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Question
Which of the following statements are correct. Answer on the basis of Figure.
(i) The two birdged hydrogen atoms and the two boron atoms lie in one plane;
(ii) Out of six B – H bonds two bonds can be described in terms of 3 centre 2-electron bonds.
(iii) Out of six B – H bonds four B – H bonds can be described in terms of 3 centre 2 electron bonds;
(iv) The four-terminal B – H bonds are two centre-two electron regular bonds.
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Solution
(i) The two birdged hydrogen atoms and the two boron atoms lie in one plane;
(ii) Out of six B – H bonds two bonds can be described in terms of 3 centre 2-electron bonds.
(iv) The four-terminal B – H bonds are two centre-two electron regular bonds.
Explanation:
Each of the two boron atoms is in sp3 – hybrid state. Of the four hybrid orbitals, three have one electron each while the fourth is empty. Two of the four orbitals of each, of the boron atom overlap with two terminal hydrogen atoms forming two normal B – H σ-bonds. One of the remaining hybrid orbitals (either empty or singly occupied) of one of the boron atoms, 15-orbital of H (bridge atom) and one of hybrid orbitals of the other boron atom overlap to form a delocalized orbital covering the three nuclei with a pair of electrons. This is three-centre two-electron bond. Similar overlapping occurs with the second hydrogen atom (bridging) forming three centre two electrons bond.
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