Find the number of process in which hybridisation of the underlined atom is changed.
(i) $B_{2} H_{6} 2 RNH_{2} 2$
(ii) $B_{2} H_{6} 2 R_{2} NH$
(iii) $B_{2} H_{6} 2 NH_{3}$
(iv) $B_{2} H_{6} 2 THF$
(v) $B_{2} H_{6} 2 R_{3} N$

Question

Find the number of process in which hybridisation of the underlined atom is changed.

(i) $B_{2} H_{6} 2 RNH_{2} 2$

(ii) $B_{2} H_{6} 2 R_{2} NH$

(iii) $B_{2} H_{6} 2 NH_{3}$

(iv) $B_{2} H_{6} 2 THF$

(v) $B_{2} H_{6} 2 R_{3} N$

TestHub · Accepted Answer

The provided solution indicates that the hybridization of boron in B₂H₆ remains unchanged in all five processes shown. In B₂H₆, each boron atom is sp³ hybridized (forming two 2 c − 2 e bonds and two 3 c − 2 e "banana" bonds). In all the resulting cleavage products with ligands like NH₂⁻, NH₃, or THF, the boron atom remains bonded to four groups, maintaining its sp³ state. Since the hybridization is sp³ both before and after these reactions, the number of processes with a change is 0.

Chemistry - CHEMICAL BONDING Question with Solution | TestHub

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