Welcome to the Molecular Orbital Theory and VSEPR Challenge. This quiz is designed to test your understanding of how atoms combine to form molecules and how we can predict their three-dimensional shapes and magnetic properties.
What you will learn:
- Application of VSEPR theory to predict molecular geometry.
- Calculation of bond order using Molecular Orbital Theory (MOT).
- Predicting paramagnetism and diamagnetism in diatomic molecules.
- Understanding the influence of lone pairs on bond angles and molecular symmetry.
Difficulty Level: Moderate
Q1. According to VSEPR theory, what is the molecular geometry of the ##SF_4## molecule?
The sulfur atom in ##SF_4## has 4 bonding pairs and 1 lone pair (total 5 electron domains). This results in a seesaw geometry to minimize electron-pair repulsion.
Q2. Using Molecular Orbital Theory, determine the magnetic behavior of the ##O_2## molecule.
Oxygen has two unpaired electrons in its ##\pi^*_{2p}## antibonding orbitals according to MOT, making it paramagnetic.
Q3. What is the molecular geometry of the triiodide ion, ##I_3^-##?
The central iodine atom in ##I_3^-## has 2 bonding pairs and 3 lone pairs. The lone pairs occupy the equatorial positions of a trigonal bipyramidal electron geometry, resulting in a linear molecular shape.
Q4. According to MOT, why is the ##B_2## molecule paramagnetic?
For ##B_2## (10 electrons), the ##\pi_{2p}## orbitals are lower in energy than the ##\sigma_{2p}## orbital. The last two electrons enter separate ##\pi_{2p}## orbitals (Hund's rule), making it paramagnetic.
Q5. Predict the molecular geometry of ##XeF_2## based on VSEPR theory.
Xenon in ##XeF_2## has 2 bonding pairs and 3 lone pairs. This arrangement leads to a linear molecular geometry.
Q6. Compare the bond stability of ##N_2## and ##N_2^+## using Molecular Orbital Theory.
##N_2## has a bond order of 3. Removing an electron to form ##N_2^+## removes it from a bonding orbital, reducing the bond order to 2.5 and making it less stable.
Q7. Why does ##NH_3## have a significantly higher dipole moment than ##NF_3##?
In ##NH_3##, the N-H bond dipoles point toward the lone pair, reinforcing the total dipole. In ##NF_3##, the highly electronegative fluorine atoms pull electron density away, opposing the lone pair dipole.
Q8. Based on Molecular Orbital Theory, what is the magnetic behavior of the ##C_2## molecule?
##C_2## (12 electrons) fills the ##\sigma_{1s}##, ##\sigma^*_{1s}##, ##\sigma_{2s}##, ##\sigma^*_{2s}##, and ##\pi_{2p}## orbitals. All electrons are paired in the ##\pi_{2p}## level, making it diamagnetic.
Q9. In the trigonal bipyramidal molecule ##PCl_5##, how do the axial and equatorial bond lengths compare?
Axial bonds in ##PCl_5## experience more repulsion from other bonding pairs (three at 90 degrees) compared to equatorial bonds, making them slightly longer and weaker.
Q10. Which statement correctly describes the stability of ##He_2^+## based on MOT?
##He_2^+## has 3 electrons. Two in the ##\sigma_{1s}## (bonding) and one in the ##\sigma^*_{1s}## (antibonding). Bond order = (2-1)/2 = 0.5, allowing it to exist.
Q11. Why is the bond angle in ##H_2O## (104.5°) smaller than the ideal tetrahedral angle (109.5°)?
According to VSEPR theory, lone pair-lone pair repulsion is stronger than lone pair-bond pair and bond pair-bond pair repulsion, which compresses the H-O-H angle.
Q12. Determine the molecular geometry of ##ClF_3## using VSEPR theory.
##ClF_3## has 3 bonding pairs and 2 lone pairs on the chlorine atom. This results in a T-shaped molecular geometry.
Q13. Predict the bond order and magnetic property of the peroxide ion, ##O_2^{2-}##.
##O_2^{2-}## has 18 electrons. Compared to ##O_2##, the two extra electrons fill the ##\pi^*_{2p}## orbitals, resulting in all electrons being paired (diamagnetic) and a bond order of (8-6)/2 = 1.
Q14. What is the bond order of the cyanide ion (##CN^-##)?
##CN^-## has 14 valence electrons (isoelectronic with ##N_2##). Its MOT configuration results in a bond order of 3.
Q15. What is the molecular geometry of ##SO_2##?
Sulfur in ##SO_2## has 2 bonding domains (double bonds) and 1 lone pair. This leads to a bent molecular geometry with an angle slightly less than 120 degrees.
Q16. According to VSEPR theory, what is the shape of the ##BrF_5## molecule?
##BrF_5## has 5 bonding pairs and 1 lone pair on the bromine atom, resulting in a square pyramidal molecular geometry.
Q17. The ##NO^+## ion is isoelectronic with which of the following molecules?
##NO^+## has 7 (N) + 8 (O) - 1 = 14 electrons. ##N_2## also has 7 + 7 = 14 electrons. Being isoelectronic, they share similar MOT configurations and bond orders.
Q18. Based on MOT, is Carbon Monoxide (##CO##) paramagnetic or diamagnetic?
##CO## has 14 electrons (isoelectronic with ##N_2##). All electrons are paired in its molecular orbitals, making it diamagnetic.
Q19. What happens to the bond order when ##O_2## is ionized to ##O_2^+##?
Ionizing ##O_2## to ##O_2^+## involves removing an electron from a ##\pi^*_{2p}## antibonding orbital. Removing an antibonding electron increases the bond order from 2 to 2.5.
Q20. Why does the ##Be_2## molecule not exist under standard conditions according to MOT?
##Be_2## has 8 electrons. The MOT configuration results in an equal number of bonding and antibonding electrons, giving a bond order of 0.
Q21. Predict the magnetic behavior of the ##N_2^{2-}## ion.
##N_2^{2-}## has 16 electrons, making it isoelectronic with ##O_2##. It has two unpaired electrons in the antibonding ##\pi^*_{2p}## orbitals and is therefore paramagnetic.
Q22. What is the molecular geometry of ##XeF_4##?
Xenon in ##XeF_4## has 4 bonding pairs and 2 lone pairs. The lone pairs occupy opposite positions to minimize repulsion, resulting in a square planar geometry.
Q23. According to MOT, what is the bond order of Nitrogen Monoxide (##NO##)?
##NO## has 15 electrons. The configuration includes one unpaired electron in a ##\pi^*_{2p}## antibonding orbital. Bond order = (10 - 5) / 2 = 2.5.
Q24. What is the bond order of the fluorine molecule (##F_2##)?
##F_2## has 18 electrons. The configuration results in 10 bonding and 8 antibonding electrons. Bond order = (10 - 8) / 2 = 1.
Q25. What is the molecular geometry of the ##PH_3## molecule?
Phosphorus in ##PH_3## has 3 bonding pairs and 1 lone pair. This arrangement results in a trigonal pyramidal molecular geometry.
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