What is the oxidation number of tin (Sn) in the compound Na2SnO2? A. -2 B. 0 C. +2 D. +3
2 answers:
Answer: Option (C) is the correct answer.
Explanation:
It is known that oxidation number of sodium is +1 and the oxidation number of oxygen is -2.
Let the oxidation number of tin (Sn) is x.
Therefore, calculate the oxidation number of tin in the given formula as follows.
= (2 \times oxidation number of Na) + Oxidation number of Sn + (2 \times oxidation number of O)
=
So, 2 + x - 4 = 0
therefore, x = +2
Thus, we can conclude that the oxidation number of tin (Sn) in the compound is +2.
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Answer:
See explanation and picture below
Explanation:
First, in the case of methyloxirane (Also known as propilene oxide) the mechanism that is taking place there is something similar to a Sn2 mechanism. Although a Sn2 mechanism is a bimolecular substitution taking place in only step, the mechanism followed here is pretty similar after the first step.
In both cases, the H atom of the HBr goes to the oxygen in the molecule. You'll have a OH⁺ in both. However, in the case of methyloxirane the next step is a Sn2 mechanism step, the bromide ion will go to the less substitued carbon, because the methyl group is exerting a steric hindrance. Not a big one but it has a little effect there, that's why the bromide will rather go to the carbon with more hydrogens. and the final product is formed.
In the case of phenyloxirane, once the OH⁺ is formed, the next step is a Sn1 mechanism. In this case, the bond C - OH⁺ is opened on the side of the phenyl to stabilize the OH. This is because that carbon is more stable than the carbon with no phenyl. (A 3° carbon is more stable than a 2° carbon). Therefore, when this bond opens, the bromide will go there in the next step, and the final product is formed. See picture below for mechanism and products.
