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Chemistry - CHEMICAL BONDING Question with Solution | TestHub

ChemistryCHEMICAL BONDINGVSEPR, Bond Angle, Bond Length, Bond EnergyMedium2 minQB

ChemistryMediumnumerical

Passage / Comprehension

The simplest boron hydride known is diborane ( $B_{2} H_{6}$ ). As implied by the mechanism of hydrolysis, diborane and many other light boron hydrides act as Lewis acids, and they are cleaved by reaction with Lewis bases. Two different cleavage patterns have been observed: symmetric cleavage and unsymmetric cleavage. In symmetric cleavage, $B_{2} H_{6}$ is broken symmetrically into two $B H_{3}$ fragments with $NM e_{3}$ . While with $N H_{3}$ , $M e N H_{2}$ , $M e N H_{2}$ results in unsymmetrical cleavage, which is a cleavage leading to an ionic product.

$3 B_{2} H_{6} + 6 NH_{3} \to$ " X " [Ionic product]

Number of bond angles present in the anionic part of one molecule of "X" is-

Answer:

6.00

Solution:

$X = [BH_{4}]^{-}$ \& $Y = B_{3} N_{3} H_{6}$

Stream:JEESubject:ChemistryTopic:CHEMICAL BONDINGSubtopic:VSEPR, Bond Angle, Bond Length, Bond Energy

⏱ 2mℹ️ Source: QB

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