Question

Oxidation state of \[Sn\] in \[Sn{O_2}\] is \[4\]. If true enter \[1\], else enter \[0\].

Answer 1

Hint:Oxidation state: It is the number of electrons lost or gained by a neutral atom while forming a chemical bond. It is also known as oxidation number. If the electrons are lost, then the value of the oxidation state will be positive whereas if an atom gains electrons, then the value of oxidation state will be negative.

Complete answer:
Rules to find oxidation state of an element in a compound:
A free element i.e., an element which is not bonded or is chemically bonded to itself will always be of zero oxidation state.
The oxidation state of a monatomic ion (ion consisting of only one atom) is equal to the charge present on the ion.
The oxidation state of oxygen atom is \[( - 2)\] with an exceptional case of peroxides. In peroxides the oxidation state of oxygen atom is \[( - 1)\].
In a molecule, the sum of oxidation states of each element present is equal to total charge on the molecule (zero in case of neutral molecules).
In \[Sn{O_2}\] i.e., Tin oxide, assume the oxidation state of tin be \[x\].
Oxidation state of oxygen atom \[ = - 2\]
As the compound is neutral so, the total charge on compound \[ = 0\]
Therefore, \[x - 2 \times 2 = 0\]
\[ \Rightarrow x - 4 = 0\]
\[ \Rightarrow x = + 4\]
Hence, the oxidation state of \[Sn\] is \[4\]. Thus, the answer is true.
So, the appropriate answer for this question is \[1\].

Note:
Tin \[(Sn)\] exists in two oxidation states i.e., \[ + 2\] and \[ + 4\]due to inert pair effect but it is more stable in \[ + 4\] oxidation state. Due to this, \[S{n^{ + 2}}\]can easily be oxidized into \[S{n^{4 + }}\]and hence acts as a strong reducing agent.