Question

Consider the following reaction 4 Na(s) + O2(g) - 2 Na2O(s) Establish oxidation numbers. Establish the number of electrons in each species. Is there a connection between the change in oxidation numbers and the change in number of electrons?

Answer 1

The given reaction is of adacs) & 02 (9) 2 Na₂O (S) Oxidation no. of Nagels O. oxidation no. of O2 (g) = 0 As, Nacs) and Ong are in elemental forum, so their oxidation no. is O. As the oxidation no. pure element is o o h Now, oxidation no. of Nao: - Here Oxygen has –2 oxidation state or no. and the Na (sodium) nas H oxidation state. s. isolat No. of é in Naiss Elbba l No. of é in O2(g) = 8 tao No. of é in wa of Na₂O) = 10 (As one è in ess, because the oxidation state of Na in Nao & H No. of è in o of Nazo (s) = 10 ( As two é & more, because the oxidation State of o s - 2 In raro) so we observed that as the oxidation state increases the no. of electrons decreases.

from the reaction & Oxidation State of "Na" changes from o to tl, ie, increased, then the no. of electrons in "Na" changed from all to 10, p.e, decreased. sinularly, oxidation state of changes from o to -2, ie, decreased, then the no. of è in 0° changed from 8 do 1o ,ie; increased. n