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  • The hybridisation of <img alt="\mathrm{\left[C u\left( NH _{3}\right)_{4}\right]^{2+}}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%5Cleft%5BC%20u%5Cleft%28%20N

The hybridisation of \mathrm{\left[C u\left( NH _{3}\right)_{4}\right]^{2+}} is

Option: 1

\mathrm{s p^{3}}


Option: 2

\mathrm{d s p^{2}}


Option: 3

\mathrm{d^{2} s p^{2}}


Option: 4

\mathrm{s p^{2} d}


Answers (1)

best_answer

[Cu(NH3)4]2+ is an exception for determining the hybridisation. On the experimental basis, it has been found that its geometry is square planar, thus atleast one d orbital is compulsory.

The electronic configuration of [Cu(NH3)4]2+ is as follows:

But in this case, the last electron can easily be removed and Cu2+ will become Cu3+, which does not exist in reality.

Thus to explain its square planar geometry, Huggins proposed sp2d hybridisation that this last unpaired electron remains in the 3d orbital and the hybridisation is done by 4s orbital, 2 4p orbitals and 1 4d orbital. The third 4p orbital does not participate in hybridisation. Thus, in this way, its hybridisation is dsp2 and the geometry is square planar.

Therefore, option(4) is correct.

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