$d^{2}sp^{3}$ hybridisation of the atomic orbitals gives

  • A
    square planar structure
  • B
    triangular structure
  • C
    tetrahedral structure
  • D
    octahedral structure

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Similar Questions

Generally,$FeCl_3 \cdot 6H_2O$ is represented as:

Match the following:
List-$I$ (Hybridisation)List-$II$ (Compound/ion)
$A. sp^3d$$I. [PtCl_4]^{2-}$
$B. sp^3d^2$$II. SF_6$
$C. dsp^2$$III. BCl_3$
$D. dsp^3$$IV. PCl_5$
$V. ClF_3$

The correct match is:

Match the following hybridization in Column $I$ with the corresponding coordination complexes in 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-}$
$(v). [Fe(CO)_5]$

In Wilkinson's catalyst,the hybridization of the central metal ion and its shape are respectively:

Which of the following complexes involves $4p$ orbitals for hybridization?

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