Question

Explain \( [Co(NH_3)_6]^{3+} \) is an inner orbital complex whereas \( [Ni(NH_3)_6]^{2+} \) is an outer orbital complex. [At. No. Co = 27, Ni = 28]

Hint:

For inner orbital complexes, low-spin configurations are typical with \( d^2sp^3 \) hybridization. For outer orbital complexes, high-spin configurations are typical with \( sp^3d^2 \) hybridization.

Answer 1

Step 1: Understanding Inner and Outer Orbital Complexes. Inner orbital complexes involve the use of \( d \)-orbitals from the inner shell (typically the \( 3d \)-orbitals in transition metals). This type of hybridization leads to low-spin complexes, where ligands coordinate using these inner \( d \)-orbitals.
Outer orbital complexes involve the use of \( d \)-orbitals from the outer shell (typically \( 4d \)-orbitals for transition metals in higher oxidation states). This type of hybridization leads to high-spin complexes, where ligands coordinate using these outer \( d \)-orbitals.

Step 2: Identifying the Hybridization in \( [Co(NH_3)_6]^{3+} \). Cobalt in the \( +3 \) oxidation state has an electronic configuration of \( 3d^6 \), and the complex \( [Co(NH_3)_6]^{3+} \) uses inner \( 3d \)-orbitals for bonding with the ligands. The \( 3d^2sp^3 \) hybridization results in an inner orbital complex.

Step 3: Identifying the Hybridization in \( [Ni(NH_3)_6]^{2+} \). Nickel in the \( +2 \) oxidation state has an electronic configuration of \( 3d^8 \), and the complex \( [Ni(NH_3)_6]^{2+} \) uses outer \( 4d \)-orbitals for bonding. The \( sp^3d^2 \) hybridization results in an outer orbital complex.
Thus, \( [Co(NH_3)_6]^{3+} \) is an inner orbital complex, while \( [Ni(NH_3)_6]^{2+} \) is an outer orbital complex.