The molecular orbital configuration of
O₂²⁻ (peroxide ion) leads to a
bond order = ½ (number of bonding electrons – number of antibonding electrons) Bonding electrons = 10, Antibonding electrons = 7 Bond order = ½ (10 – 7) =
1.5
English
20 Questions
80 Marks
24 Mins English English75 Questions300 Marks90 Mins English English30 Questions30 Marks27 Mins English| LIST I |
LIST II |
|
| A |
Starch and Inulin |
I. Animals |
| B |
Maltose |
II. Very long chain of glucose polymer |
| C |
Cellulose |
III. Plants |
| D |
Glycogen |
IV. Disaccharide of two glucose molecules |
Suggested Test Series
Suggested Test Series
