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) V3+ has 2 unpaired electrons, b) Sn4+ has 0 unpaired electrons, c) I has 5 unpaired electrons, and d) W4+ has 2 unpaired electrons.

Step-by-step solution

a) V3+ Electron Configuration and Unpaired Electrons

To find the electron configuration of \(\mathrm{V}^{3+}\), first we need to determine the ground state electron configuration of Vanadium (V). The atomic number of V is 23, so its ground state configuration is: \(1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{3}\). To obtain the configuration of \(\mathrm{V}^{3+}\), remove 3 electrons from the highest energy levels. Two electrons come from the \(4s\) orbital, and one comes from the \(3d\) orbital, leading to the electron configuration: \(1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{2}\). There are two unpaired electrons in the \(3d\) orbital.

b) Sn4+ Electron Configuration and Unpaired Electrons

To find the electron configuration of \(\mathrm{Sn}^{4+}\), first, determine the ground state electron configuration of Tin (Sn). The atomic number of Sn is 50, so its ground state configuration is: \(1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}5s^{2}4d^{10}5p^{2}\). To obtain the configuration of \(\mathrm{Sn}^{4+}\), remove 4 electrons from the highest energy levels. Two electrons come from the \(5p\) orbital, while the other two come from the \(5s\) orbital. This leads to the electron configuration: \(1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}4d^{10}\). In this configuration, all of the electrons are paired, resulting in 0 unpaired electrons.

c) I Electron Configuration and Unpaired Electrons

To find the electron configuration of Iodine (I), first, determine its atomic number. The atomic number of I is 53, so its ground state configuration is: \(1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}5s^{2}4d^{10}5p^{5}\). As Iodine is a neutral atom, we don't need to remove any electrons. The electron configuration shows that it has 5 unpaired electrons in the \(5p\) orbital.

d) W4+ Electron Configuration and Unpaired Electrons

To find the electron configuration of \(\mathrm{W}^{4+}\), first, determine the ground state electron configuration of Tungsten (W). The atomic number of W is 74, so its ground state configuration is: \(1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}5s^{2}4d^{10}5p^{6}6s^{2}4f^{14}5d^{4}\). To obtain the configuration of \(\mathrm{W}^{4+}\), remove 4 electrons from the highest energy levels. Two electrons come from the \(6s\) orbital, and two come from the \(5d\) orbital, leading to the electron configuration: \(1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}5s^{2}4d^{10}5p^{6}4f^{14}5d^{2}\). There are two unpaired electrons in the \(5d\) orbital.