$d^{2}sp^{3}$ hybridisation of the atomic orbitals gives
- Asquare planar structure
- Btriangular structure
- Ctetrahedral structure
- Doctahedral structure
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Similar Questions
Explain why $[Co(NH_{3})_{6}]^{3+}$ is an inner orbital complex whereas $[Ni(NH_{3})_{6}]^{2+}$ is an outer orbital complex.
Medium
View SolutionIn which of the following complex ions is the central metal atom's $d$-orbital devoid of electrons?
Medium
View SolutionThe complex ion $[Cu(NH_3)_4]^{2+}$ has which of the following geometries and electronic configurations?
Medium
View SolutionThe incorrect statement about $Ni(CO)_4$ is
Match List-$I$ with List-$II$:
List-$I$ (Tetrahedral Complex) List-$II$ (Electronic configuration) $A$. $TiCl_4$ $I$. $e^2, t_2^0$ $B$. $[FeO_4]^{2-}$ $II$. $e^4, t_2^3$ $C$. $[FeCl_4]^{-}$ $III$. $e^0, t_2^0$ $D$. $[CoCl_4]^{2-}$ $IV$. $e^2, t_2^3$
Choose the correct answer from the options given below:
| List-$I$ (Tetrahedral Complex) | List-$II$ (Electronic configuration) |
|---|---|
| $A$. $TiCl_4$ | $I$. $e^2, t_2^0$ |
| $B$. $[FeO_4]^{2-}$ | $II$. $e^4, t_2^3$ |
| $C$. $[FeCl_4]^{-}$ | $III$. $e^0, t_2^0$ |
| $D$. $[CoCl_4]^{2-}$ | $IV$. $e^2, t_2^3$ |
Choose the correct answer from the options given below:
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