Chemistry - Complete d-block Question with Solution | TestHub

$\sqrt{n(n+2)}=4.9$

$n(n+2)=24$

$n^2+2n-24=0$

$(n+6)(n-4)=0$

$n=4$

4 unpaired electrons

$\sqrt{l(l+1)}\frac{h}{2\pi }=\sqrt{6}\frac{h}{2\pi }$

$l=2\Rightarrow$ last subshell is d subshell, highest value of n is 3 .

$A^{+3}$ contains 4 unpaired electron . possible subshells

1s 2s 2p 3s 3p 4s 3d

unpaired in ${\mathrm{A}}^{+3}=4$

In removal of total $3{\mathrm{e}}^{-},2{\mathrm{e}}^{-}$will be removed from 4s and $1{\mathrm{e}}^{-}$from 3 d .

3d must contain 4 unpaired ${\mathrm{e}}^{-}$in ${\mathrm{A}}^{+3}$.

$A^{+3}=1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^4$

$\mathrm{A}=1{\mathrm{s}}^{2}2{\mathrm{s}}^{2}2{\mathrm{p}}^{6}3{\mathrm{s}}^{2}3{\mathrm{p}}^{6}3{\mathrm{d}}^5,4{\mathrm{s}}^2$

$\mathrm{Z}=25$

Element is Mn .