A compound of vanadium has a magnetic moment \((\mu)\) of \(1.73 \mathrm{BM}\). If the vanadium ion in the compound is present as \(\mathrm{V}^{x+}\), then, the value of \(x\) is? (a) 1 (b) 2 (c) 3 (d) 4

Determining the number of unpaired electrons in a transition metal ion is crucial for understanding its magnetic properties. Unpaired electrons are those that remain alone in an orbital and therefore contribute to the magnetic moment of the atom or ion. In physical chemistry, particularly in solving JEE problems, students often come across transition metals like vanadium which can exist in various oxidation states, influencing its number of unpaired electrons.

The step-by-step solution provided utilizes the formula \(\mu = \sqrt{n(n+2)}\) to calculate the number of unpaired electrons, where \(\mu\) is the magnetic moment in Bohr magnetons (BM) and \(n\) is the number of unpaired electrons. Starting with the magnetic moment \(\mu = 1.73 \mathrm{BM}\), one sets up the formula to solve for \(n\): \(n = \sqrt{\mu^2 -2}\). By plugging in the known value of \(\mu\), students can find the value of \(n\), which represents the number of unpaired electrons in the vanadium ion.

Understanding the calculation of the magnetic moment is a vital skill in physical chemistry and is often tested in exams such as the JEE. The magnetic moment gives us an insight into the behavior of a compound in a magnetic field and is directly related to the number of unpaired electrons in the atom or ion.

The formula \(\mu = \sqrt{n(n+2)}\) is derived from quantum mechanics and is a direct way to correlate the number of unpaired electrons to the measured magnetic moment. Here, \(\mu\) is the magnetic moment expressed in Bohr magnetons (BM). After calculating \(n\), which gives the number of unpaired electrons, you can use this number to predict magnetic properties such as diamagnetism or paramagnetism of the compound.

In the context of transition metals, like vanadium, oxidation states are integral to understanding the chemical behavior of compounds. Transition metals are known for their ability to exist in multiple oxidation states, which are in turn responsible for the metal's various colors, magnetic properties, and reactivities.

The oxidation state of an ion denotes the number of electrons lost and is indicated by a superscript plus or minus sign (e.g., \(\mathrm{V}^{3+}\) indicates that vanadium has lost three electrons). In the exercise, we examine the vanadium ion \(\mathrm{V}^{x+}\) to find its oxidation state by determining the number of unpaired electrons. For instance, vanadium has the electronic configuration \(3d^34s^2\) in its neutral state, which changes as electrons are lost to achieve a positive oxidation state.

When tackling physical chemistry problems in the Joint Entrance Examination (JEE), students face a range of questions, from calculating magnetic moment to determining oxidation states. A good grasp of core concepts and problem-solving strategies is key to success.

In the case of our problem, it pertains to a real-world application of how to deduce the oxidation state of a transition metal ion from its magnetic moment. These types of problems require a clear understanding of electronic configurations, magnetic properties of ions, and the principle of quantum numbers. Practice and familiarity with such concepts can be the difference in scoring well on these sections of the JEE.