Question

Predict the ground-state electron configurations of $Mg,Sr$and$Ba$ ?

Short answer

The ground-state electron configurations of $Mg$is$1s^{2}2s^{2}2p^{6}3s^2$,$Sr$is$1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}5s^2$and $Ba$is$1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}5s^{2}4d^{10}5p^{6}6s^2$.

Step-by-step solution

Given Information

We need to find the ground-state electron configurations of $Mg,Sr$and $Ba$.

Simplify

The electron configurations for$Mg,Sr$and $Ba$.

First, finding electron configuration for $Mg\left(Z=12\right)$:

$1s^{2}2s^{2}2p^{6}3s^2$

Total electrons is $\left(2+2+6+2=12\right)$.

Now, finding electron configuration for role="math" localid="1650270308995" $Sr\left(Z=38\right)$:

$1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}5s^2$

Total electrons is $\left(2+2+6+2+6+2+10+6+2=38\right)$

At last , finding electron configuration for $Ba(Z=56)$.

$1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}5s^{2}4d^{10}5p^{6}6s^2$

Total electrons is$\left(2+2+6+2+6+2+10+6+2+10+6+2=56\right)$