Question

Predict the formula of stable compound formed by an element with atomic number 114 and fluorine.
(A) $A{{F}_{3}}$
(B) $A{{F}_{2}}$
(C) $AF$
(D) $A{{F}_{4}}$

Answer 1

Hint: Find out the element present in the periodic table. Find out in which family it belongs to and how many electrons are required to attain noble gas configuration.

Complete step by step solution:
-Let’s begin by writing the electronic configuration of an element having atomic number 114.
-Electronic configuration = [Rn] $5{{f}^{14}}6{{d}^{10}}7{{s}^{2}}7{{p}^{2}}$
- So, its outermost shell contains 4 electrons. This implies it needs four electrons to achieve the noble gas configuration.
-Therefore, the element will form four bonds with fluorine. So, its molecular formula will be $A{{F}_{4}}$.
-The element in the periodic table containing the atomic number 114 is Flerovium which belongs to Group 14. Flerovium belongs to the Carbon family and so it will form compounds similar to that group.

Additional Information: Flerovium is a synthetically prepared element having the atomic number 114. Its highest living isotope has atomic mass number 289. It belongs to the last period of the periodic table that is, the seventh period.

Note: Write the electronic configuration of the given element first and then find out the family in which it belongs. This is an easier method to approach this question. Just find out the number of electrons required by the element to complete the inert gas configuration and that number will be the number of halogen atoms attached to the element.