Consider the boiling points for the following substances.Which of the statements
ID: 494042 • Letter: C
Question
Consider the boiling points for the following substances.Which of the statements explains the difference? 1) dimethyl ether, H3C-O-CH3, -25 degrees Celcius and ethanol, H3C-H2C-OH, 79 degrees Celcius a) The London dispersion forces in ethanol are stronger than those in diethyl ether. b) Ethanol is capable of hydrogen bonding; dimethyl ether is not c) Ethanol is a polar molecule; diethyl ether is linear about the C-O-C bond and thus nonpolar d) Bonds in ethanol are more polar than bonds in diethyl ether. 2) HF, 20 degrees Celcius and HCl, -85 degrees Celcius a) The HF molecule is larger and more polarizable; this leads to stronger London dispersion forces b) Hydrogen flouride undergoes hydrogen bonding; hydrogen chloride does not. c) Hydrogen chlordie is larger and less polarizable; it has weaker London dispersion forces. 3) n-pentane, CH3-CH2-CH2-CH2-CH3, 36.2 degrees Celcius and n-propane, Ch3-Ch2-Ch3, -42 degrees Celcius a) The smaller n-pentance molecules are less polarizable and experience weaker London dispersion forces b) The n-pentance molecules are capable of hydrogen bonding and n-propane is not. c) The C-H bonds of pentane are more polar than the C-H bonds of propane; this leads to stronger dipole-dipole interactions. d) The larger n-pentance moelcules are more polarizable and experience stronger London dispersion forces.
Explanation / Answer
Ethanol is having higher boiling point due to hydrogen bonding so answer is =b
HF is having more very polar bonds and is capable of hydrogen bonding and hcl is weak polarizable it has weaker london dispesion forces so answer = c
more no of carbon chain compounds having higher boiling points and due to large suface area of n-pentane molecules shows the strong forces so answer is = d