Question

write the ground - state electron configurations of the following transition metal ions. part 1 of 4 pd²⁺ kr4d⁸ part 2 of 4 sc³⁺ ar

Explanation:

Step1: Determine Pd's atomic number

Palladium (Pd) has atomic number 46. Kr has atomic number 36. The $4d$ sub - shell is filled after the $5s$ sub - shell in transition metals. For $Pd^{2 + }$, we start with the noble - gas core of Kr ($[Kr]=1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{10}4s^{2}4p^{6}$). Pd has 10 electrons in the $4d$ sub - shell in its neutral state. Losing 2 electrons gives $[Kr]4d^{8}$.

Step2: Determine Sc's atomic number

Scandium (Sc) has atomic number 21. Ar has atomic number 18. The electron configuration of neutral Sc is $[Ar]4s^{2}3d^{1}$. For $Sc^{3+}$, we remove 3 electrons. First, we remove the 2 electrons from the $4s$ sub - shell and then 1 electron from the $3d$ sub - shell, leaving a noble - gas core of Ar ($[Ar]=1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}$).

Answer:

Part 1 of 4: $Pd^{2+}$ has ground - state electron configuration $[Kr]4d^{8}$
Part 2 of 4: $Sc^{3+}$ has ground - state electron configuration $[Ar]$