## Chemistry: Molecular Approach (4th Edition)

(a) \begin{align} & Ni:[Ar]4{{s}^{2}}3{{d}^{8}} \\ & N{{i}^{2+}}:[Ar]3{{d}^{8}} \\ \end{align} (b) \begin{align} & Mn:[Ar]4{{s}^{2}}3{{d}^{5}} \\ & Mn:[Ar]3{{d}^{3}} \\ \end{align} (c) \begin{align} & Y:[Kr]5{{s}^{2}}4{{d}^{1}} \\ & {{Y}^{+}}:[Kr]5{{s}^{1}}4{{d}^{1}} \\ \end{align} (d) \begin{align} & Ta:[\text{Xe}]6{{s}^{2}}4{{f}^{14}}5{{d}^{3}} \\ & T{{a}^{2+}}:[\text{Xe}]4{{f}^{14}}5{{d}^{3}} \\ \end{align}
The following steps are to be used while writing the ground state electron configuration of a transition metal: 1. The noble gas preceding the element should be identified and put in square brackets. 2. The period determining the outer principal quantum level is calculated. After that 1 is subtracted from the outer quantum level to get the quantum level for the d orbital. If the element is in the third or fourth transition series, $\left( n-2 \right){{\text{f}}^{14}}$electrons are also included in the configuration. 3. The number of electrons in the neutral atom is found by calculating across the row in the period. The electrons are filled in the orbitals accordingly, and if the given species is an ion, then the required number of electrons are removed first from the s orbital and then from the d orbital. (a) Ni, Ni2+ Ni: 1. The noble gas that precedes nickel is argon $\left( Ar \right)$. 2. Ni is in the fourth period, so the orbitals to be used are $[Ar]4s3d$. 3. Ni has 10 more electrons than Ar. Therefore, the electron configuration of Ni is $[Ar]4{{s}^{2}}3{{d}^{8}}$. Ni2+: 1. The noble gas that precedes nickel is argon $\left( Ar \right)$. 2. Ni is in the fourth period, so the orbitals to be used are $[Ar]4s3d$. 3. Ni has 10 more electrons than Ar. 4. Ni2+ has lost two electrons relative to the Ni atom. Therefore, the electron configuration of $N{{i}^{2+}}$ is $[Ar]3{{d}^{8}}$. (b) Mn, Mn4+ Mn: 1. The noble gas that precedes manganese is argon $\left( Ar \right)$. 2. Mn is in the fourth period, so the orbitals to be used are $[Ar]4s3d$. 3. Mn has seven more electrons than Ar. Therefore, the electron configuration of Mn is $[Ar]4{{s}^{2}}3{{d}^{5}}$. Mn4+: 1. The noble gas that precedes manganese is argon $\left( Ar \right)$. 2. Mn is in the fourth period, so the orbitals to be used are $[Ar]4s3d$. 3. Mn has seven more electrons than Ar. 4. Mn4+ has lost four electrons relative to the Mn atom. Therefore, the electron configuration of $~M{{n}^{4+}}$ is $[Ar]3{{d}^{3}}$. (c) Y, Y+: Y: 1. The noble gas that precedes yttrium is krypton $\left( Kr \right)$. 2. Y is in the fifth period, so the orbitals to be used are $[Kr]5{{s}^{2}}4{{d}^{1}}$. 3. Y has three more electrons than Kr. Therefore, the electron configuration of Y is $[Kr]5{{s}^{2}}4{{d}^{1}}$. Y+: 1. The noble gas that precedes yttrium is krypton $\left( Kr \right)$. 2. Y is in the fifth period, so the orbitals to be used are $[Kr]5s4d$. 3. Y has three more electrons than $Kr$. 4. Y+ has lost one electron relative to the Y atom Therefore, the electron configuration of ${{Y}^{+}}$ is $[Kr]5{{s}^{1}}4{{d}^{1}}$. (d) Ta, Ta2+ Ta: 1. The noble gas that precedes tantalum is xenon $\left( Xe \right)$. 2. Ta is in the sixth period, which is the third transition series, so the $\left( n-2 \right)$${{f}^{14}}$ electrons are also included in the configuration. The orbitals to be used are $[Xe]6s4f5d$. 3. Ta has 19 more electrons than $Xe$. Therefore, the electron configuration of Ta is $[\text{Xe}]6{{s}^{2}}4{{f}^{14}}5{{d}^{3}}$. Ta2+ 1. The noble gas that precedes tantalum is xenon $\left( Xe \right)$. 2. Ta is in the sixth period, which is the third transition series, so the $\left( n-2 \right)$${{f}^{14}}$ electrons are also included in the configuration. The orbitals to be used are $[Xe]6s4f5d$. 3. Ta has 19 more electrons than $Xe$. Ta2+ has lost two electrons relative to the Ta atom. Therefore, the electron configuration of Ta2+ is $[\text{Xe}]4{{f}^{14}}5{{d}^{3}}$.