Explain the following :

(i) Low spin octahedral complexes of nickel are not known.

(ii) The $\pi$-complexes are known for transition elements only.

(iii) CO is a stronger ligand than NH3 for many metals.

Concept Videos :-

#23 | VBT & CFT: I
#24 | VBT & CFT: II
#25 | VBT & CFT: III
#26 | Colour of Coordination Compounds

Concept Questions :-

VBT & CFT, Limitation /Hybridisation

(i) The electronic configuration of Ni is [Ar] 3d8 4s2 which shows that it can only form two types of complexes i.e. square planer (dsp2) in presence of strong ligand and tetrahedral (sp3) in presence of weak ligand. There are four empty d-orbitals in Ni while octahedral complexes require six empty orbitals.

(ii) Due to presence of empty d- orbitals in transition metals, they can accept electron pairs from ligands containing $\pi$-bonding complexes.

Example : ligands like C5 H5 , C6 H6  etc.

(iii) Due to greater magnitude of ${\Delta }_{\circ }$, CO produces strong fields which cause more splitting of d- orbitals and moreover it is also able to form $\pi$ bond due to back bonding.

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