Question

What is the oxidation state of chromium in $C{r^{2 + }}$?
A) $+ 2$
B) $- 2$
C) $0$
D) $+ 1$

Answer 1

We have to remember that the oxidation state is the charge on a particular atom or element. Oxidation state tells us how many numbers of electrons are being added to or removed from a species. We also remember that the oxidation state can be positive or negative integers and also it can be zero.
For example: $N{a^ + }$has an oxidation state of $+ 1$.

Complete step by step answer:
We must remember that the oxidation can also be said that it is the total charge on a particular atom after heteronuclear bonds approximation.
$C{r^{2 + }}$has an oxidation state$= + 2$
As we know that the chromium has an atomic number which is 24 and an atomic symbol as $Cr$.
Electronic Configuration of$Cr$: $\left[ {Ar} \right]3{d^5}4{s^1}$
$C{r^{2 + }}$means that this atom has lost 2 electrons to attain this electron configuration, due to which its oxidation state comes out to be 2.

Electronic configuration of$C{r^{2 + }}$: $\left[ {Ar} \right]3{d^4}$
Option A) this is a correct option as $C{r^{2 + }}$ has an oxidation state of $+ 2$.
We can look at an example to understand this well.
$Cr{O_4}^{2 - }$, $+ 2$ is added to the oxidation states of $Cr$ and $O$ because the total charge on chromate ion $Cr{O_4}^{2 - }$is $- 2$ which is its oxidation state also.
Option B) this is an incorrect option as $+ 2$ is the correct answer.
Option C) this is an incorrect option as $0$ is the oxidation state for $Cr$ atoms and not$C{r^{2 + }}$.
Option D) this is an incorrect option.

Hence, the correct option, ‘Option A’.

Note: We have to know that the oxidation state of a neutral atom i.e. $Cr$ which has no charge on it is also known as ground state as it has $0$ oxidation state.