Question

Consider three elements A, B ,C which belongs to 2nd period of the periodic table. The element B is more metallic than C. The elements A and C respectively lose and gain one electron each to complete their octets. Examine the given information and arrange the elements in increasing order of atomic numbers ........................ ....................explain............​

Answer 1

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.

I HOPE IT HELPS..

Answer 2

Answer:

a>b>c

it is the answer that I wrote