Question

Among Co³⁺, Ti²⁺, V²⁺ and Cr²⁺ ions, one if used as a reagent cannot liberate H₂ from dilute mineral acid solution, its spin-only magnetic moment in gaseous state is________ B.M. (Nearest integer)

Answer 1

Correct Answer: 5

 To determine which ion cannot liberate H₂ from dilute mineral acid, we first consider the standard electrode potentials. H₂ is liberated if an ion can be oxidized, i.e., it has a lower or comparable standard reduction potential than H₂ (E⁰=0 V for 2H⁺+2e⁻→H₂). Let's examine the relevant cations:
 

• Co³⁺: Co³⁺ + e⁻→Co²⁺ (E⁰=+1.81 V)
• Ti²⁺:Ti²⁺ + 2e⁻→Ti (E⁰=−1.63 V)
• V²⁺:V²⁺ + 2e⁻→V (E⁰=−1.18 V)
• Cr²⁺:Cr²⁺ + 2e⁻→Cr (E⁰=−0.91 V)

Co³⁺ has the highest reduction potential among these ions, thus it cannot liberate H₂ from dilute acid.
Next, we calculate the spin-only magnetic moment, µ, for Co³⁺. As Co³⁺ is typically low spin in an octahedral field, and parameters are generally assumed unless stated otherwise due to ligand field stabilization energy:

Co (atomic number 27):	Configuration: [Ar]3d64s1
Co3+:	Configuration: [Ar]3d6

The number of unpaired electrons (n) for Co³⁺ in a low-spin state is 4. This leads to µ=√n(n+2)=√4(4+2)=√24=4.899 B.M., approximately 5 B.M. The computed value fits within the range [5,5]. Therefore, Co³⁺ has a spin-only magnetic moment of 5 B.M. in the gaseous state.