#### Answer

Both outermost electrons in Na and K experience an effective nuclear charge of $2.2+$.

#### Work Step by Step

1) $Na$:
The nuclear charge of $Na$ is $Z=11+$.
The outermost electrons of Na is in the $n=3$ shell.
Na has:
- 2 electrons in the $n=1$ shell, each of which contributes 1.00 to $S$.
- 8 electrons in the $n=2$ shell, each of which contributes 0.85 to $S$.
- 1 valence electron in the $n=3$ shell, but since Slater's rules do not account for the screening effect of 1 electron to itself, this electron does not contribute to $S$.
That means $S=2\times1+8\times0.85=8.8$
Therefore, $$Z_{eff}=Z-S=(11+)-8.8=2.2+$$
2) $K$:
The nuclear charge of $K$ is $Z=19+$.
The outermost electrons of K is in the $n=4$ shell.
K has:
- 2 electrons in the $n=1$ shell, each of which contributes 1.00 to $S$.
- 8 electrons in the $n=2$ shell, each of which contributes 1.00 to $S$.
- 8 electrons in the $n=3$ shell, each of which contributes 0.85 to $S$.
- 1 electron in the $n=4$ shell. However, Slater's rules do not account for the screenning effect of 1 electron to itself, this electron does not contribute to $S$.
That means $S=2\times1+8\times1+8\times0.85=16.8$
Therefore, $$Z_{eff}=Z-S=(19+)-16.8=2.2+$$