Answer
Verified
35.1k+ 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
To get a maximum current in an external resistance class 1 physics JEE_Main
If a wire of resistance R is stretched to double of class 12 physics JEE_Main
Let f be a twice differentiable such that fleft x rightfleft class 11 maths JEE_Main
Find the points of intersection of the tangents at class 11 maths JEE_Main
For the two circles x2+y216 and x2+y22y0 there isare class 11 maths JEE_Main
The path difference between two waves for constructive class 11 physics JEE_MAIN
Other Pages
when an object Is placed at a distance of 60 cm from class 12 physics JEE_Main
If a wire of resistance R is stretched to double of class 12 physics JEE_Main
Which of the following sets of displacements might class 11 physics JEE_Main
Oxidation state of S in H2S2O8 is A 6 B 7 C +8 D 0 class 12 chemistry JEE_Main
What is the pH of 001 M solution of HCl a 1 b 10 c class 11 chemistry JEE_Main
The correct IUPAC name of the given compound is A isopropylbenzene class 11 chemistry JEE_Main