JEE Challenger
More from Coordination Compounds
ChemMedAdv 2024 • Paper 1

Number of Metal Carbonyl Species Isoelectronic with Nickel Tetracarbonyl

Among V(CO)6\text{V(CO)}_6, Cr(CO)5\text{Cr(CO)}_5, Cu(CO)3\text{Cu(CO)}_3, Mn(CO)5\text{Mn(CO)}_5, Fe(CO)5\text{Fe(CO)}_5, [Co(CO)3]3−[\text{Co(CO)}_3]^{3-}, [Cr(CO)4]4−[\text{Cr(CO)}_4]^{4-}, and Ir(CO)3\text{Ir(CO)}_3, the total number of species isoelectronic with Ni(CO)4\text{Ni(CO)}_4 is _______.

[Given, atomic number: V=23\text{V} = 23, Cr=24\text{Cr} = 24, Mn=25\text{Mn} = 25, Fe=26\text{Fe} = 26, Co=27\text{Co} = 27, Ni=28\text{Ni} = 28, Cu=29\text{Cu} = 29, Ir=77\text{Ir} = 77]

Official Numerical Answer1

Topics & Concepts

Coordination CompoundsCoordination Compoundsd-Block Elements

Step-by-Step Solution

To determine the number of species that are isoelectronic with nickel tetracarbonyl, Ni(CO)4\text{Ni(CO)}_4, we need to find the total number of electrons present in Ni(CO)4\text{Ni(CO)}_4 and compare it with each of the given species.

1. Calculation of Total Electrons in Ni(CO)4\text{Ni(CO)}_4:

  • Atomic number of Nickel (Ni\text{Ni}) = 2828, so Ni\text{Ni} contributes 2828 electrons.
  • Each carbon monoxide (CO\text{CO}) ligand consists of 11 carbon atom (66 electrons) and 11 oxygen atom (88 electrons), giving a total of 6+8=146 + 8 = 14 electrons per CO\text{CO} ligand.

Total electrons in Ni(CO)4=28+4×14=28+56=84\text{Total electrons in } \text{Ni(CO)}_4 = 28 + 4 \times 14 = 28 + 56 = 84

2. Calculation of Total Electrons for the Given Species:

  1. V(CO)6\text{V(CO)}_6: Total electrons=23+6×14=23+84=107\text{Total electrons} = 23 + 6 \times 14 = 23 + 84 = 107

  2. Cr(CO)5\text{Cr(CO)}_5: Total electrons=24+5×14=24+70=94\text{Total electrons} = 24 + 5 \times 14 = 24 + 70 = 94

  3. Cu(CO)3\text{Cu(CO)}_3: Total electrons=29+3×14=29+42=71\text{Total electrons} = 29 + 3 \times 14 = 29 + 42 = 71

  4. Mn(CO)5\text{Mn(CO)}_5: Total electrons=25+5×14=25+70=95\text{Total electrons} = 25 + 5 \times 14 = 25 + 70 = 95

  5. Fe(CO)5\text{Fe(CO)}_5: Total electrons=26+5×14=26+70=96\text{Total electrons} = 26 + 5 \times 14 = 26 + 70 = 96

  6. [Co(CO)3]3−[\text{Co(CO)}_3]^{3-}: Total electrons=27+3×14+3=27+42+3=72\text{Total electrons} = 27 + 3 \times 14 + 3 = 27 + 42 + 3 = 72

  7. [Cr(CO)4]4−[\text{Cr(CO)}_4]^{4-}: Total electrons=24+4×14+4=24+56+4=84\text{Total electrons} = 24 + 4 \times 14 + 4 = 24 + 56 + 4 = 84

  8. Ir(CO)3\text{Ir(CO)}_3: Total electrons=77+3×14=77+42=119\text{Total electrons} = 77 + 3 \times 14 = 77 + 42 = 119

Conclusion:

Among all the given species, only [Cr(CO)4]4−[\text{Cr(CO)}_4]^{4-} has a total of 8484 electrons, which is equal to the total number of electrons in Ni(CO)4\text{Ni(CO)}_4.

Therefore, the total number of species isoelectronic with Ni(CO)4\text{Ni(CO)}_4 is 1.