I took some time off recently and went on a trip to India. Whilst wandering around the fields adjacent to the backwaters around Alappuzha (formerly Alleppey) in Kerala I came across a farmer spraying his rice crop (above). I noticed that he was not wearing any protection against the spray and wondered what chemical(s) he was actually spraying on his crops.

At the edge of the field I came across his discarded canisters (left and expanded on right). He was using the herbicide paraquat. Paraquat is a substituted bipyridine salt and has the chemical name N,N′-dimethyl-4,4′-bipyridinium dichloride. Although herbicides and pesticides were once on the old Applied Chemistry syllabus they no longer form part of the food option so IB students would not be expected to know anything about them. However paraquat could be used to illustrate a couple of useful concepts that are on the current syllabus. (Remember that future exam questions may well use unfamiliar substances such as paraquat to test the understanding of objective 3 concepts that are on the syllabus.)

One is the international aspect. Although paraquat is one of the most widely used herbicides in the world farmers and gardeners who live in the European Union cannot purchase it or use it as it has been banned since 2007. This is because it is very toxic and there is no known antidote – hence my surprise at seeing the farmer use it with no protection (and leaving the canisters lying around where possibly children could find them and play with them). The laws governing the use of certain chemicals in different countries provide a good example of the international aspects of chemistry.

The second is that paraquat provides a good example to test the understanding of hybridization for Higher Level students in Topic 14.2. Students have learned that the nitrogen atom is sp³ hybridized in both ammonia and the ammonium ion.  Ask them what the hybridization of the nitrogen is in paraquat. They should be able to deduce that as both nitrogen atoms (actually ions) are part of the aromatic ring system the hybridization is the same and the nitrogen must be sp² hybridized. Two of the p orbitals hybridize with the s orbital to give three sp² hybrid orbitals with bond angles of 120^o leaving one electron in the p orbital of each nitrogen ion to contribute to the delocalisation within the aromatic ring.