Question

a consider three elements A, B & C which belong to
the and period of the periodic table. The element B
in more metallic than C. The elements A&C
respectively lose & gain one election each to
complete their octets, examine the given information
& anange the elemente es increasing order of
atomic number​

Answer 1

Explanation:

(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

−

. 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.

Answer 2

Answer:

Here it is Given that all of three are in 2nd Period..

so in second period all have same no of orbits and when we go left to right then number of electrons increase.

let take a example, suppose these elements are in period 2 ..

E F G H I J K L

now you know that Element E is a metal which has 1 electron in outermost shell and hence it loses electron to complete its octate and Element K is a non metal which has 7 electron in outermost cell and hence it accept electron to complete its octate ..

( Here i didn't take element L because it is already stable and i have taken example only to explain you also Element L has no use in this question )

NOW , COMING TO YOUR QUESTION.

Element B is more reactive than C ( B > C )

and Element a lose 1 electron hence it is a metal and Element C gain one electron which is a non metal and in period we find that at left there is metal and when we go further ( going further also there is increase in atomic no ) then there comes non metal.

hence,

A come before C

and hence atomic no of A is less than C

Now comparing all three then ,

B>C

and A loses 1 electron hence it will have least atomic no in this period.

B is more reactive than C then it comes before C means C have the higher atomic no in this period..

HENCE, ATOMIC NO OF

A<B <C.

HOPE IT HELPS..