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Chemistry
[Cu(NH3)4]2+ is a square planar complex. The number of unpaired electrons and hybrid state ofcopper are respectively
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Q. $[Cu(NH_3)_4]^{2+}$ is a square planar complex. The number of unpaired electrons and hybrid state ofcopper are respectively
Coordination Compounds
A
$ 4, dsp^2 $
14%
B
$1 ,sp^2 $
5%
C
$1, dsp^2 $
74%
D
$4s,sp^3 $
8%
Solution:
$[Cu(NH_3)_4]^{2+}$ is square planar with one unpaired electron and $dsp^2$-hybridization of Cu