Question

10 mark for review
diagram 1: h-c-h (single bond to o with :o: lone pairs), diagram 2: h-c-h (double bond to o with :o: lone pairs)
which of the diagrams above best represents the ch₂o molecule, and why?
a diagram 1, because all bond angles are 180°.
b diagram 1, because all atoms have a formal charge of 0.
c diagram 2, because the molecule has a trigonal pyramidal shape.
d diagram 2, because all atoms have a formal charge of 0.

Explanation:

1. Analyze Diagram 1:

• For Diagram 1 (H - C - H with a single - bonded O), let's calculate formal charges. The formula for formal charge is \(FC = V - N - \frac{B}{2}\), where \(V\) is valence electrons, \(N\) is non - bonding electrons, and \(B\) is bonding electrons.
• For C: \(V = 4\), \(N = 0\), \(B = 4\) (2 single bonds with H and 1 single bond with O). \(FC=4 - 0-\frac{4}{2}=2\).
• For O: \(V = 6\), \(N = 6\) (3 lone pairs), \(B = 2\) (1 single bond with C). \(FC = 6-6 - \frac{2}{2}=- 1\). So atoms in Diagram 1 do not have formal charge 0. Also, the geometry of \(CH_2O\) (formaldehyde) is trigonal planar, not with all bond angles \(180^{\circ}\).

1. Analyze Diagram 2:

• Diagram 2 has a double bond between C and O. Let's calculate formal charges:
• For C: \(V = 4\), \(N = 0\), \(B = 8\) (2 single bonds with H and 1 double bond with O). \(FC=4 - 0-\frac{8}{2}=0\).
• For O: \(V = 6\), \(N = 4\) (2 lone pairs), \(B = 4\) (1 double bond with C). \(FC = 6 - 4-\frac{4}{2}=0\).
• For H: \(V = 1\), \(N = 0\), \(B = 2\) (1 single bond with C). \(FC=1 - 0-\frac{2}{2}=0\).
• The molecular shape of \(CH_2O\) (formaldehyde) is trigonal planar, not trigonal pyramidal. So option D is correct as all atoms in Diagram 2 have a formal charge of 0.

Answer:

D. Diagram 2, because all atoms have a formal charge of 0.