Question

1. order the following from the smallest atomic radii to the largest.

e. cl-(chlorine negative ion)
f. co²⁺(cobalt positive ion)
g. fe²⁺(iron positive ion)
h. helium

Explanation:

Step1: Recall atomic - radius trends

Atomic and ionic radii increase down a group and decrease across a period. Cations are smaller than their parent atoms and anions are larger than their parent atoms.

Step2: Analyze Helium

Helium (He) is in the first period with only one electron - shell, so it has a very small atomic radius.

Step3: Analyze positive ions

$Fe^{2 + }$ and $Co^{2+}$ are transition metals. They have lost 2 electrons. Cations have a smaller radius than their parent atoms. Among transition - metal cations, as the nuclear charge increases (atomic number increases), the ionic radius decreases slightly. But they are still larger than He as they have more electron - shells.

Step4: Analyze the anion

$Cl^-$ is an anion. It has gained an electron, increasing the electron - electron repulsion and thus increasing its radius compared to the neutral chlorine atom. It has more electron - shells than He and is larger than the cations mentioned above.

Answer:

He, $Co^{2+}$, $Fe^{2+}$, $Cl^-$