Review My Work: Arranging Atoms & Ions by Radius

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BeF2 is expected to be linear due to its two sp hybrid orbitals forming ionic bonds, while SF2 is angular because of the presence of lone pairs that distort its shape. The hybridization of one d, one s, and two p orbitals results in four dsp2 hybrid orbitals. C20H40 is solid at 25°C due to stronger intermolecular forces from more electrons, whereas C4H8 is a gas due to weaker bonds and lower boiling points. The order of increasing atomic radius is O, Cl, K+, S2-, K, with S2- being larger than Cl due to its isoelectronic configuration with Ar and smaller than K+ because of nuclear charge differences. Overall, the responses indicate a good understanding of molecular geometry, hybridization, and atomic radius concepts.
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Okay I did my work but if someone would be so kind as to look at it and see if I am understanding.
Questions
1. Explain why you would expect a BeF2 molecule to be linear but an SF2 molecule to be angular?
BeF2 forms an ionic bond with 2 single bonds. Since Be is in group IIA it forms two sp hybrid orbitals. The farthest away the pairs can be is 180 o angle, making it linear. SF2 forms two covalent bonds with two F atoms. Having four pairs of electrons around the S atom the farthest away they can be is in an angular shape.

2a) How would you designate the hybrid orbitals formed by "mixing" one d, one s and two p orbitals?
a) When mixing one d orbital, one s orbital and 2 p orbitals you get dsp2.
b) How many of the hybrid orbitals would be formed by the mixing?
b) The number you put in is the number you get as a result. So in this case we put in 4 orbitals we then get 4 hybrid orbitals.

3. Explain why C20H40 is a solid at 25degcel, while C4H8 is a gas at 25degcel?
39. C20H40 is a solid at 25o C because has more electrons. With more electrons you get more forces that can hold the lattice pattern together. C4H8 is a gas at 25o C because the bonds are weak and that gives this compound a low boiling temp.

4.Arrange the following atoms or ions in order of increasing radius:
Cl, S2-, K, K+, O
if an explanation for teh position of S2- in relation to the atom or ion in front and behind it.
OK I have:31.
O, S2-, K+, Cl ,K
------------------>
increasing radius

S2- is smaller than K+ in this case because it has a configuration the same as Ar, a noble gas. Moving horizontally in the periodic table the radius decreases in size.
S2- is bigger than Cl because?
If someone would look over my answers and see if i am correct or even close.
Thank you
 
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2 is entirely correct.1 is okay.Indeed the nonbonding pairs distort the linear structure.At 3,there's a connection between the molar mass and the the # of atoms in the hydro-carbide.

4.The order should be

O,Cl,K^{+},S^{2-},K

------------------------------>
increasing ionic/atomic radius.

1.The # of electrons and the # of shells are the main criteria.
2.For isoelectronic configurations,the largest radius corresponds to the smaller nuclear charge.

Daniel.
 
oops.

Sorry about that, I should read better before posting.. I understand it now
 
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