Which of the following elements is most likely to form an ion that will then form an ionic bond with an ion of a Group 1A element?
Be
I
Fr **my choice
Pt
Iodine is the only non-metal there, it takes a non metal to form an ionic bond with a group 1 metal.
Quick Answer: "I"
Explanation: Since you're forming an iconic bond, you want to get the total valance electrons to 8.
Elements under 1A all have 1 valance electron, so all you really need to do is find an element with 7 electrons which under VIIIA column(7th column). The only elements in that 7th column are F, Cl, Br, I, and At. Therefore, I is the correct answer in this situation.
Hopefully this helps you out! I tried my best to explain how I got the answer.
You are correct! Fr (Francium) is the most likely element to form an ion that will then form an ionic bond with an ion of a Group 1A element. Francium belongs to Group 1A (also known as Group 1 or the Alkali Metals), and Group 1A elements commonly form +1 ions by losing one electron. Therefore, Fr can readily form an ionic bond with an ion of a Group 1A element by transferring its single valence electron.
To determine which element is most likely to form an ion that will then form an ionic bond with an ion of a Group 1A element, we need to consider the electron configuration and the valence electrons of the elements.
Group 1A elements (also known as alkali metals) have one valence electron. They tend to lose this electron to achieve a stable electron configuration and form a positive ion with a charge of +1.
Now, let's look at the options:
1. Be (Beryllium) - Beryllium is in Group 2A, which means it has two valence electrons. It tends to lose both of these electrons to achieve a stable configuration. Hence, it forms a positive ion with a charge of +2. Since the Group 1A elements form ions with a charge of +1, beryllium won't form an ionic bond with them.
2. I (Iodine) - Iodine is in Group 7A, and it has seven valence electrons. It tends to gain one electron to achieve a stable octet configuration and forms a negative ion with a charge of -1. Hence, it won't form an ionic bond with Group 1A elements, which form positive ions.
3. Fr (Francium) - Francium is in Group 1A and has one valence electron. Similar to all Group 1A elements, it has a strong tendency to lose this electron and form a positive ion with a charge of +1. Therefore, francium is the element most likely to form an ion that will then form an ionic bond with an ion of a Group 1A element.
4. Pt (Platinum) - Platinum is in Group 10, and it has ten valence electrons. It typically does not form ions in ionic bonding but rather forms covalent bonds due to its d-block electron configuration.
Based on this information, the most appropriate choice would be Fr (Francium).