Solution Found!
Assign an oxidation state to each atom in each polyatomic ion(a) CrO42?(b) Cr2O72?(c)
Chapter 16, Problem 52P(choose chapter or problem)
Assign an oxidation state to each atom in each polyatomic ion.
(a) \(\mathrm{CrO_4^{~2-}}\)
(b) \(\mathrm{Cr_2O_7^{~2-}}\)
(c) \(\mathrm{PO_4^{~3-}}\)
(d) \(\mathrm{MnO_4^{~-}}\)
Equation Transcription:
Text Transcription:
CrO_4^{2-}
Cr_2O_7^{2-}
PO_4^{3-}
MnO_4^{-}
Questions & Answers
QUESTION:
Assign an oxidation state to each atom in each polyatomic ion.
(a) \(\mathrm{CrO_4^{~2-}}\)
(b) \(\mathrm{Cr_2O_7^{~2-}}\)
(c) \(\mathrm{PO_4^{~3-}}\)
(d) \(\mathrm{MnO_4^{~-}}\)
Equation Transcription:
Text Transcription:
CrO_4^{2-}
Cr_2O_7^{2-}
PO_4^{3-}
MnO_4^{-}
ANSWER:
Solution: Here, we are going to assign the oxidation state to each atom in the given polyatomic ions.
Step1:
Oxidation number denotes the oxidation state of an element in a compound ascertained according to a set of rules.
These rules are:
1. In elements, in the free or the uncombined state, each atom bears an oxidation number of zero. Evidently each atom in H2, O2, Cl2, O3, P4, S8, Na, Mg, Al has the oxidation number zero.
2. For ions composed of only one atom, the oxidation number is equal to the charge on the ion. In their compounds all alkali metals have oxidation number of +1, and all alkaline earth metals have an oxidation number of +2. Aluminium is regarded to have an oxidation number of +3 in all its compounds.
3. The oxidation number of oxygen in most compounds is –2.
4. The oxidation number of hydrogen is +1, except when it is bonded to metals in binary compounds (that is compounds containing two elements).
5. In all its compounds, fluorine has an oxidation number of –1. Other halogens (Cl, Br,and I) also have an oxidation number of –1, when they occur as halide ions in their compounds.
6. The algebraic sum of the oxidation number of all the atoms in a compound must be zero. In polyatomic ion, the algebraic sum of all the oxidation numbers of atoms of the ion must equal the charge on the ion.
Step2:
Based upon these rules, let us calculate the oxidation number of the elements in the given polyatomic ions:
- CrO42-
Here, the oxidation number of oxygen is -2
Since CrO42- is a polyatomic ion, the algebraic sum of all the oxidation numbers of atoms of the ion must equal the charge on the ion.