(b) Using crystal field theory, write the electronic configuration of central metal atom/ion of the following:

(i) [CoF₆]³⁻

(ii) [Co(NH₃)₆]³⁺ [At. No.: $\text{Co} = 27$ ]

Q: (b) Using crystal field theory, write the electronic configuration of central metal atom/ion of the following: (i) [CoF₆]³⁻ (ii) [Co(NH₃)₆]³⁺ [At. No.: $\text{Co} = 27$]

Both complexes contain **Co³⁺ ($d^6$)**, but the ligand field strength decides the configuration. **(i) [CoF₆]³⁻** - F⁻ is a **weak-field** ligand → **high-spin**. - Configuration: **$t_{2g}^4\, e_g^2$** (4 unpaired electrons → paramagnetic). **(ii) [Co(NH₃)₆]³⁺** - NH₃ is a **strong-field** ligand → **low-spin**. - Configuration: **$t_{2g}^6\, e_g^0$** (0 unpaired electrons → diamagnetic).

Q: Why is [CoF₆]³⁻ high-spin but [Co(NH₃)₆]³⁺ low-spin?

F⁻ is a weak-field ligand (small $\Delta_0 \text{pairing energy}$), so electrons pair up giving $t_{2g}^6 e_g^0$ low-spin.

Coordination Compounds

CFT electronic configuration of Co(III) complexes

Official Answer

Both complexes contain Co³⁺ ( $d^6$ ), but the ligand field strength decides the configuration.

(i) [CoF₆]³⁻

F⁻ is a weak-field ligand → high-spin.
Configuration: $t_{2g}^4\, e_g^2$ (4 unpaired electrons → paramagnetic).

(ii) [Co(NH₃)₆]³⁺

NH₃ is a strong-field ligand → low-spin.
Configuration: $t_{2g}^6\, e_g^0$ (0 unpaired electrons → diamagnetic).

Co³⁺ d⁶crystal field theoryt2g eg configurationweak field ligandstrong field ligandhigh spinlow spinfluoride ammonia

Marking Scheme

1½ mark: [CoF₆]³⁻ → $t_{2g}^4\, e_g^2$ (weak-field F⁻, high-spin).
2½ mark: [Co(NH₃)₆]³⁺ → $t_{2g}^6\, e_g^0$ (strong-field NH₃, low-spin). Full mark for both correct configurations with the high/low-spin reasoning.

Hint

Both are Co³⁺ ( $d^6$ ); decide high-spin vs low-spin from ligand strength — F⁻ weak, NH₃ strong.

Quick Oral Answer

Cobalt(III) is $d^6$ in both; with weak-field F⁻ it stays high-spin as $t_{2g}^4\, e_g^2$ with four unpaired electrons, while with strong-field NH₃ it becomes low-spin $t_{2g}^6\, e_g^0$ with no unpaired electrons.

Analysis & Explanation

Concept

Cobalt is $[\text{Ar}]3d^7 4s^2$ ; removing three electrons gives $\text{Co}^{3+} = 3d^6$ . How these six electrons distribute between the t₂g and e_g sets depends on the size of Δ₀ relative to the pairing energy P.

Weak field (F⁻): $\Delta_0 < P$ , so electrons occupy e_g before pairing → $t_{2g}^4\, e_g^2$ , high-spin, 4 unpaired.
Strong field (NH₃): $\Delta_0 > P$ , so all six pair in t₂g → $t_{2g}^6\, e_g^0$ , low-spin, 0 unpaired.

Exam trap

Write the configuration in $t_{2g}/e_g$ notation (not just $3d^6$ ). Marks are lost when students forget that F⁻ is weak-field (high-spin) while NH₃ is strong-field (low-spin).

Real-world

[Co(NH₃)₆]³⁺ is a stable, well-studied low-spin complex; its inertness made it central to Werner's pioneering coordination-chemistry experiments.

Common Mistakes

1Treating F⁻ as a strong-field ligand and wrongly pairing electrons in [CoF₆]³⁻.
2Writing the configuration simply as $3d^6$ instead of the required $t_{2g}/e_g$ split.
3Miscounting the charge: Co in both complexes is +3 ( $d^6$ ), not +2 ( $d^7$ ).

Interesting Facts

[CoF₆]³⁻ is one of the few high-spin Co(III) complexes and is paramagnetic, whereas most Co(III) complexes are low-spin and diamagnetic.

The colour difference between many Co(III) complexes arises from different $\Delta_0$ values set by the ligand in the spectrochemical series ( $F^- < H_2O < NH_3 < CN^-$ ).

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Frequently Asked Questions

Why is [CoF₆]³⁻ high-spin but [Co(NH₃)₆]³⁺ low-spin?

F⁻ is a weak-field ligand (small $\Delta_0 < \text{pairing energy}$ ), so electrons spread out giving $t_{2g}^4 e_g^2$ high-spin; NH₃ is strong-field ( $\Delta_0 > \text{pairing energy}$ ), so electrons pair up giving $t_{2g}^6 e_g^0$ low-spin.

How many unpaired electrons do these complexes have?

[CoF₆]³⁻ has 4 unpaired electrons (paramagnetic); [Co(NH₃)₆]³⁺ has 0 unpaired electrons (diamagnetic).