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(D) The hybridisation of the central atom in a compound can be calculated as $H = \frac{1}{2}[V + MA \pm 	ext{anion/cation}]$.Where,$H = 	ext{Hybrid

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(D) The hybridisation of the central atom in a compound can be calculated as $H = \frac{1}{2}[V + MA \pm 	ext{anion/cation}]$.Where,$H = 	ext{Hybridisation}$,$V = 	ext{Valency of central atom}$,$MA = 	ext{Monovalent atom}$.In $[IO_2F_5]^{2-}$,the central atom is $I$ $(V=7)$. The oxygen atoms are divalent,so they do not contribute to the $MA$ count. The fluorine atoms are monovalent $(MA=5)$. The charge is $-2$ (add $2$).$H = \frac{1}{2}[7 + 5 + 2] = \frac{14}{2} = 7$.$A$ steric number of $7$ corresponds to $sp^3d^3$ hybridisation.The shape corresponding to $sp^3d^3$ hybridisation is pentagonal bipyramidal. In $[IO_2F_5]^{2-}$,the two oxygen atoms occupy the axial positions to minimize repulsion,while the five fluorine atoms occupy the equatorial positions.

List-$I$	List-$II$
$(a)$ $SF_6$	$(i)$ $sp^3d^2$
$(b)$ $PCl_5$	(ii) $sp^3d$
$(c)$ $XeF_4$	(iii) $sp^3d^3$
$(d)$ $IF_7$	(iv) $sp^3d^2$

The hybridisation of the central atom and the shape of $[IO_2F_5]^{2-}$ ion,respectively are

Similar Questions

The change in hybridization (if any) of the $Al$ atom in the following reaction is $AlCl_3 + Cl^{-} \rightarrow AlCl_4^{-}$

Which of the following is true for the geometry of different nitrogen atoms $(N)$ present in histamine as labeled in the structure?

Match the following:
List-$I$ List-$II$
$(a)$ $SF_6$ $(i)$ $sp^3d^2$
$(b)$ $PCl_5$ (ii) $sp^3d$
$(c)$ $XeF_4$ (iii) $sp^3d^3$
$(d)$ $IF_7$ (iv) $sp^3d^2$

The correct answer is:

In $sp$ hybridisation,the shape is:

Example of $sp^2$ hybridization is

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