Hodge conjecture claims that every rational Hodge class is in the $\mathbb{Q}$-span of the image of the algebraic cycles in the cohomology. I believe the complexifed version of the conjecture should be incorrect i.e. $cl\otimes \mathbb{C}: CH^p(X)\otimes \mathbb{C}\rightarrow H^{p,p}(X,\mathbb{C})$ should not be surjective in general. This is equivalent to saying that $Hdg^p(X,\mathbb{Q})\otimes \mathbb{C}$ should not be $H^{p,p}(X, \mathbb{C})$. Is there any example showing this fails in general?
Note that here $Hdg^p(X,\mathbb{Q})$ is the intersection of $H^{2p}(X, \mathbb{Q})$ with $H^{p,p}(X, \mathbb{C})$. If Hodge conjecture is true it means that $cl: CH^p(X)\otimes \mathbb{Q}\rightarrow Hdg^p(X,\mathbb{Q})$ is surjective. This implies that $cl\otimes \mathbb{C}: CH^p(X)\otimes \mathbb{C} \rightarrow Hdg^p(X, \mathbb{Q})\otimes \mathbb{C}$ is surjective but the problem (which I am asking for an example) is that $Hdg^p(X,\mathbb{Q})\otimes \mathbb{C}\subset H^{p,p}(X, \mathbb{C})$ and this inclusion is not necessarily an isomorphism.
Any example with $H^{p,p}(X,\mathbb C)$ is larger than the complexification of $H^{p,p}(X,\mathbb C)\cap H^{2p}(X,\mathbb Z)$ will work.
There are plenty of such examples. Let's investigate $\dim X=2$ case.
For an abelian surface, its $H^{1,1}$ has dimension 4. The class $dz_1\wedge d\bar{z}_2$ has type (1,1), but does not come from complexification of an integral Hodge class.
For a K3 surface, its $H^{1,1}$ has dimension 20, but its Picard rank can range from any integer between 1 and 20. Any K3 surface with Picard rank below 20 would work.