CBSE Class 12 · 2010 · Delhi Compartment · Coordination CompoundsShort Answer

WhatsApp Email
Using the valence bond theory, predict the geometry and magnetic behaviour of [CoF]. (Atomic number of Co = 27)

Answer: Octahedral, sp³d² (outer orbital); paramagnetic — 4 unpaired electrons.

Solution

Co = 3d. F is a weak-field ligand — no pairing: te retains 4 unpaired electrons.
The ligands use 4s, three 4p and two 4d orbitals: spd — an outer-orbital (high-spin) complex.
Geometry: octahedral.
Magnetism: 4 unpaired electrons → paramagnetic, μ = √24 ≈ 4.90 BM.

CBSE Class 12 Board · 2010 · Delhi Compartment · Subjective

More board questions on Coordination Compounds

  1. The magnetic moment of [NiCl] is [Atomic number: Ni = 28]
    CBSE 2023View solution →
  2. Give the formulae of the following compounds:
    (a) Potassium tetrahydroxidozincate(II)
    (b) Hexaammineplatinum(IV) chloride
    CBSE 2020View solution →
  3. Write the IUPAC name of the complex [Pt(en)Cl]. Draw the structures of the geometrical isomers of this complex.
    CBSE 2019View solution →
  4. Using IUPAC names, write the formulae for the following:
    (i) Hexaamminecobalt(III) sulphate
    (ii) Potassium trioxalatochromate(III)
    CBSE 2019View solution →
  5. Out of [CoF] and [Co(en)], which complex is:
    (i) paramagnetic
    (ii) more stable
    (iii) an inner-orbital complex
    (iv) a high-spin complex
    (Atomic number of Co = 27)
    CBSE 2019View solution →
  6. Write the coordination number and oxidation state of platinum in the complex [Pt(en)Cl].
    CBSE 2018View solution →