Question 8.7: Draw the Lewis structure for HCN.
Draw the Lewis structure for HCN.
Hydrogen has one valence electron, carbon (Group 14) has four, and nitrogen (Group 15) has five. The total number of valence electrons is, therefore, 1 + 4 + 5 = 10. In principle, there are different ways in which we might choose to arrange the atoms. Because hydrogen can accommodate only one electron pair, it always has only one single bond associated with it. Therefore, C—H—N is an impossible arrangement. The remaining two possibilities are H—C—N and H—N—C. The first is the arrangement found experimentally. You might have guessed this because the formula is written with the atoms in this order, and carbon is less electronegative than nitrogen. Thus, we begin with the skeleton structure
H—C—N
The two bonds account for four electrons. The H atom can have only two electrons associated with it, and so we will not add any more electrons to it. If we place the remaining six electrons around N to give it an octet, we do not achieve an octet on C:
H—C—\overset{..}{\underset{..}N}\!:We therefore try a double bond between C and N, using one of the unshared pairs we placed on N. Again we end up with fewer than eight electrons on C, and so we next try a triple bond. This structure gives an octet around both C and N:
The octet rule is satisfied for the C and N atoms, and the H atom has two electrons around it. This is a correct Lewis structure.