
Structure of \[\text{ICl}_{\text{2}}^{\text{-}}\]is:
(a) Trigonal
(b) Octahedral
(c) Square planar
(d) Distorted trigonal pyramidal
Answer
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Hint: Start the question by calculating hybridization of the given compound and find the geometry. From there, you can easily identify the structure of the compound. Keep note that the geometry and structure of the compound need not be the same.
Complete step by step solution:
Hybridization is defined as, “the concept of mixing atomic orbitals into new hybrid orbitals (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 \[\text{ICl}_{\text{2}}^{\text{-}}\], Iodine (I) the central metal atom and Chlorine (Cl) is the monovalent atom. Also, it is a negatively charged molecule (i.e. the net charge is -1).
Therefore,
\[\begin{align}
& Z=\dfrac{1}{2}(7+1+2) \\
& Z=5 \\
\end{align}\]
So, the hybridization of central atom in \[\text{ICl}_{\text{2}}^{\text{-}}\]is – \[s{{p}^{3}}d\]. Looking at the table drawn above, the geometry should be trigonal bipyramidal.
But since lone pairs of electrons are present at equatorial positions, the shape is distorted trigonal bipyramidal.
Therefore, the answer is – option (d)– Structure of \[\text{ICl}_{\text{2}}^{\text{-}}\]is distorted trigonal pyramidal.
Note: Geometry and shape are not the same. Geometry means the orientation of the molecules and electron pairs, whereas shape refers to the orientation of the atoms only.
Also, 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 solution:
Hybridization is defined as, “the concept of mixing atomic orbitals into new hybrid orbitals (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 \[\text{ICl}_{\text{2}}^{\text{-}}\], Iodine (I) the central metal atom and Chlorine (Cl) is the monovalent atom. Also, it is a negatively charged molecule (i.e. the net charge is -1).
Therefore,
\[\begin{align}
& Z=\dfrac{1}{2}(7+1+2) \\
& Z=5 \\
\end{align}\]
So, the hybridization of central atom in \[\text{ICl}_{\text{2}}^{\text{-}}\]is – \[s{{p}^{3}}d\]. Looking at the table drawn above, the geometry should be trigonal bipyramidal.
But since lone pairs of electrons are present at equatorial positions, the shape is distorted trigonal bipyramidal.
Therefore, the answer is – option (d)– Structure of \[\text{ICl}_{\text{2}}^{\text{-}}\]is distorted trigonal pyramidal.
Note: Geometry and shape are not the same. Geometry means the orientation of the molecules and electron pairs, whereas shape refers to the orientation of the atoms only.
Also, hybridization can also be calculated by the formula –
z = Number of sigma bond + Number of Lone Pairs in Central Metal atom
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