Question

question 6 (1 point)
which of the molecules, co₂, nh₃ and bcl₃ will be polar?
co₂, nh₃ and bcl₃
co₂ and bcl₃
nh₃
co₂ and nh₃
bcl₃
question 7 (1 point)
place the following compounds in order of decreasing strength of intermolecular forces, starting with the strongest forces.
hf h₂ nbr₃
hf > h₂ >nbr₃
h₂ > hf >nbr₃
h₂ >nbr₃ > hf
nbr₃ > h₂ > hf
hf > nbr₃ > h₂

Explanation:

Question 6

• \(CO_2\): It has a linear geometry (\(O = C=O\)). The bond dipoles cancel each other out (\(\mu = 0\)), so it is non - polar.
• \(NH_3\): It has a trigonal pyramidal geometry due to the lone pair on nitrogen. The bond dipoles do not cancel, and there is a net dipole moment (\(\mu

eq0\)), so it is polar.

• \(BCl_3\): It has a trigonal planar geometry. The bond dipoles cancel each other (\(\mu = 0\)) (since the vectors of the \(B - Cl\) bond dipoles sum to zero), so it is non - polar.

• \(HF\): It has hydrogen bonding (a special type of dipole - dipole interaction). Hydrogen bonding occurs when \(H\) is bonded to a highly electronegative atom (\(F\) in this case).
• \(NBr_3\): It has dipole - dipole interactions. The \(N - Br\) bonds are polar, and the molecule has a net dipole moment (trigonal pyramidal geometry due to the lone pair on \(N\)).
• \(H_2\): It has only London dispersion forces. London dispersion forces are the weakest intermolecular forces and are present in all non - polar and polar molecules, but they are the only forces in non - polar \(H_2\) (\(H - H\) is a non - polar bond).

Since hydrogen bonding (\(HF\)) is stronger than dipole - dipole (\(NBr_3\)) and dipole - dipole is stronger than London dispersion forces (\(H_2\))

Answer:

\(NH_3\)

Question 7