$[Cu(NH_3)_4]^{2+}$ shows the following hybridization:
- A$dsp^2$
- B$sp^3d$
- C$dsp^3$
- D$sp^3$
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Similar Questions
Which of the following statements is correct for the complexes $Ni(CO)_4$ and $[Ni(CN)_4]^{2-}$?
Medium
View SolutionMatch List – $I$ with List – $II$:
List - $I$ (Complex) List – $II$ (Geometry) $(a)$ $[Co(NH_3)_6]^{3+}$ $(i)$ Trigonal bipyramidal $(b)$ $[NiCl_4]^{2-}$ (ii) Octahedral $(c)$ $[Ni(CN)_4]^{2-}$ (iii) Tetrahedral $(d)$ $[Fe(CO)_5]$ (iv) Square planar
Choose the correct answer from the options given below.
| List - $I$ (Complex) | List – $II$ (Geometry) |
| $(a)$ $[Co(NH_3)_6]^{3+}$ | $(i)$ Trigonal bipyramidal |
| $(b)$ $[NiCl_4]^{2-}$ | (ii) Octahedral |
| $(c)$ $[Ni(CN)_4]^{2-}$ | (iii) Tetrahedral |
| $(d)$ $[Fe(CO)_5]$ | (iv) Square planar |
Choose the correct answer from the options given below.
MediumKCET 2026
View SolutionThe complex ion $[Cu(NH_3)_4]^{2+}$ has:
Difficult
View Solution$[PdCl_2(PMe_3)_2]$ is a diamagnetic complex of $Pd(II)$. How many unpaired electrons are present in the analogous complex of $Ni(II)$?
Considering $H_2O$ as a weak field ligand,the number of unpaired electrons in $[Mn(H_2O)_6]^{2+}$ will be (Atomic number of $Mn = 25$).
MediumAIPMT 2004
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