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Coordination Compounds question

2025 · 22 Jan · Shift 2 · Q4
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  5. /2025 · 22 Jan · Shift 2 · Q4

Coordination Compounds question

2025 · 22 Jan · Shift 2 · Q4

JEE MainChemistryCoordination CompoundsMCQ+4 / −1
The correct order of the following complexes in terms of their crystal field stabilization energies is :
  1. A
    [Co(NH3)4]2+<[Co(NH3)6]2+<[Co(NH3)6]3+<[Co(en)3]3+\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_4\right]^{2+}\lt \left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]^{2+}\lt \left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]^{3+}\lt \left[\mathrm{Co}(\mathrm{en})_3\right]^{3+}[Co(NH3​)4​]2+<[Co(NH3​)6​]2+<[Co(NH3​)6​]3+<[Co(en)3​]3+
  2. B
    [Co(NH3)4]2+<[Co(NH3)6]2+<[Co(en)3]3+<[Co(NH3)6]3+\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_4\right]^{2+}\lt \left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]^{2+}\lt \left[\mathrm{Co}(\mathrm{en})_3\right]^{3+}\lt \left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]^{3+}[Co(NH3​)4​]2+<[Co(NH3​)6​]2+<[Co(en)3​]3+<[Co(NH3​)6​]3+
  3. C
    [Co(NH3)6]2+<[Co(NH3)6]3+<[Co(NH3)4]2+<[Co(en)3]3+\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]^{2+}\lt \left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]^{3+}\lt \left[\mathrm{Co}\left(\mathrm{NH}_3\right)_4\right]^{2+}\lt \left[\mathrm{Co}(\mathrm{en})_3\right]^{3+}[Co(NH3​)6​]2+<[Co(NH3​)6​]3+<[Co(NH3​)4​]2+<[Co(en)3​]3+
  4. D
    [Co(en)3]3+<[Co(NH3)6]3+<[Co(NH3)6]2+<[Co(NH3)4]2+\left[\mathrm{Co}(\mathrm{en})_3\right]^{3+}\lt \left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]^{3+}\lt \left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]^{2+}\lt \left[\mathrm{Co}\left(\mathrm{NH}_3\right)_4\right]^{2+}[Co(en)3​]3+<[Co(NH3​)6​]3+<[Co(NH3​)6​]2+<[Co(NH3​)4​]2+
View written solutionFree

Correct answer: A

  1. Find oxidation state and d-electron count of Co in each complex
  • In [Co(NH3)4]2+\left[\mathrm{Co}(\mathrm{NH}_3)_4\right]^{2+}[Co(NH3​)4​]2+, NH3\mathrm{NH}_3NH3​ is neutral, so Co is +2+2+2.
  • In [Co(NH3)6]2+\left[\mathrm{Co}(\mathrm{NH}_3)_6\right]^{2+}[Co(NH3​)6​]2+, Co is also +2+2+2.
  • In [Co(NH3)6]3+\left[\mathrm{Co}(\mathrm{NH}_3)_6\right]^{3+}[Co(NH3​)6​]3+, Co is +3+3+3.
  • In [Co(en)3]3+\left[\mathrm{Co}(\mathrm{en})_3\right]^{3+}[Co(en)3​]3+, en is neutral, so Co is +3+3+3.

Now,

  • Co2+:3d7\mathrm{Co}^{2+} : 3d^7Co2+:3d7
  • Co3+:3d6\mathrm{Co}^{3+} : 3d^6Co3+:3d6
  1. Determine geometry and spin state
  • [Co(NH3)4]2+\left[\mathrm{Co}(\mathrm{NH}_3)_4\right]^{2+}[Co(NH3​)4​]2+ is tetrahedral (4-coordinate Co(II) with NH3_33​ commonly forms tetrahedral complex).
  • [Co(NH3)6]2+\left[\mathrm{Co}(\mathrm{NH}_3)_6\right]^{2+}[Co(NH3​)6​]2+ is octahedral, d7d^7d7.
  • [Co(NH3)6]3+\left[\mathrm{Co}(\mathrm{NH}_3)_6\right]^{3+}[Co(NH3​)6​]3+ is octahedral, d6d^6d6.
  • [Co(en)3]3+\left[\mathrm{Co}(\mathrm{en})_3\right]^{3+}[Co(en)3​]3+ is octahedral, d6d^6d6.

For Co(III), Δo\Delta_oΔo​ is large, so these are low-spin d6d^6d6 complexes.

Also, en is a stronger field ligand than NH3_33​, so

Δo([Co(en)3]3+)>Δo([Co(NH3)6]3+)\Delta_o\big([\mathrm{Co}(\mathrm{en})_3]^{3+}\big) > \Delta_o\big([\mathrm{Co}(\mathrm{NH}_3)_6]^{3+}\big)Δo​([Co(en)3​]3+)>Δo​([Co(NH3​)6​]3+)
  1. Calculate/compare CFSE

(i) [Co(NH3)4]2+\left[\mathrm{Co}(\mathrm{NH}_3)_4\right]^{2+}[Co(NH3​)4​]2+

Tetrahedral, d7d^7d7.

For tetrahedral splitting:

  • eee lower: −0.6Δt-0.6\Delta_t−0.6Δt​
  • t2t_2t2​ upper: +0.4Δt+0.4\Delta_t+0.4Δt​

Configuration: e4t23e^4 t_2^3e4t23​

So,

CFSE=4(−0.6Δt)+3(+0.4Δt)=−2.4Δt+1.2Δt=−1.2Δt\mathrm{CFSE} = 4(-0.6\Delta_t) + 3(+0.4\Delta_t) = -2.4\Delta_t + 1.2\Delta_t = -1.2\Delta_tCFSE=4(−0.6Δt​)+3(+0.4Δt​)=−2.4Δt​+1.2Δt​=−1.2Δt​

Magnitude =1.2Δt=1.2\Delta_t=1.2Δt​.

Since Δt≈49Δo\Delta_t \approx \frac{4}{9}\Delta_oΔt​≈94​Δo​, this is relatively small.

(ii) [Co(NH3)6]2+\left[\mathrm{Co}(\mathrm{NH}_3)_6\right]^{2+}[Co(NH3​)6​]2+

Octahedral, d7d^7d7.

With NH3_33​ and Co(II), this is generally high-spin:

t2g5eg2t_{2g}^5 e_g^2t2g5​eg2​

Hence,

CFSE=5(−0.4Δo)+2(+0.6Δo)=−2.0Δo+1.2Δo=−0.8Δo\mathrm{CFSE} = 5(-0.4\Delta_o) + 2(+0.6\Delta_o) = -2.0\Delta_o + 1.2\Delta_o = -0.8\Delta_oCFSE=5(−0.4Δo​)+2(+0.6Δo​)=−2.0Δo​+1.2Δo​=−0.8Δo​

Magnitude =0.8Δo=0.8\Delta_o=0.8Δo​.

This is greater than tetrahedral d7d^7d7 stabilization.

(iii) [Co(NH3)6]3+\left[\mathrm{Co}(\mathrm{NH}_3)_6\right]^{3+}[Co(NH3​)6​]3+

Octahedral, low-spin d6d^6d6:

t2g6eg0t_{2g}^6 e_g^0t2g6​eg0​

So,

CFSE=6(−0.4Δo)=−2.4Δo\mathrm{CFSE} = 6(-0.4\Delta_o) = -2.4\Delta_oCFSE=6(−0.4Δo​)=−2.4Δo​

Magnitude =2.4Δo=2.4\Delta_o=2.4Δo​.

(iv) [Co(en)3]3+\left[\mathrm{Co}(\mathrm{en})_3\right]^{3+}[Co(en)3​]3+

Octahedral, low-spin d6d^6d6:

t2g6eg0t_{2g}^6 e_g^0t2g6​eg0​

Again,

CFSE=−2.4Δo\mathrm{CFSE} = -2.4\Delta_oCFSE=−2.4Δo​

But since en is a stronger field ligand than NH3_33​, its Δo\Delta_oΔo​ is larger, so its CFSE magnitude is larger.

Therefore,

[Co(NH3)6]3+<[Co(en)3]3+[\mathrm{Co}(\mathrm{NH}_3)_6]^{3+} < [\mathrm{Co}(\mathrm{en})_3]^{3+}[Co(NH3​)6​]3+<[Co(en)3​]3+

in terms of increasing stabilization energy.

  1. Arrange in increasing order of CFSE

From the above:

  • smallest: [Co(NH3)4]2+\left[\mathrm{Co}(\mathrm{NH}_3)_4\right]^{2+}[Co(NH3​)4​]2+
  • then: [Co(NH3)6]2+\left[\mathrm{Co}(\mathrm{NH}_3)_6\right]^{2+}[Co(NH3​)6​]2+
  • then: [Co(NH3)6]3+\left[\mathrm{Co}(\mathrm{NH}_3)_6\right]^{3+}[Co(NH3​)6​]3+
  • largest: [Co(en)3]3+\left[\mathrm{Co}(\mathrm{en})_3\right]^{3+}[Co(en)3​]3+

Thus,

[Co(NH3)4]2+<[Co(NH3)6]2+<[Co(NH3)6]3+<[Co(en)3]3+\left[\mathrm{Co}(\mathrm{NH}_3)_4\right]^{2+} < \left[\mathrm{Co}(\mathrm{NH}_3)_6\right]^{2+} < \left[\mathrm{Co}(\mathrm{NH}_3)_6\right]^{3+} < \left[\mathrm{Co}(\mathrm{en})_3\right]^{3+}[Co(NH3​)4​]2+<[Co(NH3​)6​]2+<[Co(NH3​)6​]3+<[Co(en)3​]3+

So the correct option is A.

  1. Comparison with stored answer

Stored correct answer: A

My derived answer: A

They agree.

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