What does the name chromium $(II)$ oxide mean in terms of electrons lost and gained?
Answer
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Hint: Chromium has multiple states in which it exists due to its unique electronic configuration \[[Ar]\,3{d^5}4{s^1}\] which in interaction highly electronegative p block elements. The oxidation states it possesses are $ + 2,\, + 3\,and\, + 6$.
Complete answer:
Chromium is the d block element and exists in the sixth period and having a electronic configuration of \[[Ar]\,3{d^5}4{s^1}\], which states that it has 5 unpaired electrons in its d orbital and 1 unpaired electron in its s orbital. Now for chromium exits like this because for an atom the most stable configuration is where all are paired or all are unpaired, rest all configurations of electrons are less stable than the above mentioned.
Now as oxygen is the atom more electronegative than chromium and oxygen except for peroxide doesn’t exist in any other stage except $ - 2$.
Taking out the oxidation state of chromium keeping oxidation state of oxygen as $ - 2$
Let $x$ be the oxidation state of chromium in chromium oxide and whole charge on chromium oxide is zero, then:
$
x - 2 = 0 \\
\Rightarrow x = 2 \\
$
Therefore, chromium exists here in $ + 2$ state and by looking at the oxidation states of chromium and oxygen in chromium oxide we can say that two electrons were lost by chromium and 2 electrons were gained by oxygen.
The name chromium$(II)$ oxide means two electrons were lost and gained by chromium and oxygen respectively to form a stable compound.
Note:
Oxygen only exists in two states that is $ - 2,\, + 1\,and\, - 1$. $ + 1$ oxidation state of oxygen exists only in the presence of higher electronegative atoms being halogens which form peroxide.
Complete answer:
Chromium is the d block element and exists in the sixth period and having a electronic configuration of \[[Ar]\,3{d^5}4{s^1}\], which states that it has 5 unpaired electrons in its d orbital and 1 unpaired electron in its s orbital. Now for chromium exits like this because for an atom the most stable configuration is where all are paired or all are unpaired, rest all configurations of electrons are less stable than the above mentioned.
Now as oxygen is the atom more electronegative than chromium and oxygen except for peroxide doesn’t exist in any other stage except $ - 2$.
Taking out the oxidation state of chromium keeping oxidation state of oxygen as $ - 2$
Let $x$ be the oxidation state of chromium in chromium oxide and whole charge on chromium oxide is zero, then:
$
x - 2 = 0 \\
\Rightarrow x = 2 \\
$
Therefore, chromium exists here in $ + 2$ state and by looking at the oxidation states of chromium and oxygen in chromium oxide we can say that two electrons were lost by chromium and 2 electrons were gained by oxygen.
The name chromium$(II)$ oxide means two electrons were lost and gained by chromium and oxygen respectively to form a stable compound.
Note:
Oxygen only exists in two states that is $ - 2,\, + 1\,and\, - 1$. $ + 1$ oxidation state of oxygen exists only in the presence of higher electronegative atoms being halogens which form peroxide.
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