Step 1: Understanding Inner and Outer Orbital Complexes
- In coordination chemistry, inner and outer orbital complexes refer to hybridization of central metal ions in coordination compounds.
- Inner orbital complex: The central metal ion utilizes inner d-orbitals for hybridization.
- Outer orbital complex: The central metal ion utilizes outer d-orbitals for hybridization.
Step 2: Analysis of \( [Co(NH_3)_6]^{3+ \) and \( [Ni(NH_3)_6]^{3+} \)}
- \( [Co(NH_3)_6]^{3+} \):
- \( Co^{3+} \) has an electronic configuration \( d^6 \).
- With strong ligand \( NH_3 \), it forms low spin inner orbital complex (hybridization: \( d^2sp^3 \)).
- \( [Ni(NH_3)_6]^{3+} \):
- \( Ni^{3+} \) has an electronic configuration \( d^7 \).
- It forms an outer orbital complex (hybridization: \( sp^3d^2 \)).
Step 3: Verifying Other Options
- Option (1) \( Al_2O_3 \) with 5\% \( Cr^{3+} \) is incorrect because Ruby is \( Al_2O_3 \) with trace amounts of \( Cr^{3+} \), but not exactly 5\%.
- Option (2) \( Mn_2(CO)_{10} \) does not contain two bridged carbonyl groups, so it's incorrect.
- Option (4) \( [Ni(CN)_4]^{2-} \) and \( [NiCl_4]^{2-} \) do not both have tetrahedral geometry.
- \( [Ni(CN)_4]^{2-} \) is square planar due to strong field ligand \( CN^- \).
- \( [NiCl_4]^{2-} \) is tetrahedral due to weak field ligand \( Cl^- \).
Step 4: Conclusion
- The correct statement is Option (3): \( [Co(NH_3)_6]^{3+} \) is an inner orbital complex, whereas \( [Ni(NH_3)_6]^{3+} \) is an outer orbital complex.
Match the following: