Structure of \[\text{ICl}_{\text{2}}^{\text{-}}\]is:
(a) Trigonal
(b) Octahedral
(c) Square planar
(d) Distorted trigonal pyramidal
Answer
Verified
123k+ views
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
Recently Updated Pages
The hybridization and shape of NH2 ion are a sp2 and class 11 chemistry JEE_Main
Total number of orbitals associated with the 3rd shell class 11 chemistry JEE_Main
Which of the following has the lowest boiling point class 11 chemistry JEE_Main
Which of the following compounds has zero dipole moment class 11 chemistry JEE_Main
Number of g of oxygen in 322 g Na2SO410H2O is Molwt class 11 chemistry JEE_Main
In the neutralization process of H3PO4 and NaOH the class 11 chemistry JEE_Main
Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility & More
JEE Main Login 2045: Step-by-Step Instructions and Details
JEE Main Chemistry Question Paper with Answer Keys and Solutions
JEE Main Exam Marking Scheme: Detailed Breakdown of Marks and Negative Marking
JEE Main 2023 January 24 Shift 2 Question Paper with Answer Keys & Solutions
JEE Main Chemistry Exam Pattern 2025
Other Pages
NCERT Solutions for Class 11 Chemistry Chapter 7 Redox Reaction
NCERT Solutions for Class 11 Chemistry Chapter 5 Thermodynamics
NCERT Solutions for Class 11 Chemistry Chapter 8 Organic Chemistry
NCERT Solutions for Class 11 Chemistry Chapter 6 Equilibrium
NCERT Solutions for Class 11 Chemistry Chapter 9 Hydrocarbons
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs