Question

Consider the bonding in an \(\mathrm{MgH}_{2}\) molecule. (a) Draw a Lewis structure for the molecule, and predict its molecular geometry. (b) What hybridization scheme is used in \(\mathrm{MgH}_{2}\) ? (c) Sketch one of the two-electron bonds between an \(\mathrm{Mg}\) hybrid orbital and an \(\mathrm{H} 1 \mathrm{~s}\) atomic orbital.

Short answer

(a) The Lewis structure for $\mathrm{MgH}_{2}$ has a central Mg atom with two H atoms bonded linearly. The molecular geometry is linear with bond angles of 180 degrees. (b) The hybridization scheme in $\mathrm{MgH}_{2}$ for the Mg atom is sp, as there are 2 sigma bonds and no lone pairs. (c) The bonding between the Mg and H orbitals involves one of the Mg's hybrid sp orbitals overlapping with the H's 1s atomic orbital to form a two-electron bond. A sketch of this bonding can be found in chemistry textbooks or online resources.

Step-by-step solution

Part (a): Lewis Structure and Molecular Geometry

Step 1: Find the total number of valence electrons in MgH2. Mg has 2 valence electrons and H has 1 valence electron. So, the total number of valence electrons = 2 + (2 × 1) = 4. Step 2: Arrange the atoms and distribute the electrons. Place Mg in the center of the molecule as it has the lowest electronegativity. Now, distribute two hydrogen atoms around Mg and place the 4 valence electrons as bonding pairs between each Mg and H. The Lewis structure for MgH2 would look like this: H | Mg | H Step 3: Predict the molecular geometry. Since there are only two bonding pairs around the Mg atom, it has a linear molecular geometry with bond angles of 180 degrees.

Part (b): Hybridization Scheme in MgH2

Step 1: Find the number of sigma bonds and lone pairs. Mg has two sigma bonds (one with each hydrogen atom) and no lone pairs of electrons. Step 2: Calculate the hybridization scheme. Hybridization is determined by the number of sigma bonds and lone pairs around the central atom. The number of hybrid orbitals is equal to the sum of the number of sigma bonds and lone pairs. In this case, there are 2 sigma bonds and 0 lone pairs around the Mg atom. Therefore, 2 hybrid orbitals are required for Mg. Step 3: Determine the hybridization scheme. Since there are 2 hybrid orbitals required, the hybridization scheme is s + (n-1)p, where n is the number of hybrid orbitals. Here, n=2, so the hybridization scheme is s + p = sp. Therefore, Mg has an sp hybridization scheme in MgH2.

Part (c): Sketch the Bonding between Mg and H Orbitals

Step 1: Identify the orbitals involved in bonding. The bonding between the magnesium and hydrogen atoms involves one of the magnesium's hybrid sp orbitals and the hydrogen's 1s atomic orbital. Step 2: Sketch the bonding. To sketch the bonding between the orbitals, represent the Mg sp hybrid orbital as a combination of the s and p orbitals. Draw the 1s orbital for the hydrogen atom and overlap it with the Mg sp hybrid orbital to form a two-electron bond. [Note: Creating a sketch of orbitals is beyond the capabilities of text-based response. In order to visualize the sketch, please refer to any chemistry textbook or online resources that provide orbital diagrams for this type of bonding.]