The given compound is \( K_2[Ni(CN)_4] \), where the oxidation state of Ni is +2. On reacting with excess \( K/\text{liq. NH}_3 \), a yellow compound is formed. This typically suggests a reduction of Ni(II) to Ni(0) since yellow complexes are often associated with nickel in the zero oxidation state. Thus, the oxidation state of Ni in compound X is 0.
Final Answer:
\[
\boxed{0}
\]