$A$ square planar complex is formed by the hybridization of which atomic orbitals?
- A$s, p_x, p_y, d_{yz}$
- B$s, p_x, p_y, d_{x^2 - y^2}$
- C$s, p_x, p_y, d_{z^2}$
- D$s, p_y, p_z, d_{xy}$
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Similar Questions
In which of the following complex ions is the central metal atom's $d$-orbital devoid of electrons?
Medium
View SolutionThe Jahn-Teller effect is not observed in high spin complexes of:
MediumNEET 2016
View SolutionThe hybridization involved in the complex $[Ni(CN)_4]^{2-}$ is (Atomic number of $Ni = 28$):
MediumNEET 2015
View SolutionMatch List-$I$ with List-$II$.
List-$I$
List-$II$
$A$. $[MnBr_4]^{2-}$
$I$. $d^2sp^3$ and diamagnetic
$B$. $[FeF_6]^{3-}$
$II$. $sp^3d^2$ and paramagnetic
$C$. $[Co(C_2O_4)_3]^{3-}$
$III$. $sp^3$ and diamagnetic
$D$. $[Ni(CO)_4]$
$IV$. $sp^3$ and paramagnetic
Choose the correct answer from the options given below:
| List-$I$ | List-$II$ |
|---|---|
| $A$. $[MnBr_4]^{2-}$ | $I$. $d^2sp^3$ and diamagnetic |
| $B$. $[FeF_6]^{3-}$ | $II$. $sp^3d^2$ and paramagnetic |
| $C$. $[Co(C_2O_4)_3]^{3-}$ | $III$. $sp^3$ and diamagnetic |
| $D$. $[Ni(CO)_4]$ | $IV$. $sp^3$ and paramagnetic |
Choose the correct answer from the options given below:
The hybridization of the central atom in $NH_3, [PtCl_4]^{2-}, PCl_5$,and $BCl_3$ respectively is:
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