Which member of each pair has the greater polarizability-i.e. which member of ea
ID: 990438 • Letter: W
Question
Which member of each pair has the greater polarizability-i.e. which member of each pair will most form stronger dispersion forces? Briefly explain your reasoning. Ca2+ or Ca CH_3CH_3 or CH_3CH_3CH_3 CCI_4 or CF_4 Circle the member of each pair that has the greater boiling point? Briefly explain your reasoning. Your explanation must include the nature of the intermolecular forces involved. CH_3CH_2OH or CH_3CH_2CH_3 NO or N_2 H_2S or H_2Te Which would you expect to have the lower boiling point, CH_3CH_2CH_2CH_3 or cyclobutane (C_4H_8)? Circle your choice and briefly explain your reasoning. Your explanation should include the nature of the intermolecular forces involved.Explanation / Answer
11. (a) Ca2+
Due more effective nuclear charge in the Ca2+, it will be more polar.
11. (b) CH3CH2CH3
The surface area of the molecule propane will be higher than ethane due to large size. Thus, the greater van der Waals forces between adjacent molecules make it more polarizable.
11. (c) CCl4
Because the net dipole in both the case will be zero due to tetrahedral geometry. but the bigger size of CCl4 will be accountable for higher dispersion force.
12. (a) CH3CH2OH
Ethanol will be stabilized by inter molecular hydrogen bonding and consequently will be having higher b.p. than CH3CH2CH3.
12. (b) NO
Because N2 is a neutral homo diatomic molecule with no dipole. But NO is a hetero diatomic radical and have dipole.
12. (C) H2Te
The size of the TeH2 molecules is bigger compared to H2S which gives more interactions
13. Cyclobutane have much higher boiling point compared to butane. Because the cyclic molecules have less surface area, but they might enable more productive interactions over larger parts of their surface area due to their fixed conformations.
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