Coordination Compounds Test - 11

dsp² type of hybridisation is seen specially in case of transition metal ions. The orbitals involved in this type of hybridisation are dx² - y², s and two p. The four dsp² hybrid orbitals adopt a square planar geometry. For example, the complex ion [Ni(CN)₄]^2- involves dsp² hybridisation.

Oxidation state of Co in the given complex is +3.
Co in ground state: 3d⁷ 4s²
So, Co⁺³: 3d⁶ 4s⁰
There will be no unpaired electrons in this coordination compound because NH₃ provides a strong ligand field and is able to pair up electrons of the given compound, forming d²sp³ hybridisation. Hence, the given complex is diamagnetic.

Cr exists as Cr(II) ion and is in d⁴ high spin configuration (for which the number of unpaired electrons is 4). Hence, the spin only magnetic moment is √24 BM. In the other complex, Fe(II) ion is in d⁶ high spin configuration (for which the number of unpaired electrons is 4). Hence, the spin only magnetic moment is √24 BM.

CO is a neutral ligand.
So, X + 4(0) = 0
X = 0

A ligand should be negatively charged or must possess at least one lone pair of electrons so as to form dative bonds with the central metal ion. NH₄⁺ has no lone pair of electrons and thus, cannot act as a ligand in the formation of complexes.

Oxidation state of Cr:
x + 3(-2) = -3
x – 6 = - 3
x = -3 + 6 = + 3

The coordination number is the number of ligands attached to the central ion (more specifically, the number of donor atoms). As C₂O₄^2- is a bidentate ligand, so coordination number of Cr in this compound is 6.