If the principal quantum number n = 6, the correct sequence of filling of electrons will be:

The question asks for the correct sequence of filling electrons when the principal quantum number $$ n = 6 $$. According to the Aufbau principle, electrons fill subshells in order of increasing energy, determined by the $$ (n + l) $$ rule, where $$ n $$ is the principal quantum number and $$ l $$ is the azimuthal quantum number. If two subshells have the same $$ (n + l) $$ value, the subshell with the lower $$ n $$ is filled first.

For $$ n = 6 $$, the relevant subshells and their notations are:

• $$ ns $$ corresponds to $$ 6s $$
• $$ (n-1)d $$ corresponds to $$ (6-1)d = 5d $$
• $$ (n-2)f $$ corresponds to $$ (6-2)f = 4f $$
• $$ np $$ corresponds to $$ 6p $$

Now, calculate the $$ (n + l) $$ values for each subshell:

• For $$ 6s $$: $$ n = 6 $$, $$ l = 0 $$ (since s orbital has $$ l = 0 $$), so $$ n + l = 6 + 0 = 6 $$.
• For $$ 4f $$: $$ n = 4 $$, $$ l = 3 $$ (since f orbital has $$ l = 3 $$), so $$ n + l = 4 + 3 = 7 $$.
• For $$ 5d $$: $$ n = 5 $$, $$ l = 2 $$ (since d orbital has $$ l = 2 $$), so $$ n + l = 5 + 2 = 7 $$.
• For $$ 6p $$: $$ n = 6 $$, $$ l = 1 $$ (since p orbital has $$ l = 1 $$), so $$ n + l = 6 + 1 = 7 $$.

Comparing the $$ (n + l) $$ values:

• The $$ 6s $$ subshell has the lowest $$ (n + l) = 6 $$, so it is filled first.
• The subshells $$ 4f $$, $$ 5d $$, and $$ 6p $$ all have $$ (n + l) = 7 $$. Since they have the same $$ (n + l) $$ value, we compare their $$ n $$ values: $$ 4f $$ has $$ n = 4 $$, $$ 5d $$ has $$ n = 5 $$, and $$ 6p $$ has $$ n = 6 $$. The subshell with the smallest $$ n $$ is filled first, so $$ 4f $$ is filled before $$ 5d $$, and $$ 5d $$ is filled before $$ 6p $$.

Therefore, the sequence of filling is: $$ 6s \rightarrow 4f \rightarrow 5d \rightarrow 6p $$.

In terms of the given notations:

• $$ 6s $$ is $$ ns $$
• $$ 4f $$ is $$ (n-2)f $$
• $$ 5d $$ is $$ (n-1)d $$
• $$ 6p $$ is $$ np $$

So the sequence becomes: $$ ns \rightarrow (n-2)f \rightarrow (n-1)d \rightarrow np $$.

Now, comparing this with the options:

• Option A: $$ ns \rightarrow (n-1)d \rightarrow (n-2)f \rightarrow np $$ corresponds to $$ 6s \rightarrow 5d \rightarrow 4f \rightarrow 6p $$, which is incorrect because $$ 4f $$ should come before $$ 5d $$.
• Option B: $$ ns \rightarrow np \rightarrow (n-1)d \rightarrow (n-2)f $$ corresponds to $$ 6s \rightarrow 6p \rightarrow 5d \rightarrow 4f $$, which is incorrect because $$ 6p $$ should be last.
• Option C: $$ ns \rightarrow (n-2)f \rightarrow np \rightarrow (n-1)d $$ corresponds to $$ 6s \rightarrow 4f \rightarrow 6p \rightarrow 5d $$, which is incorrect because $$ 5d $$ should come before $$ 6p $$.
• Option D: $$ ns \rightarrow (n-2)f \rightarrow (n-1)d \rightarrow np $$ corresponds to $$ 6s \rightarrow 4f \rightarrow 5d \rightarrow 6p $$, which matches the correct sequence.

Hence, the correct answer is Option D.