Q17 of 26 Page 1

(a) Write the hybridisation and shape of the following complexes

i)


ii)


(Atomic number: Co = 27, Ni = 28)


b) Out of and CO, which ligand forms a more stable complex with a transition metal and why?


a i)




Fluoride is a weak ligand and it is unable to bind the inner d orbitals. Cobalt ion has 6 electrons in its 3d orbital in which 4 are unpaired. The Flouride ion pair with the 3d electrons by utilising 1 4s, 3 4p and 2 4d electrons for the hybridisation. Thus, the hybridisation is .


Shape is Octahedral


IUPAC name is hexafluorocobalt(3-)


ii)




The electronic configuration of Nickel ion is . Each hybrid orbital is occupied by a pair of electrons from each cyanide molecule. Thus, the hybridisation is . Shape is Square planar IUPAC name is Tetracyanonickelate(II)


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i) What is the principle behind the zone refining of metals?

ii) What is the role of Silica in the extraction of Copper?


iii) How is ‘cast ion’ different from ‘pig iron’?


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Give reasons for the following:

i) is less reactive at room temperature.


ii) is the strongest reducing agent amongst all the hydrides of Group 16 elements.


iii) Helium is used in diving apparatus as diluent for Oxygen.


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Write the structure of the major product in each of the following reactions:





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Give reasons for the following:

i) Phenol is more acidic than ethanol


ii) Boiling point of ethanol is higher in comparison to methoxymethane


iii) on reaction with HI gives and as the main products not and