Exams › JEE Main › Chemistry

For the following coordination species, identify the correct sequence of spin-only paramagnetic moments: I. [FeF6]3− II. [CrF6]3− III. [V(H2O)6]3+ IV. [Ti(H2O)6]3+

1. I > II > III > IV
2. I < II < III < IV
3. I > III > IV > II
4. I > II > IV > III

Correct answer: I > III > IV > II

Solution

The correct sequence reflects the number of unpaired electrons in each complex, which determines their paramagnetic moments. [FeF6]3− has the highest number of unpaired electrons due to its d5 configuration in a weak field, followed by [V(H2O)6]3+ and [Ti(H2O)6]3+, both of which have fewer unpaired electrons, while [CrF6]3− has the least due to its strong field ligands leading to a low-spin state.

Related JEE Main Chemistry questions

• When sodium hydroxide is introduced into an aqueous Zn2+ solution, a white precipitate forms, and this precipitate dissolves when more NaOH is added. In such a solution, zinc is present in the:
• When excess ammonia is added to a copper sulphate solution, a deep blue colour appears because of the formation of which species?
• In the nitroprusside ion, iron and NO are present as Fe$^{2+}$ and NO$^+$ instead of Fe$^{3+}$ and NO. These two formulations can be distinguished by
• In the brown ring test for nitrite and nitrate ions, the characteristic brown ring is produced because a complex ion is formed with which formula?
• What is the spin-only magnetic moment, expressed in Bohr magnetons (μB), for Ni²⁺ in aqueous solution? (Atomic number of Ni = 28)
• Silver chloride dissolves in aqueous ammonium hydroxide because it leads to the formation of which species?

⚔️ Practice JEE Main Chemistry free + battle 1v1 →