Question

Write the ground state electron configuration for (a) \(\mathrm{Mg}, \mathrm{Mg}^{2+}\) (b) \(\mathrm{N}, \mathrm{N}^{3-}\) (c) \(\mathrm{Ti}, \mathrm{Ti}^{4+}\) (d) \(\mathrm{Sn}^{2+}, \mathrm{Sn}^{4+}\)

Short answer

Question: Write the electron configurations for the following species: 1. Mg and Mg²⁺ 2. N and N³⁻ 3. Ti and Ti⁴⁺ 4. Sn²⁺ and Sn⁴⁺ Answer: 1. Mg: [1s² 2s² 2p⁶ 3s²] ; Mg²⁺: [1s² 2s² 2p⁶] 2. N: [1s² 2s² 2p³] ; N³⁻: [1s² 2s² 2p⁶] 3. Ti: [1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d²] ; Ti⁴⁺: [1s² 2s² 2p⁶ 3s² 3p⁶] 4. Sn²⁺: [1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s² 4p⁶ 4d¹⁰ 5s² 4d¹⁰]; Sn⁴⁺: [1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s² 4p⁶ 4d⁸]

Step-by-step solution

(a) Writing Electron Configuration for Mg and Mg²⁺)

To write the electron configuration of magnesium (Mg), we first need to locate Mg in the periodic table. Mg has an atomic number of 12. Looking at the periodic table, Mg is in period 3 and group 2. So its electron configuration follows that of Ne. The electron configuration of Ne is \([1s^2 2s^2 2p^6]\), and we need to add 2 more electrons for Mg. The electron configuration for Mg is: Mg: \([1s^2 2s^2 2p^6 3s^2]\) For \(\mathrm{Mg}^{2+}\), since it has lost 2 electrons, the \(\mathrm{3s^2}\) orbital will be empty: \(\mathrm{Mg}^{2+}\): \([1s^2 2s^2 2p^6]\)

(b) Writing Electron Configuration for N and N³⁻)

Nitrogen (N) has atomic number 7. Nitrogen is located in period 2, group 15 in the periodic table. Its electron configuration follows that of He. The electron configuration of He is \([1s^2]\), and we need to add 5 more electrons for N. The electron configuration for N is: N: \([1s^2 2s^2 2p^3]\) For \(\mathrm{N}^{3-}\), since it has gained 3 electrons, they will completely fill the \(\mathrm{2p}\) orbitals: \(\mathrm{N}^{3-}\): \([1s^2 2s^2 2p^6]\)

(c) Writing Electron Configuration for Ti and Ti⁴⁺)

Titanium (Ti) has atomic number 22. Ti is located in period 4 and group 4. Its electron configuration follows that of Ar. The electron configuration of Ar is \([1s^2 2s^2 2p^6 3s^2 3p^6]\), and we need to add 4 more electrons for Ti. The electron configuration for Ti is: Ti: \([1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^2]\) For \(\mathrm{Ti}^{4+}\), since it has lost 4 electrons, the \(\mathrm{4s^2}\) and two \(\mathrm{3d}\) electrons are removed: \(\mathrm{Ti}^{4+}\): \([1s^2 2s^2 2p^6 3s^2 3p^6]\)

(d) Writing Electron Configuration for Sn²⁺ and Sn⁴⁺)

For Tin (Sn), we first look at the neutral species. Sn has atomic number 50. Looking at the periodic table, Sn is in period 5 and group 14. So its electron configuration follows that of Kr. The electron configuration of Kr is \([1s^2 2s^2 2p^6 3s^2 3p^6 3d^{10} 4s^2 4p^6 4d^{10}]\), and we need to add 6 more electrons for Sn. The electron configuration for Sn is: Sn: \([1s^2 2s^2 2p^6 3s^2 3p^6 3d^{10} 4s^2 4p^6 4d^{10} 5s^2 4d^{10} 5p^2]\) For \(\mathrm{Sn}^{2+}\), we remove 2 electrons from 5p orbital: \(\mathrm{Sn}^{2+}\): \([1s^2 2s^2 2p^6 3s^2 3p^6 3d^{10} 4s^2 4p^6 4d^{10} 5s^2 4d^{10}]\) For \(\mathrm{Sn}^{4+}\), we remove 2 more electrons from 5s and the remaining from 4d orbitals: \(\mathrm{Sn}^{4+}\): \([1s^2 2s^2 2p^6 3s^2 3p^6 3d^{10} 4s^2 4p^6 4d^{8}]\)