Its All about Pj Problem Strings (SiPjAjk) - 7 Spaces Of Interest (Si) and their associated Basic Sequences; 7 Pj Problems of Interest (PPI) and their Alleles (Ajk)
Precise molecular shapes are determined by the bond lengths and bond angles established by the bonding atoms. However, the Valence Shell Electron Pair Repulsion (VSEPR) model predicts molecular shapes fairly well. Here, we examine VSEPR as it relates to the prediction of molecular shapes of AB molecules.
1. Define the following: (a) ABn molecules (b) Lewis Electron-Dot Diagram (c) Electron Domain.
2. Determine the Lewis structure for the following molecules: (a) CO2 (b) H2O (c) BF3 (d) XeF4 (e) CCl4 (f) NH3 (g) PCl5 (h) SF6
3. Use the Valence-Shell Electron-Pair Reduction (VSEPR) model to predict the molecular shapes and bond angles of the molecules indicated in problems 2a, 2c, 2e, 2g and 2h.
4. Predict the molecular geometries of the molecules of 2b, 2d and 2f, from the electron-domain geometries of problem 2. Will the bond angles of these molecules be greater or smaller than the bond angles indicated in the electron-domain geometries from which their molecular geometries were derived?
5. Write a general representative string for the VSEPR model.
1(a) ABn molecules are molecules with a central atom (A) and n similar atoms (B) bonded to A. For example CO2 (where carbon is the A atom and oxygen the B atoms with n =2).
1(b) The Lewis Dot Diagram is the representation of an atom, ion, or molecule such that the element's symbol represents the nucleus and all inner shell electrons while the dots that surround the symbol represent the valence electrons. The diagram is also called the Lewis structure of the atom, ion or molecule.
1(c) Electron domain refers to the region occupied by non-bonding pair of electrons, a single covalent bond, or multiple covalent bonds around the central atom of an ABn molecule.
2. The Lewis structures for the molecules are:
3. The general steps for using the VSEPR model is as follows:
(i). Determine the Lewis structure of the molecule or ion, then count the total electron-domains around the central atom. Each nonbonding pair, single bond, or multiple bond around the central atom counts as an electron-domain.
(ii). Determine the electron-domain geometry by arranging the electron-domains about the central atom in a manner that minimizes the electron repulsion among them.
(iii). Use the arrangement of the bonded atoms to determine the molecular geometry of the molecules.
Electron domains = 2; bonding domains = 2; nonbonding domains = 0; electron geometry = linear; molecular geometry = linear; bonding angle = 180o
Electron domains = 3; bonding domains = 3; nonbonding domains = 0; electron geometry = trigonal planar; molecular geometry = trigonal planar; bonding angle =120o. In the trigonal planar, all the atoms are on the same triangular plane of an equilateral triangle.
Electron domains = 4; bonding domains = 4; nonbonding domains = 0; electron geometry = tetrahedral; molecular geometry = tetrahedral; bonding angle = 109.5o. The tetrahedron has 4 vertices and four faces all of which are equilateral triangles. The vertices of the equilateral base holds three atoms. The central atom is the center of the tetrahedron which is above the plane of the base. The fourth atom is at the fourth vertex which is above both the base of the tetrahedron and the central plane.
Electron domains = 5; bonding domains = 5; nonbonding domains = 0; electron geometry = trigonal bipyramidal; molecular geometry = trigonal bipyramidal; bonding angle between axial and equatorial bond = 90o; bond angle between equatorial bonds = 120o. The trigonal bipyramid has 5 vertices. Three of which are the vertices of an equilateral triangle. The three atoms at the vertices of the equilateral triangle are equatorial atoms. The central atom A, is at the center of the equilateral triangle. The two atoms at the vertices above and below the plane of the equilateral triangle are axial atoms.
Electron domains = 6; bonding domains = 6; nonbonding domains = 0; electron geometry = octahedral; molecular geometry = octahedral; bonding angle between axial and equatorial bond = 90o; bond angle between equatorial bonds = 90o. The octahedral has six vertices and 8 surfaces which are all equilateral triangles. Four equatorial atoms are at the vertices of a square. The central atom A, is at the center of the square. The axial atoms are at the vertices above and below the plane of the square.
electron domains = 4; bonding domains = 2; nonbonding domains = 2; electron geometry = tetrahedral; molecular geometry = bent (tetrahedral minus 2 electron domains); bond angle will be less than 120o because of the effect of the nonbonding domains.
electron domains = 6; bonding domains = 4; nonbonding domains = 2; electron geometry = octahedral; molecular geometry = square planar (ocahedral minus 2 axial electron domains); bond angle 90o. No axial bonds.
electron domains = 4; bonding domains = 3; nonbonding domains = 1; electron geometry = tetrahedral; molecular geometry = trigonal pyramidal (tetraahedral minus 1 electron domain); bond angle will be less than 120o because of the effect of the nonbonding domain.
5. The VSEPR model is based on repulsion so the general representative string is: