Problem 117 - A Complete Resource Book in Chemistry for JEE Main

Question

Which of the following are diamagnetic? (a) \(\left[\mathrm{Ni}\left(\mathrm{H}_{2} \mathrm{O}\right)\right]^{2+}\) (b) \(\left[\mathrm{Zn}\left(\mathrm{NH}_{3}\right)_{4}\right]^{2+}\) (c) \(\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{2-}\) (d) \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right]^{3+}\)

Step-by-Step Solution

Answer

Diamagnetic: (b) \([\mathrm{Zn}(\mathrm{NH}_3)_4]^{2+}\), (c) \([\mathrm{Ni}(\mathrm{CN})_4]^{2-}\), and (d) \([\mathrm{Co}(\mathrm{NH}_3)_6]^{3+}\).

Step 1: Determine the Electronic Configuration

Determine the electronic configuration of the central metal ions in each complex taking into account their oxidation state.

Step 2: Identify Oxidation States

(a) In \([\mathrm{Ni}(\mathrm{H}_2\mathrm{O})_6]^{2+}\), Ni is in a +2 oxidation state, thus configuration is [Ar] 3d\(^8\).(b) In \([\mathrm{Zn}(\mathrm{NH}_3)_4]^{2+}\), Zn is in a +2 oxidation state, thus configuration is [Ar] 3d\(^{10}\).(c) In \([\mathrm{Ni}(\mathrm{CN})_4]^{2-}\), Ni is in a +2 oxidation state, thus configuration is [Ar] 3d\(^8\).(d) In \([\mathrm{Co}(\mathrm{NH}_3)_6]^{3+}\), Co is in a +3 oxidation state, thus configuration is [Ar] 3d\(^6\).

Step 3: Check the Number of Unpaired Electrons in Each Complex

A complex is diamagnetic if all its electrons are paired.(a) \([\mathrm{Ni}(\mathrm{H}_2\mathrm{O})_6]^{2+}\) has 2 unpaired electrons (high-spin d\(^8\)).(b) \([\mathrm{Zn}(\mathrm{NH}_3)_4]^{2+}\) has 0 unpaired electrons (d\(^{10}\)).(c) \([\mathrm{Ni}(\mathrm{CN})_4]^{2-}\) also typically forms a low-spin square planar complex with 0 unpaired electrons.(d) \([\mathrm{Co}(\mathrm{NH}_3)_6]^{3+}\) is a low-spin d\(^6\) complex, having 0 unpaired electrons.

Step 4: Determine Diamagnetic Compounds

Based on the number of unpaired electrons, (b) \([\mathrm{Zn}(\mathrm{NH}_3)_4]^{2+}\), (c) \([\mathrm{Ni}(\mathrm{CN})_4]^{2-}\), and (d) \([\mathrm{Co}(\mathrm{NH}_3)_6]^{3+}\) are diamagnetic because they do not have any unpaired electrons.

Key Concepts

Electronic Configuration

The electronic configuration of an atom or ion describes the distribution of electrons in its atomic orbitals. Understanding the electronic configuration is crucial in predicting the chemical behavior of an element or compound. To determine the electronic configuration of a metal ion in a complex, you first need to note the number of electrons in the neutral atom of the element. Next, you consider the oxidation state of the metal ion, which tells you how many electrons have been lost.

For example, neutral nickel (Ni) has an electronic configuration of [Ar] 3d\(^8\) 4s\(^2\). When nickel forms the ion \ extit{Ni}^{2+}\, it loses two electrons from the 4s orbital. Thus, the electronic configuration becomes [Ar] 3d\(^8\). Knowing the electronic configuration of the metal within a complex allows you to predict other properties, particularly magnetic ones.

• Look at the periodic table placement to identify the neutral electron count.
• Consider the oxidation state for electron removal.
• Re-arrange the electrons among orbitals to find the configuration of the ion.

Oxidation States

An oxidation state indicates the degree of oxidation (loss of electrons) an atom has undergone in a compound. Oxidation states help identify how many electrons have been removed when forming an ion from a neutral element. It is crucial to know an element's oxidation state in order to determine its electronic configuration and other properties like magnetism.

In coordination chemistry, ligands can influence the metal center's oxidation state. For instance, in the complex \ extit{[Ni(H}_2\(\textit{O)}\)_6\text{]}^{2+}\, nickel has an oxidation state of +2, meaning it has lost two electrons compared to its elemental state. This knowledge helps us further dissect its electronic configuration and magnetic characteristics.

• The sum of oxidation states in a complex equals the overall charge.
• Common oxidation states for transition metals are often +1, +2, or +3.
• Changes in oxidation states can affect bonding and magnetic properties.

Unpaired Electrons

The presence of unpaired electrons in a substance contributes significantly to its magnetic properties. Specifically, complexes can be labeled as either diamagnetic or paramagnetic based on the presence of unpaired electrons. Diamagnetic materials have all electrons paired and do not exhibit magnetic properties in the presence of an external magnetic field, while paramagnetic materials do have unpaired electrons and are attracted to magnetic fields.

For a complex to be diamagnetic, like \ extit{[Zn(NH}_3\text{)}_4\text{]}^{2+}\, all the electrons must be paired, a characteristic of a completely filled d-orbital or a low-spin configuration. In contrast, if a complex, such as \ extit{[Ni(H}_2\(\textit{O)}\)_6\text{]}^{2+}\, has unpaired electrons, it will be paramagnetic. The spin state and type of ligands in a complex can influence whether electrons pair or remain unpaired.

• Unpaired electrons lead to paramagnetic properties.
• Paired electrons signify diamagnetic properties.
• Magnetism is dependent on electron pairing and ligand field strength.