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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Vanillin is the common name for 4-hydroxy-3-methoxy-benzaldehyde.
See attached figure for the structure.
Vanillin have 3 functional groups:
1) aldehyde group: R-HC=O, in which the carbon is double bonded to oxygen
2) phenolic hydroxide group: R-OH, were the hydroxyl group is bounded to a carbon from the benzene ring
3) ether group: R-O-R, were hydrogen is bounded through sigma bonds to carbons
Now for the hybridization we have:
The carbon atoms involved in the benzene ring and the red carbon atom (from the aldehyde group) have a <u>sp²</u> hybridization because they are involved in double bonds.
The carbon atom from the methoxy group (R-O-CH₃) and the blue oxygen's have a <u>sp³</u> hybridization because they are involved only in single bonds.
