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Coordination Compounds question

2018 · Shift 2 · Q5
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  5. /2018 · Shift 2 · Q5

Coordination Compounds question

2018 · Shift 2 · Q5

JEE AdvancedChemistryCoordination CompoundsMultiple correct+4 / −1
The correct option(s) regarding the complex [Co(en)(NH3)3(H2O)]3+  (en=H2NCH2CH2NH2){\left[ {Co\left( {en} \right){{\left( {N{H_3}} \right)}_3}\left( {{H_2}O} \right)} \right]^{3 + }}\,\,\left( {en = {H_2}NC{H_2}C{H_2}N{H_2}} \right)[Co(en)(NH3​)3​(H2​O)]3+(en=H2​NCH2​CH2​NH2​) is (are)
  1. A
    It has two geometrical isomers
  2. B
    It will have three geometrical isomers if bidentate 'en' is replaced by two cyanide ligands
  3. C
    It is paramagnetic
  4. D
    It absorbs light at longer wavelength as compared to [Co(en)(NH3)4]3+{\left[ {Co\left( {en} \right){{\left( {N{H_3}} \right)}_4}} \right]^{3 + }}[Co(en)(NH3​)4​]3+
View written solutionFree

Correct answer: A, B, D

  1. Identify the oxidation state and electronic configuration of Co

In

[Co(en)(NH3)3(H2O)]3+\left[\mathrm{Co}(en)(NH_3)_3(H_2O)\right]^{3+}[Co(en)(NH3​)3​(H2​O)]3+

all ligands are neutral:

  • enenen is neutral
  • NH3NH_3NH3​ is neutral
  • H2OH_2OH2​O is neutral

So, oxidation state of Co is +3+3+3.

Cobalt: Z=27Z=27Z=27

Co:[Ar]3d74s2\mathrm{Co}: [Ar]3d^74s^2Co:[Ar]3d74s2

Thus,

Co3+:3d6\mathrm{Co}^{3+}: 3d^6Co3+:3d6

Since Co(III) usually forms low-spin octahedral complexes with ligands like enenen, NH3NH_3NH3​, and even H2OH_2OH2​O in such mixed complexes, the configuration is:

t2g6eg0t_{2g}^6 e_g^0t2g6​eg0​

Hence all electrons are paired.

So the complex is diamagnetic, not paramagnetic.

Therefore, Option C is false.

  1. Check geometrical isomerism of
[Co(en)(NH3)3(H2O)]3+\left[\mathrm{Co}(en)(NH_3)_3(H_2O)\right]^{3+}[Co(en)(NH3​)3​(H2​O)]3+

This is an octahedral complex of type:

[M(AA)B3C][M(AA)B_3C][M(AA)B3​C]

where:

  • AA=enAA = enAA=en (bidentate)
  • B=NH3B = NH_3B=NH3​
  • C=H2OC = H_2OC=H2​O

For octahedral complexes of type [M(AA)B3C][M(AA)B_3C][M(AA)B3​C], there are two geometrical isomers.

Reason: the monodentate ligand CCC can be placed either:

  • trans to one donor position associated relative to the chelate arrangement, or
  • cis in the other non-equivalent arrangement.

Thus Option A is correct.

  1. If en is replaced by two cyanide ligands

Then the complex becomes:

[Co(CN)2(NH3)3(H2O)]+\left[\mathrm{Co}(CN)_2(NH_3)_3(H_2O)\right]^+[Co(CN)2​(NH3​)3​(H2​O)]+

Check charge:

  • 2×CN−=−22\times CN^- = -22×CN−=−2
  • NH3NH_3NH3​ and H2OH_2OH2​O are neutral
  • Co is still +3+3+3 So overall charge is +1+1+1.

This is an octahedral complex of type:

[MA2B3C][M A_2 B_3 C][MA2​B3​C]

with:

  • A=CN−A = CN^-A=CN−
  • B=NH3B = NH_3B=NH3​
  • C=H2OC = H_2OC=H2​O

Now count geometrical isomers.

For [MA2B3C][MA_2B_3C][MA2​B3​C], the two identical AAA ligands can be:

  1. trans to each other → gives one isomer
  2. cis to each other → then CCC can occupy positions giving two non-equivalent arrangements

Hence total geometrical isomers:

1+2=31+2=31+2=3

Therefore, Option B is correct.

  1. Compare absorption wavelength with
[Co(en)(NH3)4]3+\left[\mathrm{Co}(en)(NH_3)_4\right]^{3+}[Co(en)(NH3​)4​]3+

We compare ligand field strength.

Ligands involved:

  • enenen and NH3NH_3NH3​ are relatively stronger-field ligands
  • H2OH_2OH2​O is weaker than NH3NH_3NH3​

So in

[Co(en)(NH3)3(H2O)]3+\left[\mathrm{Co}(en)(NH_3)_3(H_2O)\right]^{3+}[Co(en)(NH3​)3​(H2​O)]3+

one NH3NH_3NH3​ is replaced by weaker ligand H2OH_2OH2​O compared to

[Co(en)(NH3)4]3+\left[\mathrm{Co}(en)(NH_3)_4\right]^{3+}[Co(en)(NH3​)4​]3+

Therefore crystal field splitting Δo\Delta_oΔo​ decreases.

Since absorbed energy is:

E=hν=hcλE = h\nu = \frac{hc}{\lambda}E=hν=λhc​

smaller Δo\Delta_oΔo​ means lower energy absorption, hence larger wavelength absorbed.

So the first complex absorbs light of longer wavelength.

Therefore, Option D is correct.

  1. Final evaluation of options
  • A: True
  • B: True
  • C: False
  • D: True

So the correct options are:

A, B, D\boxed{A,\ B,\ D}A, B, D​
  1. Comparison with stored correct answer

Stored correct answer: A, B, D

This matches the derived answer exactly.

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