MHTCET 17 Apr 2026 Shift 1 Paper

Concept:A monodentate ligand coordinates to the metal ion through only one donor atom.Explanation:In (a) tetracyanonickelate(II), $\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{2-}$, $\mathrm{CN}^-$ binds through one carbon atom, so it is monodentate.In (b) potassium hexafluoroaluminate(III), $\mathrm{K}_{3}\left[\mathrm{AlF}_{6}\right]$, $\mathrm{F}^-$ binds through one fluorine atom, so it is monodentate.In (c) tetraamminecopper(II), $\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\right]^{2+}$, $\mathrm{NH}_3$ binds through one nitrogen atom, so it is monodentate.In (d) potassium trioxalatoaluminate(III), $\mathrm{K}_{3}\left[\mathrm{Al}\left(\mathrm{C}_2\mathrm{O}_4\right)_{3}\right]$, oxalate $\left(\mathrm{C}_2\mathrm{O}_4^{2-}\right)$ binds through two oxygen atoms, so it is bidentate and not monodentate.Thus, complexes (a), (b), and (c) contain monodentate ligands, giving a total of $3$.Hence, the correct option is B.Answer:Option B, 3