Suppose $i_Z \hookrightarrow X$ be a closed immersion, with $Z$ and $X$ being smooth varieties over $\mathbb{C}$, and $c, d$ are the dimensions of $Z$ and $X$ respectively.

$\textbf{Question}:$ Is it true that $L_n i^*_{Z} \mathcal{F} = 0$ for any $n>d-c$ and $\mathcal{F}\in Coh(X)$ ?

The reason I ask is the following inconsistency I encounter (suppose $d=2$ and $c=1$ for simplicity), \begin{equation} E_2^{p,q}:= Ext_Z^p(L_q i^*_{Z}\mathcal{F},\mathcal{O}_Z) \Longrightarrow E^{p+q}_{\infty} = Ext_Z^{p+q}(L i^*_{Z}\mathcal{F},\mathcal{O}_Z)=Ext_X^{p+q}(\mathcal{F},i_{Z *} \mathcal{O}_Z). \end{equation}

Note that $E_2^{1,2}$ survives to infinity, so $Ext_X^{3}(\mathcal{F},i_{Z *} \mathcal{O}_Z) \ne 0$, this is absurd because $X$ is a surface and $Ext_X^3 = 0$. But this must be true for any curve $Z\subset X$, and any sheaf over that (not just the structure sheaf $\mathcal{O}_Z$). So I got the question above.

This is puzzling for me because in general there is a resolution by locally free sheaves (again I just assume $d=2$),

\begin{equation} 0 \longrightarrow V^{2}\longrightarrow V^1 \longrightarrow V^0 \longrightarrow \mathcal{F}\longrightarrow 0, \end{equation} and $L_n i_Z^* \mathcal{F} = \mathcal{H}^{-n} i_Z^* (V^{\bullet})$, so why if $n=2$ the corresponding cohomology should be zero?