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The number of unpaired electrons calculated in \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right]^{3+}\) and \(\left[\mathrm{CoF}_{6}\right]^{3-}\) are : \(\quad\) [V.I.T.E.E.E. 2008] (a) 4 and 4 \(\square\) (b) 0 and 2 (c) 2 and 4 \(\square\) (d) 0 and 4

Short Answer

Expert verified
Option (d) 0 and 4.

Step by step solution

01

Determine the Oxidation State of Cobalt

For \([\mathrm{Co}(\mathrm{NH}_{3})_{6}]^{3+}\) and \([\mathrm{CoF}_{6}]^{3-}\), identify the oxidation state of cobalt. In both cases, the overall charge affects the oxidation state. For \([\mathrm{Co}(\mathrm{NH}_{3})_{6}]^{3+}\), each NH鈧 is neutral, so cobalt must be +3 to account for the +3 complex charge. Similarly, in \([\mathrm{CoF}_{6}]^{3-}\), each F ligand is -1, so to balance the charge, cobalt also has an oxidation state of +3.
02

Determine Electronic Configuration

Cobalt in its elemental form has the electronic configuration \([\mathrm{Ar}]\, 3d^7\,4s^2\). In a +3 oxidation state, Co loses three electrons, primarily from the 4s and then the 3d orbitals, so its configuration becomes \([\mathrm{Ar}]\, 3d^6\).
03

Analyze Ligands and Field Strength

\([\mathrm{Co}(\mathrm{NH}_{3})_{6}]^{3+}\) features NH鈧, a strong field ligand that typically causes a pairing of electrons in the lower energy orbitals, leading to a low-spin configuration. \([\mathrm{CoF}_{6}]^{3-}\) features F鈦, a weak field ligand that does not cause pairing of the electrons, resulting in a high-spin configuration.
04

Determine the Number of Unpaired Electrons

For \([\mathrm{Co}(\mathrm{NH}_{3})_{6}]^{3+}\), the t鈧俫 orbitals (lower energy) are filled before the eg orbitals with strong-field ligands. Therefore, all six 3d electrons pair up, resulting in 0 unpaired electrons. For \([\mathrm{CoF}_{6}]^{3-}\), orbitals are filled with high-spin configurations, allowing four electrons to remain unpaired in the 3d orbitals.
05

Compare Against Options

Based on the calculations above, \([\mathrm{Co}(\mathrm{NH}_{3})_{6}]^{3+}\) has 0 unpaired electrons, and \([\mathrm{CoF}_{6}]^{3-}\) has 4 unpaired electrons. This matches with option (d).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Cobalt complexes
Cobalt complexes are fascinating because they illustrate how transition metals, like cobalt, can form stable compounds with various ligands. Cobalt, a transition metal, can bond with different ions or molecules, known as ligands, to form what we call coordination compounds.
These ligands donate electrons to the cobalt metal center, which plays an essential part in determining the characteristics of the complex. Depending on the ligands attached, cobalt can exhibit different chemical behaviors. This is inherently linked to the concept of oxidation state and electron configuration. In our exercise,
  • the oxidation state of cobalt is +3 in both complexes studied:
    • \([Co(NH_3)_6]^{3+}\) and \([CoF_6]^{3-}\)
This consistent oxidation state impacts the subsequent steps in determining the number of unpaired electrons.
Unpaired electrons
Unpaired electrons play a critical role in determining the magnetism and other properties of the cobalt complexes. In coordination chemistry, electron configuration is key to understanding these phenomena. For cobalt in the +3 oxidation state, we first remove three electrons from its atomic configuration, \([Ar] 3d^7 4s^2\), leading to a new configuration of \([Ar] 3d^6\). The question of how many electrons remain unpaired hinges on the electron arrangement within the 3d orbitals.
Two main scenarios arise based on field strength of the ligands:
  • Strong field ligands, like NH鈧, cause electron pairing within the t鈧俫 orbitals, leading to a low-spin state and no unpaired electrons.
  • Weak field ligands, such as F鈦, allow for higher energy configurations (high-spin), resulting in more unpaired electrons. In \([CoF_6]^{3-}\), four electrons are unpaired.
The net result strongly influences the magnetic properties, turning some complexes into paramagnetic materials.
Ligand field strength
Ligand field strength is a critical concept in Crystal Field Theory, as it determines how electrons are distributed in the d orbitals of the metal center. Ligands around a metal split the degenerate d orbitals into two sets: t鈧俫 and eg. The strength of the ligand directly affects this splitting.
Strong field ligands, such as NH鈧, induce a large separation between these two sets. This makes it energetically favorable for electrons to pair up in the lower energy t鈧俫 orbitals, resulting in a low-spin configuration. Hence, in \([Co(NH_3)_6]^{3+}\), all electrons are paired due to NH鈧's strong field nature.
On the other hand, weak field ligands like F鈦 create a smaller energy gap between t鈧俫 and eg. Thus, electrons occupy the higher energy eg orbitals first, leading to a high-spin configuration and more unpaired electrons, as seen in \([CoF_6]^{3-}\).
Recognizing the ligand field strength helps in predicting electronic structures and understanding the magnetic and optical properties of the complexes.

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Most popular questions from this chapter

Optical isomerism is not shown by the complex: a) \(\left[\mathrm{Cr}(\mathrm{ox})_{3}\right]^{3-}\) (b) \(\left[\mathrm{Co}(\mathrm{en})_{2} \mathrm{Cl}_{2}\right]^{+}(\) cis-form \()\) (c) \(\left[\mathrm{Co}(\mathrm{en})_{2} \mathrm{Cl}_{2}\right]^{+}\) (trans- form) (d) \(\left[\mathrm{Cr}(\mathrm{en})_{3}\right]^{3+}\) ox = oxalate; en=ethylenediamine

Among \(\left[\mathrm{Ni}(\mathrm{CO})_{4}\right],\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{2-}\) and \(\left[\mathrm{NiCl}_{4}\right]^{2-}\) species, the hybridization states of the Ni-atom are respectively: (a) \(s p^{3}, d s p^{2}, d s p^{2}\) (c) \(s p^{3}, s p^{3}, d s p^{2}\) [C.B.S.E. (P.M.T.) 2004] \(\square\) (b) \(s p^{3}, d s p_{3}^{2}, s p^{3}\). \(\square\) (d) \(d s p^{2}, s p^{3}, s p^{3}\)

Which one of the following complexes will have four isomers? (a) \(\left[\mathrm{Co}(\mathrm{en})\left(\mathrm{NH}_{3}\right)_{2} \mathrm{Cl}_{2}\right] \mathrm{Cl}\) (b) \(\left[\mathrm{Co}\left(\mathrm{PPh}_{3}\right)_{2}\left(\mathrm{NH}_{3}\right)_{2} \mathrm{Cl}_{2}\right] \mathrm{Cl}\) (c) \(\left[\mathrm{Co}(\mathrm{en})_{3}\right] \mathrm{Cl}_{3}\) (d) \(\left[\mathrm{Co}(\mathrm{en})_{2} \mathrm{Cl}_{2}\right] \mathrm{Br}\)

What is the shape of \(\mathrm{Fe}(\mathrm{CO})_{5}\) molecule? \(\quad\) [C.B.S.E. 2000] (a) Tetrahedral \(\square\) (b) Octahedral (c) Trigonal bipyramidal \(\square\) (d) Square pyramidal -

One mole of the complex compound \(\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{Cl}_{3}\), gives 3 moles- of ions on dissolution in water. One-mole-of the same complex reacts with two moles of \(\mathrm{AgNO}_{3}\) solution to yield two moles of \(\mathrm{AgCl}(s)\). The structure of the complex is: [A.I.E.E.E. 2003] (a) \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{Cl}\right] \mathrm{Cl}_{2}\) (b) \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{3} \mathrm{Cl}_{3}\right] \cdot 2 \mathrm{NH}_{3}\) (c) \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{2}\right] \mathrm{Cl} \cdot \mathrm{NH}_{3}\) (d) \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}\right] \mathrm{Cl}_{2} \cdot \mathrm{NH}_{3}\)

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