Question

Ratio of carbon and oxygen in carbon dioxide, molecule.
(A) $ 2:1 $
(B) $ 3:2 $
(C) $ 4:2 $
(D) $ 1:2 $

Answer 1

Hint :We know that the Carbon has a valency of $ 4. $ So it forms the compound by sharing the electrons as it requires a large amount of energy in completing its octet by gaining or losing electrons. We know that carbon and oxygen generally together form two compounds namely carbon monoxide and carbon dioxide.

Complete Step By Step Answer:
We know that carbon Dioxide always has a formula of $ C{{O}_{2}} $ , independent of its source. So, we can easily calculate the ratio by mass of the two elements that constitute this compound, which are carbon and oxygen. This is in agreement with the Law of Constant proportion or Law of Definite Proportion which was proposed by Joseph Proust. Since each molecule has one Carbon atom and two Oxygen atoms, we can easily calculate the ratio by mass of these two elements if we know the individual masses of the atoms of these two elements.
Electron configuration of $ C:2,4. $ Each $ C $ atom contributes $ 4 $ electrons for sharing. Electronic configuration of $ O:2,6~ $ Each O atom contributes $ 2 $ electrons for sharing. Therefore two $ O $ atoms are needed to contribute a total of $ 4 $ electrons for sharing.
Therefore, the ratio of carbon and oxygen in carbon dioxide molecules is $ 1:2. $
So, correct answer is option D.

Additional Information:
There is another law that is concerned with mass ratios that is the law of Multiple Proportions. According to this law, if two elements combine to form more than one compound, then, the ratios of the masses of the second element which combine with a fixed mass of the first element are always the ratios of small whole numbers.

Note :
Remember that the different reactions which occur are classified based on the balanced chemical equation so be careful while balancing them. Thus, the Law of multiple proportions states that the masses of one element combined with a fixed mass of the other element are in a ratio of whole numbers.