<u>Answer:</u> The formula of the compound formed between rubidium and fluorine is RbF
<u>Explanation:</u>
Ionic bond is defined as the bond which is formed by complete transfer of electrons from one atom to another atom.
The atom which looses the electron is known as electropositive atom and the atom which gains the electron is known as electronegative atom. This bond is usually formed between a metal and a non-metal.
Rubidium is the 37th element of the periodic table having electronic configuration of 
This will loose 1 electron to form
ion
Fluoride is the 9th element of the periodic table having electronic configuration of 
This will gain 1 electron to form
ion
To form
compound, 1 rubidium ion is needed to neutralize the charge on fluoride ion
The formation of the given compounds is shown in the image below.
From the periodic table, beryllium has an atomic number 4. This means that beryllium has 4 electrons.
Now, the first energy level can hold only 2 electrons (which will occupy the s sublevel) wile the remaining two electrons will occupy the second energy level (also in the s sublevel).
Based on the above, the electronic configuration of beryllium will be as follows:
1s2 2s2
K, ca, sc is the right answer. Take a look at table S of your chemistry reference table.
Do I need to solve 12.046 x 10^25 to get the answer