairs, so it uses two hybrid orbitals (sp) to form the sigma bonds. The remaining p orbitals participate in the pi bonds, and the two regions arrange 180° apart, giving a linear geometry for the molecule.

Methane has four regions of electron density around carbon (four C–H bonds) and no lone pairs, so it uses four equivalent sp3 hybrids to form the four sigma bonds. These four orbitals point toward the corners of a tetrahedron, producing a tetrahedral geometry with bond angles near 109.5°.

So the typical descriptions are: carbon dioxide is sp hybridized with linear geometry, and methane is sp3 hybridized with tetrahedral geometry.