90.
If n =6, the correct sequence for filling of electrons will be
-
ns → (n-1) d → ( n - 2)f → np
-
ns - (n-2)f → np → (n-1)d
-
ns - np → (n-1)d → (n-2)f
-
ns - np → (n-1)d → (n-2)f
D.
ns - np → (n-1)d → (n-2)f
6s → 4f → 5d → 6p for n = 6
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Text Solution
42106
Answer : B
Solution : n=3, l=1, the resultant subshell is 3p. Thus, number of orbitals having n=3, l=1 and m=-1 is one and the maximum number of electrons that can be accommodated in this orbital is two.