Short answer: no! The first known exception to occur in the periodic table is in period 5: niobium’s ground state electronic configuration is:

Nb: [Kr] 5s¹ 4d⁴

which is not warranted by the “usual” rules for determining electronic configuration. So, the “half-full rule” is not sufficient in periods 5 and higher.

Let’s look at Wikipedia’s list of exceptions to Madelung’s rule. In period 4, the only exceptions are Cr and Cu, which are accounted for by the “half-full rule”. (Don’t bother too much about the dispute for Ni.)

In period 5, the exceptions to Madelung’s rule are classified in three groups:

• Mo and Ag: they are the analogues of Cu and Cr
• Pd is 5s⁰ 4d¹⁰: a different type of application of the explanation that “fully filled shells are particularly stable”, because in this case two electrons from the s shell were pulled into the d shell. Not what you usually learn.
• Nb is 5s¹ 4d⁴, and Ru is 5s¹ 4d⁷: these cannot be explained by simple rules, and one has to perform complex quantum chemistry calculations to understand these electronic configurations.

Period 6 contains further inconsistencies, because of the introduction of f orbitals into the mix.

Short answer: no! The first known exception to occur in the periodic table is in period 5: niobium’s ground state electronic configuration is

$\ce{Nb: [Kr] 5s^1 4d^4}$

which is not warranted by the “usual” rules for determining electronic configuration. So, the “half-full rule” is not sufficient in periods 5 and higher.

Let’s look at Wikipedia’s list of exceptions to Madelung’s rule. In period 4, the only exceptions are $\ce{Cr}$ and $\ce{Cu}$, which are accounted for by the “half-full rule”. (Don’t bother too much about the dispute for Ni.)

In period 5, the exceptions to Madelung’s rule are classified in three groups:

• $\ce{Mo}$ and $\ce{Ag}$: they are the analogues of Cu and Cr
• $\ce{Pd}$ is $\ce{5s^0 4d^10}$: a different type of application of the explanation that “fully filled shells are particularly stable”, because in this case two electrons from the s shell were pulled into the d shell. Not what you usually learn.
• $\ce{Nb}$ is $\ce{5s^1 4d^4}$, and $\ce{Ru}$ is $\ce{5s^1 4d^7}$: these cannot be explained by simple rules, and one has to perform complex quantum chemistry calculations to understand these electronic configurations.

Period 6 contains further inconsistencies, because of the introduction of $\ce{f}$ orbitals into the mix.