Question

If the principal quantum number $\text{n} = 6 ,$ the correct sequence of filling of electrons will be:

• A. $ns \rightarrow \left(n - 1\right) \, d \rightarrow \left(n - 2\right) \, f \rightarrow np$
• B. $ns \rightarrow np \rightarrow \left(n - 1\right)d \rightarrow \left(n - 2\right)f$
• C. $ns \rightarrow \left(n - 2\right)f \rightarrow np \rightarrow \left(n - 1\right)d$
• D. $ns \rightarrow \left(n - 2\right)f \rightarrow \left(n - 1\right)d \rightarrow np$

Answer 1

Answer: (D)

$ns \rightarrow \left(n - 2\right)f \rightarrow \left(n - 1\right)d \rightarrow np$

Answer 2

As per $\left(\text{n} + \text{l}\right)$ rule electrons fill first in that orbital which have least $\left(\text{n} + \text{l}\right)$ value. When $\left(\text{n} + \text{l}\right)$ values are same, then electron fills that orbital which have lowest $\text{n}$ value. when $\text{n} = 6$ as per $\left(\text{n} + \text{l}\right)$ rule when $\text{n} = 6$ $\text{ns}$ subshell $\Rightarrow 6+0=6$ $\left(\text{n} - 1\right) \text{d}$ subshell $\Rightarrow 5+2=7$ $\left(\text{n} - 2\right) \text{f}$ subshell $\Rightarrow 4+3=7$ $\text{np}$ subshell $\Rightarrow 6+1=7$ $ns \, , \, \left(n - 2\right)f, \, \left(n - 1\right)d,np$ $\left(n + l\right) \, values \, \Rightarrow 7, \, 7, \, 7 \,$ $n \, value \, \Rightarrow 4, \, 5, \, 6$