Paramagnetic substances are those which attract the external magnetic field or which move from a weaker to stronger part of the magnetic field. Paramagnetic substances contain unpaired electrons. Diamagnetic substances are those which repel the external magnetic field or which tend to move away from stronger part to weaker part of the magnetic field. Diamagnetic substances are those which contain all paired electrons. Ferromagnetic substances are those which are attracted very strongly into the applied magnetic field. In ferromagnetic substances, the spins of all unpaired electrons and thus their magnetic moments are aligned in the same direction. Ferromagnetism is an extreme form of paramagnetism. Iron, Cobalt and Nickel are ferromagnetic substances. The magnetic property of a substance is due the spin angular momentum and orbital angular momentum of the electron and can be calculated using the formula

\[\mu_{S + L}= \sqrt{4S(S+1)+L(L+1)}BM\]

where S is the sum of spin quantum numbers of all electrons and L is the sum of orbital angular momentum quantum number. The unit of magnetic moment is Bohr magnetons (BM).

\[1 BM = \frac{eh}{4\pi m_e} = 9.237 × 10^{-21} \text{erg Gauss}^{-1}\]

In many compounds of 3d series metals, the magnetic moment due to orbital movement of electron is neglected as the orbital movement of these electrons is quenched by the surrounding species in compound or solution. The spin only magnetic moment can be calculated by using the spin only formula

\[\mu_s = \sqrt{n(n + 2)}BM\]

The magnetic moment of an ion in its +3 oxidation state is 3.85 BM. The number of unpaired electrons in it is

3

The correct answer is option 3. 3.

To determine the number of unpaired electrons in an ion with a given magnetic moment, we use the relationship between the magnetic moment and the number of unpaired electrons for transition metal ions in different oxidation states. The magnetic moment (\( \mu \)) of a transition metal ion in units of Bohr magnetons (BM) is related to the number of unpaired electrons (\( n \)) by the formula:

\( \mu = \sqrt{n(n+2)}\)

Given:

\(\mu = 3.85 \text{ BM}\)

We need to solve for \( n \):

\(3.85 = \sqrt{n(n+2)}\)

Squaring both sides:

\((3.85)^2 = n(n+2)\)
\(14.8225 = n^2 + 2n\)

Now, solve this quadratic equation:

\(n^2 + 2n - 14.8225 = 0\)

Using the quadratic formula \( n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = 2 \), and \( c = -14.8225 \):

\(n = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 1 \cdot (-14.8225)}}{2 \cdot 1}\)

\(n = \frac{-2 \pm \sqrt{4 + 59.29}}{2}\)

\(n = \frac{-2 \pm \sqrt{63.29}}{2}\)

\(n = \frac{-2 \pm 7.96}{2}\)

Calculating the roots:

\(n = \frac{5.96}{2} = 2.98\approx 3 \quad \text{or} \quad n = \frac{-9.96}{2} = -4.98\)

Since \( n \) represents the number of unpaired electrons and must be a positive integer, we take:

\(n = 3\)

Therefore, the number of unpaired electrons in the ion in its +3 oxidation state, given the magnetic moment of 3.85 BM, is 3.