Observe the structures of the sodium salts of various carboxylic...

Heating a sodium salt of a carboxylic acid with soda lime ($NaOH$ and $CaO$) results in decarboxylation, yielding an alkane with one carbon atom less than the original salt.

The general reaction is:
$R-COONa \xrightarrow{NaOH + CaO, \Delta} R-H + Na_2CO_3$

We need the product to be 2-methylpropane, which has the structure $CH_3-CH(CH_3)-CH_3$ or $(CH_3)_3CH$.

Let us analyze the decarboxylation products of the given salts:

For I: $CH_3-CH_2-CH(CH_3)-COONa \xrightarrow{\Delta} CH_3-CH_2-CH_2-CH_3$ (n-butane)

For II: $(CH_3)_3C-COONa \xrightarrow{\Delta} (CH_3)_3CH$ (2-methylpropane)

For III: $CH_3-CH(CH_3)-CH_2-COONa \xrightarrow{\Delta} CH_3-CH(CH_3)-CH_3$ (2-methylpropane)

For IV: $CH_3-CH_2-CH_2-CH_2-COONa \xrightarrow{\Delta} CH_3-CH_2-CH_2-CH_3$ (n-butane)

Thus, both salts II and III yield 2-methylpropane as the major organic product.

Answer: II and III only