
Concept explainers
Give the oxidation numbers for the underlined atoms in the following molecules and ions:

Interpretation:
The oxidation number of the underlined atoms isto be determined.
Concept introduction:
The oxidation number can be assigned using the following rules:
In free state, an element is always in
For a neutral compound, oxidation numbers of atoms add upto
In a polyatomic ion, oxidation number of atoms add up to the charge on the ion.
The oxidation number of monoatomic ions is same as its charge.
Oxygen is usually in
Hydrogen is usually in
Group 1 elements are always in
Answer to Problem 45QP
Solution:
(a)
(b) +7
(c)
(d)
(e)
(f)
(g)
(h)
(i)
(j)
(k)
(l)
(m)
(n)
Explanation of Solution
(a)
Let the oxidation number of Clbe
Hence, the oxidation number ofCl can be calculated as follows:
Thus, the oxidation number of Clis
(b)
In
Hence, the oxidation number ofiodine can be calculated as follows:
Thus, the oxidation number of iodine is
(c)
Let the oxidation number of carbonbe
Hence, the oxidation number ofcarbon can be calculated as follows:
Thus, the oxidation number of carbonis
(d)
Let the oxidation number of carbonbe
Hence, the oxidation number ofcarbon can be calculated as follows:
Thus, the oxidation number of carbon is
(e)
Let the oxidation number of carbonbe
Hence, the oxidation number ofcarbon can be calculated as follows:
Thus, the oxidation number of carbon is
(f)
Let the oxidation number of chromium (Cr)be
Hence, the oxidation number ofchromium can be calculated as follows:
Thus, the oxidation number of chromium is
(g)
Let the oxidation number of chromium be
Hence, the oxidation number ofchromium can be calculated as follows:
Thus, the oxidation number of chromium is
(h)
Let the oxidation number of manganese be
Hence, the oxidation number ofmanganese can be calculated as follows:
Thus, the oxidation number of manganese is
(i)
Let the oxidation number of carbonbe
Hence, the oxidation number ofcarbon can be calculated as follows:
Thus, the oxidation number of carbon is
(j)
The oxidation number of any atom in its elemental form is zero.
Hence, the oxidation number of lithiumis
(k)
Let the oxidation number of iodinebe
Hence, the oxidation number ofiodine can be calculated as follows:
Thus, the oxidation number of iodine is
(l)
Let the oxidation number of oxygenbe
Hence, the oxidation number ofoxygen can be calculated as follows:
Thus, the oxidation number of oxygen is
(m)
Let the oxidation number of phosphorus be
Hence, the oxidation number ofphosphorus can be calculated as follows:
Thus, the oxidation number of phosphorus is
(n)
Let the oxidation number of Aube
Hence, the oxidation number of Au can be calculated as follows:
Thus, the oxidation number of Au is
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