Exams › JEE Main › Chemistry

The metal atom present in the complex MABXL (where A, B, X and L are unidentate ligands and M is metal) involves sp3 hybridization. The number of geometrical isomers exhibited by the complex is: (1) 4 (2) 0 (3) 2 (4) 3

1. 4
2. 0
3. 2
4. 3

Correct answer: 0

Solution

The complex MABXL has a coordination number of 4, which typically leads to tetrahedral geometry. Tetrahedral complexes do not exhibit geometrical isomerism due to their symmetrical arrangement, resulting in zero geometrical isomers.

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 →