He resonance structures of carbon monoxide are shown below. show how each structure can be converted into the other using the cu
rved-arrow notation. to draw an arrow, click on the structure to enter the edit mode, then click on the curved arrow icon.
1 answer:
The two resonating structures of Carbon Monoxide are shown below.
The movement of electrons are shown by arrow.
Structure A:
In structure a A the formal charges of Carbon and Oxygen are zero. As Formal Charge is calculated as,
= # of valence electron - electrons in lone pairs + 1/2 bonding electrons electrons
For C:
= 4 - 2 + 4/2
= 4- 4
= 0
For O:
= 6 - 4 + 4/2
= 6- 6
= 0
Structure B:
Formal Charge on C:
= 4 - 2 + 6/2
= 4- 5
= -1
Formal Charge on O:
= 6 - 2 + 6/2
= 6 - 5
= +1
The movement of electrons are shown by arrow.
Structure A:
In structure a A the formal charges of Carbon and Oxygen are zero. As Formal Charge is calculated as,
= # of valence electron - electrons in lone pairs + 1/2 bonding electrons electrons
For C:
= 4 - 2 + 4/2
= 4- 4
= 0
For O:
= 6 - 4 + 4/2
= 6- 6
= 0
Structure B:
Formal Charge on C:
= 4 - 2 + 6/2
= 4- 5
= -1
Formal Charge on O:
= 6 - 2 + 6/2
= 6 - 5
= +1
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Answer: (E) 300 bq
Explanation:
Half life is the amount of time taken by a radioactive material to decay to half of its original value.
Radioactive decay process is a type of process in which a less stable nuclei decomposes to a stable nuclei by releasing some radiations or particles like alpha, beta particles or gamma-radiations. The radioactive decay follows first order kinetics.
Half life is represented by
Half life of Thallium-208 = 3.053 min
Thus after 9 minutes , three half lives will be passed, after ist half life, the activity would be reduced to half of original i.e. , after second half life, the activity would be reduced to half of 1200 i.e.
, and after third half life, the activity would be reduced to half of 600 i.e.
,
Thus the activity 9 minutes later is 300 bq.