Which of the following is a low-spin (spin-paired) complex? — Coordination Compounds Chemistry Question

💡 Solution & Explanation

Step 1: Check if there is a distinction between low-spin and high-spin configurations. This distinction only exists for d^4 to d^7 octahedral systems. Step 2: [Ti(H2O)6]^3+ is d^1 and [Cr(NH3)6]^3+ is d^3, so they have no spin-pairing options. [Ni(NH3)6]^2+ is d^8 and must have 2 unpaired electrons. [Fe(NH3)6]^3+ contains Fe^3+ (3d^5). Step 3: In [Fe(NH3)6]^3+, the strong-field NH3 ligands force the 5 d-electrons to pair up in the t_2g orbitals (t_2g^5 e_g^0), making it a low-spin (spin-paired) complex.