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Coordination Compounds question

2020 · 4 Sep · Shift 2 · Q1
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Coordination Compounds question

2020 · 4 Sep · Shift 2 · Q1

JEE MainChemistryCoordination CompoundsMCQ+4 / −1
The one that can exhibit highest paramagnetic behaviour among the following is : gly = glycinato; bpy = 2, 2'-bipyridine
  1. A
    [Fe(en)(bpy)(NH3)2]2+[Fe(en)(bpy)(NH_3)_2]^{2+}[Fe(en)(bpy)(NH3​)2​]2+
  2. B
    [Pd(gly)2][Pd(gly)_2][Pd(gly)2​]
  3. C
    [Co(OX)2(OH)2]−[Co(OX)_2(OH)_2]^-[Co(OX)2​(OH)2​]− (Δ\DeltaΔ 0 > P)
  4. D
    [Ti(NH3)6]3+[Ti(NH_3)_6]^{3+}[Ti(NH3​)6​]3+
View written solutionFree

Correct answer: C

  1. Goal: Find the complex with the highest paramagnetic behaviour, i.e. the one with the maximum number of unpaired electrons.

  2. Check each complex one by one.


Option A: [Fe(en)(bpy)(NH3)2]2+[Fe(en)(bpy)(NH_3)_2]^{2+}[Fe(en)(bpy)(NH3​)2​]2+

  • Oxidation state of Fe:

    • enenen, bpybpybpy, NH3NH_3NH3​ are all neutral ligands.
    • Therefore, Fe is in +2+2+2 oxidation state.
  • Electronic configuration: Fe2+:3d6Fe^{2+} : 3d^6Fe2+:3d6

  • Nature of ligands:

    • enenen and bpybpybpy are strong field ligands.
    • NH3NH_3NH3​ is also reasonably strong compared to weak ligands.

So this is expected to be a low-spin octahedral complex: t2g6eg0t_{2g}^6 e_g^0t2g6​eg0​

  • Number of unpaired electrons =0=0=0.

So, A is diamagnetic.


Option B: [Pd(gly)2][Pd(gly)_2][Pd(gly)2​]

  • Glycinato (gly−)(gly^-)(gly−) is a mononegative bidentate ligand.

  • Two gly−gly^-gly− ligands contribute −2-2−2, and the complex is neutral.

  • Hence oxidation state of Pd is: x−2=0⇒x=+2x-2=0 \Rightarrow x=+2x−2=0⇒x=+2

  • Electronic configuration: Pd2+:4d8Pd^{2+} : 4d^8Pd2+:4d8

  • For 4d4d4d metals, pairing is generally strong; Pd(II)Pd(II)Pd(II) commonly forms square planar complexes.

  • Square planar d8d^8d8 complexes are typically diamagnetic.

  • Number of unpaired electrons =0=0=0.

So, B is diamagnetic.


Option C: [Co(OX)2(OH)2]−[Co(OX)_2(OH)_2]^-[Co(OX)2​(OH)2​]− with (Δo>P)(\Delta_o > P)(Δo​>P)

  • Let oxidation state of Co be xxx.
  • Oxalate (OX2−)(OX^{2-})(OX2−): two ligands give −4-4−4.
  • Two OH−OH^-OH− give −2-2−2.
  • Overall charge is −1-1−1.

Thus, x−4−2=−1x-4-2=-1x−4−2=−1 x=+5x=+5x=+5

So cobalt is Co5+Co^{5+}Co5+.

  • Electronic configuration:
    • Co: [Ar]3d74s2[Ar]3d^7 4s^2[Ar]3d74s2
    • Co5+:3d4Co^{5+} : 3d^4Co5+:3d4

Given Δo>P\Delta_o > PΔo​>P, it is a low-spin octahedral complex.

For low-spin d4d^4d4 octahedral: t2g4eg0t_{2g}^4 e_g^0t2g4​eg0​

This gives 2 unpaired electrons.

So, C is paramagnetic with 2 unpaired electrons.


Option D: [Ti(NH3)6]3+[Ti(NH_3)_6]^{3+}[Ti(NH3​)6​]3+

  • NH3NH_3NH3​ is neutral, so Ti is in +3+3+3 oxidation state.

  • Electronic configuration: Ti3+:3d1Ti^{3+} : 3d^1Ti3+:3d1

In octahedral field: t2g1eg0t_{2g}^1 e_g^0t2g1​eg0​

  • Number of unpaired electrons =1=1=1.

So, D is paramagnetic with 1 unpaired electron.


  1. Compare number of unpaired electrons:
  • A: 000
  • B: 000
  • C: 222
  • D: 111

Hence, the complex showing the highest paramagnetic behaviour is: [Co(OX)2(OH)2]−\boxed{[Co(OX)_2(OH)_2]^-}[Co(OX)2​(OH)2​]−​

So, Option C is correct.

  1. Comparison with stored answer:
  • Stored correct answer: C
  • Derived answer: C

They agree.

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