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The order of increasing sizes of atomic radii among the elements O, S, Se and As is :
To determine the order of increasing atomic radii for the elements Oxygen (O), Sulfur (S), Selenium (Se), and Arsenic (As), we need to recall the periodic trends. Atomic radius generally decreases from left to right across a period and increases from top to bottom down a group.
First, locate the elements in the periodic table. Oxygen (O) is in period 2, group 16. Sulfur (S) is in period 3, group 16. Selenium (Se) is in period 4, group 16. Arsenic (As) is in period 4, group 15.
Since O, S, and Se are all in group 16, they share the same group. Moving down the group, atomic radius increases because each successive element has an additional electron shell. Therefore, the atomic radius order among them is: $$ \text{O} < \text{S} < \text{Se} $$.
Now, consider Arsenic (As) and Selenium (Se). Both are in period 4. As is in group 15, and Se is in group 16. Since atomic radius decreases from left to right across a period, As (being to the left of Se) has a larger atomic radius than Se. So, $$ \text{Se} < \text{As} $$.
Next, compare Sulfur (S) and Arsenic (As). S is in period 3, group 16, while As is in period 4, group 15. Moving down a period adds an electron shell, so As has a larger atomic radius than S. Thus, $$ \text{S} < \text{As} $$.
Also, since Se is in period 4 and S is in period 3, Se has a larger atomic radius than S, which we already have as $$ \text{S} < \text{Se} $$.
Combining all the comparisons:
Therefore, the increasing order of atomic radii is: $$ \text{O} < \text{S} < \text{Se} < \text{As} $$.
Now, comparing with the options:
Hence, the correct answer is Option C.
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