The common positive oxidation states for an element with atomic number 24, are :
Solution
<sub>24</sub>Cr = [Ar]3d<sup>5</sup>4s<sup>1</sup><br>
<p>The element with atomic number 24 is Chromium (Cr). Chromium commonly exhibits several positive oxidation states, which are due to the loss of electrons from both the 4s and 3d orbitals.</p>
<p>Here are the common oxidation states for Chromium :</p>
<ol>
<li>+2 (as in ${Cr}^{2+} $) : This is observed where the electron configuration is $[{Ar}] 3d^4$</li>
<br/><li>+3 (as in ${Cr}^{3+} )$ : This is the most stable state, with the electron configuration $[ {Ar} ] 3d^3$</li>
<br/><li>+6 (as in ${CrO}_4^{2-} $ and ${Cr}_2\text{O}_7^{2-} )$ : In these cases, chromium exhibits an oxidation state of +6.</li>
</ol>
<p>Hence, the common positive oxidation states for Chromium are +2, +3, and +6. <br/><br/>Therefore, the correct option is :</p>
<p>Option B : +2 to +6.</p>
About this question
Subject: Chemistry · Chapter: d and f Block Elements · Topic: Properties of Transition Metals
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