Question

both the questions plzzzzzzzzzzz​

Answer 1

Answer:

for question 68.

Z atomic no is 30

Explanation:

write electronic configuration.

3p and 4s has 6 and 2 electrons respectively

Answer 2

67)U already calculated values of n1 and n2 in exm probably n1=2 and n2=6.Now u need to know the energy diff. btwn n1 nd n2.U can easily find it if u remember the values for diffm energy levels.Rememer till 4th or 5th enrgy level and higher ones u can calculate using -13.6z^2/n^2.Now equate this enrgy diff. u calcluted as delE=nhc/lambda but here is the point....u hv to convert the delE u calculated into JOULE as nhc/lambda is in si units.U will get the answer!!!

68)isko ese kr skti ho.We need only those elec. which have n+l=4 so,

n + l

0 4 (in Zn there is no morr than d

subshell i.e. l=2 so 1st option

wrong)

1 3 (again same reason)

2 2 (this time l is 2 but n is also 2

and ther is no 2d but only 3d

so not possible)

3 1 ( this is possible as 3p is

available nd has 6 elec.)

4 0 (this is also possible as 4s is

is available nd has 2 electrons)

So n+l=4 is applicable for a total of 6+2= 8 electrons.Hope i made it clear!!!