Consider that d? metal ion (M²?) forms a complex with aqua ligands, and the spin only magnetic moment of the complex is 4.90BM. The geometry and the crystal field stabilization energy of the complex is:

Option 1 - <p>tetrahedral and -1.6Δt + 1P<br>&lt;!-- [if !supportLineBreakNewLine]--&gt;<br>&lt;!--[endif]--&gt;</p>
Option 2 - <p>tetrahedral and -0.6Δt<br>&lt;!-- [if !supportLineBreakNewLine]--&gt;<br>&lt;!--[endif]--&gt;</p>
Option 3 - <p>octahedral and -2.4Δo + 2P</p>
Option 4 - <p>octahedral and -1.6Δo</p>
5 Views|Posted 7 months ago
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1 Answer
R
7 months ago
Correct Option - 2
Detailed Solution:

Since spin only magnetic moment is 4.90BM so number of unpaired electrons must be 4, so If the complex is octahedral, then it has to be high spin complex with configuration t? g? e²g¹ in that case
CFSE = 4* (-0.4Δ? ) + 2*0.6Δ? = -0.4Δ?
If the complex is tetrahedral then its electronic configuration w

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Chemistry Coordination Compounds 2025

Chemistry Coordination Compounds 2025

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