Advertisements
Advertisements
Question
Explain, why [Co(NH3)6]3+ ion is low spin? Calculate number of unpaired electrons and write the geometry of [Co(NH3)6]3+.
Advertisements
Solution
NH3 is a strong field ligand. Strong field ligands cause larger splitting of d orbitals and pairing of electrons is favoured. Hence, [Co(NH3)6]3+ is a low spin complex.
- In [Co(NH3)6]3+ ion, oxidation state of cobalt is +3.
Valence shell electronic configuration of Co3+ is:
- Number of ammine ligands is 6. Therefore, the number of vacant metal ion orbitals required for bonding with ligands must be six. Complex is low spin, so the pairing of electrons will take place prior to hybridisation.
Electronic configuration after pairing would be:
- Six orbitals available for hybridisation are two 3d, one 4s, three 4p orbitals.
The orbitals for hybridization are decided from the number of ammine ligands which is six. Here, (n−1)d orbitals participate in hybridization since it is the low spin complex. - Electronic configuration after complex formation is:
- Number of unpaired electrons = 0
- Geometry of the complex ion = Octahedral
APPEARS IN
RELATED QUESTIONS
What are the high-spin complexes?
Answer in brief.
[CoCl4]2− is a tetrahedral complex. Draw its box orbital diagram. State which orbitals participate in hybridization.
Answer in brief.
[CoCl4]2- is a tetrahedral complex. Draw its box orbital diagram. State which orbitals participate in hybridization.
Answer in brief.
With the help of the crystal field, the energy-level diagram explains why the complex [Cr(en)3]3⊕ is coloured.
Answer the following question.
Draw a qualitatively energy-level diagram showing d-orbital splitting in the octahedral environment. Predict the number of unpaired electrons in the complex [Fe(CN)6]4-. Is the complex diamagnetic or paramagnetic? Is it coloured? Explain.
Answer the following with respect to [CoF6]3– ion
- Type of hybridization
- Number of unpaired electrons
- Geometry of complex ion
- Magnetic property
Identify the number of donor groups present in EDTA.
The number of unpaired electrons in the complex ion [CoF6]3− is ____________.
The number of unpaired electrons in the complex ion [NiCl4]2− is ____________.
What is the value of primary valence of Co in CoCl3?
What is the type of magnetic behavior and geometry respectively in Cuproammonium sulphate (Atomic number of Cu = 29)?
Identify the increasing order of effective magnetic moment of the following elements in their +2 oxidation state.
[Fe (Z = 26), Co (Z = 27), Ni (Z = 28), Cu (Z = 29)]
Which of the following types of square planar complexes can show geometrical isomerism [M = metal, a, b, = monodentate ligand]?
Explain the formation of [CoF6]3Θ complex with respect to
- Hybridisation
- Magnetic properties
- Inner/outer complex
- Geometry
A compound forms hcp structure. What is the number of (a) octahedral voids, (b) tetrahedral voids, (c) total voids formed in 0.2 mol of the compound?
Give VBT description of the bonding in a square planar complex [Cu(NH3) 4]2+. Show orbital diagrams for free metal ion and metal ion in the complex. Which hybrid orbitals are used by metal for bonding? State magnetic nature of the complex.
A compound forms a hep structure. Calculate the number of octahedral voids in 0.4 mol. (NA = 6.022 × 1023).
Octahedral complex have ______ hybridisation.
Give the limitations of VBT.
A compound forms hexagonal close packed (hcp) structure. What is the number of (i) Octahedral voids, (ii) Tetrahedral voids, (iii) Total voids formed in 0.7 mol of it.
Mention the type of hybridization in [Co(NH3)6]3+ complex.
Mention the number of unpaired electrons and geometry of the following complex:
\[\ce{[Ni(Cl)4]^{2-}}\]
