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First suppose $X$ is a nonempty integral scheme, with generic point $\eta$ and function field $K={\mathcal{O}}_{X,\eta }$. Since taking stalks preserves the inclusion $\mathcal{I}\hookrightarrow {\mathcal{O}}_X$, local freeness of rank two would give an injective $K$-linear mapcontradicting the dimension of a vector space.

For a nonempty Noetherian scheme that is reduced, there are finitely many irreducible components. Choose one, and remove the union of the others. The resulting nonempty open subscheme is irreducible and reduced, hence integral. Restricting the proposed locally free sheaf to it gives the contradiction above. Therefore no locally free ideal of rank two exists on a nonempty reduced Noetherian scheme.

The rank bound for locally free ideals on reduced schemes in fact removes the Noetherian assumption: on an affine open where the rank is fixed, localization at a minimal prime ideal gives a field, so a free ideal has rank at most one. Nonemptiness is necessary for the literal statement: on the empty scheme the zero sheaf is vacuously locally free of every stipulated rank. The question is interpreted with this usual nonempty hypothesis.

A finite locally free sheaf of ideals on a reduced scheme has rank at most one at every point. On a nonempty affine open subscheme where its rank is $r$, localize its inclusion into the structure sheaf at a minimal prime ideal. The resulting local ring is a field $K$, giving an injection $K^r\hookrightarrow K$, hence $r\le 1$. This works without a Noetherian assumption. The empty scheme is a vacuous exception to claims phrased as nonexistence of a sheaf of a prescribed rank.