Question

Write the electron configuration for (a) the \(\mathrm{Ni}^{2+}\) ion and (b) the \(\mathrm{Sn}^{2+}\) ion. How many unpaired electrons does each contain?

Short answer

The electron configurations for Ni²⁺ and Sn²⁺ ions are: Ni²⁺: \( 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^6 \) and Sn²⁺: \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10} \). Ni²⁺ has 2 unpaired electrons in the 3d orbital, and Sn²⁺ has 0 unpaired electrons, as all its orbitals are fully occupied.

Step-by-step solution

Identify the atomic numbers of Ni and Sn

Look up the atomic numbers of Nickel (Ni) and Tin (Sn) on the periodic table. Ni has an atomic number of 28 and Sn has an atomic number of 50.

Write the ground-state electron configuration for neutral atoms

Using the periodic table or an electron configuration chart, write the ground-state electron configurations for Ni and Sn. Ni: \( 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^8 \) Sn: \( 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10} 5p^2 \)

Write the electron configurations for Ni²⁺ and Sn²⁺

Remove 2 electrons from the neutral electron configurations to show the electron configurations of Ni⁺² and Sn⁺² ions. Ni⁺²: \( 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^6 \) Sn⁺²: \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10} \) Note that in the case of transition metals like Ni, it's common to remove electrons from the ns orbital (here 4s) before removing from the (n-1)d orbital (here 3d).

Identify the number of unpaired electrons

Inspect the electron configuration of each ion to determine the number of unpaired electrons. Ni²⁺: \( 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^6 \) - 2 unpaired electrons in the 3d orbital Sn²⁺: \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10} \) - No unpaired electrons, as all the orbitals are fully occupied.

Final Answer

Electron configurations: Ni²⁺: \( 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^6 \) - 2 unpaired electrons Sn²⁺: \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10} \) - 0 unpaired electrons