using VSEPR THEORY explain why

1)BF4- is a tetrahedral molecule
2)SF3+ is a Trigonal pyramidal molecule
3)ICI4- IS A SQUARE PLANER molecule
4IF5 IS A SQUARE pyramidal molecule
5)PF5 IS A TRIGNAL BIPYRAMIDAl molecule help Dr bob222

I don't know what you want in the way of an explanation. You count the number of regions of high electron density. For BF4^-, that is 4 and since the groups bond to the central atom arrange themselves so they are as far apart as possible, then tetrahedral is the shape it takes.

yes that is what I wanted can you do this for every one Dr Bob 222 thanks

1) To determine the molecular shape of BF4-, we first look at the central atom, which is Boron. Boron has three valence electrons and can form three bonds. Fluoride (F) has seven valence electrons and needs one more electron to complete its octet. Hence, each F atom forms a single bond with Boron. This accounts for four electrons, leaving one additional electron from Boron. The lone pair and bond pairs repel each other equally, resulting in a tetrahedral shape for BF4-.

2) For SF3+, we again focus on the central atom, which is Sulfur. Sulfur has six valence electrons and can form six bonds. Fluoride (F) has seven valence electrons and needs one more electron to complete its octet. Therefore, three F atoms form single bonds with Sulfur. This accounts for three of the valence electrons. One of Sulfur's lone pairs repels the bond pairs, creating a trigonal pyramidal molecular shape.

3) In ICI4-, Iodine (I) is the central atom. Iodine has seven valence electrons and can form seven bonds. Chlorine (Cl) has seven valence electrons and needs one more electron to complete its octet. Hence, four Cl atoms form single bonds with Iodine. This accounts for four of the valence electrons. There are no lone pairs on the central Iodine atom, leading to a square planar molecular shape.

4) In IF5, Iodine (I) is the central atom, which has seven valence electrons and can form seven bonds. Fluoride (F) has seven valence electrons and needs one more electron to complete its octet. Therefore, five fluorine atoms form single bonds with Iodine. This accounts for five of the valence electrons. One of Iodine's lone pairs repels the bond pairs to give a square pyramidal molecular shape.

5) Lastly, for PF5, Phosphorus (P) is the central atom with five valence electrons and can form five bonds. Fluoride (F) has seven valence electrons and needs one more electron to complete its octet. Hence, five F atoms form single bonds with Phosphorus. This accounts for five of the valence electrons. There are no lone pairs on the central Phosphorus atom, resulting in a trigonal bipyramidal molecular shape.

Sure! VSEPR theory (Valence Shell Electron Pair Repulsion theory) helps us predict the shape and geometry of molecules based on the repulsion between electron pairs (both bonding and non-bonding) around the central atom. Let's apply this theory to each of the given molecules:

1) BF4- (Boron tetrafluoride anion):
To determine the shape of BF4-, we start by considering the central atom (Boron) and the surrounding atoms (Fluorine). Boron has 3 valence electrons, and each Fluorine contributes 7 valence electrons, making a total of 4 × (7) + 3 = 32 valence electrons.

Since BF4- is an anion, we need to add an additional electron to the electron count, giving us a total of 33 electrons. These electrons are then arranged in the molecule to minimize repulsion as much as possible.

To achieve this, one Fluorine atom forms a single bond with Boron, which uses two electrons. The remaining three Fluorine atoms contribute an additional one electron each as lone pairs. This gives us two electron domains around the central atom.

According to VSEPR theory, electron domains push away from each other and try to position themselves as far apart as possible. This results in a tetrahedral electron geometry, with the lone pairs occupying three of the four vertices and the bonded pairs occupying the fourth vertex. Therefore, BF4- is a tetrahedral molecule.

2) SF3+ (Sulfur trifluoride cation):
For SF3+, Sulfur is again the central atom, and each Fluorine contributes 7 valence electrons. Sulfur has 6 valence electrons, and because SF3+ is a cation, we need to subtract one electron, giving us a total of 6 + 3 × (7) - 1 = 26 valence electrons.

In SF3+, two Flourine atoms form single bonds with Sulfur, which uses four electrons. The remaining Fluorine atom contributes an additional lone pair. This gives us four electron domains.

Following VSEPR theory, these electron domains need to position themselves as far apart as possible. The best arrangement is trigonal pyramidal, with the three bonded pairs and one lone pair occupying the four vertices. Therefore, SF3+ is a trigonal pyramidal molecule.

3) ICl4- (Iodine tetrachloride anion):
In ICl4-, Iodine is the central atom, and each Chlorine contributes 7 valence electrons. Iodine has 7 valence electrons, and since ICl4- is an anion, we need to add an extra electron, giving us a total of 7 + 4 × (7) + 1 = 36 valence electrons.

In ICl4-, each Chlorine forms a single bond with Iodine, which uses eight electrons. This gives us four electron domains around the central atom.

To minimize repulsion, these electron domains arrange themselves as far apart as possible, resulting in a square planar electron geometry. Therefore, ICl4- is a square planar molecule.

4) IF5 (Iodine pentafluoride):
For IF5, Iodine is once again the central atom, and each Fluorine contributes 7 valence electrons. Iodine has 7 valence electrons, so the total electron count is 7 + 5 × (7) = 42 valence electrons.

In IF5, each Fluorine atom forms a single bond with Iodine, using five electrons. This gives us five electron domains around the central atom.

According to VSEPR theory, these electron domains position themselves as far apart as possible, resulting in a square pyramidal electron geometry. Therefore, IF5 is a square pyramidal molecule.

5) PF5 (Phosphorus pentafluoride):
In PF5, Phosphorus is the central atom, and each Fluorine contributes 7 valence electrons. Phosphorus has 5 valence electrons, so the total electron count is 5 + 5 × (7) = 40 valence electrons.

In PF5, each Fluorine atom forms a single bond with Phosphorus, using five electrons. This gives us five electron domains around the central atom.

Following VSEPR theory, these electron domains arrange themselves as far apart as possible, resulting in a trigonal bipyramidal electron geometry. Therefore, PF5 is a trigonal bipyramidal molecule.