The Hodge diamond \begin{array}{ccccc}&&1&&\\&0&&0&\\a&&1&&a\\&0&&0&\\&&1&&\end{array} is naively realisable.
Suppose $M$ is a compact Kähler surface with the given Hodge diamond with $a \geq 2$. As $h^{2,0}(M) > 1$, the Kodaira dimension of $M$ is either $1$ or $2$. Note that $$c_1(M)^2 = 2\chi(M) + 3\sigma(M) = 2(2a + 3) + 3(2a + 1) = 10a + 9.$$ It follows from the Kodaira-Enriques classification that $c_1(M)^2 \leq 0$ for a surface of Kodaira dimension $1$, so we must have $\kappa(M) = 2$. Note however that $$c_1(M)^2 = 10a + 9 > 6a + 9 = 3(2a + 3) = 3c_2(M)$$ which contradicts the Bogomolov-Miyaoka-Yau inequality. Therefore, the above Hodge diamond with $a \geq 2$ is not realisable.
Finally, the corresponding Betti sequence is realisable as the blowup of $\mathbb{CP}^2$ at $2a$ points demonstrates.
The same arguments apply to the Hodge diamond \begin{array}{ccccc}&&1&&\\&0&&0&\\a&&b&&a\\&0&&0&\\&&1&&\end{array} with $a \geq b \geq 1$ and $a \geq 2$.
The case $a = b = 1$ provides another example. The same arguments as above can be used to rule out Kodaira dimensions $-\infty$, $1$, and $2$. To rule out $\kappa(M) = 0$, note that $M$ must be minimal and hence must be finitely covered by a K3 surface or a complex torus. As $\chi(M) = 5$ and the Euler characteristic is multiplicative under finite covers, this is impossible.