A divalent ion of 'V' (Atomic no. 23) in aqueous solution is:
√15 BM
Analysing the Magnetic Moment of a Divalent Vanadium Ion
The question asks us to determine the magnetic moment of a divalent ion of Vanadium (V) in aqueous solution. The atomic number of Vanadium (V) is given as 23. A divalent ion means it carries a charge of +2, so we are looking at the V2+ ion.
Understanding Vanadium's Electronic Configuration
First, let's write the electronic configuration of a neutral Vanadium atom (Z=23). Vanadium is a transition metal in the first series. Its electronic configuration is:
\(V: [Ar] 3d^3 4s^2\)
Forming the Divalent V$^{2+}$ Ion
When a transition metal atom forms a positive ion, it typically loses electrons first from the outermost shell, which is the 4s orbital, before losing electrons from the 3d orbitals. To form the V$^{2+}$ ion, the Vanadium atom loses 2 electrons. These two electrons are removed from the 4s orbital.
So, the electronic configuration of the V$^{2+}$ ion is:
\(V^{2+}: [Ar] 3d^3\)
Calculating Unpaired Electrons in V$^{2+}$
The magnetic properties of transition metal ions are largely determined by the number of unpaired electrons in the d orbitals. The 3d subshell has 5 orbitals. According to Hund's rule, electrons will singly occupy each orbital in a subshell before pairing up.
For the \(3d^3\) configuration of V$^{2+}$, we have 3 electrons to place in the 5 d orbitals. These three electrons will occupy three different d orbitals individually, each with the same spin.
Thus, the number of unpaired electrons (n) in V$^{2+}$ is 3.
Spin-Only Magnetic Moment Calculation
The magnetic moment (\(\mu\)) of a transition metal ion due to electron spin can be calculated using the spin-only formula:
\(\mu_s = \sqrt{n(n+2)}\) BM
where:
- \(\mu_s\) is the spin-only magnetic moment.
- \(n\) is the number of unpaired electrons.
- BM stands for Bohr Magnetons, the unit of magnetic moment.
In the case of V$^{2+}$, we found that the number of unpaired electrons \(n = 3\). Now, let's plug this value into the formula:
\(\mu_s = \sqrt{3(3+2)}\) BM
\(\mu_s = \sqrt{3(5)}\) BM
\(\mu_s = \sqrt{15}\) BM
Conclusion on V$^{2+}$ Magnetic Moment
The calculated spin-only magnetic moment for the divalent Vanadium ion, V$^{2+}$, in aqueous solution is \(\sqrt{15}\) BM. This value corresponds to one of the given options.
Revision Table: Magnetic Moment of Ions
| Ion | Electronic Configuration | Number of Unpaired Electrons (n) | Spin-Only Magnetic Moment (\(\sqrt{n(n+2)}\) BM) |
|---|---|---|---|
| V (Z=23) | \([Ar] 3d^3 4s^2\) | N/A (Atom) | N/A |
| V$^{2+}$ | \([Ar] 3d^3\) | 3 | \(\sqrt{3(3+2)} = \sqrt{15}\) BM |
| V$^{3+}$ | \([Ar] 3d^2\) | 2 | \(\sqrt{2(2+2)} = \sqrt{8}\) BM |
| V$^{4+}$ | \([Ar] 3d^1\) | 1 | \(\sqrt{1(1+2)} = \sqrt{3}\) BM |
| V$^{5+}$ | \([Ar]\) | 0 | \(\sqrt{0(0+2)} = 0\) BM |
Additional Information: Transition Metal Magnetic Moments
Transition metal ions in aqueous solution often exist as complex ions, with water molecules acting as ligands. These ligands can affect the splitting of the d orbitals, which can sometimes influence the number of unpaired electrons (in cases of strong field ligands causing pairing). However, for first transition series ions with 3 or fewer d electrons (\(d^1, d^2, d^3\)) or 8 or more d electrons (\(d^8, d^9, d^{10}\)), the number of unpaired electrons is usually the same regardless of whether the ligand field is strong or weak. For \(d^4, d^5, d^6, d^7\) configurations, the ligand field strength can be crucial.
The question specifies an "aqueous solution," implying the presence of water ligands. Water is typically a weak field ligand. However, since V$^{2+}$ has a \(3d^3\) configuration, the number of unpaired electrons (n=3) is the same for both weak and strong field cases, as the three electrons will occupy separate orbitals before pairing occurs, regardless of splitting magnitude.
Therefore, the spin-only formula provides a good approximation for the magnetic moment in this case.
Important Questions from p-block Elements
Second most abundant element in alloy misch metal is:
Match List-I with List-II:
List-I List-II (A) Gel (I) Hair cream (B) Foam (II) Dust (C) Emulsion (III) Cheese (D) Aerosol (IV) Whipped cream Choose the correct answer from the options given below:
Rate of a reaction changes from 2.48 × 10⁻³ mol⁻¹ sec⁻¹ to 4.96 × 10⁻³ mol⁻¹ sec⁻¹ when concentration of reactant is changed from 0.6 M to 2.4 M respectively, the order of reaction is:
Degree of dissociation, when molar conductivity of X at its concentration C is 24.14 and its limiting molar conductivity is 48.28 will be:
Which of the following sols are correctly matched with their corresponding charges?
(A) Cr(OH)₃ sol : negatively charged sol
(B) TiO₂ sol : positively charged sol
(C) CdS sol : positively charged sol
(D) Gum : negatively charged sol
(E) Silver sol : positively charged sol
Choose the correct answer from the options given below:
