A) \[Si{{F}_{4}}\]
B) \[C{{H}_{4}}\]
C) \[B{{F}_{3}}\]
D) \[{{H}_{2}}C{{O}_{3}}\]
Correct Answer: D
Solution :
\[{{H}_{2}}C{{O}_{3}}\]has\[s{{p}^{2}}\]hybridisation as: Number of hybrid orbitals \[=\frac{1}{2}\] (number of electrons in valence shell of atom + number of monovalent atoms - charge on cation + charge on anion. \[=\frac{1}{2}[4+2-0+0]\] \[=3\] \[\therefore \]hybridisation is\[s{{p}^{2}}\] \[{{H}_{2}}C{{O}_{3}}\]is a polar compound due to the difference in the electronegativity?s of the elemens.
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