VSEPR is easy if you follow the right steps. 1. Find the number of valence (outer) electrons around the central atom (remember the charge) 2. Work out how many of these are involved in bonding (remember you may have single or double bonds) 3. Remember the lone pairs. 4. Work out the shape from the number of atoms bonded to the central atom and the number of lone pairs. So for SeO2, Se is in group 6, and has 6 valence electrons. Se=O is a double bond, so there are 2 of Se\'s electrons in a double bond There are two Se=O bonds, so 4 electrons are involved in Se=O bonds. This leaves another 2 electrons which are a lone pair. So overall you have three groups around the central atom (2 Se=O bonds and a lone pair) so it is trigonal planar if you drew in the lone pair. Removing the lone pair gives you a bent molecule. For PF4+: P has 5 valence electrons, but it is a positive ion, so you knock off an electron to get 4 P-F bonds are single bonds, and there are 4 of them, which accounts for all the electrons. Therefore PF4+ is a tetrahedral shape. Finally, ICl3: I has 7 valence electrons. I-Cl bonds are single bonds, so the I uses one electron in each bond. Therefore 3 of the 7 electrons are involved in I-Cl bonds, the other 4 must be in 2 lone pairs. This gives 5 electron groups around the central ion, so the shape is trigonal bipyramid if you draw in the lone pairs (trigonal plane with 2 groups above and below the plane. You can determine that these two groups must be the lone pairs since they repel more than the I-Cl bonds). Removing the lone pairs gives you a trigonal planar molecule. The number of groups around the central ion determines the shape. 2 is linear 3 is trigonal planar Solution VSEPR is easy if you follow the right steps. 1. Find the number of valence (outer) electrons around the central atom (remember the charge) 2. Work out how many of these are involved in bonding (remember you may have single or double bonds) 3. Remember the lone pairs. 4. Work out the shape from the number of atoms bonded to the central atom and the number of lone pairs. So for SeO2, Se is in group 6, and has 6 valence electrons. Se=O is a double bond, so there are 2 of Se\'s electrons in a double bond There are two Se=O bonds, so 4 electrons are involved in Se=O bonds. This leaves another 2 electrons which are a lone pair. So overall you have three groups around the central atom (2 Se=O bonds and a lone pair) so it is trigonal planar if you drew in the lone pair. Removing the lone pair gives you a bent molecule. For PF4+: P has 5 valence electrons, but it is a positive ion, so you knock off an electron to get 4 P-F bonds are single bonds, and there are 4 of them, which accounts for all the electrons. Therefore PF4+ is a tetrahedral shape. Finally, ICl3: I has 7 valence electrons. I-Cl bonds are single bonds, so the I uses one electron in each bond. Therefore 3 of the 7 electrons are involved in I-Cl bonds, the other 4 must be in 2 lone pairs. This gives 5 electro.

1. VSEPR is easy if you follow the right steps. 1. Find the number of valence (outer)
electrons around the central atom (remember the charge) 2. Work out how many of these are
involved in bonding (remember you may have single or double bonds) 3. Remember the lone
pairs. 4. Work out the shape from the number of atoms bonded to the central atom and the
number of lone pairs. So for SeO2, Se is in group 6, and has 6 valence electrons. Se=O is a
double bond, so there are 2 of Se's electrons in a double bond There are two Se=O bonds, so 4
electrons are involved in Se=O bonds. This leaves another 2 electrons which are a lone pair. So
overall you have three groups around the central atom (2 Se=O bonds and a lone pair) so it is
trigonal planar if you drew in the lone pair. Removing the lone pair gives you a bent molecule.
For PF4+: P has 5 valence electrons, but it is a positive ion, so you knock off an electron to get 4
P-F bonds are single bonds, and there are 4 of them, which accounts for all the electrons.
Therefore PF4+ is a tetrahedral shape. Finally, ICl3: I has 7 valence electrons. I-Cl bonds are
single bonds, so the I uses one electron in each bond. Therefore 3 of the 7 electrons are involved
in I-Cl bonds, the other 4 must be in 2 lone pairs. This gives 5 electron groups around the central
ion, so the shape is trigonal bipyramid if you draw in the lone pairs (trigonal plane with 2 groups
above and below the plane. You can determine that these two groups must be the lone pairs since
they repel more than the I-Cl bonds). Removing the lone pairs gives you a trigonal planar
molecule. The number of groups around the central ion determines the shape. 2 is linear 3 is
trigonal planar
Solution
VSEPR is easy if you follow the right steps. 1. Find the number of valence (outer)
electrons around the central atom (remember the charge) 2. Work out how many of these are
involved in bonding (remember you may have single or double bonds) 3. Remember the lone
pairs. 4. Work out the shape from the number of atoms bonded to the central atom and the
number of lone pairs. So for SeO2, Se is in group 6, and has 6 valence electrons. Se=O is a
double bond, so there are 2 of Se's electrons in a double bond There are two Se=O bonds, so 4
electrons are involved in Se=O bonds. This leaves another 2 electrons which are a lone pair. So
overall you have three groups around the central atom (2 Se=O bonds and a lone pair) so it is
trigonal planar if you drew in the lone pair. Removing the lone pair gives you a bent molecule.
For PF4+: P has 5 valence electrons, but it is a positive ion, so you knock off an electron to get 4
P-F bonds are single bonds, and there are 4 of them, which accounts for all the electrons.
Therefore PF4+ is a tetrahedral shape. Finally, ICl3: I has 7 valence electrons. I-Cl bonds are
single bonds, so the I uses one electron in each bond. Therefore 3 of the 7 electrons are involved
in I-Cl bonds, the other 4 must be in 2 lone pairs. This gives 5 electron groups around the central
ion, so the shape is trigonal bipyramid if you draw in the lone pairs (trigonal plane with 2 groups

2. above and below the plane. You can determine that these two groups must be the lone pairs since
they repel more than the I-Cl bonds). Removing the lone pairs gives you a trigonal planar
molecule. The number of groups around the central ion determines the shape. 2 is linear 3 is
trigonal planar