Find the number of process in which hybridisation of the underlined atom is changed. (i) B 2 H 6 2 RNH 2 (ii) B 2 H 6 2 R 2 NH (iii) B 2 H 6 2 NH 3 (iv) B 2 H 6 2 T

Question

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

(i) $\underline{B}_{2} H_{6} 2 RNH_{2}$

(ii) $\underline{B}_{2} H_{6} 2 R_{2} NH$

(iii) $\underline{B}_{2} H_{6} 2 NH_{3}$

(iv) $\underline{B}_{2} H_{6} 2 THF$

(v) $\underline{B}_{2} H_{6} 2 R_{3} N$

(vi) $\underline{B}_{2} H_{6}$ $T > 200^{\circ} C ⟶ N H_{3}$

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.

In last reaction the product is (BN)_x in which B is sp² hybridized

Chemistry - CHEMICAL BONDING Question with Solution | TestHub

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JEE Advanced 2027 (Class 12+)

JEE Advanced 2027 (Class 12)