| (i) \[{{\operatorname{NH}}_{3}}\] |
| (ii) \[{{\operatorname{PH}}_{3}}\] |
| (iii) \[{{\operatorname{AsH}}_{3}}\] |
| (iv) \[{{\operatorname{SbH}}_{3}}\] |
| (v) \[{{\operatorname{H}}_{2}}0\] |
A) (v) > (iv) > (i) > (iii) > (ii)
B) (v) > (i) > (ii) > (iii) > (iv)
C) (ii) > (iv) > (iii) > (i) > (v)
D) (iv) > (iii) > (i) > (ii) > (v)
Correct Answer: A
Solution :
v > iv > i > iii > ii \[{{\operatorname{H}}_{2}}O\]- Contain hydrogen bonds due to fact that some extra energy is needed to break these bonds, so it has high boiling point. \[{{\operatorname{SbH}}_{3}}\]has also H-bonding. \[{{\operatorname{NH}}_{3}}\]it has higher boiling point tan \[{{\operatorname{AsH}}_{3}}\]and\[{{\operatorname{PH}}_{3}}\]. This is again because there is hydrogen bonding in \[{{\operatorname{NH}}_{3}}\] and not in \[{{\operatorname{PH}}_{3}}\]. \[\therefore \]Compounds having hydrogen bonding show abnormally high boiling and melting points.
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