Why does chromium form \[C{r^{ + 2}}\] ion?
Answer
510.3k+ views
Hint: In the case of transition elements, the $\left( { - d} \right)$ orbitals start filling successively. Common electronic configuration of transition elements is $\left( {n - 1} \right){d^{1 - 10}}n{s^{1 - 2}}$. Chromium elements are transition elements and have transition metals properties.
Complete answer:
As we know the atomic number of chromium elements is $24$ and electronic configuration is $\left[ {Ar} \right]3{d^5}4{s^1}$. From the electronic configuration of chromium, we can see that it has a half filled $\left( { - d} \right)$ orbital which impart additional stability to the chromium atom.
Transition metals have the characteristic to show more than one oxidation state due to participation of inner $\left( {n - 1} \right)d$ electrons. The most common oxidation state exhibited by transition metals of the first row is $\left( { + 2} \right)$ which is shown due to loss of a pair of electrons or two electrons present in $4s$ orbital. As we move from scandium to titanium the strength of $3d$ orbitals increases and becomes more stable due to this reason electrons are removed from $4s$ orbital to give an oxidation number of $\left( { + 2} \right)$.
The most common oxidation states shown by chromium elements are $ + 2, + 3, + 4, + 5, + 6$ due to availability of $4s$ and $\left( {n - 1} \right)d$ electrons. Out of all these oxidation numbers, only a few compounds of oxidation number $ + 1, + 4, + 5$ are found to be stable in nature. Chromium atoms usually form chromate and dichromate ions.
$ \Rightarrow $ Due to higher stability of $3d$ orbital of chromium due to half-filled orbital, electrons are removed from its $4s$ orbital therefore, it easily form stable \[C{r^{ + 2}}\] ions.
Note:
Chromium element has exceptionally high magnitude of ionization enthalpy due to extra stability of half filled $\left( { - d} \right)$ orbital. Higher oxidation state of any transition metal is calculated by adding its outer $\left( s \right)$ and $\left( {n - 1} \right)d$ electrons. Compounds of chromium are also used as a catalyst during many chemical reactions.
Complete answer:
As we know the atomic number of chromium elements is $24$ and electronic configuration is $\left[ {Ar} \right]3{d^5}4{s^1}$. From the electronic configuration of chromium, we can see that it has a half filled $\left( { - d} \right)$ orbital which impart additional stability to the chromium atom.
Transition metals have the characteristic to show more than one oxidation state due to participation of inner $\left( {n - 1} \right)d$ electrons. The most common oxidation state exhibited by transition metals of the first row is $\left( { + 2} \right)$ which is shown due to loss of a pair of electrons or two electrons present in $4s$ orbital. As we move from scandium to titanium the strength of $3d$ orbitals increases and becomes more stable due to this reason electrons are removed from $4s$ orbital to give an oxidation number of $\left( { + 2} \right)$.
The most common oxidation states shown by chromium elements are $ + 2, + 3, + 4, + 5, + 6$ due to availability of $4s$ and $\left( {n - 1} \right)d$ electrons. Out of all these oxidation numbers, only a few compounds of oxidation number $ + 1, + 4, + 5$ are found to be stable in nature. Chromium atoms usually form chromate and dichromate ions.
$ \Rightarrow $ Due to higher stability of $3d$ orbital of chromium due to half-filled orbital, electrons are removed from its $4s$ orbital therefore, it easily form stable \[C{r^{ + 2}}\] ions.
Note:
Chromium element has exceptionally high magnitude of ionization enthalpy due to extra stability of half filled $\left( { - d} \right)$ orbital. Higher oxidation state of any transition metal is calculated by adding its outer $\left( s \right)$ and $\left( {n - 1} \right)d$ electrons. Compounds of chromium are also used as a catalyst during many chemical reactions.
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