(1)
[Co(NH3)6]3+Co3+⇒3 d64 s0
=[−0.4×6+0.6×(0)]Δ0=−2.4Δ0(2)
<br/><br>[Mn(SCN)6]4Mn2+⇒3 d54 s0<br/>
<br/><br>μ=35 B.M. =5.96 B.M. CFSE=(−0.4×3+0.6×2)Δ0<br/>So
Δ0=0(3)
<br/>[Fe(CN)6]−4Fe2+⇒3 d64 s0<br/>
CFSE=−2.4Δ0(4) \left[\mathrm{FeF}_6\right]^{4-}
<br/>\mathrm{Fe}^{2+} \Rightarrow 3 \mathrm{~d}^6 4 \mathrm{~s}^0
<imgsrc="https://cdn−question−pool.getmarks.app/pyq/jeemain/zqvcznomxHPLkpqdaC4cOWvkldF0f22bBKDvUFo9−cY.original.fullsize.png"/><br/><br/><br>μ=24 B.M. =4.89 B.M. CFSE =(−0.4×4+0.6×2)Δ0=−1.2Δ0<br/>$