To determine which of these ions have a noble gas configuration, we must consider what the phrase "noble gas configuration" means. Atoms or ions with a noble gas configuration have completely filled electron shells, similar to the electron configuration of the noble gases, which are the elements found in Group 18 of the periodic table. Noble gases are helium (He), neon (Ne), argon (Ar), krypton (Kr), xenon (Xe), and radon (Rn).

Let's examine each ion given:

• $$\mathrm{Sr}^{2+}(z=38)$$
  : Strontium has atomic number 38 and loses 2 electrons to form a
  $$\mathrm{Sr}^{2+}$$
  ion, giving it the same electron configuration as krypton (Kr), with atomic number 36. Hence, it has a noble gas configuration.

${\left[\mathrm{Sr}^{2+}\right]=[\mathrm{Kr}]}$


• $$\mathrm{Cs}^{+}(z=55)$$
  : Cesium has atomic number 55, and by losing 1 electron to form
  $$\mathrm{Cs}^{+}$$
  , it has the electron configuration of xenon (Xe), with atomic number 54. Thus, it also has a noble gas configuration.

${\left[\mathrm{Cs}^{+}\right]=[\mathrm{Xe}]}$


• $$\mathrm{La}^{2+}(z=57)$$
  : Lanthanum has atomic number 57. If it lost 2 electrons to form
  $$\mathrm{La}^{2+}$$
  , it would not have a noble gas configuration because it would have one electron more than xenon (Xe), which means it would not fully match any noble gas electron configuration.

${\left[\mathrm{La}^{2+}\right]=[\mathrm{Xe}] 5 \mathrm{~d}^1}$


• $$\mathrm{Pb}^{2+}(z=82)$$
  : Lead has an atomic number of 82, so when it loses 2 electrons to form
  $$\mathrm{Pb}^{2+}$$
  , it has 80 electrons, which is the same electron number as mercury (Hg) and not a noble gas. Therefore,
  $$\mathrm{Pb}^{2+}$$
  does not have a noble gas configuration.


${\left[\mathrm{~Pb}^{2+}\right]=[\mathrm{Xe}] 4 \mathrm{f}^{14} 5 \mathrm{~d}^{10} 6 \mathrm{~s}^2}$


• $$\mathrm{Yb}^{2+}(z=70)$$
  : Ytterbium has an atomic number of 70. By losing 2 electrons to form
  $$\mathrm{Yb}^{2+}$$
  , it has 68 electrons, aligning with the electron configuration of Erbium (Er).
  $$\mathrm{Yb}^{2+}$$
  and not a noble gas.

${\left[\mathrm{Yb}^{2+}\right]=[Xe]4f^{14}}$


• $$\mathrm{Fe}^{2+}(z=26)$$
  : Iron has an atomic number of 26. When it becomes
  $$\mathrm{Fe}^{2+}$$
  by losing 2 electrons, it has 24 electrons. This does not correspond to any noble gas configuration, as the nearest noble gas is argon (Ar) with 18 electrons.

${\left[\mathrm{Fe}^{2+}\right]=[\mathrm{Ar}] 3 \mathrm{~d}^6}$

Therefore, the ions

and

$$\mathrm{Cs}^{+}$$

have a noble gas configuration. Altogether, there are 2 ions with a noble gas configuration.