What is vsepr shape

For some molecules in the Table, we note that there is more than one possible shape that would satisfy the VSEPR rules. For example, the XeF 2 molecule has a steric number of five and a trigonal bipyramidal geometry.

The observed geometry of XeF 2 is linear, which can be rationalized by considering the orbitals that are used to make bonds or lone pairs in the axial and equatorial positions.

There are four available orbitals, s, p x , p y , and p z. If we choose the z-axis as the axial direction, we can see that the p x and p y orbitals lie in the equatorial plane. We assume that the spherical s orbital is shared equally by the five electron domains in the molecule, the two axial bonds share the p z orbital, and the three equatorial bonds share the p x and p y orbitals.

We can then calculate the bond orders to axial and equatorial F atoms as follows:. Because fluorine is more electronegative than a lone pair, it prefers the axial site where it will have more negative formal charge. In general, by this reasoning, lone pairs and electropositive ligands such as CH 3 will always prefer the equatorial sites in the trigonal bipyramidal geometry.

Electronegative ligands such as F will always go to the axial sites. In the case of the BrF 4 - anion, which is isoelectronic with XeF 4 in the Table, the electronic geometry is octahedral and there are two possible isomers in which the two lone pairs are cis or trans to each other. In this case, lone pair - lone pair repulsion dominates and we obtain the trans arrangement of lone pairs, giving a square planar molecular geometry.

The observation of molecules in the various electronic shapes shown above is, at first blush, in conflict with our picture of atomic orbitals. For an atom such as oxygen, we know that the 2s orbital is spherical, and that the 2p x , 2p y , and 2p z orbitals are dumbell-shaped and point along the Cartesian axes. Given the relative orientations of the atomic orbitals, how do we arrive at angles between electron domains of To understand this we will need to learn a little bit about the quantum mechanics of electrons in atoms and molecules.

By analogy with classical mechanics, the Hamiltonian is commonly expressed as the sum of operators corresponding to the kinetic and potential energies of a system in the form. Because we can't locate the nonbonding electrons with any precision, this prediction can't be tested directly. But the results of the VSEPR theory can be used to predict the positions of the nuclei in these molecules, which can be tested experimentally.

If we focus on the positions of the nuclei in ammonia, we predict that the NH 3 molecule should have a shape best described as trigonal pyramidal , with the nitrogen at the top of the pyramid. Water, on the other hand, should have a shape that can be described as bent , or angular.

Both of these predictions have been shown to be correct, which reinforces our faith in the VSEPR theory. Click here to check your answer to Practice Problem 6. Use the Lewis structure of the NO 2 molecule shown in the figure below to predict the shape of this molecule. Click here to check your answer to Practice Problem 7.

When we extend the VSEPR theory to molecules in which the electrons are distributed toward the corners of a trigonal bipyramid, we run into the question of whether nonbonding electrons should be placed in equatorial or axial positions.

Experimentally we find that nonbonding electrons usually occupy equatorial positions in a trigonal bipyramid. To understand why, we have to recognize that nonbonding electrons take up more space than bonding electrons. Nonbonding electrons need to be close to only one nucleus, and there is a considerable amount of space in which nonbonding electrons can reside and still be near the nucleus of the atom. Bonding electrons, however, must be simultaneously close to two nuclei, and only a small region of space between the nuclei satisfies this restriction.

Because they occupy more space, the force of repulsion between pairs of nonbonding electrons is relatively large. The lone electron pairs on the O atom are omitted for clarity. The molecule will not be a perfect equilateral triangle because the C—O double bond is different from the two C—H bonds, but both planar and triangular describe the appropriate approximate shape of this molecule. The first step is to draw the Lewis structure of the molecule.

The lone electron pairs on the Cl atoms are omitted for clarity. The P atom has four electron groups with three of them bonded to surrounding atoms, so the molecular shape is trigonal pyramidal. The N atom has three electron groups on it, two of which are bonded to other atoms. The molecular shape is bent. What is the approximate molecular shape of CH 2 Cl 2?

Ethylene C 2 H 4 has two central atoms. Determine the geometry around each central atom and the shape of the overall molecule. Hint, hydrogen is a terminal atom. The approximate shape of a molecule can be predicted from the number of electron groups and the number of surrounding atoms.

The orbitals containing the various bonding and nonbonding pairs in the valence shell will extend out from the central atom in directions that minimize their mutual repulsions. Coordination number refers to the number of electron pairs that surround a given atom, often referred to as the central atom.

The geometries of molecules with lone pairs will differ from those without lone pairs, because the lone pair looks like empty space in a molecule. Both classes of geometry are named after the shapes of the imaginary geometric figures mostly regular solid polygons that would be centered on the central atom and have an electron pair at each vertex.

In the water molecule AX 2 E 2 , the central atom is O, and the Lewis electron dot formula predicts that there will be two pairs of nonbonding electrons. The oxygen atom will therefore be tetrahedrally coordinated, meaning that it sits at the center of the tetrahedron. Two of the coordination positions are occupied by the shared electron-pairs that constitute the O—H bonds, and the other two by the non-bonding pairs.

Therefore, although the oxygen atom is tetrahedrally coordinated, the bonding geometry shape of the H 2 O molecule is described as bent.

The effect of the lone pair on water : Although the oxygen atom is tetrahedrally coordinated, the bonding geometry shape of the H2O molecule is described as bent. There is an important difference between bonding and non-bonding electron orbitals.

Because a nonbonding orbital has no atomic nucleus at its far end to draw the electron cloud toward it, the charge in such an orbital will be concentrated closer to the central atom; as a consequence, nonbonding orbitals exert more repulsion on other orbitals than do bonding orbitals. In H 2 O, the two nonbonding orbitals push the bonding orbitals closer together, making the H—O—H angle The electron-dot structure of NH 3 places one pair of nonbonding electrons in the valence shell of the nitrogen atom.

This means that there are three bonded atoms and one lone pair for a coordination number of four around the nitrogen, the same as occurs in H 2 O. The Lewis dot structure for ammonia, NH3. We can therefore predict that the three hydrogen atoms will lie at the corners of a tetrahedron centered on the nitrogen atom.

The lone pair orbital will point toward the fourth corner of the tetrahedron, but since that position will be vacant, the NH 3 molecule itself cannot be tetrahedral; instead, it assumes a pyramidal shape, more specifically, that of a trigonal pyramid a pyramid with a triangular base. The hydrogen atoms are all in the same plane, with the nitrogen outside of the plane.