Question

Predict the total spin in \(\mathrm{Ni}^{2+}\) ion (1) \(\pm \frac{5}{2}\) (2) \(\pm \frac{3}{2}\) (3) \(\pm \frac{1}{2}\) (4) \(\pm 1\)

Short answer

Option (4), \( \text{\textpm 1} \).

Step-by-step solution

Determine the Electron Configuration of Neutral Nickel

Nickel (Ni) has an atomic number of 28. The electron configuration for a neutral Ni atom is \[ \text{[Ar]} 3d^8 4s^2 \].

Find the Electron Configuration of \(\text{Ni}^{2+}\) Ion

When a neutral Ni atom loses two electrons to form \(\text{Ni}^{2+}\), the two 4s electrons are removed first, giving the electron configuration: \[ \text{[Ar]} 3d^8 \].

Calculate the Number of Unpaired Electrons

In the 3d \(3d^8\) subshell, there are 8 electrons. To determine the number of unpaired electrons, sketch or visualize the placement in the five d orbitals: two orbitals will be fully filled (2 electrons each), and the remaining three will each contain 1 unpaired electron. Therefore, there are 2 unpaired electrons.

Determine the Total Spin

Each unpaired electron contributes a spin of \( + \frac{1}{2} \) or \( - \frac{1}{2} \). Since there are 2 unpaired electrons, the total spin, denoted as \(\text{S}\), will be: \[ S = \frac{1}{2} + \frac{1}{2} = 1 \].

Final Step: Evaluate the Options

Based on the total spin calculation, the total spin for \(\text{Ni}^{2+}\) ion is \( \text{S} = 1 \), which corresponds to option (4): \(\text{\textpm 1}\).

Key concepts

Electron Configuration

Electrons are arranged around the nucleus of an atom in specific orbitals. These arrangements form what we call the electron configuration of an element. Each orbital can hold a certain number of electrons, following the pattern:

• s orbital: 2 electrons
• p orbital: 6 electrons
• d orbital: 10 electrons

For neutral Nickel (Ni), which has an atomic number of 28, the electron configuration is \[ \text{[Ar]} 3d^8 4s^2 \]. This means that after the electrons fill the Argon core ([Ar]), they occupy the 3d and 4s orbitals. When Nickel becomes a \(\text{Ni}^{2+}\) ion, it loses two electrons—specifically, the two electrons in the 4s orbital. This results in an electron configuration of \[ \text{[Ar]} 3d^8 \]. Understanding this is crucial for determining further properties, like unpaired electrons and total spin.

Unpaired Electrons

Unpaired electrons are those that are not paired with another electron in the same orbital. In the case of the \(\text{Ni}^{2+}\) ion, after losing the two 4s electrons, it has a 3d^8 configuration. The placement of electrons in the d orbitals can be visualized as: a fully filled orbital can hold 2 electrons. In the 3d orbital, there are five possible orbitals, and 8 electrons to place:

• Two orbitals will be fully filled, having 2 electrons each.
• The remaining six electrons will be distributed among the remaining three orbitals.

When visualizing this, two electrons will pair up in two orbitals, leaving three electrons unpaired. Thus, there are 2 unpaired electrons for the \(\text{Ni}^{2+}\) ion.

Spin Quantum Number

The Spin Quantum Number describes the intrinsic angular momentum of an electron within an atom. Its values can be either \(+\frac{1}{2}\) or \(-\frac{1}{2}\), often called 'spin-up' and 'spin-down'. For each unpaired electron, you must account for its spin. In \(\text{Ni}^{2+}\), we have identified 2 unpaired electrons. Each unpaired electron contributes a spin of \(+\frac{1}{2}\) or \(-\frac{1}{2}\). When summing these spins, we aim to find the total spin (\text{S}). For the two unpaired electrons in \(\text{Ni}^{2+}\): \[ S = \frac{1}{2} + \frac{1}{2} = 1 \] Thus, the total spin of the \(\text{Ni}^{2+}\) ion is \(\text{S} = 1\). This calculation is used to understand and predict magnetic properties and other quantum characteristics of ions and atoms.