Question

Match the following: List - I \quad (Type of hydride) \quad List - II \quad (Example) A) Electron precise \quad i) SiH$_4$ B) Saline \quad ii) H$_2$O C) Electron deficient \quad iii) MgH$_2$ D) Electron rich \quad iv) B$_2$H$_6$

• A. A-(ii), B-(iii), C-(i), D-(iv)
• B. A-(i), B-(iii), C-(iv), D-(ii)
• C. A-(iv), B-(ii), C-(iii), D-(i)
• D. A-(ii), B-(i), C-(iv), D-(iii)

Hint:

\textbf{Electron precise:} Group 14 hydrides (e.g., CH$_4$, SiH$_4$). Central atom has exactly enough electrons for bonding to achieve an octet (no lone pairs, no deficiency).
\textbf{Saline/Ionic:} Group 1 and most Group 2 hydrides (e.g., LiH, CaH$_2$, MgH$_2$). Contain H$^-$ ions.
\textbf{Electron deficient:} Group 13 hydrides (e.g., B$_2$H$_6$, AlH$_3$). Central atom has fewer electrons than needed for an octet through normal covalent bonds.
\textbf{Electron rich:} Group 15-17 hydrides (e.g., NH$_3$, H$_2$O, HF). Central atom has lone pairs after forming bonds.

Answer 1

The Correct Option is B

Let's classify each example hydride from List-II according to the types in List-I. Types of Hydrides: \begin{itemize} \item Electron precise hydrides: These are typically formed by Group 14 elements. The central atom has exactly the number of electrons required to form normal covalent bonds (octet satisfied with no lone pairs or electron deficiency). Example: CH$_4$, SiH$_4$. \item Saline hydrides (or Ionic hydrides): These are formed by s-block elements (Group 1 and Group 2, except Be and sometimes Mg due to covalent character). They are ionic compounds containing H$^-$ ions. Example: NaH, CaH$_2$, MgH$_2$. \item Electron deficient hydrides: These are typically formed by Group 13 elements. The central atom has fewer electrons than required to form normal covalent bonds and achieve an octet. They often form polymeric structures or involve multi-center bonds. Example: B$_2$H$_6$ (diborane). \item Electron rich hydrides: These are typically formed by Group 15, 16, and 17 elements. The central atom has one or more lone pairs of electrons after forming normal covalent bonds. Example: NH$_3$, H$_2$O, HF. \end{itemize} Now let's match the examples from List-II: \begin{itemize} \item i) SiH$_4$ (Silane): Silicon is in Group 14. SiH$_4$ has a tetrahedral structure with Si forming four single covalent bonds with H atoms. Silicon completes its octet. This is an electron precise hydride. So, A - Electron precise matches with i) SiH$_4$. \item ii) H$_2$O (Water): Oxygen is in Group 16. In H$_2$O, oxygen forms two single covalent bonds with H atoms and has two lone pairs of electrons. It has more electrons (lone pairs) than needed for bonding. This is an electron rich hydride. So, D - Electron rich matches with ii) H$_2$O. \item iii) MgH$_2$ (Magnesium hydride): Magnesium is in Group 2 (alkaline earth metal). MgH$_2$ is generally considered an ionic or saline hydride, containing Mg$^{2+}$ and H$^-$ ions, although it has some covalent character. Among the given categories, "Saline" is the best fit. So, B - Saline matches with iii) MgH$_2$. \item iv) B$_2$H$_6$ (Diborane): Boron is in Group 13. In B$_2$H$_6$, each boron atom does not have enough electrons to form normal two-center-two-electron bonds with all surrounding atoms. It features banana bonds (three-center-two-electron bonds). This is an electron deficient hydride. So, C - Electron deficient matches with iv) B$_2$H$_6$. \end{itemize} Summarizing the matches: A - (i) B - (iii) C - (iv) D - (ii) This set of matches corresponds to option (b). \[ \boxed{\text{A-(i), B-(iii), C-(iv), D-(ii)}} \]