Problem 9 - A Complete Resource Book in Chemistry for JEE Main

Question

Which of the following order is correct for the size of \(\mathrm{Fe}^{3+}, \mathrm{Fe}\) and \(\mathrm{Fe}^{2+} ?\) (a) \(\mathrm{Fe}^{3+}<\mathrm{Fe}^{2+}<\mathrm{Fe}\) (b) \(\mathrm{Fe}<\mathrm{Fe}^{3+}<\mathrm{Fe}^{2+}\) (c) \(\mathrm{Fe}<\mathrm{Fe}^{2+}<\mathrm{Fe}^{3+}\) (d) \(\mathrm{Fe}^{2+}<\mathrm{Fe}^{3+}<\mathrm{Fe}\)

Step-by-Step Solution

Answer

The correct order is (a) Fe^{3+}

Step 1: Understand the Concept

The size of an atom or ion is determined by its electron configuration and the number of electrons it contains. The more electrons an atom or ion has, the larger it tends to be due to increased electron-electron repulsion, unless the additional protons in the nucleus pull the outer electrons more tightly.

Step 2: Analyze Neutral Iron (Fe)

Neutral iron (Fe) has an atomic number of 26, which means it has 26 electrons. The electron configuration can lead to establishing the baseline size of the atom. Since it has no charge, it is the largest among the options provided, assuming no loss of electron layers.

Step 3: Consider Iron(II) Ion (Fe^{2+})

The Fe^{2+} ion is formed when Fe loses two electrons. Fewer electrons result in a reduced electron-electron repulsion leading to a decrease in size. However, it still has more electrons compared to Fe^{3+}.

Step 4: Evaluate Iron(III) Ion (Fe^{3+})

The Fe^{3+} ion is formed when Fe loses three electrons. This means it has 23 electrons. With even fewer electrons than Fe^{2+}, the pull from the protons in the nucleus is more effective at reducing the size of the ion.

Step 5: Determine the Order

Based on the analysis, Fe^{3+} is the smallest due to the highest positive charge drawing electrons closer, followed by Fe^{2+}, and then neutral Fe being the largest. Thus, the order is Fe^{3+}

Key Concepts

Electron Configuration

Electron configuration is a way of understanding how electrons are distributed in an atom. Each element has a unique arrangement of electrons in its atomic orbitals. The electron configuration of an atom can be determined based on the principle of filling orbitals in order of their energy levels, starting with the lowest energy (1s) moving upwards. For example, the neutral iron atom (Fe) has 26 electrons and its configuration is written as \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^6\). This notation indicates that iron's electrons fill up the 1s, 2s, 2p, 3s, 3p, 4s, and 3d orbitals in that order.

Neutral atoms are larger because their outer electrons are not as tightly bound by the attraction of the protons in the nucleus compared to ions. When atoms lose electrons to become cations, as in \(\text{Fe}^{2+}\) and \(\text{Fe}^{3+}\), their electron configuration changes, typically resulting in reduced sizes. This is due to fewer electrons in outer orbitals, which lessens repulsion among electrons, allowing the remaining electrons to be pulled closer to the nucleus, thus affecting atomic size.

Iron Oxidation States

Iron exhibits various oxidation states, with \(+2\) and \(+3\) being the most common ones. This indicates how many electrons an atom has lost. For iron, transitioning from the neutral state to \(\text{Fe}^{2+}\) means losing two electrons, while going to \(\text{Fe}^{3+}\) means losing three electrons. Each removal of an electron results in a positive charge, reducing the size of the ion because there are fewer electrons repelling each other.

When an ion has fewer electrons, the electrostatic pull from protons in the nucleus becomes more pronounced, further decreasing the atomic radius. Therefore, while \(\text{Fe}^{3+}\) is smaller than \(\text{Fe}^{2+}\), neutral \(\text{Fe}\) being the largest due to having the full number of electrons before any oxidation occurs. In terms of sizes for these oxidation states, it follows the trend: \(\text{Fe}^{3+} < \text{Fe}^{2+} < \text{Fe}\). This highlights how variations in electron count influence atomic size and ionic character.

Ionization and Ionic Radius

The concept of ionization refers to the process by which an atom or molecule acquires a charge by gaining or losing electrons. This process directly influences an ion's radius, also known as the ionic radius. When an atom loses an electron, forming a cation, it generally becomes smaller. This size reduction occurs because the loss of negative charge means fewer electrons are available to repel each other, allowing the positive nuclear charge to draw the remaining electrons closer.

For \(\text{Fe}^{2+}\) and \(\text{Fe}^{3+}\), the radius decreases further as the number of electrons decreases. With \(\text{Fe}^{2+}\), there are 24 electrons compared to 23 in \(\text{Fe}^{3+}\), resulting in \(\text{Fe}^{3+}\) being even smaller. Additionally, more electrostatic pulling power from the protons in \(\text{Fe}^{3+}\) huddles the electrons more effectively, leading to a smaller ionic radius than \(\text{Fe}^{2+}\), which in turn is smaller than the neutral \(\text{Fe}\).

Understanding ionization and resulting changes in ionic radius is crucial for explaining how ionic size varies with the number of electrons retained in ions.

Previous

Problem 8

Next

Problem 10

Other exercises in this chapter

Problem 7

Misch metal is (a) an alloy of copper (b) an alloy of lanthanoid metal (c) an alloy of aluminium (d) a mixture of chromium and lead chromate

View solution

Problem 8

Which one of the following forms, with an excess of \(\mathrm{CN}^{-}\), a complex having coordination number two? (a) \(\mathrm{Ni}^{2+}\) (b) \(\mathrm{Cu}^{+

View solution

Problem 10

Which of the following pairs of ions is colourless? (a) \(\mathrm{Ti}^{-3+}, \mathrm{Cu}^{2+}\) (b) \(\mathrm{Sc}^{3+}, \mathrm{Zn}^{2+}\) (c) \(\mathrm{Co}^{2+

View solution

Problem 11

Which of the following pairs of ions have same paramagnetic moment? (a) \(\mathrm{Cu}^{2+}, \mathrm{Ti}^{3+}\) (b) \(\mathrm{Ti}^{3+}, \mathrm{Ni}^{2+}\) (c) \(

View solution

Key Concepts

Electron Configuration

Electron configuration is a way of understanding how electrons are distributed in an atom. Each element has a unique arrangement of electrons in its atomic orbitals. The electron configuration of an atom can be determined based on the principle of filling orbitals in order of their energy levels, starting with the lowest energy (1s) moving upwards. For example, the neutral iron atom (Fe) has 26 electrons and its configuration is written as \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^6\). This notation indicates that iron's electrons fill up the 1s, 2s, 2p, 3s, 3p, 4s, and 3d orbitals in that order.

Neutral atoms are larger because their outer electrons are not as tightly bound by the attraction of the protons in the nucleus compared to ions. When atoms lose electrons to become cations, as in \(\text{Fe}^{2+}\) and \(\text{Fe}^{3+}\), their electron configuration changes, typically resulting in reduced sizes. This is due to fewer electrons in outer orbitals, which lessens repulsion among electrons, allowing the remaining electrons to be pulled closer to the nucleus, thus affecting atomic size.

Iron Oxidation States

Iron exhibits various oxidation states, with \(+2\) and \(+3\) being the most common ones. This indicates how many electrons an atom has lost. For iron, transitioning from the neutral state to \(\text{Fe}^{2+}\) means losing two electrons, while going to \(\text{Fe}^{3+}\) means losing three electrons. Each removal of an electron results in a positive charge, reducing the size of the ion because there are fewer electrons repelling each other.

When an ion has fewer electrons, the electrostatic pull from protons in the nucleus becomes more pronounced, further decreasing the atomic radius. Therefore, while \(\text{Fe}^{3+}\) is smaller than \(\text{Fe}^{2+}\), neutral \(\text{Fe}\) being the largest due to having the full number of electrons before any oxidation occurs. In terms of sizes for these oxidation states, it follows the trend: \(\text{Fe}^{3+} < \text{Fe}^{2+} < \text{Fe}\). This highlights how variations in electron count influence atomic size and ionic character.

Ionization and Ionic Radius

The concept of ionization refers to the process by which an atom or molecule acquires a charge by gaining or losing electrons. This process directly influences an ion's radius, also known as the ionic radius. When an atom loses an electron, forming a cation, it generally becomes smaller. This size reduction occurs because the loss of negative charge means fewer electrons are available to repel each other, allowing the positive nuclear charge to draw the remaining electrons closer.

For \(\text{Fe}^{2+}\) and \(\text{Fe}^{3+}\), the radius decreases further as the number of electrons decreases. With \(\text{Fe}^{2+}\), there are 24 electrons compared to 23 in \(\text{Fe}^{3+}\), resulting in \(\text{Fe}^{3+}\) being even smaller. Additionally, more electrostatic pulling power from the protons in \(\text{Fe}^{3+}\) huddles the electrons more effectively, leading to a smaller ionic radius than \(\text{Fe}^{2+}\), which in turn is smaller than the neutral \(\text{Fe}\).

Understanding ionization and resulting changes in ionic radius is crucial for explaining how ionic size varies with the number of electrons retained in ions.