Doubtify – JEE Ki Taiyaari, Simple Bhasha Mein

  ❓ Question The number of paramagnetic metal complex species among the following — that have the same number of unpaired electrons — is asked for: [ Co(NH 3 ) 6 ] 3 + ,    [ Co(C 2 O 4 ) 3 ] 3 − ,    [ MnCl 6 ] 3 − ,    [ Mn(CN) 6 ] 3 − ,    [ CoF 6 ] 3 − ,    [ Fe(CN) 6 ] 3 − ,    [ FeF 6 ] 3 − We must find which of these are paramagnetic and then see how many of them share the same number of unpaired electrons. 🖼️ Question Image ✍️ Short Solution (stepwise) We analyse each complex: General steps: Determine metal oxidation state from complex charge. Write the d-electron count. Decide whether ligand is strong-field (low-spin) or weak-field (high-spin). Count unpaired electrons in octahedral splitting. 1. [ Co(NH 3 ) 6 ] 3 + [ \text{Co(NH}_3)_6 ]^{3+} Co oxidation: + 3 +3  → d 6 d^6 . NH₃ is a borderline/relatively strong ligand for Co³⁺ → typically low-spin . Low-spin d 6 d^6 d 6 → t 2 g 6 e g 0 t_{2g}^6 e_g^0  → 0 unpaired → di...

  ❓ Question The number of unpaired electrons responsible for the paramagnetic nature of the following complex ions are, respectively: [ Fe(CN) 6 ] 3 − [ FeF 6 ] 3 − [ \text{FeF}_6 ]^{3-} [ CoF 6 ] 3 − [ \text{CoF}_6 ]^{3-} [ Mn(CN) 6 ] 3 − [ \text{Mn(CN)}_6 ]^{3-} 🖼️ Question Image ✍️ Short Solution We analyze each complex using: Oxidation state Electronic configuration Nature of ligand (strong field or weak field) High-spin vs low-spin Count unpaired electrons ✅ (1) [ Fe(CN) 6 ] 3 − [ \text{Fe(CN)}_6 ]^{3-} Step-1: CN⁻ is a strong-field ligand → large Δ₀ → low-spin . Oxidation state: Fe + 6 ( − 1 ) = − 3 ⇒ Fe 3 + \text{Fe} + 6(-1) = -3 \Rightarrow \text{Fe}^{3+} Fe 3 + : 3 d 5 \text{Fe}^{3+} : 3d^5 In low-spin d⁵ , electrons pair as much as possible: Configuration: t 2 g 5 e g 0 t_{2g}^5 e_g^0 ​ Unpaired electrons = 1 ✅ (2) [ FeF 6 ] 3 − [ \text{FeF}_6 ]^{3-} F⁻ is a weak-field ligand → small Δ₀ → high-spin . Oxidation state o...