Hybrid Atomic Orbitals Practice Questions

Why is the concept of hybridization required in valence bond theory?

Solution:
Hybridization is introduced to explain the geometry of bonding orbitals in valance bond theory.

Give the shape that describes each hybrid orbital set:

(a) sp2

(b) sp3d

(c) sp

(d) sp3d2

Explain why a carbon atom cannot form five bonds using sp3d hybrid orbitals.

Solution:
There are no d orbitals in the valence shell of carbon.

What is the hybridization of the central atom in each of the following?

(a) BeH2

(b) SF6

(c) $PO_4^{3−}$

(d) PCl5

A molecule with the formula AB3 could have one of four different shapes. Give the shape and the hybridization of the central A atom for each.

Solution:
trigonal planar, sp2; trigonal pyramidal (one lone pair on A) sp3; T-shaped (two lone pairs on A sp3d, or (three lone pairs on A) sp3d2

Methionine, CH3SCH2CH2CH(NH2)CO2H, is an amino acid found in proteins. The Lewis structure of this compound is shown below. What is the hybridization type of each carbon, oxygen, the nitrogen, and the sulfur?

Sulfuric acid is manufactured by a series of reactions represented by the following equations:
$$S_8(s)+8O_2(g)⟶8SO_2(g)$$ $$2SO_2(g)+O_2(g)⟶2SO_3(g)$$ $$SO_3(g)+H_2O(l)⟶H_2SO_4(l)$$

Draw a Lewis structure, predict the molecular geometry by VSEPR, and determine the hybridization of sulfur for the following:

(a) circular S8 molecule

(b) SO2 molecule

(c) SO3 molecule

(d) H2SO4 molecule (the hydrogen atoms are bonded to oxygen atoms) Solution

(a) Each S has a bent (109°) geometry, sp3



(b) Bent (120°), sp2



(c) Trigonal planar, sp2



(d) Tetrahedral, sp3

Two important industrial chemicals, ethene, C2H4, and propene, C3H6, are produced by the steam (or thermal) cracking process:
$$2C_3H_8(g)⟶C_2H_4(g)+C_3H_6(g)+CH_4(g)+H_2(g)$$

For each of the four carbon compounds, do the following:

(a) Draw a Lewis structure.

(b) Predict the geometry about the carbon atom.

(c) Determine the hybridization of each type of carbon atom.

Analysis of a compound indicates that it contains 77.55% Xe and 22.45% F by mass.

(a) What is the empirical formula for this compound? (Assume this is also the molecular formula in responding to the remaining parts of this exercise).

(b) Write a Lewis structure for the compound.

(c) Predict the shape of the molecules of the compound.

(d) What hybridization is consistent with the shape you predicted? Solution

(a) XeF2
(b)



(c) linear (d) sp3d

Consider nitrous acid, HNO2 (HONO).

(a) Write a Lewis structure.

(b) What are the electron pair and molecular geometries of the internal oxygen and nitrogen atoms in the HNO2 molecule?

(c) What is the hybridization on the internal oxygen and nitrogen atoms in HNO2?

Strike-anywhere matches contain a layer of KClO3 and a layer of P4S3. The heat produced by the friction of striking the match causes these two compounds to react vigorously, which sets fire to the wooden stem of the match. KClO3 contains the $ClO_3^−$ ion. P4S3 is an unusual molecule with the skeletal structure.

(a) Write Lewis structures for P4S3 and the $ClO_3^–$ ion.

(b) Describe the geometry about the P atoms, the S atom, and the Cl atom in these species.

(c) Assign a hybridization to the P atoms, the S atom, and the Cl atom in these species.

(d) Determine the oxidation states and formal charge of the atoms in P4S3 and the $ClO_3^–$ ion.

Solution
(a)



(b) P atoms, trigonal pyramidal; S atoms, bent, with two lone pairs; Cl atoms, trigonal pyramidal; (c) Hybridization about P, S, and Cl is, in all cases, sp3; (d) Oxidation states P +1, S $−1\frac{1}{3}$, Cl +5, O –2. Formal charges: P 0; S 0; Cl +2: O –1

Identify the hybridization of each carbon atom in the following molecule. (The arrangement of atoms is given; you need to determine how many bonds connect each pair of atoms.)

Write Lewis structures for NF3 and PF5. On the basis of hybrid orbitals, explain the fact that NF3, PF3, and PF5 are stable molecules, but NF5 does not exist. Solution

Phosphorus and nitrogen can form sp3 hybrids to form three bonds and hold one lone pair in PF3 and NF3, respectively. However, nitrogen has no valence d orbitals, so it cannot form a set of sp3d hybrid orbitals to bind five fluorine atoms in NF5. Phosphorus has d orbitals and can bind five fluorine atoms with sp3d hybrid orbitals in PF5.

In addition to NF3, two other fluoro derivatives of nitrogen are known: N2F4 and N2F2. What shapes do you predict for these two molecules? What is the hybridization for the nitrogen in each molecule?