Looking for someone to check my chem. questions.

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SUMMARY

The discussion centers on molecular geometry and hybridization in chemistry, specifically addressing why BeF2 is linear while SF2 is angular. Beryllium fluoride exhibits linearity due to its two bonding pairs and absence of lone pairs, resulting in a 180-degree bond angle. In contrast, sulfur difluoride has lone pairs that create repulsion, leading to an angular shape with a bond angle of approximately 109 degrees. Additionally, the hybridization of orbitals is clarified, with dsp2 hybridization resulting from the mixing of one d, one s, and two p orbitals, yielding four hybrid orbitals.

PREREQUISITES
  • Understanding of valence electron-pair repulsion theory
  • Knowledge of molecular geometry and bond angles
  • Familiarity with hybridization concepts in chemistry
  • Basic comprehension of intermolecular forces and their effects on states of matter
NEXT STEPS
  • Study the principles of VSEPR theory for predicting molecular shapes
  • Learn about hybridization types and their applications in molecular bonding
  • Investigate the impact of molecular size on physical states and boiling points
  • Explore intermolecular forces and their role in determining the properties of substances
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Chemistry students, educators, and anyone interested in molecular structure and bonding theories will benefit from this discussion.

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1Q. Explain why you would expect a BeF2 molecule to be linear but a SF2 molecule to be angular?

A. Beryllium fluoride would be linear because each of the two valence electrons in a beryllium atom is shared with an electron from a fluorine atom. Using the valence electron-pair repulsion theory applied to two electron pairs, we would perdict that the angle between the fluorine atoms to be 180deg. The central atom uses all of its bonding electrons to bond to two identical atoms.
SF2, when two atoms are bonded to a third, the resulting molecules are angular if unbonded outer electrons are present in the central atom. The two fluorine atoms would repel at a 109deg.

2aQ. How would you designate the hybrid orbitals formed by "mixing" one d, one s, and two p orbitals?

A. The hybrid orbitals formed would be dsp2.

b) How many hybrid orbitals would be formed by the "mixing"?

A. The number of hybrid orbitals formed would be 4. The total number of orbitals you put in are equal to the number of hybrid orbitals you get.

3Q. Explain why C20H40 is a solid @ 25degcel, while C4H8 is a gas @ 25degcel?

A. C20H40 is a larger compound having more atoms than C4H8 therefore having more electrons. The forces in C20H40 are a lot greater that C4H8 making C20H40 harder to break down resulting in a higher boiling point that C4H8.

Am I on the right track here? Any help would be greatly appreciated.
 
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peck said:
C20H40 is a larger compound having more atoms than C4H8 therefore having more electrons. The forces in C20H40 are a lot greater that C4H8 making C20H40 harder to break down resulting in a higher boiling point that C4H8.

Am I on the right track here? Any help would be greatly appreciated.

Melting is not about breaking down, it is about intermolecular forces.

For the first question think in terms of hybrydization.
 

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