The compound that does not produce nitrogen gas by thermal decomposition is:

We start by recalling that “thermal decomposition” simply means heating a substance until it breaks up into simpler products. Our task is to check, one by one, whether nitrogen gas, $$N_2$$, appears among those products.

First we take the azide salt $$Ba(N_3)_2$$. A standard rule from coordination and main-group chemistry is that a metal azide decomposes as

(Thermal decomposition of a metal azide) $$M(N_3)_n \xrightarrow{\Delta} M + \dfrac{3n}{2}\,N_2$$

Putting $$M = Ba$$ and $$n = 2$$ gives

$$Ba(N_3)_2 \;\xrightarrow{\Delta}\; Ba + 3\,N_2.$$

We clearly see molecular nitrogen, so $$Ba(N_3)_2$$ does produce $$N_2$$.

Next we examine ammonium dichromate, $$(NH_4)_2Cr_2O_7$$. Textbook and laboratory demonstrations of the “ammonium-dichromate volcano” are based on the reaction

$$ (NH_4)_2Cr_2O_7 \;\xrightarrow{\Delta}\; Cr_2O_3 + N_2 + 4\,H_2O.$$

Nitrogen gas again appears explicitly on the right-hand side, so $$(NH_4)_2Cr_2O_7$$ gives $$N_2$$ when heated.

For ammonium nitrite, $$NH_4NO_2$$, we use the well-known self-oxidation-reduction (disproportionation) that occurs on heating:

$$ NH_4NO_2 \;\xrightarrow{\Delta}\; N_2 + 2\,H_2O.$$

Thus $$NH_4NO_2$$ definitely releases $$N_2$$.

Finally, we consider ammonium sulfate, $$(NH_4)_2SO_4$$. When heated, this salt first loses ammonia:

$$ (NH_4)_2SO_4 \;\xrightarrow{\Delta}\; NH_3 \;+\; NH_4HSO_4.$$

With stronger heating, further stepwise reactions produce sulfur dioxide, water vapour and additional ammonia or hydrogen sulfide, but no balanced thermal sequence for $$(NH_4)_2SO_4$$ ever contains molecular $$N_2$$. The nitrogen atoms remain in the form of ammonia or other nitrogen-containing ions; they do not couple to give $$N_2$$.

Summarising the four cases:

$$Ba(N_3)_2 \to N_2$$ (yes)

$$(NH_4)_2Cr_2O_7 \to N_2$$ (yes)

$$NH_4NO_2 \to N_2$$ (yes)

$$(NH_4)_2SO_4 \not\to N_2$$ (no)

Hence, the correct answer is Option A.