Question

How many unpaired electrons are there in the following ions? (a) \(\mathrm{V}^{3+}\) (b) \(\mathrm{Sn}^{4+}\) (c) I \(^{-}\) (d) \(\mathrm{W}^{4+}\)

Short answer

Answer: (a) 3 unpaired electrons, (b) 0 unpaired electrons, (c) 0 unpaired electrons, (d) 2 unpaired electrons.

Step-by-step solution

Find the electron configurations of the ions

To find the electron configurations, you can use the periodic table. For each ion, identify its atomic number, which tells you the number of electrons in the neutral atom. Then, add or subtract electrons to account for the charge of the ion. Finally, use the aufbau principle to fill in the electron configuration for the ion with the correct number of electrons. (a) V\(^{3+}\): The atomic number of V (vanadium) is 23, so the neutral atom has 23 electrons. Since the ion has a charge of +3, it has lost 3 electrons, giving it a total of 20 electrons. The electron configuration of V\(^{3+}\) is [Ar]3d\(^3\). (b) Sn\(^{4+}\): The atomic number of Sn (tin) is 50, so the neutral atom has 50 electrons. Since the ion has a charge of +4, it has lost 4 electrons, giving it a total of 46 electrons. The electron configuration of Sn\(^{4+}\) is [Kr]4d\(^{10}\)5s\(^2\). (c) I\(^-\): The atomic number of I (iodine) is 53, so the neutral atom has 53 electrons. Since the ion has a charge of -1, it has gained 1 electron, giving it a total of 54 electrons. The electron configuration of I\(^-\) is [Kr]4d\(^{10}\)5s\(^2\)5p\(^6\). (d) W\(^{4+}\): The atomic number of W (tungsten) is 74, so the neutral atom has 74 electrons. Since the ion has a charge of +4, it has lost 4 electrons, giving it a total of 70 electrons. The electron configuration of W\(^{4+}\) is [Xe]4f\(^{14}\)5d\(^2\)6s\(^2\).

Count the unpaired electrons

Now, look at the electron configurations to count the number of unpaired electrons. (a) V\(^{3+}\): In the 3d orbitals, there are 3 electrons that are unpaired. So, there are 3 unpaired electrons in V\(^{3+}\). (b) Sn\(^{4+}\): All electrons in the 4d and 5s orbitals are paired. So, there are 0 unpaired electrons in Sn\(^{4+}\). (c) I\(^-\): All electrons in the 4d, 5s, and 5p orbitals are paired. So, there are 0 unpaired electrons in I\(^-\). (d) W\(^{4+}\): In the 5d orbitals, there are 2 electrons that are unpaired. So, there are 2 unpaired electrons in W\(^{4+}\).