Answer
See the explanation
Work Step by Step
a. Neat hypothetical - if
$m_s$ had three possible values (instead of the usual two), each orbital could hold three electrons. From that everything else follows:.
Number of orbitals in shell: $n=n^2$
Electrons per orbital (hypothetical) $ = 3$
So capacity of shell $n=3n^2$
How many electrons would an orbital be able to hold?
$3$ electrons (one for each allowed $m_s$)
b. The first period will contain:
$3(1^2)=3$ elements
The second period will contain:
$3(2^2)=12$ elements
c. Let's count the elements in the first transition-metal series:
$l=2\Rightarrow 5$ orbitals $\Rightarrow 5\times 3=15$ elements
d. Let's count the number of electrons the set of 4f orbitals would hold:
$l=3\Rightarrow 7$ orbitals $\Rightarrow 7\times 3=21$ electrons
