Question

what is ths hybridization of cu in cu[(nh3)4]2+

Answer 1

There is a lots of concept is in this question .

Because if u see the complex u might think about that it will show sp3 hybridisation or dsp2 whatever u r thinking is wrong

The metal atom in the complex shows sp2d hybridisation .you might not ever heard about the sp2d hybridisation but it is really true.

Let us see how it is,

1.the electronic configuration of copper is Ar(3d104s1). Now copper has oxidation state of 2+ .so it configuration will be 3d9.means it has empty4 s orbital and 4porbital and 4d orbital and 3d orbital contain one unpaired electron .Now NH3 is a strong ligand so it should pair up all the electron but 3d orbital contains only one unpaired electron so there is no chance to pair up. There for NH3 should donate its lone pair in empty s and p subshell . therefore it should have sp3hybridisation.

But if this is true then it should have tetrahedral geometry but it is practically found that it has square plannar geometry . So it is not having sp3 hybridisation.

2. So there is another concept ,the concept says that the unpaired electron present in 3d subshell is shifted to 4p orbital.Now ,as usual again NH3will donate its lone pair of electron .Note that in this case it will donate 1pair of electron in 3d orbital and one in4 s orbital and remaining two pairs in 4px and 4py orbital .so it should have dsp2 hybridisation .

Although the above statement satisfy the condition that it has square plannar geometry ,still a problem in this statement .

The problem is unpaired electron present in 4p orbital will experience electronic repulsion so it will be easily lost means if it would have dsp2hybridisation then cu2+ will get oxidised to cu3+.this means complex ion is not stable .But this complex is stable also .

So the point is it should show square plannar geometry and the complex should be stable also.

4. in order to explain the stability and square plannar geometry of complex Huggins suggested that the unpaired electron will remain in 3d orbital and NH3will donate its one pair of electron in s orbital and one pair of electron in px orbital and one pair in py orbital .pz orbital will remain empty and it will donate its remaining lone pair of electron in 4d orbital .so the hybridisation is sp2d.

This is the correct explanation of this question.

I hope it will clear your concept .

So the hybridisation is sp2d not sp3 or dsp2.

Answer 2

Answer:

The hybridization of Cu in $\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\right]^{2+}$ is dsp².

Explanation:

• The oxidation state of COPPER$\mathrm{Cu} in \left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\right]^{2+} is +2 . The electronic configuration of \mathrm{Cu}^{2}+ is 3 \mathrm{~d}^{9} 4 s^{0} .$
• $d s p^{2} hybridisation of one d , one s and two p orbitals result in square planar geometry.$
• c. Coordinate bonds are formed by overlap of metal dsp² hybrid orbitals.
• refer figure.