
Oxidation number in sucrose is:
(A) 0
(B) +12
(C) -12
(D) +7
Answer
588.3k+ views
Hint: To answer this question we must first recall the concept of oxidation number. We would also have to consider that the molecule as a whole is neutral and we have to assign the general oxidation number to hydrogen and oxygen.
Complete step by step solution:
Oxidation number or oxidation state of any atom is the total number of electrons it either gains or loses while forming a chemical bond with another atom. It is important to note that this plays an essential role in systematic nomenclature of organic compounds.
Let us first see the formula of Glucose to find out how many number of atoms of each element is present:
\[{C_6}{H_{12}}{O_6}\]
As we can see, there are 6 molecules of C present along with 12 atoms of H and 6 of O.
We know that the common oxidation state of hydrogen is +1 and that of oxygen is -2. Let us now assume that the oxidation state of carbon is X.
If we assume that the total charge on the molecule is 0, we can formulate the following equation:
\[6(X) + 12( + 1) + 6( - 2) = 0\]
Or, \[6X + 12 - 12 = 0\]
Therefore, \[X = 0\]
Hence, the oxidation state of carbon in glucose is Option (A) 0.
Note: We must remember that although oxidation number most of the time is equal to the number of valence electrons of an atom, this is not the case for carbon. We consider carbon to have a valency of 4 but its oxidation number remains 0 most of the time. We must not confuse this with formal charge which determines the likelihood of a molecule existing.
Complete step by step solution:
Oxidation number or oxidation state of any atom is the total number of electrons it either gains or loses while forming a chemical bond with another atom. It is important to note that this plays an essential role in systematic nomenclature of organic compounds.
Let us first see the formula of Glucose to find out how many number of atoms of each element is present:
\[{C_6}{H_{12}}{O_6}\]
As we can see, there are 6 molecules of C present along with 12 atoms of H and 6 of O.
We know that the common oxidation state of hydrogen is +1 and that of oxygen is -2. Let us now assume that the oxidation state of carbon is X.
If we assume that the total charge on the molecule is 0, we can formulate the following equation:
\[6(X) + 12( + 1) + 6( - 2) = 0\]
Or, \[6X + 12 - 12 = 0\]
Therefore, \[X = 0\]
Hence, the oxidation state of carbon in glucose is Option (A) 0.
Note: We must remember that although oxidation number most of the time is equal to the number of valence electrons of an atom, this is not the case for carbon. We consider carbon to have a valency of 4 but its oxidation number remains 0 most of the time. We must not confuse this with formal charge which determines the likelihood of a molecule existing.
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