1) Compare nitrogen (N) and Chlorine (Cl). Select the correct statement and explain your choice. a) N has larger atomic radii than Cl and a higher electronegativity value than Cl. b) N has larger atomic radii than Cl and a lower electronegativity value than Cl. c) N has smaller atomic radii than Cl and a higher electronegativity value than Cl. d) N has smaller atomic radii than Cl and a lower electronegativity value than Cl.

Question

1) Compare nitrogen (N)  and Chlorine (Cl). Select the correct statement and explain your choice. a) N has larger atomic radii than Cl and a higher electronegativity value than Cl. b) N has larger atomic radii than Cl and a lower electronegativity value than Cl. c) N has smaller atomic radii than Cl and a higher electronegativity value than Cl. d) N has smaller atomic radii than Cl and a lower electronegativity value than Cl.

anonymous · Answer

Which of the following best explains the forces acting upon polar molecules?
a)They have ionic bonds which are acted upon by strong dipole forces.

b)They have high electronegativity values which account for strong London dispersion forces.

c)They have small atomic radii which make for greater forces of attraction.

d)They have partial charges and are bound together by dipole-dipole intermolecular forces.

anonymous · Answer

These are the radii trend and electronegativity trend. Hope it will help you understand the problem better.

$N$ is in period 2 while $Cl$ is in period 3; that means $N$'s radii is less than $Cl$'s radii.

About the negativity, we cannot compare $N$'s and $Cl$'s electronegativities directly. So we have to check the table. $Cl$'s electronegativity is slightly higher than $N$'s.

anonymous · Answer

So for the first question, D is correct.

anonymous · Answer

For the second question, D is correct. The bound type is dipole-dipole forces. You can find out more intermolecular forces here: http://chemed.chem.purdue.edu/genchem/topicreview/bp/intermol/intermol.html

anonymous · Answer

@tinybookworm, thank you for your assistance, I appreciate it. :)

anonymous · Answer

@tinybookworm, can you please help me with one more? 


What type of bonding, molecular geometry (shape), and intermolecular forces for HF and HBr. Which substance would experience a stronger attraction between its molecules, and why?


(Chemistry isn't my strongest subject...)

anonymous · Answer

We use the difference between the electronegativities of elements ($\Delta \lambda$) to decide the type of bonding of a molecule.
    + If the difference range is >1.8, the bond is ionic bond.
    + If the difference range is 0.5-1.7, the bond is polar covalent bond.
    + If the difference range is 0.0-0.4, then the bond is nonpolar covalent bond.

The electronegativities of H, F, and Br are 2.2, 3.98, and 2.96.

$\Delta \lambda$ of HF is $3.98-2.2=1.78$. Based on the rule, the bond of HF is very polar and nearly ionic. In conclusion, the bond of HF is polar covalent bond.
$\Delta \lambda$ of HBr is $2.96-2.2=0.76$. Therefore, the bond of HBr is polar covalent bond.

For the geometric shape, both HF and HBr are made of two atoms. Therefore, the molecular geometry for both molecule is linear.

HF and HBr are polar. Therefore, the intermolecular force between the molecules is dipole-dipole.

Finally, we also use $\Delta \lambda$ to compare the attraction between molecules of a compound. The higher $\Delta \lambda$ is, the stronger the attraction is. $\Delta \lambda _{HF} ~(1.78)>\Delta \lambda _{HBr} ~(0.76)$ so I believe you can give your conclusion for the question "Which substance would experience a stronger attraction between its molecules?"

Sorry for the long explanation, I hope you will get the idea

anonymous · Answer

@tinybookworm thank you for all your help :)

anonymous · Answer

You are very welcome