Which of the four elements is most likely to form cations in its reactions?
A. Nh
B. Ts
C. Mc
D. Og
The Correct answer and Explanation is:
The element most likely to form cations in its reactions is C. Mc (Moscovium).
Explanation:
To understand why Moscovium (Mc) is most likely to form cations compared to the other elements listed (Nh – Nihonium, Ts – Tennessine, Og – Oganesson), we need to consider their positions on the periodic table and their properties.
- Position on the Periodic Table:
- Moscovium (Mc) is located in Group 13, which is also known as the thallium group. Elements in this group typically have three valence electrons. They tend to lose these electrons in reactions to form cations, specifically a +1 oxidation state or sometimes a +3 state in certain compounds.
- Nihonium (Nh), located in the same group as Mc, also has three valence electrons. However, due to relativistic effects and instability, its behavior may not follow the trends observed in lighter elements of the group.
- Tennessine (Ts) is in Group 17, which is known for forming anions rather than cations, as these elements typically gain electrons to achieve a stable octet configuration.
- Oganesson (Og), a noble gas, is found in Group 18. Noble gases are characterized by their full valence electron shells, making them very stable and unlikely to form cations.
- Reactivity and Ionization Energy:
- Moscovium, although predicted to be relatively unstable, would have lower ionization energy compared to heavier noble gases and halogens. This property allows it to more readily lose its valence electrons, facilitating the formation of cations.
- The other elements either have a stronger tendency to gain electrons (like Ts) or have very high ionization energies (like Og), making cation formation less favorable.
In summary, among the choices given, Moscovium (Mc) is the element most likely to form cations due to its group position, predicted reactivity, and electron configuration.