Q.2.The position of three elements A, B and C in the Periodic Table is shown below: Giving reasons, explain the following: (a) Element A is a non-metal. Answer= (b) Element B has a larger atomic size than element C. (c) Element C has a valency of 1.
And this is the table
![\\ \\ \\ \\ \\ \left[\begin{array}{ccc}group 16&group 17\\--&--\\--&A\\--&--\\B&C\\\end{array}\right]](https://tex.z-dn.net/?f=%5C%5C+%5C%5C+%5C%5C+%5C%5C+%5C%5C+%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dgroup+16%26amp%3Bgroup+17%5C%5C--%26amp%3B--%5C%5C--%26amp%3BA%5C%5C--%26amp%3B--%5C%5CB%26amp%3BC%5C%5C%5Cend%7Barray%7D%5Cright%5D "\\ \\ \\ \\ \\ \left[\begin{array}{ccc}group 16&group 17\\--&--\\--&A\\--&--\\B&C\\\end{array}\right]")
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(a) A belongs to group 17. There are 7 electrons in its valance shell. Therefore it tend to accept one electron to complete octet. Hence, it is a non-metal.
(b) C is less reactive than A. As both elements belong to same group and going down to group size of element increases. A is a smaller than C therefore the cloud of electron density surrounding the nucleus is more compact. The stability of completing the octet for A is greater than for C as A is more keen to obtain an electron than C.
(c) C is smaller than B. C and B belongs to same period. Atomic radius decreases from left to right within a period. This is caused by the increase in the number of protons and electrons across a period. One proton has a greater effect than one electron; thus, electrons are pulled towards the nucleus, resulting in a smaller radius.
(d) A will form anion A
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. A belongs to group 17. There are 7 electrons in its valance shell. Therefore it tend to accept one electron to complete octet and forms anion. ....
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