Question

Among $VO_2^+$, $MnO_4^-$, and $Cr_2O_7^{2-}$, the spin-only magnetic moment value of the species with least oxidising ability is _____________ BM (Nearest integer).(Given atomic number $V = 23$, $Mn = 25$, $Cr = 24$)

Answer 1

Correct Answer: 0

The problem asks for the spin-only magnetic moment of the species with the least oxidizing ability among the given ions: \(VO_2^+\), \(MnO_4^-\), and \(Cr_2O_7^{2-}\).

Concept Used:

1. Oxidation State: The first step is to determine the oxidation state of the central transition metal atom in each of the given oxoanions/cations.

2. Oxidizing Ability: An oxidizing agent is a substance that has a tendency to accept electrons and get reduced. For transition metal oxoanions, the oxidizing ability generally increases with the increase in the oxidation state of the central metal atom. The species with the metal in the lowest oxidation state will be the weakest oxidizing agent.

3. Spin-only Magnetic Moment (\(\mu_s\)): The magnetic moment of a transition metal ion is primarily due to the spin of its unpaired electrons. The spin-only magnetic moment is calculated using the formula:

\[ \mu_s = \sqrt{n(n+2)} \, \text{B.M.} \]

where \(n\) is the number of unpaired electrons in the d-orbitals of the metal ion, and B.M. stands for Bohr Magneton.

Step-by-Step Solution:

Step 1: Determine the oxidation state of the central metal atom in each species.

• For \(VO_2^+\): Let the oxidation state of Vanadium (V) be \(x\). The oxidation state of oxygen is -2. \[ x + 2(-2) = +1 \implies x - 4 = +1 \implies x = +5 \] The oxidation state of V is +5.
• For \(MnO_4^-\): Let the oxidation state of Manganese (Mn) be \(y\). \[ y + 4(-2) = -1 \implies y - 8 = -1 \implies y = +7 \] The oxidation state of Mn is +7.
• For \(Cr_2O_7^{2-}\): Let the oxidation state of Chromium (Cr) be \(z\). \[ 2z + 7(-2) = -2 \implies 2z - 14 = -2 \implies 2z = 12 \implies z = +6 \] The oxidation state of Cr is +6.

Step 2: Identify the species with the least oxidizing ability.

The oxidation states are V(+5), Cr(+6), and Mn(+7). Since a higher oxidation state generally corresponds to a stronger oxidizing ability, the species with the lowest oxidation state will be the weakest oxidizing agent.

Comparing the oxidation states: \(+5 < +6 < +7\).

Therefore, \(VO_2^+\) is the species with the least oxidizing ability.

Step 3: Determine the number of unpaired electrons (\(n\)) in the central atom of the weakest oxidizing agent.

The weakest oxidizing agent is \(VO_2^+\), where the central atom is \(V^{5+}\).

The atomic number of Vanadium (V) is 23. Its ground-state electronic configuration is:

\[ V: [Ar] \, 3d^3 4s^2 \]

To form the \(V^{5+}\) ion, five electrons are removed (two from the 4s orbital and three from the 3d orbital).

The electronic configuration of \(V^{5+}\) is:

\[ V^{5+}: [Ar] \, 3d^0 4s^0 \]

Since the 3d subshell is empty, there are no unpaired electrons. Thus, \(n = 0\).

Final Computation & Result:

Step 4: Calculate the spin-only magnetic moment using the value of \(n\).

Using the formula \(\mu_s = \sqrt{n(n+2)}\) with \(n = 0\):

\[ \mu_s = \sqrt{0(0+2)} \] \[ \mu_s = \sqrt{0} = 0 \, \text{B.M.} \]

The calculated spin-only magnetic moment is 0 B.M. The value to the nearest integer is 0.

The spin-only magnetic moment value of the species with the least oxidising ability is 0 BM.

Answer 2

For 3d transition series:

\[ \text{Oxidising power: } V^{+5} < Cr^{+6} < Mn^{+7} \]

For \( V^{+5} \):

\[ [\text{Ar}] 4s^0 3d^0 \]

Number of unpaired electrons = 0

\[ \mu = 0 \]