B_2H_6 + 2NaH to compound P. Identify the hyb | vedclass

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(B) The reaction between diborane $(B_2H_6)$ and sodium hydride $(NaH)$ in an ether solvent yields sodium borohydride $(NaBH_4)$: $B_2H_6 + 2NaH 	o 2

Vedclass · Accepted Answer

$sp^3$

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(B) The reaction between diborane $(B_2H_6)$ and sodium hydride $(NaH)$ in an ether solvent yields sodium borohydride $(NaBH_4)$: $B_2H_6 + 2NaH 	o 2NaBH_4$. In the borohydride ion $([BH_4]^-)$,the Boron atom is bonded to four Hydrogen atoms via single covalent bonds. Since the Boron atom has a steric number of $4$ (four sigma bonds and zero lone pairs),its hybridisation is $sp^3$.

List-$I$	List-$II$
$A$. Electron-deficient hydride	$I$. $CH_4$
$B$. Electron-precise hydride	$II$. $B_2H_6$
$C$. Electron-rich hydride	$III$. $CaH_2$
$D$. Saline hydride	$IV$. $NiH_{0.6}$
	$V$. $PH_3$

$B_2H_6 + 2NaH \to$ compound $P$. Identify the hybridisation of Boron in compound $P$.

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Match the following:
List-$I$ List-$II$
$A$. Electron-deficient hydride $I$. $CH_4$
$B$. Electron-precise hydride $II$. $B_2H_6$
$C$. Electron-rich hydride $III$. $CaH_2$
$D$. Saline hydride $IV$. $NiH_{0.6}$
$V$. $PH_3$

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