If $Z$ is independent of outcome $Y$, can we still use it as an instrument?

Question

In a standard instrumental variable estimation setting, consider treatment $X$, outcome $Y$, instrument $Z$ and unobserved confounder $H$. For it to be a valid instrument, $Z$ should satisfy the following properties:

1. Relevance: $Z \not{\perp \!\!\! \perp} X$
2. Exclusion Restriction: $Z {\perp \!\!\! \perp} Y \mid X$
3. Non-Confoundedness: $Z {\perp \!\!\! \perp} H$

My question is the following: Generally, if $Z {\perp \!\!\! \perp} Y$ (without conditioning on $X$), is it still a valid instrument? Since

$$Z {\perp \!\!\! \perp} Y \;\;\;\Longrightarrow\;\;\; Z {\perp \!\!\! \perp} Y \mid X\;,$$

but

$$Z {\perp \!\!\! \perp} Y \;\;\;\not\Longleftarrow\;\;\; Z {\perp \!\!\! \perp} Y \mid X\;,$$

can we still say that exclusion restriction holds and therefore $Z$ is a valid instrument?

Answer 1

No. If $Y$ and $Z$ are marginally independent, it means that either $Z$ has no effect on $X$ (violates assumption $1$), or $X$ has no effect on $Y$.