Question

The highest oxidation state of $\mathrm{Cr}$ will be

• A. $2$
• B. $3$
• C. $4$
• D. $6$

Answer 1

The correct option is D

$6$

Explanation for correct option

(D)$6$

Oxidation state

• "Oxidation number denotes the oxidation state of an element in a compound ascertained according to a set of rules formulated on the basis that electron pair in a covalent bond belongs entirely to a more electronegative element."

Oxidation state of $\mathrm{Cr}$

• The atomic number of $\mathrm{Cr}$ is $24$.
• The electronic configuration of $\mathrm{Cr}$ is $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{1}3d^5$.
• $\mathrm{Cr}$ can lose five electrons from $d$ orbitals and one electron from $s$ orbital.
• The possible oxidation state of $\mathrm{Cr}$ is $+2,+3,+4,+5,+6$.
• So, the highest oxidation state of $\mathrm{Cr}$ will be $+6$.

Explanation for incorrect option

The highest oxidation state of $\mathrm{Cr}$ will be $+6$. Hence, options (A), (B), and (C) are incorrect.

Hence, option (D) is correct. The highest oxidation state of $\mathrm{Cr}$ will be $+6$.