Molecular Geometry And Hybridization Quiz #3

Molecules of the general formula AB can have n greater than four only when A is an element from the third period or below. This means that if A is an element from the third period or below in the periodic table, the molecule AB can have more than four atoms. If A is an element from a higher period, the molecule AB will not have more than four atoms.

The central atom utilizes d orbitals to form hybrid orbitals in molecules that have an expanded octet or have more than 8 valence electrons. In this case, (iii) TeCl, (iv) XeF, and (v) SF are the only molecules that fit this criteria. PCl and CCl do not have an expanded octet and do not require the use of d orbitals for hybridization.

The BeCl molecule consists of two atoms, Be and Cl, with a linear arrangement. Since there are no lone pairs on the central atom (Be), the molecule is symmetrical and the dipole moments of the two Be-Cl bonds cancel each other out, resulting in a nonpolar molecule.

The spd atomic hybrid orbital set consists of three orbitals: one s orbital, three p orbitals, and five d orbitals. Each orbital can accommodate a maximum of two electrons. Therefore, when all the orbitals in the spd set are considered, a total of six electron domains can be accommodated.

The hybridization of iodine in IF is sp^3d, while the hybridization of iodine in IF7 is sp^3d^2.

The XeF molecule is formed by the bonding of xenon (Xe) and fluorine (F) atoms. In order to form bonds, Xe utilizes hybrid orbitals. The sp^3d^2 hybrid orbitals are used by Xe in the XeF molecule. This hybridization allows Xe to form six bonds, as it has six electron pairs in its valence shell. Therefore, the correct answer is sp^3d^2.

BF3 is the only nonpolar molecule among the given options. This is because BF3 has a trigonal planar molecular geometry with the boron atom in the center and three fluorine atoms surrounding it. The fluorine atoms are all identical and have the same electronegativity, resulting in a symmetrical distribution of charge around the boron atom. Since there are no lone pairs of electrons on the central atom and the bond dipoles cancel out each other, BF3 is nonpolar. In contrast, NF3, IF3, PBr3, and BrCl3 have polar bonds or lone pairs, leading to an overall dipole moment and making them polar molecules.

AsH3 (arsine) is the only polar molecule among the given options. This is because it has a trigonal pyramidal molecular geometry with a lone pair of electrons on the central atom (arsenic). The asymmetry in the molecule leads to an uneven distribution of charge, resulting in a polar molecule. The other molecules listed (SbF5, I2, SF6, CH4) have symmetrical molecular geometries or no lone pairs, making them nonpolar.

CO3^-2 is the only molecule among the given options that has sp hybridization of the central atom. In CO3^-2, the central atom is carbon, which forms three sigma bonds with three oxygen atoms. To accommodate these bonds, carbon undergoes sp hybridization, where one s orbital and two p orbitals hybridize to form three sp hybrid orbitals. These orbitals are then used to form sigma bonds with the oxygen atoms, resulting in a trigonal planar geometry around the carbon atom. The other molecules in the options do not have sp hybridization of the central atom.

SF6 only is the correct answer because SF6 is an octahedral molecule with six fluorine atoms surrounding the central sulfur atom. The fluorine atoms are arranged symmetrically around the sulfur atom, resulting in a symmetrical distribution of electron density. As a result, SF6 is a nonpolar molecule. On the other hand, SF2 and SF4 have a bent and seesaw molecular geometry respectively, which leads to an asymmetrical distribution of electron density and makes them polar molecules.