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 objective is to determine the spin-only magnetic moment for the ion exhibiting the weakest oxidizing power among \(VO_2^+\), \(MnO_4^-\), and \(Cr_2O_7^{2-}\).

Concept Used:

1. Oxidation State Determination: Ascertain the oxidation state of the central transition metal in each given oxoanion/cation.

2. Oxidizing Ability Correlation: Oxidizing agents readily accept electrons and undergo reduction. For transition metal oxoanions, oxidizing strength generally escalates with the metal's oxidation state. The species with the lowest metal oxidation state will possess the weakest oxidizing ability.

3. Spin-only Magnetic Moment (\(\mu_s\)): The magnetic moment of a transition metal ion arises primarily from the spin of its unpaired electrons. This moment is quantified by the formula:

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

where \(n\) represents the count of unpaired electrons in the metal ion's d-orbitals, and B.M. denotes Bohr Magneton.

Step-by-Step Solution:

Step 1: Calculate the oxidation state of the central metal atom for each species.

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

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

The oxidation states are V(+5), Cr(+6), and Mn(+7). A higher oxidation state correlates with stronger oxidizing power; therefore, the species with the lowest oxidation state is the weakest oxidizing agent.

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

Consequently, \(VO_2^+\) exhibits 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 \(VO_2^+\), containing \(V^{5+}\).

Vanadium (V) has atomic number 23 and an electronic configuration of:

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

Removal of five electrons to form \(V^{5+}\) (two from 4s and three from 3d) yields:

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

With an empty 3d subshell, there are no unpaired electrons, so \(n = 0\).

Final Computation & Result:

Step 4: Compute the spin-only magnetic moment using \(n\).

Applying 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 nearest integer value is 0.

The spin-only magnetic moment for the species with the least oxidizing ability is 0 BM.