Which of the following sets of atomic orbitals results in a square planar complex?
- A$s, p_x, p_y, p_z$
- B$s, p_x, p_y, d_{x^2-y^2}$
- C$s, p_x, p_y, d_{z^2}$
- D$s, p_x, p_z, d_{zy}$
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Similar Questions
$A$ tetrahedral complex ion is formed due to $........$ hybridization.
Easy
View SolutionMatch the coordination number and type of hybridisation with distribution of hybrid orbitals in space based on Valence bond theory.
Coordination number and type of hybridisation Distribution of hybrid orbitals in space $(a)$. $4, sp^3$ $(i)$. trigonal bipyramidal $(b)$. $3, sp^2$ $(ii)$. octahedral $(c)$. $5, sp^3 d$ $(iii)$. tetrahedral $(d)$. $6, sp^3 d^2$ $(iv)$. trigonal planar
Select the correct option $:$
| Coordination number and type of hybridisation | Distribution of hybrid orbitals in space |
|---|---|
| $(a)$. $4, sp^3$ | $(i)$. trigonal bipyramidal |
| $(b)$. $3, sp^2$ | $(ii)$. octahedral |
| $(c)$. $5, sp^3 d$ | $(iii)$. tetrahedral |
| $(d)$. $6, sp^3 d^2$ | $(iv)$. trigonal planar |
Select the correct option $:$
Medium
View SolutionWhich of the following matches is incorrect?
Complex compounds $-$ Type of hybridization
Complex compounds $-$ Type of hybridization
Difficult
View Solution$[Co(C_2O_4)_3]^{3-}$ is a :
MediumAIIMS 2019
View SolutionMatch the hybridisation in Column $I$ with the complexes in Column $II$. The options represent the matches for $(A), (B), (C), (D)$ respectively.
Column $I$ Column $II$ $(A)$ $sp^3$ $(i)$ $[Co(NH_3)_6]^{3+}$ $(B)$ $dsp^2$ $(ii)$ $[Ni(CO)_4]$ $(C)$ $sp^3d^2$ $(iii)$ $[Pt(NH_3)_2Cl_2]$ $(D)$ $d^2sp^3$ $(iv)$ $[CoF_6]^{3-}$ $(v)$ $[Fe(CO)_5]$
| Column $I$ | Column $II$ |
|---|---|
| $(A)$ $sp^3$ | $(i)$ $[Co(NH_3)_6]^{3+}$ |
| $(B)$ $dsp^2$ | $(ii)$ $[Ni(CO)_4]$ |
| $(C)$ $sp^3d^2$ | $(iii)$ $[Pt(NH_3)_2Cl_2]$ |
| $(D)$ $d^2sp^3$ | $(iv)$ $[CoF_6]^{3-}$ |
| $(v)$ $[Fe(CO)_5]$ |
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