The hybridisation of the central atom and the | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

Vedclass · Answer

(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.

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

Similar Questions

In the formation of a certain compound,the central atom uses its hybrid orbitals and has a bond angle around it of $105^o$. The percentage of $s-$ character in the hybrid orbitals is likely to be

The geometry of the ammonia molecule and the hybridization of the nitrogen atom involved in it are:

The order of electronegativity of carbon in $sp$,$sp^{2}$ and $sp^{3}$ hybridised states follows:

Which molecule has $sp^3$ hybridization of the central atom?

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module