VSEPR and Bond Hybridization - AP Chemistry

when a central atom of a molecule has 6 electron domains coming off of it (none of which are lone pairs of electrons), it is considered octahedral

SF4 has 6 electron domains coming off of it- 4 F molecules and 2 lone pairs of e–. This is an example of see-saw shape.

COH2:

:O:

||

H—C—H

This is a trigonal planar molecule, which only has bond angles of 120o

Each p subshell can have a maximum of 6 electrons in it. Note that there are three p orbitals, denoted px, py, and pz. These orbitals represent the plane in space in which they reside.

NH3 is trigonal pyramidal because it has 4 electron domains, one of them being a lone pair of electrons and the other three being H atoms. When this is arranged in a three-dimensional space, it is trigonal pyramidal in shape.

CO2H is trigonal planar, so the bond angles will be 120 only.

the sulfur has 2 atoms bonded (the H) and 2 pairs of electrons, which is a total of 4 electron domains. The hybridization of the sulfur therefore is sp3

(a) Octahedral: six charge clouds and six atoms (b) Square planar: six charge clouds and four atoms (c) Trigonal pyramidal: four charge clouds and three atoms (d) Tetrahedral: four charge clouds and four atoms (e) Trigonal bipyramidal: five charge clouds and five atoms

The answer to this question can be determined by drawing a Lewis structure of the molecule. The S atom is central, with an O atom on each side. Adding in double bonds to each O atom and placing lone pairs on the O atoms give them each an octet, which leaves 2 more lone pairs of electrons to go on the S atom, creating a bent shape.

Molecules with three atoms around a central atom such as BF3 are trigonal planar because electron repulsion is minimized by positioning the three attachments toward the corners of an equilateral triangle.

Multiple bonds, although they may be stronger, are not necessarily favored in a Lewis structure. It is dependent on the atoms involved. Thus, this is not a criterion used in determining a preferred Lewis structure.

There are three bonding electron pairs around the central atom and one non-bonding pair, for a total of 4 electron groups. They arrange spontaneously to be furthest apart according to VSEPR theory, which corresponds to when they form a trigonal pyramidal shape.

It is important to note that electronic geometry includes the orientation fo lone pairs, while molecular geometry considers only the geometry of the atoms present in the compound.

Methane, or CH4, has four hydrogen atoms bound to a central carbon, resulting in a tetrahedral geometry.

BF3 has three fluorine atoms bound to a central boron, resulting in a trigonal planar geometry.

Be(Cl)2 has two chlorine atoms bound to a central beryllium, resulting in a linear geometry.

Water, or H2O, has two hydrogen atoms bound to a central oxygen atom with two lone pairs. The molecular geometry is bent, but the electronic geometry is tetrahedral.

Ammonia, or NH3, has three hydrogens bound to a central nitrogen atom with one lone pair. The moelcular geometry is trigonal pyramidal, but the electronic geometry is tetrahedral.

Ammonia, water, and metahne all have the same electronic geometries (tetrahedral), giving us our final answer.

CO_2 has two bonding regions and no lone pairs around the central atom, so it has a linear geometry, so its bond angle is $180^{circ}$.

CH_4 has four bonding regions and no lone pairs around the central atom, so it has a tetrahedral geometry with bond angles of $109.5^{circ}$.

SF_6 has six bonding regions and no lone pairs around the central atom, so it has an octahedral geometry with bond angles of $90^{circ}$ and $180^{circ}$. For the sake of ranking, we will only consider the smaller angle.

NH_3 has three bonding regions and one lone pair around the central atom, so it has a trigonal pyramidal geometry with bond angles of $107^{circ}$.

BF_3 has three bonding regions and no lone pairs around the central atom, so it has a trigonal planar geometry with bond angles of $120^{circ}$.

Trigonal bipyramidal geometries arise from the hybridization of one s orbital, three p orbitals, and one d orbital, hence the name $sp^3d$.

CH_4 has four bonding regions and no lone pairs, so it has a tetrahedral geometry. H_2O has two bonding regions, but it also has two lone pairs, so it has a bent geometry.

Nitrogen is a Group 5 element, so when it is bound to three hydrogen atoms, it will have a lone pair.

NH_3 has four areas of electron density (three bonds and one lone pair), so its bonds are $sp^3$ hybridized.

Because of the lone pair, NH_3 has a trigonal pyramidal geometry.

CH_2Cl_2 has four bonding regions and no lone pairs, so it has a tetrahedral geometry. The bond angles between all of the bonded atoms is $109.4^{circ}$.

In the reactant, , the carbon is bound to three hydrogens and one oxygen, and it has no lone pairs, so it has a tetrahedral geometry. In the product, , the carbon is bound to two hydrogens and one oxygen, so it must be double bonded to the oxygen in order to have a complete octet. Therefore, it has a trigonal planar geometry.

Ammonia exhibits a tetrahedral electron pair geometry. It has three bonded pairs (between nitrogen and each hydrogen), and one lone pair (on nitrogen). This combination forms a trigonal pyramidal molecular geometry.