I'm having difficulty on drawing lewis dot structures for certain compounds and figuring out the molecular shape. I can do the simple ones like IO4- which is tetrahedral...but i'm stumped on ICl4-. When i drew the structure I thought it was octahedral, but my textbook says it's square planar. help?
I'm having similar problems with ICl2-. I put trigonal planar, when it was linear.
Try this and see if it helps. Repost if you still have trouble and tell us, if you can, what is giving you the problem.
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NH2Cl
ICl2- is Linear because the Cl bind at the top and bottom, and around the center there are 3 sets of Lone Pair Electrons. there is a negative sign, so you must add a set of LP electrons. So yes, Its Trigonal Bipyrmidal- if you go by the total electron density locations, but, its linear because the Cl atoms stay in linear formation with the 3 sets of LP electrons around center of the central I atom.
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Drawing Lewis dot structures and determining molecular shapes can be challenging, but there are some general guidelines that can help. Let's start by addressing ICl4- and ICl2-.
To draw the Lewis dot structure, follow these steps:
1. Determine the total number of valence electrons for the compound by adding the valence electrons of each element involved. For ICl4-, iodine (I) has 7 valence electrons, and chlorine (Cl) has 7 valence electrons each.
Total valence electrons = 7 (I) + 7 (Cl) x 4 - 1 (negative charge of ICl4-) = 32.
2. Place the central atom in the middle. In this case, iodine (I) is the central atom.
3. Connect the peripheral atoms (chlorine atoms) to the central atom with a single bond. Each single bond represents 2 electrons.
4. Distribute the remaining electrons around the atoms, starting with the peripheral atoms, until each atom has a full octet (except for hydrogen, which needs only 2 electrons).
5. If you still have remaining electrons after filling the octets, place them on the central atom as lone pairs until you run out.
Applying these steps to ICl4-:
- Iodine is the central atom because it is less electronegative than chlorine.
- Connect each chlorine atom to the iodine atom with a single bond. This uses 8 electrons (4 bonds * 2 electrons = 8 electrons).
- Distribute the remaining 24 electrons around the atoms (8 electrons on iodine + 6 electrons on each chlorine atom). This gives each atom a full octet.
- The Lewis dot structure for ICl4- should look like this:
Cl Cl
| |
Cl - I - Cl
|
Cl
Now, let's determine the molecular shape of ICl4-:
- The central atom (iodine) has four peripheral atoms (chlorine atoms) bonded to it.
- The presence of four peripheral atoms in a plane around the central atom gives the molecule a square planar shape.
- Therefore, square planar is the correct molecular shape for ICl4-.
Moving on to ICl2-:
Following the same steps as before, you would find:
- The total valence electrons = 7 (I) + 7 (Cl) x 2 - 1 (negative charge of ICl2-) = 20.
- Iodine (I) is the central atom.
- Connect the two chlorine atoms to the iodine atom with single bonds.
- Distribute the remaining 14 electrons around the atoms.
- The Lewis dot structure for ICl2- should look like this:
Cl
|
Cl - I - Cl
|
Now, let's determine the molecular shape of ICl2-:
- The central atom (iodine) has two peripheral atoms (chlorine atoms) bonded to it.
- The presence of two peripheral atoms in a straight line gives the molecule a linear shape.
- Therefore, linear is the correct molecular shape for ICl2-.
Remember, drawing Lewis dot structures and determining molecular shapes are based on following the rules of valence electrons and octet configuration. It may take practice to become more skilled at it.