Question:
The common positive oxidation states for an element with atomic number 24, are :
Correct Option: 1
Solution:
$\operatorname{Cr}(Z=24)$
$[\mathrm{Ar}] 4 \mathrm{~s}^{1} 3 \mathrm{~d}^{5} \mathrm{Cr}$ shows common oxidation states
starting from $+2$ to $+6$.