Q47E
Question
Determine the oxidation number of each element in each of the following compounds:
(a) \({\bf{HCN}}\)
(b) \(O{F_2}\)
(c) \(AsC{l_3}\)
Step-by-Step Solution
Verified(a)The oxidation number of hydrogen is\( + 1\), carbon is\( + 2\), and nitrogen is \( - 3\).
(b) The oxidation number of oxygen is\( + 2\)and fluorine is\( - 1\).
(c) The oxidation number of arsenic is\( + 3\)and chlorine is\( - 1\).
In simple words, anoxidation number is the number assigned to the factors in a chemical combination.
Anoxidation number is the total number of electrons that can participate, lose, or acquire while establishing chemical relations with another element.
- The oxidation number of an element present in a certain molecule tells us how many electrons that element has lost or obtained.
- One element can have more than one oxidation number, depending on the molecule in which it is placed.
- In the following molecules, the total sum of the oxidation numbers is zero.
(a) HCN
In HCN, the oxidation number of hydrogen is\( + 1\), the oxidation number of carbon is\( + 2\), while the oxidation number of nitrogen is \( - 3\).
(b)\(O{F_4}\)
In \(O{F_4}\), the oxidation number of oxygen is\( + 2\), while the oxidation number of fluorine is\( - 1\).
(c) \({\rm{AsC}}{{\rm{l}}_3}\)
In \({\rm{AsC}}{{\rm{l}}_3}\), the oxidation number of arsenic is\( + 3\), while the oxidation number of chlorine is\( - 1\).