Consider the relation shown for the radius of the species indicated. Which choice presents a valid comparison?
a) Na > Na+
b) Ca^+2 > Ca+
c) Cl > Cl-
d) S- > S^2-
I think that the answer is "a"...but not 100% certain.
pls help.
a is correct.
Of course, I'm here to help, even if I have to clown around a bit! Let's compare these ions like circus performers on a tightrope:
a) Na > Na+: Well, it seems like the sodium ion (Na+) is missing an electron, which makes it smaller than the neutral sodium atom (Na). So, their radius relationship is backward. Na+ is actually smaller than Na.
b) Ca^+2 > Ca+: Here, the calcium ion (Ca^+2) is missing two electrons, resulting in a smaller size compared to the singly charged calcium ion (Ca+). So, this comparison is also incorrect.
c) Cl > Cl-: In this case, the negatively charged chloride ion (Cl-) has gained an electron, making it larger than the neutral chlorine atom (Cl). So, this comparison is correct! Cl- is indeed larger than Cl.
d) S- > S^2-: The sulfur atom (S) is larger than the sulfide ion (S^2-), as the latter has gained two extra electrons. Oops, this comparison is reversed!
So, after taking a spin around the circus, we can see that the correct choice is c) Cl > Cl-. Hope that clarifies things for you!
To determine which choice presents a valid comparison, let's analyze each option:
a) Na > Na+: This statement is comparing sodium (Na) to sodium ion (Na+). In this case, sodium has a larger atomic radius than the sodium ion, so the statement is correct. Thus, Option a) is valid.
b) Ca^+2 > Ca+: This statement compares calcium ion with a +2 charge (Ca^+2) to calcium ion with a +1 charge (Ca+). Since both ions have the same number of protons but different numbers of electrons, the comparison is not valid. Thus, Option b) is incorrect.
c) Cl > Cl-: This statement compares chlorine (Cl) to chloride ion (Cl-). When an atom loses an electron to become an ion, its radius generally decreases. Thus, the statement is incorrect. Thus, Option c) is incorrect.
d) S- > S^2-: This statement compares sulfide ion (S-) to sulfide ion with a -2 charge (S^2-). The statement is comparing two ions with the same element but different charges. When an atom gains more electrons, it will increase in size. Thus, the statement is correct. Thus, Option d) is valid.
Therefore, the valid comparison is Option a) Na > Na+.
To determine the validity of the comparisons, we first need to understand the concept of atomic radius and how it relates to the ionization of atoms.
Atomic radius refers to the size of an atom, which can vary based on the number of electron shells and the effective nuclear charge. When an atom loses or gains electrons, it forms an ion, and the size of the ion may differ from the original atom.
In general, cations (ions with a positive charge) have a smaller radius compared to their respective parent atoms because they have lost one or more electrons, reducing the electron-electron repulsion and causing the remaining electrons to be pulled closer to the nucleus.
On the other hand, anions (ions with a negative charge) have a larger radius compared to their respective parent atoms because they have gained one or more electrons. The additional electron-electron repulsion increases the electron cloud's size, leading to an increased atomic radius.
Now let's analyze the given choices:
a) Na > Na+
This comparison suggests that the radius of a sodium ion (Na+) is greater than the radius of a sodium atom (Na). However, as mentioned earlier, when an atom loses an electron and becomes a cation, its radius generally decreases. Therefore, this comparison is invalid.
b) Ca^+2 > Ca+
This comparison suggests that the radius of a calcium ion with a 2+ charge (Ca^+2) is greater than the radius of a calcium ion with a positive charge (Ca+). Similar to the explanation above, both ions have lost one electron, so the radius of Ca+2 would be smaller than Ca+. Therefore, this comparison is also invalid.
c) Cl > Cl-
This comparison suggests that the radius of a chlorine atom (Cl) is greater than the radius of a chloride ion (Cl-). Since chlorine gains one electron to become Cl-, the chloride ion's radius should be larger than that of the chlorine atom. Therefore, this comparison is valid.
d) S- > S^2-
This comparison suggests that the radius of a sulfur ion with a 1- charge (S-) is greater than the radius of a sulfur ion with a 2- charge (S^2-). Since both ions have gained electrons, increasing the electron-electron repulsion, the S- ion should have a greater radius than the S^2- ion. Therefore, this comparison is valid.
In conclusion, the correct answer is c) Cl > Cl-.