Problem 49 - Chemistry: The Central Science

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?

Step-by-Step Solution

Answer

The electron configurations for the \(\mathrm{Ni}^{2+}\) and \(\mathrm{Sn}^{2+}\) ions are as follows: \(\mathrm{Ni^{2+}}: 1s^2 2s^2 2p^6 3s^2 3p^6 3d^8\) \(\mathrm{Sn^{2+}}: 1s^2 2s^2 2p^6 3s^2 3p^6 3d^{10} 4s^2 4p^6 4d^{10} 5s^2\) The \(\mathrm{Ni}^{2+}\) ion has 2 unpaired electrons while the \(\mathrm{Sn}^{2+}\) ion has 0 unpaired electrons.

Step 1: Identify the ground state electron configurations of the neutral Ni and Sn atoms

Using the periodic table, we can find that Ni (Nickel) has an atomic number of 28 and Sn (Tin) has an atomic number of 50. For Ni: \(1s^2 2s^2 2p^6 3s^2 3p^6 3d^8 4s^2\) For Sn: \(1s^2 2s^2 2p^6 3s^2 3p^6 3d^{10} 4s^2 4p^6 4d^{10} 5s^2 5p^2\)

Step 2: Remove electrons to account for the ion charge

For the \(\mathrm{Ni}^{2+}\) ion, we need to remove 2 electrons from the neutral Ni atom: Ni: \(1s^2 2s^2 2p^6 3s^2 3p^6 3d^8 4s^2\) becomes \(\mathrm{Ni^{2+}}: 1s^2 2s^2 2p^6 3s^2 3p^6 3d^8\) For the \(\mathrm{Sn}^{2+}\) ion, we need to remove 2 electrons from the neutral Sn atom: Sn: \(1s^2 2s^2 2p^6 3s^2 3p^6 3d^{10} 4s^2 4p^6 4d^{10} 5s^2 5p^2\) becomes \(\mathrm{Sn^{2+}}: 1s^2 2s^2 2p^6 3s^2 3p^6 3d^{10} 4s^2 4p^6 4d^{10} 5s^2\)

Step 3: Determine the number of unpaired electrons

For the \(\mathrm{Ni}^{2+}\) ion, we can see that the electron configuration is \(1s^2 2s^2 2p^6 3s^2 3p^6 3d^8\). There are 8 electrons in the 3d orbitals, which means that 2 of the orbitals are fully occupied, and 4 electrons are unpaired. For the \(\mathrm{Sn}^{2+}\) ion, we can see that the electron configuration is \(1s^2 2s^2 2p^6 3s^2 3p^6 3d^{10} 4s^2 4p^6 4d^{10} 5s^2\). There are no unpaired electrons in this ion since all the orbitals are fully occupied. So, the \(\mathrm{Ni}^{2+}\) ion has 2 unpaired electrons, and the \(\mathrm{Sn}^{2+}\) ion has 0 unpaired electrons.

Key Concepts

Transition Metals

Transition metals are the elements found in groups 3 to 12 on the periodic table, characterized by their ability to form compounds with variable oxidation states, colored compounds, and to act as catalysts. These elements have d-electrons that are not in the outermost shell, an attribute leading to a rich chemistry and the formation of complexes.

For example, nickel (Ni) is a transition metal that commonly forms a +2 ionic charge, known as the Ni^2+ ion. This is because transition metals tend to lose the s-electrons of their highest electron shell first, before losing d-electrons. In the case of nickel, the 4s electrons are removed before the 3d electrons when forming ions.

Ionic Charge

Ionic charge refers to the electric charge that an atom acquires when it has gained or lost electrons to achieve a fuller outer shell, thereby becoming more stable. In transition metals, such as Ni and Sn, the common charges are often +2 or +3 or higher depending on the element concerned, which correlates with the fact that their d orbitals are initially only partially filled.

The process of losing electrons to form a positive ion is called oxidation. Contrastingly, when an atom gains electrons, forming a negative ion, it is known as reduction. For instance, when nickel loses two electrons, it gets oxidized to form Ni^2+. It is important to recognize that when transition metals form ions, they generally lose their s orbital electrons before any d orbital electrons.

Unpaired Electrons

Unpaired electrons are those that do not have a paired electron with opposite spin in the same orbital. Having unpaired electrons makes an atom or ion paramagnetic, which means it is attracted to magnetic fields. This property is particularly important in transition metals, which often have several unpaired electrons.

For example, neutral nickel (Ni) has the electron configuration [Ar] 3d^8 4s^2, but when it forms a Ni^2+ ion, it loses the 4s electrons, resulting in the electron configuration [Ar] 3d^8. In this configuration, nickel has 2 unpaired electrons. These unpaired d-electrons are responsible for many of the magnetic and chemical behaviors of transition metal ions. In contrast, tin (Sn) in the Sn^2+ ion has no unpaired electrons, hence it is diamagnetic and not attracted to magnetic fields.