[Cu(NH3)4]2+ is a square planar complex. The number of unpaired electrons and hybrid state ofcopper are respectively

Question

[Cu(NH3)4]2+ is a square planar complex. The number of unpaired electrons and hybrid state ofcopper are respectively (A)  4, dsp2  (B) 1 ,sp2  (C) 1, dsp2  (D) 4s,sp3

Tardigrade Admin · Accepted Answer

[Cu(NH3)4]2+ is square planar with one unpaired electron and dsp2-hybridization of Cu

Q. $[C u (N H_{3})_{4}]^{2 +}$ is a square planar complex. The number of unpaired electrons and hybrid state ofcopper are respectively

$4, d s p^{2}$

$1, s p^{2}$

$1, d s p^{2}$

$4 s, s p^{3}$

$[C u (N H_{3})_{4}]^{2 +}$ is square planar with one unpaired electron and $d s p^{2}$ -hybridization of Cu

Q. [Cu(NH3​)4​]2+ is a square planar complex. The number of unpaired electrons and hybrid state ofcopper are respectively

4,dsp2

1,sp2

1,dsp2

4s,sp3