• # question_answer 80)   Explain why $PC{{l}_{5}}$ is trigonal bipyramidal whereas $I{{F}_{5}}$ is square pyramidal?

P-atom has the configuration : $1{{s}^{2}},2{{s}^{2}}2{{p}^{6}},3{{s}^{2}}3p_{x}^{1}3p_{y}^{1}3p_{z}^{1}$ To get pentavalency, $3s$-orbital is unpaired and electron is shifted to 3d. In this excited state, the five orbitals undergo hybridisation ($s{{p}^{3}}d$) giving rise to trigonal bipyramidal structure. All the five hybrid orbitals overlap with five chlorine atoms and form sigma bonds. $I{{F}_{5}}$ is square pyramidal. $I$-atom have ${{s}^{2}}{{p}^{5}}$ outer most configuration. Two $p$-orbitals are unpaired and electrons are shifted to $5d$-orbitals. In the excited state $s{{p}^{3}}{{d}^{2}}-$hybridisation takes place giving birth to six hybrid orbitals, i.e., octahedral structure. Five orbitals overlap with five fluorine atoms and one orbital is occupied by lone pair, i.e., square pyramidal structure comes into existence.