<u>Answer:</u> The ion that is expected to have a larger radius than the corresponding atom is chlorine.
<u>Explanation:</u>
There are two types of ions:
- <u>Cations:</u> They are formed when an atom looses its valence electrons. They are positive ions.
- <u>Anions:</u> They are formed when an atom gain electrons in its outermost shell. They are negative ions.
For positive ions, the removal of electron increases the nuclear charge for an outermost electron because the outermost electrons are more strongly attracted by the nucleus. So, the effective nuclear charge increases for cations and thus, the size of the cation will be smaller than that of the corresponding atom.
For negative ions, the addition of electron decreases the nuclear charge for an outermost electron because the outermost electrons are less strongly attracted by the nucleus. So, the effective nuclear charge decreases for anions and thus, the size of the anion will be larger than that of the corresponding atom.
For the given options:
<u>Option a:</u> Chlorine
Chlorine gains 1 electron and form
ion
<u>Option b:</u> Sodium
Sodium looses 1 electron and form
ion
<u>Option c:</u> Copper
Copper looses 2 electrons and form
ion
<u>Option d:</u> Strontium
Strontium looses 2 electrons and form
ion
Hence, the ion that is expected to have a larger radius than the corresponding atom is chlorine.
Answer:
I believe it would be D, to enhance the survival rate.
Explanation:
Both of you are overlooking a pretty big component of the question...the Group I cation isn't being dissociated into water. We're testing the solubility of the cation when mixed with HCl. And this IS a legitimate question, seeing as our lab manual is the one asking.
<span>By the way, the answer you're looking for is "Because Group I cations have insoluble chlorides". </span>
<span>"In order...to distinguish cation Group I, one adds HCl to a sample. If a Group I cation is present in the sample, a precipitate will form." </span>
Answer:
we need to know which atom you're talking about and then you need to say what the mass number is then we can tell how many electrons there are.
(I think)