Question

Х Lab 11.pdf Lab 11.pdf INTRODUCTION The chemical and physical properties of substances are influenced by the way in which valence electrons are distributed and by the three-dimensional shape of the molecule or ion. X- ray and neutron diffraction, infrared, Raman and microwave absorption spectroscopy, as well as dipole moment measurements, are used to ascertain the three-dimensional structure of a substance. Thus, techniques are available which can identify the geometry of substances Results of these measurements generally agree well with the geometries predicted from theory. In this experiment, models will be constructed for several simple molecules and polyatomic ions Most molecules and most complex ions have a "central atom". When the distribution of electrons and bonded atoms around this central atom is determined, the electron geometry. the molecular geometry, the hybridization, and the polarity can be predicted Molecular models will be constructed using a standard model kit. The instructor will demonstrate formation of: (1) single bond (2) lone pair (3) double bond (4) triple bond. LEWIS DOT FORMULA The key to predicting the electronic geometry of a molecule is writing the appropriate dot formula. Many (but not all) examples used will obey the "odet rule". These rules are to be followed: 1. Determine the total number of valence electrons in the formula For each atom, its valence electrons are determined by its Group Number. Thus, carbon in Group IV-A (group 14) has four valence electrons, and chlorine Group VII-A (group 17) has seven valence electrons. In the case of ions, add one electron for each negative charge and subtract one electron for each positive charge 2. Determine the central atom. It is usually the least electronegative atom (except hydrogen). 3. Surround the central atom with the outer atoms. 4. Form one bond (two electrons) from each outer atom to the central atom. 5. Place electron pairs on the outer atoms so that each has an octet (except H which should have only two electrons). 6. If any electrons remain, these are to be placed on the central atom as lone pairs 7. At this point, the central atom should have eight or more electrons surrounding it (except for Be and B which can have fewer than eight). If necessary, move a lone pair of electrons from one of the outer atoms so that it is shared between the outer atom and the central atom (creating a double bond). If the central atom still does not have an octet, repeat the process creating a second double bond or a triple bond. 8. If the central atom has five or more electron groups, it does NOT obey the octet rule. In these cases, multiple bonding is not necessary. Often, when one or more double bonds are created in the formation of a dot structure, the lone pair used to form the double bond may be obtained from one of several atoms. When such choice is available, each option is a possibility and represents a "resonance structure" for the molecule or ion

please I need help with this

Answer 1

As per rules the first four will be solved. Please repost the rest.

1. BeI2 Valence electrons -16 i.e 2 valence electrons from Be and 7 valence electrons from both the I atoms. Electron geometry = Linear Molecular geometry = Linear . Be Be Lewis dot formula I is more electronegative than Be. So the bond moments are directed towards I. Since they run in opposite directions they cancel each other i.e dipole moment = 0. Hence Bel2 is non-polar molecular. 2. CO2 Valence electrons = 16 ie 4 valence electrons from C and 6 valence electrons from each of O atoms. Electron geometry = Linear Molecular geometry = Linear 10: :c: :: =o O is more electronegative than C. So the bond moments are directed towards O. Since they run in opposite directions, they cancel out each other i.e dipole moment = 0. Hence CO2 is non-polar molecule

3. BF3 Valence electrons = 24 ie 3 valence electrons of B and 7 valence electrons each from the three F atoms. Electron geometry = Trigonal planar Molecular geometry = Trigonal planar F is more electronegative than B. So the bond moments -B are directed towards F. The bond moment cancel out each other i.e dipole moment = 0. Hence BF3 is a non- polar molecule. 4. SiC14 Valence electrons = 32 ie 4 valence electrons of Si and 7 valence electrons each from the four Ci atoms. Electron geometry = Tetrahedral Molecular geometry = Tetrahedral :01: Cl is more electronegative than Si So the bond moments are directed towards C1. The bond moments cancel out each other i.e dipole Ta Ci moment = 0. Hence SiC14 is a non-polar :ci: moelcule. S; c1