I'm playing around with products $M = \Bbb S^{n_1} \times \Bbb S^{n_2}$, and a quick computation using the Künneth formula tells us that if $(n_1,n_2)$ is not $(1,1)$ or $(2,4)$, $M$ is not symplectic (WLOG $1 \leq n_1 \leq n_2$, of course). The $(1,1)$ case is obviously symplectic, but I couldn't decide about the $(2,4)$ case. So:

Is $\Bbb S^2 \times \Bbb S^4$ symplectic?

I'm playing around with products $M = \Bbb S^{n_1} \times \Bbb S^{n_2}$, and a quick computation using the Künneth formula tells us that if $(n_1,n_2)$ is not $(1,1)$ or $(2,4)$, $M$ is not symplectic (WLOG $1 \leq n_1 \leq n_2$, of course). The $(1,1)$ case is obviously symplectic, but I couldn't decide about the $(2,4)$ case. So:

Is $\Bbb S^2 \times \Bbb S^4$ symplectic?

Edit: after some time I came back to those calculations. I had missed the obvious case $(n_1,n_2) = (2,2)$. The proof given in the answers can be adapted to show that products of the form $\Bbb S^2 \times \Bbb S^{n_2}$ for even $n_2>2$ are not symplectic. The conclusion of what happened here is the

Theorem: Let $1 \leq n_1 \leq n_2$ be natural numbers. Then $\Bbb S^{n_1}\times \Bbb S^{n_2}$ is symplectic if and only if $n_1=n_2=1$ or $n_1=n_2=2$.

I'm playing around with products $M = \Bbb S^{n_1} \times \Bbb S^{n_2}$, and a quick computation using the Künneth formula tells us that unless $(n_1,n_2) = (1,1)$ or $(2,4)$, $M$ is not symplectic (WLOG $1 \leq n_1 \leq n_2$, of course). The $(1,1)$ case is obviously symplectic, but I couldn't decide about the $(2,4)$ case. So:

Is $\Bbb S^2 \times \Bbb S^4$ symplectic?

I'm playing around with products $M = \Bbb S^{n_1} \times \Bbb S^{n_2}$, and a quick computation using the Künneth formula tells us that if $(n_1,n_2)$ is not $(1,1)$ or $(2,4)$, $M$ is not symplectic (WLOG $1 \leq n_1 \leq n_2$, of course). The $(1,1)$ case is obviously symplectic, but I couldn't decide about the $(2,4)$ case. So:

Is $\Bbb S^2 \times \Bbb S^4$ symplectic?