Answer
Verified
39k+ views
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.
Complete step by step answer:
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 –
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 – \[s{{p}^{3}}{{d}^{3}}\].
Additional information:
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.
Note: Hybridization can also be calculated by the formula –
z = Number of sigma bond + Number of Lone Pairs in Central Metal atom
Complete step by step answer:
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 – \[s{{p}^{3}}{{d}^{3}}\].
Additional information:
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.
Note: Hybridization can also be calculated by the formula –
z = Number of sigma bond + Number of Lone Pairs in Central Metal atom
Recently Updated Pages
silver wire has diameter 04mm and resistivity 16 times class 12 physics JEE_Main
A parallel plate capacitor has a capacitance C When class 12 physics JEE_Main
A series combination of n1 capacitors each of value class 12 physics JEE_Main
When propyne is treated with aqueous H2SO4 in presence class 12 chemistry JEE_Main
Which of the following is not true in the case of motion class 12 physics JEE_Main
The length of a potentiometer wire is 10m The distance class 12 physics JEE_MAIN
Other Pages
A closed organ pipe and an open organ pipe are tuned class 11 physics JEE_Main
A convex lens is dipped in a liquid whose refractive class 12 physics JEE_Main
Iodoform can be prepared from all except A Acetaldehyde class 12 chemistry JEE_Main
The resultant of vec A and vec B is perpendicular to class 11 physics JEE_Main
448 litres of methane at NTP corresponds to A 12times class 11 chemistry JEE_Main
Dissolving 120g of urea molwt60 in 1000g of water gave class 11 chemistry JEE_Main