The trans-alkenes are formed by the reduction of alkynes with:

We have to find out which reagent converts an alkyne ($$R-C \equiv C-R'$$) into a trans (that is, $$E$$-) alkene. In organic chemistry two standard partial‐reduction methods are famous:

1. The Lindlar catalyst, written as $$H_{2} / Pd-CaCO_{3} \ or\ Pd/BaSO_{4} \ (poisoned)$$, gives syn addition of hydrogen. Because both hydrogens add from the same face, the product is the cis (or $$Z$$-) alkene.

2. The dissolving-metal reduction, usually $$Na / liq. NH_{3}$$ (or $$Li / liq. NH_{3}$$), proceeds through radicals and delivers the two hydrogens from opposite faces (anti-addition). Anti-addition produces the trans (or $$E$$-) alkene.

Let us write the key steps for the dissolving-metal (Birch) reduction of an internal alkyne:

$$ R-C \equiv C-R' \;\overset{Na,\ liq.\ NH_{3}}{\underset{\text{-33 °C}}{\rightarrow}} R{-}\underset{\text{trans}}{\text{CH}=\text{CH}}{-}R' $$

Step-wise mechanism (shown only to keep every algebraic/chemical step explicit):

1. Electron addition from $$Na$$ gives a radical anion: $$R-C \equiv C-R' + e^- \to R-C \equiv C-R'^{\bullet-}$$

2. Protonation by liquid ammonia ($$NH_{3}$$ is the proton source) forms a vinyl radical: $$R-C \equiv C-R'^{\bullet- + NH3 \to R-C(H)=C^{\bullet}-R' + NH2^-}$$

3. A second electron addition gives a vinyl anion: $$R-C(H)=C^{\bullet-R' + e^- \to R-CH= C^{-}-R'}$$

4. Final protonation furnishes the trans-alkene: $$R-CH= C^{--R' + NH3 \to R-CH=CH-R' + NH2^-}$$

Because the two protonation steps take place from opposite sides of the molecule, the overall result is anti-addition, giving the trans product.

Now we test each option given in the question:

A. $$Sn/HCl$$ - this reagent is used for reducing nitro groups to amines, not for selective alkyne reduction. It does not give trans-alkenes.

B. $$H_{2}-Pd/C, BaSO_{4}$$ - this is exactly the Lindlar (poisoned) catalyst discussed above; it furnishes cis alkenes, not trans.

C. $$NaBH_{4}$$ - sodium borohydride is a mild hydride donor; it reduces aldehydes and ketones, but it does not reduce carbon-carbon triple bonds at all.

D. $$Na / liq. NH_{3}$$ - as explained, this is the dissolving-metal (Birch) reduction that converts an alkyne specifically into a trans alkene.

So, the only reagent that satisfies the requirement is Option D.

Hence, the correct answer is Option D.