Question

write the ground - state electron configurations of the following transition metal ions. part 1 of 4 fe³⁺ part 2 of 4 cu⁺ part 3 of 4 ni³⁺

Explanation:

Step1: Recall Fe neutral - atom electrons

Fe (iron) has atomic number 26, so neutral Fe has 26 electrons. Its electron - configuration is $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{6}$.

Step2: Determine electrons in $Fe^{3 + }$

$Fe^{3 + }$ has lost 3 electrons. First, remove 2 electrons from 4s and 1 electron from 3d. So the electron - configuration of $Fe^{3 + }$ is $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{5}$.

Step3: Recall Cu neutral - atom electrons

Cu (copper) has atomic number 29, so neutral Cu has 29 electrons. Its electron - configuration is $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{1}3d^{10}$ (due to the stability of half - filled and fully - filled sub - shells).

Step4: Determine electrons in $Cu^{+}$

$Cu^{+}$ has lost 1 electron. Remove 1 electron from 4s. So the electron - configuration of $Cu^{+}$ is $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{10}$.

Step5: Recall Ni neutral - atom electrons

Ni (nickel) has atomic number 28, so neutral Ni has 28 electrons. Its electron - configuration is $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{8}$.

Step6: Determine electrons in $Ni^{3 + }$

$Ni^{3 + }$ has lost 3 electrons. First, remove 2 electrons from 4s and 1 electron from 3d. So the electron - configuration of $Ni^{3 + }$ is $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{7}$.

Answer:

Part 1 of 4: $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{5}$
Part 2 of 4: $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{10}$
Part 3 of 4: $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{7}$