question_answer

Select correct statements and choose the correct code out of the following. I. \[{{\operatorname{CH}}_{3}} -C\,-C{{H}_{3}}\] angle in \[{{(C{{H}_{3}})}_{2}}C\,\,=\,\,C{{H}_{2}}\] is smaller and the \[{{\operatorname{CH}}_{3}} -\,\,C=C{{H}_{2}}\]; angle is larger than trigonal \[120{}^\circ \]. II. \[O-I-F\] angle is less than \[90{}^\circ  \left( 89{}^\circ  \right) in IOF_{4}^{\Theta }\]. III. \[SeOC{{l}_{2}}, Cl\,\,-Se\,\,-\,\,Cl\] angle is less than the \[\operatorname{Cl}\,-Se-O\] angle. IV. \[\operatorname{POC}{{l}_{3}}\] is tetrahedral with a double bond between P and O, there is no lone pair on central atom.

A)  I and II are correct        

B)  II and III are correct     

C)  I, II and III are correct   

D)  All are correct

Correct Answer: D

Solution :

This is due to wide space occupied by ?electron density which pushes both \[\operatorname{C}\,-C{{H}_{3}}\] bond to reduce \[\operatorname{Cl}-C-C{{H}_{3}}\] bond angle. \[O-I-F\] bond angle is also less than \[90{}^\circ \] in \[IOF_{4}^{\Theta }\] due to lone pair on I-atom. \[\operatorname{SeOC}{{l}_{2}}\] has \[\operatorname{Cl}-Se-Cl\] bond angle less than \[\operatorname{Cl}-Se-O\] due to re-electron. \[\operatorname{POC}{{l}_{3}}\] has tetrahedral structure due to presence of \[\pi -bond\] because if this \[\pi -bond\] is not.