JEE MAIN - Chemistry (2023 - 11th April Morning Shift - No. 12)
The set which does not have ambidentate ligand(s) is :
$$\mathrm{C}_{2} \mathrm{O}_{4}{ }^{2-}, \mathrm{NO}_{2}{ }^{-}, \mathrm{NCS}^{-}$$
$$\mathrm{C}_{2} \mathrm{O}_{4}{ }^{2-}$$, ethylene diammine, $$\mathrm{H}_{2} \mathrm{O}$$
$$\mathrm{NO}_{2}^{-}, \mathrm{C}_{2} \mathrm{O}_{4}{ }^{2-}$$, EDTA $$^{4-}$$
$$\mathrm{EDTA}^{4-}, \mathrm{NCS}^{-}, \mathrm{C}_{2} \mathrm{O}_{4}{ }^{2-}$$
Explanation
Option A :
- $$\mathrm{C}_{2} \mathrm{O}_{4}{ }^{2-}$$ (Oxalate) is a bidentate ligand. It always binds through its two oxygen atoms.
- $$\mathrm{NO}_{2}{ }^{-}$$ (Nitrite) is an ambidentate ligand. It can coordinate either through nitrogen or oxygen.
- $$\mathrm{NCS}^{-}$$ (Thiocyanate) can also bind through either nitrogen or sulfur, making it an ambidentate ligand.
Option B :
- $$\mathrm{C}_{2} \mathrm{O}_{4}{ }^{2-}$$ (Oxalate) as mentioned earlier, is a bidentate ligand.
- Ethylene diamine (en) is also a bidentate ligand, as it always binds through its two nitrogen atoms.
- $$\mathrm{H}_{2} \mathrm{O}$$ (Water) is a monodentate ligand, binding only through one oxygen atom.
Option C :
- $$\mathrm{NO}_{2}^{-}$$ as mentioned, is an ambidentate ligand.
- $$\mathrm{C}_{2} \mathrm{O}_{4}{ }^{2-}$$ as mentioned, is a bidentate ligand.
- $$\mathrm{EDTA}^{4-}$$ is a hexadentate ligand. It always binds through six donor atoms and is not ambidentate.
Option D :
- $$\mathrm{EDTA}^{4-}$$ as mentioned, is a hexadentate ligand.
- $$\mathrm{NCS}^{-}$$ as mentioned, is an ambidentate ligand.
- $$\mathrm{C}_{2} \mathrm{O}_{4}{ }^{2-}$$ as mentioned, is a bidentate ligand.
From the above analysis, it's clear that Option B is the one that does not contain any ambidentate ligands. So, the correct answer is Option B.
- $$\mathrm{C}_{2} \mathrm{O}_{4}{ }^{2-}$$ (Oxalate) is a bidentate ligand. It always binds through its two oxygen atoms.
- $$\mathrm{NO}_{2}{ }^{-}$$ (Nitrite) is an ambidentate ligand. It can coordinate either through nitrogen or oxygen.
- $$\mathrm{NCS}^{-}$$ (Thiocyanate) can also bind through either nitrogen or sulfur, making it an ambidentate ligand.
Option B :
- $$\mathrm{C}_{2} \mathrm{O}_{4}{ }^{2-}$$ (Oxalate) as mentioned earlier, is a bidentate ligand.
- Ethylene diamine (en) is also a bidentate ligand, as it always binds through its two nitrogen atoms.
- $$\mathrm{H}_{2} \mathrm{O}$$ (Water) is a monodentate ligand, binding only through one oxygen atom.
Option C :
- $$\mathrm{NO}_{2}^{-}$$ as mentioned, is an ambidentate ligand.
- $$\mathrm{C}_{2} \mathrm{O}_{4}{ }^{2-}$$ as mentioned, is a bidentate ligand.
- $$\mathrm{EDTA}^{4-}$$ is a hexadentate ligand. It always binds through six donor atoms and is not ambidentate.
Option D :
- $$\mathrm{EDTA}^{4-}$$ as mentioned, is a hexadentate ligand.
- $$\mathrm{NCS}^{-}$$ as mentioned, is an ambidentate ligand.
- $$\mathrm{C}_{2} \mathrm{O}_{4}{ }^{2-}$$ as mentioned, is a bidentate ligand.
From the above analysis, it's clear that Option B is the one that does not contain any ambidentate ligands. So, the correct answer is Option B.
Comments (0)
