Chemical Bonding II (Geometry) Practice Questions & Answers

The VSEPR model assumes that electron pairs in the valence shell of a central atom will adopt an arrangement that minimizes repulsions between these electron pairs by maximizing the distance between them.[1][2] Source: OpenStax

Lone pairs occupy more space than bonding pairs, leading to stronger repulsions.[1][2] The order is: lone pair-lone pair > lone pair-bonding pair > bonding pair-bonding pair. Source: OpenStax

We count the number of regions of electron density (lone pairs and bonds) around the central atom.[1][2][3] A single, double, or triple bond counts as one region of electron density.[1][2][4] Source: OpenStax

The bond angles are slightly smaller than 109.5° because the lone pair-bonding pair repulsion is greater than the bonding pair-bonding pair repulsion, taking up more space.[1][2][3][4][5] Source: OpenStax

With two bonds and no lone pairs of electrons on the central atom, the bonds are as far apart as possible (180°), adopting a linear structure.[1][2] Source: OpenStax

$BCl_3$BCl3​ contains three bonds and no lone pairs.[1][2][3] The arrangement of three regions of high electron density gives a trigonal planar geometry.[1][2][3] Source: OpenStax

Methane has four bonding pairs around the central carbon atom; the electron-pair geometry is tetrahedral, as is the molecular structure. Source: OpenStax

Ammonia has four electron pairs (tetrahedral geometry), but one is a lone pair.[1][2][3] The molecular structure is trigonal pyramidal.[1][2] Source: OpenStax

Water has four regions of electron density (two bonds, two lone pairs).[1][2][3] The electron geometry is tetrahedral, but the molecular structure is bent.[1][2] Source: OpenStax

Any molecule with five electron pairs around the central atom including no lone pairs will be trigonal bipyramidal.[1][2] $PF_5$PF5​ is a common example.[1][2] Source: OpenStax