Molecular Geometry And Hybridization Quiz #1

According to VSEPR theory, if there are four electron domains in the valence shell of an atom, they will be arranged in a tetrahedral geometry. In this arrangement, the four electron domains are positioned as far apart as possible, creating a three-dimensional shape with a central atom and four surrounding atoms or lone pairs. This geometry is commonly observed in molecules like methane (CH4), where the central carbon atom is surrounded by four hydrogen atoms.

The molecular shape of a molecule with the formula AB can be either linear or bent. This is because the central atom (A) can have two or three bonding pairs of electrons, resulting in either a linear shape if there are two bonding pairs, or a bent shape if there are three bonding pairs. The actual shape depends on the arrangement of the bonding and non-bonding electron pairs around the central atom.

The hybridization of orbitals on the central atom in a molecule is sp. This means that the central atom has two hybridized orbitals and two unhybridized p orbitals. In a linear electron-domain geometry, the central atom is surrounded by two electron domains, which corresponds to the sp hybridization. Therefore, the correct answer is linear.

XeF4 is the correct answer because it has a square planar molecular geometry. In XeF4, xenon (Xe) is surrounded by four fluorine (F) atoms in a square planar arrangement. This arrangement allows for the bond angles to be 90 degrees, resulting in a square shape. The other options, CCl4, PH3, XeF2, and ICl3, have different molecular geometries and do not have a square planar shape.

The electron-domain geometry of a sulfur-centered compound is trigonal bipyramidal, which means that there are five electron domains around the central sulfur atom. In order to accommodate these five electron domains, the central sulfur atom must undergo hybridization. The sp^3d hybridization allows the central sulfur atom to form five hybrid orbitals, which are arranged in a trigonal bipyramidal shape. Therefore, the correct hybridization of the central sulfur atom in this compound is sp^3d.

The F-B-F bond angle in the BF molecule is 120 degrees. This is because the BF molecule has a trigonal planar molecular geometry, with the central boron atom bonded to three fluorine atoms. In a trigonal planar arrangement, the bond angles between the atoms are all equal and measure 120 degrees.

The HO ion consists of one oxygen atom bonded to one hydrogen atom. The oxygen atom has two lone pairs of electrons, causing the molecule to have a trigonal pyramidal molecular geometry. In this geometry, the oxygen atom is at the center, and the three atoms (two lone pairs and one hydrogen) are arranged in a pyramid shape around it.

In valence bond theory, the formation of a bond involves the overlapping of atomic orbitals. In the case of bromine (Br), the correct answer is 4p. This is because bromine has an electron configuration of [Ar] 3d10 4s2 4p5. The valence electrons in bromine are in the 4p orbital, and during bond formation, these orbitals overlap with orbitals from other atoms to create the bond. Therefore, the 4p orbitals on bromine atoms overlap in the formation of the bond in Br.

The electron-domain geometry of iodine trichloride is trigonal bipyramidal because it has five electron domains around the central iodine atom. The molecular geometry is T-shaped because it has three bonded atoms and two lone pairs of electrons, resulting in a T-shaped arrangement.

The O-S-O bond angle in SO is slightly less than 120. This is because the molecule has a bent or V-shaped geometry due to the presence of two lone pairs of electrons on the central sulfur atom. These lone pairs repel the bonding pairs, causing the bond angle to be slightly less than the ideal tetrahedral angle of 109.5 degrees.