Match List-I with List-II List-I Type of | vedclass

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Question

(A) $B_2H_6$ is an electron-deficient hydride because it has fewer electrons than required for normal covalent bonding.$HF$ is an electron-rich hydrid

Vedclass · Accepted Answer

$A-III, B-II, C-IV, D-I$

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(A) $B_2H_6$ is an electron-deficient hydride because it has fewer electrons than required for normal covalent bonding.$HF$ is an electron-rich hydride because it has lone pairs of electrons on the central atom.$CH_4$ is an electron-precise hydride because it has the exact number of electrons required for covalent bonding.$MgH_2$ is a saline (ionic) hydride formed by the reaction of $Mg$ with $H_2$.Therefore,the correct matching is $A-III, B-II, C-IV, D-I$.

List-$I$ Type of Hydride	List-$II$ Example
$A$. Electron deficient hydride	$I$. $MgH_2$
$B$. Electron rich hydride	$II$. $HF$
$C$. Electron precise hydride	$III$. $B_2H_6$
$D$. Saline hydride	$IV$. $CH_4$

Match List-$I$ with List-$II$

List-$I$ Type of Hydride List-$II$ Example

$A$. Electron deficient hydride $I$. $MgH_2$

$B$. Electron rich hydride $II$. $HF$

$C$. Electron precise hydride $III$. $B_2H_6$

$D$. Saline hydride $IV$. $CH_4$

Choose the correct answer from the options given below:

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Match List-$I$ with List-$II$ List-$I$ Type of Hydride List-$II$ Example $A$. Electron deficient hydride $I$. $MgH_2$ $B$. Electron rich hydride $II$. $HF$ $C$. Electron precise hydride $III$. $B_2H_6$ $D$. Saline hydride $IV$. $CH_4$ Choose the correct answer from the options given below: