Why can't methanol, ch3oh, be used as a solvent for sodium amide, nanh2? sodium amide is nonpolar and methanol is polar. sodium amide is polar and methanol is nonpolar. sodium amide undergoes an acid-base reaction with methanol. there would be no ion-dipole attractive forces between the two compounds.

sodium amide undergoes an acid -base reaction

Explanation:

sodium amide is a ionic compound and basically exists as sodium cation and amide anion. Amide anion is highly basic in nature and hence as soon as  there is amide anion generated in the solution , Due to its very pronounced acidity it very quickly abstracts the slightly acidic proton available on methanol.

This leads to formation of ammonia and sodium methoxide.

Hence sodium amide reacts with methanol and abstracts its only acidic proton and form ammonia and sodium Methoxide.

Hence the 3rd statement is a corrects statement.

So we cannot use methanol for sodium amide because sodium amide itself would react with methanol and the inherent molecular natur of sodium amide would then change.

The 1st and 2nd statements both are incorrect because both the compounds methanol as well as sodium amide have dipole moments and hence are polar molecules.

The 4th statement is also incorrect as both the molecules have dipole moment and hence there would be ion-dipole forces operating between them.

The following reaction occurs:

NaNH₂+CH₃OH→NH₃+CH₃ONa

i'm assuming that the correct answer is indeed the first one/ "a"..

the correct answer is b. a strong acid completely dissociates to its ions when dissolved in water. completely dissociates to and ions. a is incorrect because does not fully ionize to and ions and some remain as the compound . c and d are incorrect because, and do not fall into the category of bronsted lowry acids. a bronsted lowry acid is a proton donor and since both molecules do not have a proton to donate they are not bronsted lowry acids. likewise,   if we use the lewis acid definition, they cannot be acids. a lewis acid is an electron pair acceptor. both and have a complete octet so can not accept any electrons which disqualifies them as acids.