Write the Lewis structure for ICl4- and identify its shape and hybridization scheme
The Correct Answer and Explanation is:
To draw the Lewis structure for ICl₄⁻ and determine its shape and hybridization, let’s follow these steps:
Step 1: Count the Total Valence Electrons
- Iodine (I) is in Group 17 and has 7 valence electrons.
- Chlorine (Cl) is also in Group 17 and each Cl atom contributes 7 valence electrons.
- The negative charge (⁻) indicates an extra electron.
Thus, the total number of valence electrons:
7 (from I)+4×7 (from Cl)+1 (extra from the negative charge)=36 valence electrons7 \, (\text{from I}) + 4 \times 7 \, (\text{from Cl}) + 1 \, (\text{extra from the negative charge}) = 36 \, \text{valence electrons}7(from I)+4×7(from Cl)+1(extra from the negative charge)=36valence electrons
Step 2: Draw the Skeleton Structure
- Iodine (I) will be the central atom because it is less electronegative than chlorine (Cl).
- The four chlorine (Cl) atoms will be arranged around iodine.
Step 3: Distribute Electrons
- Place single bonds between iodine and each chlorine atom. Each bond uses 2 electrons, and 4 bonds will use 8 electrons, leaving us with:
36−8=28 electrons left36 – 8 = 28 \, \text{electrons left}36−8=28electrons left - Now, complete the octets for each chlorine atom by adding three lone pairs around each chlorine. This uses 24 electrons:
28−24=4 electrons left28 – 24 = 4 \, \text{electrons left}28−24=4electrons left - Place the remaining 4 electrons as a lone pair on the iodine atom.
Step 4: Check the Structure
- Each chlorine atom has 8 electrons around it (6 from lone pairs and 2 from the bonding pair), which satisfies the octet rule.
- Iodine has 8 electrons from the bonds and 2 from the lone pair, so it satisfies its expanded octet (Iodine can have more than 8 electrons in its valence shell because it’s in period 5).
Thus, the Lewis structure for ICl₄⁻ is:
lessCopyEdit Cl
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Cl - I - Cl
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Cl
With a lone pair on iodine.
Step 5: Determine the Shape and Hybridization
- Shape: The molecule has 5 regions of electron density (4 single bonds and 1 lone pair). This gives a trigonal bipyramidal geometry.
- Lone Pair Effect: The lone pair occupies an equatorial position to minimize electron repulsion, leading to a seesaw shape.
- Hybridization: Since there are 5 regions of electron density, the hybridization is sp³d.
Conclusion:
- The Lewis structure of ICl₄⁻ shows iodine at the center bonded to four chlorine atoms with a lone pair on iodine.
- The shape of the molecule is seesaw.
- The hybridization scheme is sp³d.
