It would just be Aluminum with four chlorines attached. The negative charge comes from the Al which will have one extra valence electron, from the 4 bonds, compared to it's normal 3 valence electrons.
SiI4 2- is not the formula of any real species. RnF2: Rn has 8 outer electrons (from position in periodic table), F needs 8 round it and has 7. So each F shares a pair of electrons with Rn, one from each atom, making two Rn:F (Lewis pair) bonds. Rn has 3 lone pairs left over; so we have 5 pairs round Rn; two bonds and 3 lone pairs. The 5 spread out as much as possible. That gives you a trigonal pyramid (the shape should be in your textbook). Because lone pairs are more repulsive than bond pairs, they occupy the three equatorial positions, while the bonds to F are top and bottom. KrO3: O makes a double bond. So of the 8 electrons round Kr, 6 are involved in bonds and you have one lone pair. This gives you 4 "things" round the Kr, which spread out as much as possible, roughly into a tetrahedron. Lone pair in one corner, Kr in the middle, O atoms in each of the other 3. Principles: set up your bonds. Count your total things rond the central atom (bonds and lone pairs). Spread out. Lone pairs take up more room than bonds. It's not difficult but it DOES take practice. I hope this helps.
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It would just be Aluminum with four chlorines attached. The negative charge comes from the Al which will have one extra valence electron, from the 4 bonds, compared to it's normal 3 valence electrons.
Alcl4 Lewis Structure
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SiI4 2- is not the formula of any real species. RnF2: Rn has 8 outer electrons (from position in periodic table), F needs 8 round it and has 7. So each F shares a pair of electrons with Rn, one from each atom, making two Rn:F (Lewis pair) bonds. Rn has 3 lone pairs left over; so we have 5 pairs round Rn; two bonds and 3 lone pairs. The 5 spread out as much as possible. That gives you a trigonal pyramid (the shape should be in your textbook). Because lone pairs are more repulsive than bond pairs, they occupy the three equatorial positions, while the bonds to F are top and bottom. KrO3: O makes a double bond. So of the 8 electrons round Kr, 6 are involved in bonds and you have one lone pair. This gives you 4 "things" round the Kr, which spread out as much as possible, roughly into a tetrahedron. Lone pair in one corner, Kr in the middle, O atoms in each of the other 3. Principles: set up your bonds. Count your total things rond the central atom (bonds and lone pairs). Spread out. Lone pairs take up more room than bonds. It's not difficult but it DOES take practice. I hope this helps.
formal charge: # of valence electrons - # of lone pair electrons - (# of covalent bonds / 2) i suggest you refer to the shapes of the molecules to determine their hybridization. molecular shape helps a lot when determining hybrid orbitals. from Wikipedia: AX1 (e.g., LiH): no hybridisation; trivially linear shape AX2 (e.g., BeCl2): sp hybridisation; linear or diagonal shape; bond angles = 180° AX2E (e.g., GeF2): bent/V shape, < 120° AX3 (e.g., BCl3): sp2 hybridisation; trigonal planar shape; bond angles = 120° AX3E (e.g.,NH3): trigonal pyramidal, 107° AX4 (e.g., CCl4): sp3 hybridisation; tetrahedral shape; bond angles ≈ 109.5° AX5 (e.g., PCl5): sp3d hybridisation; trigonal bipyramidal shape AX6 (e.g., SF6): sp3d2 hybridisation; octahedral (or square bipyramidal) shape