Consider the following Lie algebra:

${\cal G}=Vect\{a,a',b,b',c,c',d,d'\}$ $[a,a']=[b,b']=c$,

$[a,b]=[a',b]=[a,b']=[a',b']=[c,a]=[c,b]=[c,a']=[c,b']=[c',a]=[c',a']=[c',b]=[c',b']=[c,c']=[d,a]=[d,a']=[d,b]=[d,b']=[d,c]=[d,c']=[d',a]=[d',a']=[d,b]=[d',b']=[d',c]=[d',c']=[d,d]=0.$

${\cal G}$ is a 2-nilpotent algebra and its center is $Vect\{c,c',d,d'\}$. Its derived ideal $[{\cal G},{\cal G}]=Vect\{c\}$.

Suppose that there exists a symplectic form $\omega$ on ${\cal G}$, you have:

$\omega(c,c')=\omega([a,a'],c')$, since the form is closed, you have:

$\omega([a,a'],c)+\omega([c,a],a')+\omega([a',c],a)=0=\omega([a,a'],c)=\omega(c,c')$ since $c$ is in the center.

$\omega(c,d)=\omega([a,a'],d)$.

$\omega([a,a'],d)+\omega([d,a],a')+\omega([a',d],a)=0=\omega([a,a'],d)=\omega(c,d)=0$ since $d$ is in the center of ${\cal G}$

$\omega(c,d')=\omega([a,a'],d')$.

$\omega([a,a'],d')+\omega([d',a],a')+\omega([a',d'],a)=0=\omega([a,a'],d')=\omega(c,d')=0$ since $d'$ is in the center of ${\cal G}$

$\omega(c,a)=\omega([b,b'],a)$, we have:

$\omega([b,b'],a)+\omega([a,b],b')+\omega([b',a],b)=0$, since $[a,b]=[b',a]=0$, we deduce that $\omega([b,b'],a)=\omega(c,a)=0$.

$\omega(c,a')=\omega([b,b'],a')$

$\omega([b,b'],a')+\omega([a',b],b')+\omega([b',a'],b)=0$, since $[a',b]=[b',a']=0$, we deduce that $\omega([b,b'],a')=\omega(c,a')=0$.

$\omega(c,b)=\omega([a,a'],b)$

we have:
$\omega([a,a'],b)+\omega([b,a],a')+\omega([a',b],a)=0$, since $[b,a]=[a',b]=0$, we deduce that $\omega([a,a'],b)=\omega(c,b)=0$.

$\omega(c,b')=\omega([a,a'],b')$

we have:
$\omega([a,a'],b')+\omega([b',a],a')+\omega([a',b'],a)=0$, since $[b',a]=[a',b']=0$, we deduce that $\omega([a,a'],b')=\omega(c,b')=0$.

Since $\omega(c,c)=0$, we deduce that for every $x\in {\cal G}, \omega(c,x)=0$. Contradiction since a symplectic form is not degenerated.
There does not exist a symplectic form on the 2-nilpotent Lie algebra ${\cal G}$.