Question 107 - In N + 1 Rule in Quantum Chemistry, whenever a spin 1 / 2 nucleus is adjacent to N other nuclei, it is split into N + 1 distinct peaks. In 1 peak or singlet, there is only 1 magnitude. In 2 peaks or doublet, the ratio of magnitude of each peak is 1 : 1. In 3 peaks or triplet, the ratio of magnitude of each peak is 1 : 2 : 1. In 4 peaks or quartet, the ratio of magnitude of each peak is 1 : 3 : 3 : 1. In 5 peaks or quintet, the ratio of magnitude of each peak is 1 : 4 : 6 : 4 : 1. (a) By using binomial coefficients or Triangle of Pascal find the ratio of magnitude of each peak if 6 peaks exists. (b) How many adjacent nuclei are available in a spin 1 / 2 nucleus in such situation of 6 peaks?

Answer 107 - (a) 1 : 5 : 10 : 10 : 5 : 1, since (0 + 1) : (1 + 4) : (4 + 6) : (6 + 4) : (4 + 1) : (1 + 0) with reference to the ratio of magnitude of each peak in quintet. (b) Number of adjacent nuclei = Number of peaks - 1 = 6 - 1 = 5. The answer is given by Kang Chuen Tat; PO Box 6263, Dandenong, Victoria VIC 3175, Australia; SMS +61405421706; chuentat@hotmail.com; http://kangchuentat.wordpress.com.

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