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 $(s)$ $H_2O$ $(t)$ $B_2H_6$
| 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 $(1-None, 2-q, 3-r, 4-s)$.
$(1)$ 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 $(1-None, 2-q, 3-r, 4-s)$.
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