Match the correct option from Column-$A$ and Column-$B$.

Column-$A$	Column-$B$
$(1)$ Ionic hydride	$(p)$ $BeH_2$
$(2)$ Interstitial hydride	$(q)$ $TiH$
$(3)$ Electron-precise hydride	$(r)$ $CH_4$
$(4)$ Electron-rich hydride	$(s)$ $H_2O$
	$(t)$ $B_2H_6$

(A) The correct matches are as follows:
$(1)$ Ionic hydride: These are formed by $s$-block elements. However,$BeH_2$ and $MgH_2$ are polymeric and covalent. The question implies a standard classification where ionic hydrides are typically $s$-block (except $Be, Mg$). Given the options,let's re-evaluate:
$(1)$ Ionic hydride: $NaH$ (not listed).
$(2)$ Interstitial hydride: $TiH$ (metallic/interstitial).
$(3)$ Electron-precise hydride: $CH_4$ (Group $14$ elements).
$(4)$ Electron-rich hydride: $H_2O$ (Group $15-17$ elements).
Note: $BeH_2$ is polymeric covalent,$B_2H_6$ is electron-deficient. Based on standard classification: $(1)$ Ionic hydride: None of the options perfectly fit,but often $s$-block hydrides are categorized here. $(2)$ Interstitial: $TiH$. $(3)$ Electron-precise: $CH_4$. $(4)$ Electron-rich: $H_2O$. The provided solution $1-p$ is technically incorrect as $BeH_2$ is covalent. Correct matching is $(1-None, 2-q, 3-r, 4-s)$.