Molecule # [ TeO2 ] Draw the Lewis Structure below. Calculate the number of valence electrons. Draw the three-dimensional structure of the molecule using the wedge and dash notation. The molecular geometry is: This molecule is polar/nonpolar (Circle your choice). Our logic for this choice is: The formal charge on the central Te atom is. The hybridization of Te is. The formal charge on the terminal O atom is. Estimate the O-Te-O bond angle.
The Correct Answer and Explanation is:
Here are the answers for the worksheet on the tellurite ion, TeO3^2-.
Calculate the number of valence electrons: 26
Draw the Lewis Structure below:
The Lewis structure shows a central Tellurium (Te) atom. The Te atom has one lone pair of electrons. It is bonded to three Oxygen (O) atoms. One oxygen atom is connected by a double bond and has two lone pairs. The other two oxygen atoms are connected by single bonds and each has three lone pairs. The entire structure is enclosed in square brackets with a 2- charge on the outside. (Note: This is one of three equivalent resonance structures. In reality, the three Te-O bonds are identical, with a bond order of 4/3).
Draw the three-dimensional structure of the molecule using the “wedge and dash” notation:
The 3D structure has the Tellurium (Te) atom at the top center. Below it, one Te-O bond is drawn with a solid line (in the plane of the page), one Te-O bond is drawn with a solid wedge (coming out of the page), and one Te-O bond is drawn with a dashed line (going into the page). This arrangement shows a trigonal pyramidal shape. The invisible lone pair on the Te atom occupies the fourth position of a tetrahedral electron geometry, pushing the three oxygen atoms down into the pyramid base.
The molecular geometry is: Trigonal pyramidal
This molecule is polar nonpolar: polar (The word “polar” should be circled).
Our logic for this choice is: The tellurite ion is polar because of its asymmetrical molecular geometry. The central tellurium atom has three bonding pairs and one lone pair, resulting in a trigonal pyramidal shape. The bonds between tellurium and the more electronegative oxygen atoms are polar. Due to the asymmetrical arrangement, the individual bond dipoles do not cancel each other out. The lone pair on the central atom creates an uneven distribution of electron density, leading to a net molecular dipole moment and making the ion polar.
The formal charge on the central Te atom is 0
on the terminal O atom is -1 (for the single-bonded oxygens)
The hybridization of Te is sp3
Estimate O-Te-O bond angle < 109.5°
Explanation
To analyze the tellurite ion (TeO3^2-), we first calculate the total number of valence electrons. Tellurium (Te) and Oxygen (O) are both in Group 16, so they each contribute 6 valence electrons. The 2- charge adds two more electrons. The total is (6 for Te) + (3 × 6 for O) + 2 = 26 valence electrons.
For the Lewis structure, tellurium is the central atom as it is the least electronegative. After connecting the three oxygen atoms to tellurium with single bonds (using 6 electrons), we distribute the remaining 20 electrons. We place three lone pairs on each oxygen atom to satisfy their octets (using 18 electrons). The final 2 electrons are placed on the central tellurium atom as a lone pair. This initial structure gives tellurium a formal charge of +1 and each oxygen a formal charge of -1. To minimize formal charges, we can form a double bond between tellurium and one oxygen, creating a resonance structure. In this more stable structure, the double-bonded oxygen has a formal charge of 0, the two single-bonded oxygens have a formal charge of -1, and the central tellurium atom has a formal charge of 0. The overall charge remains -2.
The VSEPR model is used to predict the geometry. The central tellurium atom has four electron domains: three bonding domains (to the O atoms) and one lone pair domain. This corresponds to an AX3E1 arrangement. The electron geometry is tetrahedral, but the molecular geometry, which describes the arrangement of atoms, is trigonal pyramidal.
This trigonal pyramidal geometry is inherently asymmetrical. Because oxygen is more electronegative than tellurium, the Te-O bonds are polar. The asymmetrical shape prevents these polar bond dipoles from canceling out, resulting in a net dipole moment and making the ion polar.
The hybridization of the central tellurium atom is determined by its four electron domains, which requires four hybrid orbitals. This corresponds to sp3 hybridization. The ideal bond angle for a tetrahedral arrangement is 109.5°. However, the lone pair on the tellurium atom exerts greater repulsion than the bonding pairs, compressing the O-Te-O bond angles to be less than 109.5°.
