If n=6what is the correct sequence for filling of elrctrons will be
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QUESTION ==
If n = 6, the correct sequence of filling of electrons will be
(1) ns → np(n -1)d → (n - 2)f
(2) ns → n(n - 2)f → (n -1)d → np
(3) ns → (n -1)d → (n - 2)f → np
(4) ns → (n - 2)f → np → (n -1)d
ANSWER ==
Aufban principle (n+l rule):Electron will fill orbit by sum of quantum numbers n and l. Orbitals with equal values of n+l will fill with the lower n values first.Since n=6 :ns⇒6s, n+l=6+0=6ns⇒6s,n+l=6+0=6np⇒6p, n+l=6+1=7np⇒6p,n+l=6+1=7(n−1)d⇒5d, n+l=5+2=7(n−2)f⇒4f, n+l=4+3=7Since in last three value of n+l value is same. Therefore we will with the lower n values first.So correct order,ns→(n−2)f→(n−1)d→np
I hope it will help you☺️
#yahyaahmad# ✌️
QUESTION ==
If n = 6, the correct sequence of filling of electrons will be
(1) ns → np(n -1)d → (n - 2)f
(2) ns → n(n - 2)f → (n -1)d → np
(3) ns → (n -1)d → (n - 2)f → np
(4) ns → (n - 2)f → np → (n -1)d
ANSWER ==
Aufban principle (n+l rule):Electron will fill orbit by sum of quantum numbers n and l. Orbitals with equal values of n+l will fill with the lower n values first.Since n=6 :ns⇒6s, n+l=6+0=6ns⇒6s,n+l=6+0=6np⇒6p, n+l=6+1=7np⇒6p,n+l=6+1=7(n−1)d⇒5d, n+l=5+2=7(n−2)f⇒4f, n+l=4+3=7Since in last three value of n+l value is same. Therefore we will with the lower n values first.So correct order,ns→(n−2)f→(n−1)d→np
I hope it will help you☺️
#yahyaahmad# ✌️
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