write the ground - state electron configurations of the following transition metal ions. part 1 of 4 fe³⁺…

Explanation:

Step1: Recall electron - configuration rules

For transition - metal ions, electrons are removed from the outermost s - orbital first and then from the d - orbitals.

Step2: Determine Fe³⁺ electron configuration

Fe has an atomic number of 26, with ground - state electron configuration $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{6}$. For Fe³⁺, three electrons are removed. First, two from the 4s orbital and then one from the 3d orbital, resulting in $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{5}$.

Step3: Determine Cu⁺ electron configuration

Cu has an atomic number of 29, with ground - state electron configuration $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{1}3d^{10}$ (due to the stability of a half - filled 4s and a filled 3d). For Cu⁺, one electron is removed from the 4s orbital, giving $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{10}$.

Step4: Determine Ni³⁺ electron configuration

Ni has an atomic number of 28, with ground - state electron configuration $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{8}$. For Ni³⁺, three electrons are removed. First, two from the 4s orbital and then one from the 3d orbital, resulting in $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{7}$.

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: $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{7}$