CHAPTER 1: Question 4

 

2D Explanation

A boron atom has three electrons in its outer shell. In BH3, each of these is combined with the electron from one of the hydrogen atoms to form the three B-H bonds. Thus, the boron atom is surrounded by three identical bond pairs of electrons, which will distribute themselves as far apart as possible, and thus will give a trigonal planar geometry

The bonding situation in B2H6 is slightly more complex. There are two different types of hydrogen atom in this molecule, which are shown in red and blue in the diagram below. Each bond between a boron atom and a red hydrogen atom is a conventional two electron bond to which the boron and hydrogen atoms both provide one electron. This uses two of the three electrons in the outer shell of each boron atom. The remaining two hydrogens (shown in blue) are shared between the two boron atoms. However, each pair of B-H-B bonds contains only two electrons (one from the hydrogen atom and one from one of the boron atoms), and is called a three centre two electron bond. Thus each boron atom is surrounded by two conventional bond pairs and two three centre two electron bond pairs. The basic structure will thus be tetrahedral, but since each B-(blue H) bond contains on average only one electron, the structure will be distorted so that the (red H)-B-(red H) bond angle > (Red H)-B-(blue H) bond angle > (blue H)-B-(blue H) bond angle.

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