NEET Chemistry Question (2026) — Solution

Step-by-step solution

The spin-only magnetic moment is given by = n(n+2) BM, where n is the number of unpaired electrons. We determine n for each complex: For [ Ti(H _2 O )_6]^ 3+ : Ti is in +3 oxidation state. The electronic configuration of Ti^ 3+ is [ Ar ] 3d^1. Number of unpaired electrons, n = 1. = 1(1+2) = 3 BM. For [ Mn(CN) _6]^ 3- : Mn is in +3 oxidation state. The electronic configuration of Mn^ 3+ is [ Ar ] 3d^4. Since CN^- is a strong field ligand, pairing of electrons takes place. The configuration in the octahedral field is t_ 2g ^4 e_g^0. Number of unpaired electrons, n = 2. = 2(2+2) = 8 BM. For [ Fe(CN) _6]^ 3- : Fe is in +3 oxidation state. The electronic configuration of Fe^ 3+ is [ Ar ] 3d^5. CN^- is a strong field ligand, causing pairing. The configuration is t_ 2g ^5 e_g^0. Number of unpaired electrons, n = 1. = 1(1+2) = 3 BM. For [ Co(NH _3)_6]^ 3+ : Co is in +3 oxidation state. The electronic configuration of Co^ 3+ is [ Ar ] 3d^6. NH_3 acts as a strong field ligand for Co^ 3+ , causing complete pairing. The configuration is t_ 2g ^6 e_g^0. Number of unpaired electrons, n = 0. = 0 BM. Comparing the values, [ Mn(CN) _6]^ 3- has the highest number of unpaired electrons (n=2) and thus the highest spin-only magnetic moment. Answer: [ Mn(CN) _6]^ 3-