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QJEEM24Ch13c2: JEE Main 2024 Chemistry - Class 12 JEEM24Ch13c2 Solution: Consider the following complex ions The correct order of the…

Consider the following complex ions

amp;P=[FeF6]3−amp;Q=[V(H2O)6]2+amp;R=[Fe(H2O)6]2+\begin{aligned}& \mathrm{P}=\left[\mathrm{FeF}_6\right]^{3-} \\& \mathrm{Q}=\left[\mathrm{V}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+} \\& \mathrm{R}=\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}\end{aligned}

The correct order of the complex ions, according to their spin only magnetic moment values (in B.M.) is :

Multiple Choice Options

  • A R < Q < P
  • B R < P < Q
  • C Q < R < P
  • D Q < P < R

Step-by-Step Verified Solution

[FeF6]3−:Fe+3:[Ar]3 d5\left[\mathrm{FeF}_6\right]^{3-}: \mathrm{Fe}^{+3}:[\mathrm{Ar}] 3 \mathrm{~d}^5

F : Weak field Ligand

JEE Main 2024 (Online) 27th January Morning Shift Chemistry - Coordination Compounds Question 61 English Explanation 1

No. of unpaired electron's =5=5

amp;μ=5(5+2)amp;μ=35 BMamp;[V(H2O)6]+2:V+2:3 d3\begin{aligned}&amp; \mu=\sqrt{5(5+2)} \\&amp; \mu=\sqrt{35} \mathrm{~BM} \\&amp; {\left[\mathrm{V}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{+2}: \mathrm{V}^{+2}: 3 \mathrm{~d}^3}\end{aligned}

JEE Main 2024 (Online) 27th January Morning Shift Chemistry - Coordination Compounds Question 61 English Explanation 2

No. of unpaired electron's =3=3

amp;μ=3(3+2)amp;μ=15 BMamp;[Fe(H2O)6]+2:Fe+2:3 d6\begin{aligned}&amp; \mu=\sqrt{3(3+2)} \\&amp; \mu=\sqrt{15} \mathrm{~BM} \\&amp; {\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{+2}: \mathrm{Fe}^{+2}: 3 \mathrm{~d}^6}\end{aligned}

H 2_2 O : Weak field Ligand

JEE Main 2024 (Online) 27th January Morning Shift Chemistry - Coordination Compounds Question 61 English Explanation 3

No. of unpaired electron's =4=4

amp;μ=4(4+2)amp;μ=24 BM\begin{aligned}&amp; \mu=\sqrt{4(4+2)} \\&amp; \mu=\sqrt{24} \mathrm{~BM}\end{aligned}

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