Problem 63 - A Complete Resource Book in Chemistry for JEE Main

Question

Among the following complex ions, the one with the highest paramagnetism is (a) \(\left[\mathrm{FeF}_{6}\right]^{2+}\) (b) \(\left[\mathrm{Cu}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}\) (c) \(\left[\mathrm{Zn}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}\) (d) \(\left[\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{6}\right]^{3+}\)

Step-by-Step Solution

Answer

The complex ion \([\mathrm{FeF}_{6}]^{2+}\) has the highest paramagnetism.

Step 1: Determine Oxidation States

For each complex ion, determine the oxidation state of the central metal atom. This will help identify the electronic configuration.- In \([\mathrm{FeF}_{6}]^{2+},\) iron (Fe) is in the +4 oxidation state.- In \([\mathrm{Cu(H}_{2}\mathrm{O})_{6}]^{2+},\) copper (Cu) is in the +2 oxidation state.- In \([\mathrm{Zn(H}_{2}\mathrm{O})_{6}]^{2+},\) zinc (Zn) is in the +2 oxidation state.- In \([\mathrm{Cr(NH}_{3})_{6}]^{3+},\) chromium (Cr) is in the +3 oxidation state.

Step 2: Determine Electronic Configuration

Determine the electronic configuration of the central metal ion in its respective oxidation state to find out the number of unpaired electrons.- Fe in +4 state: \([\mathrm{Ar}]\ 3d^{4}\).- Cu in +2 state: \([\mathrm{Ar}]\ 3d^{9}\).- Zn in +2 state: \([\mathrm{Ar}]\ 3d^{10}\).- Cr in +3 state: \([\mathrm{Ar}]\ 3d^{3}\).

Step 3: Count Unpaired Electrons

Now count the number of unpaired electrons for each metal ion based on their electronic configuration. - Fe: 4 unpaired electrons. - Cu: 1 unpaired electron. - Zn: 0 unpaired electrons. - Cr: 3 unpaired electrons.

Step 4: Identify the Most Paramagnetic Ion

Paramagnetism is directly proportional to the number of unpaired electrons. The more unpaired electrons, the higher the paramagnetism.- \([\mathrm{FeF}_{6}]^{2+}\) has 4 unpaired electrons (most).- \([\mathrm{Cu(H}_{2}\mathrm{O})_{6}]^{2+}\) has 1 unpaired electron.- \([\mathrm{Zn(H}_{2}\mathrm{O})_{6}]^{2+}\) has 0 unpaired electrons.- \([\mathrm{Cr(NH}_{3})_{6}]^{3+}\) has 3 unpaired electrons.So, \([\mathrm{FeF}_{6}]^{2+}\) is the most paramagnetic.

Key Concepts

Oxidation States

The oxidation state of an atom in a compound tells us how its electrons are behaving and how it is interacting with nearby atoms. To determine the oxidation state, we examine the charge of the ion and the atoms it's bonded to. For transition metals, knowing the oxidation state is crucial because it affects their electronic configuration and how many electrons are unpaired.
- **Fe in \([\mathrm{FeF}_{6}]^{2+}\)**: To find the oxidation state of iron, note that fluoride (F) has a -1 charge. Six fluoride ions contribute a total charge of -6. Since the complex ion has an overall charge of +2, we can set up the equation: \(x + (-6) = +2\), where \(x\) is the oxidation state of iron, giving us \(x = +4\).- **Cu in \([\mathrm{Cu(H}_{2}\mathrm{O})_{6}]^{2+}\)**: Each water molecule is neutral, so the copper's oxidation state is simply the same as the charge on the complex, which is +2.- **Zn in \([\mathrm{Zn(H}_{2}\mathrm{O})_{6}]^{2+}\)**: Similarly, zinc in this complex also has an oxidation state corresponding to the complex's charge: +2.- **Cr in \([\mathrm{Cr(NH}_{3})_{6}]^{3+}\)**: Ammonia (NH₃) is neutral, so the oxidation state of chromium is +3. By determining these oxidation states, we set the stage for understanding how these metals will interact with their surrounding electrons and how they affect paramagnetism.

Electronic Configuration

The electronic configuration of an ion helps us understand the distribution of its electrons in various energy levels or orbitals. Electronic configuration dictates how electrons are arranged in an atom and is key to determining its properties, such as magnetism.
When an atom ionizes, electrons are removed starting from the outermost orbitals. For transition metals, we primarily focus on the d orbitals:- **Fe in the +4 state**: The ground state configuration of Fe is \([\mathrm{Ar}] \, 3d^6\). Loss of 4 electrons results in \([\mathrm{Ar}] \, 3d^4\), leaving us with four d electrons.- **Cu in the +2 state**: Copper in its elemental form is \([\mathrm{Ar}] \, 3d^{10} 4s^1\). After losing two electrons, we find \([\mathrm{Ar}] \, 3d^9\).- **Zn in the +2 state**: Zinc's ground state is \([\mathrm{Ar}] \, 3d^{10} 4s^2\). Upon losing two electrons, it becomes \([\mathrm{Ar} ] \, 3d^{10}\), maintaining a full d orbital.- **Cr in the +3 state**: Chromium starts with \([\mathrm{Ar}] \, 3d^5 4s^1\) in its ground state. By losing three electrons, it results in \([\mathrm{Ar} ] \, 3d^3\), which directly influences its magnetic properties.Understanding the electron configurations tells us how these atoms will behave. It helps us predict their interactions in the magnetic field, based on how their electrons are distributed.

Unpaired Electrons

The concept of unpaired electrons is crucial in determining the magnetic properties of a complex ion. Unpaired electrons serve as tiny magnets that contribute to the overall magnetism of a substance, particularly paramagnetism.
- **Iron (Fe):** With an electron configuration of \([\mathrm{Ar}]3d^4\), in the +4 oxidation state, Fe has four unpaired electrons. This results in significant paramagnetism, as the unpaired electrons enhance the magnetic field of the ion.- **Copper (Cu):** In the +2 state, with \([\mathrm{Ar}]3d^9\), Cu has only one unpaired electron, leading to weaker paramagnetism compared to Fe.- **Zinc (Zn):** In the +2 state, Zn boasts a complete 3d orbital (\mathrm{3d^{10}}), meaning all the electrons are paired. As a result, it shows no paramagnetism.- **Chromium (Cr):** The \([\mathrm{Ar}]3d^3\) configuration for the +3 state hints that Cr has three unpaired electrons, making it paramagnetic, but less so than Fe.The more unpaired electrons present, the stronger the paramagnetism. Hence, \([\mathrm{FeF}_{6}]^{2+}\) exhibits the most paramagnetism among the given ions, because it has the maximum number of unpaired electrons.