Verified Solution • 2025–26 Session moderate

QJEEM24Chb8fd: JEE Main 2024 Chemistry - Class 12 JEEM24Chb8fd Solution: Which of the following complex is homoleptic?

Which of the following complex is homoleptic?

Multiple Choice Options

  • A [Ni(NH3)2Cl2]\left[\mathrm{Ni}\left(\mathrm{NH}_3\right)_2 \mathrm{Cl}_2\right]
  • B [Co(NH3)4Cl2]+\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_4 \mathrm{Cl}_2\right]^{+}
  • C [Fe(NH3)4Cl2]+\left[\mathrm{Fe}\left(\mathrm{NH}_3\right)_4 \mathrm{Cl}_2\right]^{+}
  • D [Ni(CN)4]2−\left[\mathrm{Ni}(\mathrm{CN})_4\right]^{2-}

Step-by-Step Verified Solution

A homoleptic complex is one in which a central metal atom or ion is surrounded by only one kind of donor groups, ligands or ions. On the other hand, a heteroleptic complex contains a central metal surrounded by more than one kind of donor groups, ligands or ions.

Let's examine the given options :

Option A: [Ni(NH3)2Cl2]\left[\mathrm{Ni}\left(\mathrm{NH}_3\right)_2 \mathrm{Cl}_2\right]

This complex contains two different ligands, ammonia ( NH3\mathrm{NH}_3 ) and chloride (Cl). Therefore, it is a heteroleptic complex.

Option B: [Co(NH3)4Cl2]+\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_4 \mathrm{Cl}_2\right]^{+}

Just like Option A, this complex has two types of ligands, ammonia ( NH3\mathrm{NH}_3 ) and chloride (Cl), making it a heteroleptic complex.

Option C: [Fe(NH3)4Cl2]+\left[\mathrm{Fe}\left(\mathrm{NH}_3\right)_4 \mathrm{Cl}_2\right]^{+}

Again, this complex has ammonia ( NH3\mathrm{NH}_3 ) and chloride (Cl) ligands, so it is a heteroleptic complex.

Option D: [Ni(CN)4]2−\left[\mathrm{Ni}(\mathrm{CN})_4\right]^{2-}

This complex has only cyanide (CN) ligands surrounding the nickel ion, with no other types of ligands present. Thus, this is a homoleptic complex because it is coordinated by a single type of ligand.

Therefore, Option D is the correct answer as it represents a homoleptic complex.

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