The change in hybridisation of the Al atom in the reaction of AlCl3+Cl−→AlCl−4
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AlCl3 is sp2 hybridised while AlCl4- is sp3 hybridised.
No of hybrid orbitals = bp + lp
In AlCl3 there are 3 bp and 0 lp
Thus hybrid orbitals = 3
Possible hybridisation is sp2
In AlCl4-, there are 4 bp and 0 lp
Thus hybrid orbitals = 4
Possible hybridisation is sp3 and dsp2 ( but dsp2 can be ruled out since AlCl3 already was sp2 hybridised.
No of hybrid orbitals = bp + lp
In AlCl3 there are 3 bp and 0 lp
Thus hybrid orbitals = 3
Possible hybridisation is sp2
In AlCl4-, there are 4 bp and 0 lp
Thus hybrid orbitals = 4
Possible hybridisation is sp3 and dsp2 ( but dsp2 can be ruled out since AlCl3 already was sp2 hybridised.
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