Easy
If the period is $6$,then $n = 6$ and $l = 0, 1, 2, 3, 4$. Why is the number of elements in this period still $32$?
(N/A) For $n = 6$,the subshells available are $6s, 6p, 5d, 4f$. The $5g$ and $6g$ subshells are not filled in the ground state of known elements because their energy is significantly higher than the $7s$ orbital.
| $n=6$ subshells | $6s, 6p, 5d, 4f$ |
| Number of orbitals | $1 (6s) + 3 (6p) + 5 (5d) + 7 (4f) = 16$ orbitals |
| Total elements | $16 \times 2 = 32$ elements |
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