Answer:
(a) Boron has the
electronic configuration:
\[1{{s}^{2}},2{{s}^{2}}2p_{x}^{1}2p_{y}^{0}2p_{z}^{0}\]
Under excited state 2s-orbital
is unpaired and one electron is shifted to p-orbital. Now, the hybridisation
occurs between one s- and three p-orbitals to give a tetrahedral configuration.
\[\underbrace{1{{s}^{2}}2{{s}^{1}}2p_{x}^{1}2p_{y}^{1}2p_{z}^{0}}_{s{{p}^{3}}-hybridisation}\]
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