An example of the nature as in title has asked in mathstack; but here I am trying to do it by a different way.
Consider ring $\mathcal{O}=\mathbb{Z}[\sqrt{5}]$ and ideal $I=(2,1+\sqrt{5})$. Claim is that this is not invertible.
(1) Consider $(\mathcal{O}:I)=\{x\in K: xI\subset \mathcal{O}\}$ where $K$ is field of fractions of $\mathcal{O}$.
(2) Then $K=\mathbb{Q}(\sqrt{5})$ and $(\mathcal{O}:I)=\{ \frac{a+b\sqrt{5}}{2}: a,b\in\mathbb{Z}, a\equiv b\pmod{2}\}$.
(3) Then $I(\mathcal{O}:I)=\{a+b\sqrt{5} : a,b\in\mathbb{Z}, a\equiv b\pmod{2}\}$; so this is not equal to $\mathbb{Z}[\sqrt{5}]$.
Hence $I$ is not invertible ideal in $\mathbb{Z}[\sqrt{5}]$. Is this computation correct?
If we consider in general $\mathbb{Z}[\sqrt{d}]$ where $d$ is a square-free integer and $d\equiv 1\pmod{4}$ then still $(2,1+\sqrt{d})$ works as non-invertible ideal in this ring, am I right?