The complex ion $[Cu(NH_3)_4]^{2+}$ has:
- ATetrahedral geometry with one unpaired electron
- BSquare planar geometry with one unpaired electron
- CTetrahedral geometry with all paired electrons
- DSquare planar geometry with all paired electrons
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Similar Questions
Match the List-$I$ with List-$II$List-$I$ (Molecules/ion) List-$II$ (Hybridisation of central atom) $A. PF_5$ $I. dsp^2$ $B. SF_6$ $II. sp^3d$ $C. Ni(CO)_4$ $III. sp^3d^2$ $D. [PtCl_4]^{2-}$ $IV. sp^3$
Choose the correct answer from the options given below $:$
| List-$I$ (Molecules/ion) | List-$II$ (Hybridisation of central atom) |
| $A. PF_5$ | $I. dsp^2$ |
| $B. SF_6$ | $II. sp^3d$ |
| $C. Ni(CO)_4$ | $III. sp^3d^2$ |
| $D. [PtCl_4]^{2-}$ | $IV. sp^3$ |
Consider the reaction $CuSO_4 + KCN \to CuCN + Y \xrightarrow{\text{excess } K^{+}CN^{-}} Z$. Identify the correct statement regarding the products $Y$ and $Z$.
Difficult
View SolutionAmong $[Co(CN)_4]^{4-}, [Co(CO)_3(NO)], XeF_4, [PCl_4]^{+}, [PdCl_4]^{2-}, [ICl_4]^{-}, [Cu(CN)_4]^{3-}$ and $P_4$,the total number of species with tetrahedral geometry is:
MediumIIT 2024
View SolutionMatch the coordination number and type of hybridisation with the distribution of hybrid orbitals in space based on Valence Bond Theory.
Coordination number and type of hybridisation Distribution of hybrid orbitals in space $a. 4, sp^3$ $i. \text{trigonal bipyramidal}$ $b. 4, dsp^2$ $ii. \text{octahedral}$ $c. 5, sp^3d$ $iii. \text{tetrahedral}$ $d. 6, d^2sp^3$ $iv. \text{square planar}$
Select the correct option.
| Coordination number and type of hybridisation | Distribution of hybrid orbitals in space |
|---|---|
| $a. 4, sp^3$ | $i. \text{trigonal bipyramidal}$ |
| $b. 4, dsp^2$ | $ii. \text{octahedral}$ |
| $c. 5, sp^3d$ | $iii. \text{tetrahedral}$ |
| $d. 6, d^2sp^3$ | $iv. \text{square planar}$ |
Select the correct option.
MediumNEET 2020
View SolutionConsidering $H_2O$ 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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