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Match List-I with List-II.
\(\begin{array}{|c|l|c|l|}
\hline & \text { List - I } & & \text { List - II } \\
\hline & \text { (Hydride) } & & \text { (Type of hydride) } \\
\hline(A) & \mathrm{NaH} & \text { (I) } & \text { Electron precise } \\
\hline \text { (B) } & \mathrm{PH}_3 & \text { (II) } & \text { Saline } \\
\hline \text { (C) } & \mathrm{GeH}_4 & \text { (III) } & \text { Metallic } \\
\hline \text { (D) } & \mathrm{LaH}_{2.87} & \text { (IV) } & \text { Electron rich } \\
\hline
\end{array}\)
Choose the correct answer from the options given below:
Options:
\(\begin{array}{|c|l|c|l|}
\hline & \text { List - I } & & \text { List - II } \\
\hline & \text { (Hydride) } & & \text { (Type of hydride) } \\
\hline(A) & \mathrm{NaH} & \text { (I) } & \text { Electron precise } \\
\hline \text { (B) } & \mathrm{PH}_3 & \text { (II) } & \text { Saline } \\
\hline \text { (C) } & \mathrm{GeH}_4 & \text { (III) } & \text { Metallic } \\
\hline \text { (D) } & \mathrm{LaH}_{2.87} & \text { (IV) } & \text { Electron rich } \\
\hline
\end{array}\)
Choose the correct answer from the options given below:
Solution:
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Verified Answer
The correct answer is:
(A)-(II), (B)-(IV), (C)-(I), (D)-(III)
$\begin{array}{ll}\text { Hydride } & \text { Type of Hydride } \\ \mathrm{NaH} & \text { Saline or lonic hydride (Generally formed by s-block elements) } \\ \mathrm{PH}_3 & \text { Electron rich hydride (Elements of group 15-17 form such hydrides) } \\ \mathrm{GeH}_4 & \text { Electron precise (Elements of group } 14 \text { form such hydrides) } \\ \mathrm{LaH}_{2.87} & \text { Metallic or Non-stoichiometric hydrides (Formed by f block element) }\end{array}$
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