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JEE Main 20254 AprilMorning Shift
Exam
JEE Main
Year
2025
Date
4 April
Shift
Morning Shift
Subject
Chemistry
Chapter
Coordination Compounds

JEE Main 2025 · 4 April · Morning Shift · Chemistry · Coordination Compounds

Which one of the following complexes will have Δ_0=0 and μ=5.96...

Which one of the following complexes will have Δ0=0\Delta_0=0 and μ=5.96\mu=5.96 B.M.?Which one of the following complexes will have Δ_0=0 and μ=5.96 B.M.?Which one of the following complexes will have Δ_0=0 and μ=5.96 B.M.?Which one of the following complexes will have Δ_0=0 and μ=5.96 B.M.?Which one of the following complexes will have Δ_0=0 and μ=5.96 B.M.?

Options

  1. A[Fe(CN)6]4\left[\mathrm{Fe}(\mathrm{CN})_6\right]^4
  2. B\left[\mathrm{CO}\left(\mathrm{NH}_3\right)_6\right]^{3+}
  3. C[FeF6]4\left[\mathrm{FeF}_6\right]^4
  4. D[Mn(SCN)6]4\left[\mathrm{Mn}(\mathrm{SCN})_6\right]^4

Correct answer

D. [Mn(SCN)6]4\left[\mathrm{Mn}(\mathrm{SCN})_6\right]^4

Step-by-step solution

(1) [Co(NH3)6]3+Co3+⇒3 d64 s0\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]^{3+} \quad \mathrm{Co}^{3+} \Rightarrow 3 \mathrm{~d}^6 4 \mathrm{~s}^0
Which one of the following complexes will have Δ_0=0 and μ=5.96 B.M.?
=[−0.4×6+0.6×(0)]Δ0=−2.4Δ0=[-0.4 \times 6+0.6 \times(0)] \Delta_0=-2.4 \Delta_0
(2)
<br/>[Mn(SCN)6]4<br>Mn2+⇒3 d54 s0<br/>\begin{aligned}<br />& {\left[\mathrm{Mn}(\mathrm{SCN})_6\right]^4} \\<br>& \mathrm{Mn}^{2+} \Rightarrow 3 \mathrm{~d}^5 4 \mathrm{~s}^0<br />\end{aligned}Which one of the following complexes will have Δ_0=0 and μ=5.96 B.M.?
<br/>μ=35 B.M. =5.96 B.M. <br>CFSE=(−0.4×3+0.6×2)Δ0<br/>\begin{aligned}<br />& \mu=\sqrt{35} \text{ B.M. }=5.96 \text{ B.M. } \\<br>& \mathrm{CFSE}=(-0.4 \times 3+0.6 \times 2) \Delta_0<br />\end{aligned}
So Δ0=0\Delta_0=0
(3)
<br/>[Fe(CN)6]−4Fe2+⇒3 d64 s0<br/>\begin{aligned}<br />& {\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{-4} \quad \mathrm{Fe}^{2+} \Rightarrow 3 \mathrm{~d}^6 4 \mathrm{~s}^0}<br />\end{aligned}Which one of the following complexes will have Δ_0=0 and μ=5.96 B.M.?
CFSE=−2.4Δ0\mathrm{CFSE}=-2.4 \Delta_0
(4) \left[\mathrm{FeF}_6\right]^{4-}<br/><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/><img src="https://cdn-question-pool.Quantrex.app/pyq/jee_main/zqvcznomxHPLkpqdaC4cOWvkldF0f22bBKDvUFo9-cY.original.fullsize.png"/><br /><br/>μ=24 B.M. =4.89 B.M. <br> CFSE =(−0.4×4+0.6×2)Δ0=−1.2Δ0<br/>\begin{aligned}<br />& \mu=\sqrt{24} \text{ B.M. }=4.89 \text{ B.M. } \\<br>& \text{ CFSE }=(-0.4 \times 4+0.6 \times 2) \Delta_0=-1.2 \Delta_0<br />\end{aligned}$
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