The electrons in an atom's outermost energy level are called: Electron dots. Quantum electrons. Valence electrons. Noble-gas electrons.

The valence electrons are located in an atom's outermost energy level

Further explanation  

In an atom, there are levels of energy in the shell and subshell  

This energy level is expressed in the form of electron configurations.  

Electrons fill the orbitals based on the Aufbau principle, namely the filling of electrons starts from the lowest energy level

Electron dots

Usually used in describing Lewis structures, placed around the element

Quantum electrons

Each orbital in an atom consists of 4 quantum numbers  : n, l, ml, ms

Each electron has a different quantum number and this conforms to Pauli's rule: no electrons have the same 4 quantum numbers

Valence electrons

Valence electrons are electrons used in a chemical bond , and  in an atom's outermost energy level

Noble-gas electrons

Noble gas has a stable electron configuration of 8 valence electrons which guides the formation of chemical bonds called the octet rule

Answerthis solution will not form a precipitateexplanationsolubility equilbrium:         pbf₂(s)  ⇄ pb²⁺(aq) + 2 f⁻(aq)the solubility product expression is    (we exclude solids from the equilibrium expression).now suppose we have the trial ion product q, which we can define as    then, we can interpret  q as the product of the concentration of the ions that we have and ksp as the product of the ion concentrations that are needed to form a saturated solution.if q > ksp, then that implies that there will be a surplus of ions that exceed the amount that is required to form a saturated solution. a precipitate will then form (the precipitate continually  forms until the surplus ions are removed and q is decreased to ksp's value).if q = ksp, the saturated solution that forms is barely saturated and the precipitate will be minimal.if q < ksp, then the ion concentration will not be enough to form a saturated solution (so no precipitate).using our information, we calculate the trial ion product:     since q = 2.7  × 10⁻⁸ < ksp =  4  ×  10⁻⁸, a precipitate will not form.