because Second electron is removed from stable

configuration $3 \mathrm{~d}^5$

$\mathrm{Cr}^{+}:[\mathrm{Ar}] 3 \mathrm{~d}^5 4 \mathrm{~s}^0$

$\therefore$ No of unpaired $\mathrm{e}^{-}$in $\mathrm{Cr}^{+}$is $5, \mathrm{n}=5$

$\text { So, Magnetic moment } =\sqrt{\mathrm{n}(\mathrm{n}+2)} \text { B.M }$

$=\sqrt{5(5+2)}=5.92 \mathrm{BM} \approx 6$