The total number of orbitals associated with the principal quantum number 5 is:

The question asks for the total number of orbitals that correspond to the principal quantum number $$n = 5$$. To find this, we will recall how quantum numbers determine orbitals and then add up every possibility one by one.

First, remember that the principal quantum number $$n$$ fixes the value of the azimuthal (or orbital‐angular‐momentum) quantum number $$l$$. The rule is:

$$l = 0,1,2,\dots,(n-1).$$

So, when $$n = 5$$ we have the following possible $$l$$ values:

$$l = 0,\;1,\;2,\;3,\;4.$$

Now, for each value of $$l$$ there is a magnetic quantum number $$m_l$$ that can take on integer values from $$-l$$ to $$+l$$ in steps of one. The total number of possible $$m_l$$ values (and hence the number of orbitals) for a given $$l$$ equals $$2l + 1$$. We will list them and add them all up.

For $$l = 0$$ (this is the 5s subshell):

Number of orbitals $$= 2(0) + 1 = 1.$$

For $$l = 1$$ (this is the 5p subshell):

Number of orbitals $$= 2(1) + 1 = 3.$$

For $$l = 2$$ (this is the 5d subshell):

Number of orbitals $$= 2(2) + 1 = 5.$$

For $$l = 3$$ (this is the 5f subshell):

Number of orbitals $$= 2(3) + 1 = 7.$$

For $$l = 4$$ (this is the 5g subshell):

Number of orbitals $$= 2(4) + 1 = 9.$$

Now we add them all together to get the total number of orbitals for the shell where $$n = 5$$:

$$1 + 3 + 5 + 7 + 9 = 25.$$

There is another quicker formula that gives the same result directly: the total number of orbitals in any shell with principal quantum number $$n$$ is $$n^2$$. Substituting $$n = 5$$ gives

$$n^2 = 5^2 = 25,$$

which matches the detailed counting above.

Hence, the correct answer is Option B.