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# Explain hybridization of the central atom in: $I{{F}_{7}}$.

Last updated date: 01st Mar 2024
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Hint: Hybridisation can be calculated by taking into account the bond pairs and lone pairs in the molecule. Try relating the hybridization with Z formula.

Hybridization is defined as, “the concept of mixing atomic orbital’s into new hybrid orbital’s (with different energies, shapes, etc.) suitable for the pairing of electrons to form chemical bonds in valence bond theory”.
Let us define a term Z to calculate hybridisation. They can be related as –

 Z Hybridization Geometry 2 $sp$ Linear 3 $s{{p}^{2}}$ Trigonal planar 4 $s{{p}^{3}}$ Tetrahedral 5 $s{{p}^{3}}d$ Trigonal bipyramidal 6 $s{{p}^{3}}{{d}^{2}}$ Octahedral 7 $s{{p}^{3}}{{d}^{3}}$ Pentagonal bipyramidal

The formula for Z is given as –
Z = $\dfrac{1}{2}\left[ v\text{ }+\text{ }n\text{ }-\text{ }p\text{ }+m \right]$
Where, v = Number of valence electrons on central atom
n = negative charge
p = positive charge
m = number of monovalent atoms (e.g. – H, F, Cl, Br)
In the case of $I{{F}_{7}}$, Iodine (I) the central metal atom and Fluorine (F) is the monovalent atom. Also, it is a neutral molecule (i.e. the negative and positive charge is zero).
Therefore,
\begin{align} & Z=\dfrac{1}{2}(7+7) \\ & Z=7 \\ \end{align}
So, the hybridization of central atom in $I{{F}_{7}}$ is – $s{{p}^{3}}{{d}^{3}}$.

The geometry of $I{{F}_{7}}$ is pentagonal bipyramidal. This means that five fluorine’s are placed at equatorial position and two are placed at axial position.