Question

Hey give me answer fast urgent

Answer 1

Writing Configuration Of Chromium

1s2 2s2 2p6 3s2 3p6 3d5 4s1

since the last shell is 4, therefore n=4,

as there is only one orbital in the s sub shell l=0

since we're still in the s sub shell only m=0

So, P = n - l + m

P = 4 - 0 - 0

P = 4