A) \[s{{p}^{3}},\,\,s{{p}^{3}}d,\,\,s{{p}^{3}}{{d}^{2}}\]
B) \[s{{p}^{3}}d,\,\,s{{p}^{3}},\,\,s{{p}^{2}}{{d}^{2}}\]
C) \[s{{p}^{3}},\,\,s{{p}^{3}}{{d}^{2}},\,\,s{{p}^{3}}\]
D) \[s{{p}^{3}}{{d}^{2}},\,\,s{{p}^{3}},\,\,s{{p}^{3}}d\]
Correct Answer: B
Solution :
In\[PC{{l}_{5}}\]No.\[{{e}^{-}}\]pairs around central atom\[P=5\] No of lone pairs = 0 and hybridisation-\[s{{p}^{3}}d\] Shape = trigonal bipyramidal. In\[PCl_{4}^{+}-\]number, of\[{{e}^{-}}\]pairs\[=4\], lone pairs\[=0\], hybridisation\[=s{{p}^{3}}\]and shape - tetrahedral In\[PCl_{6}^{-}-\]number of \[{{e}^{-}}\] pairs = 6, lone pairs = 0, hybridisation \[=s{{p}^{3}}{{d}^{2}}\] and shape is octahedral
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