This answer is based on the electron configuration.
And you can use Aufbau's rule to predict the atomic number of the next elements.
Radon, Rn is the element number 86.
Following Aufbau's rules, the electron configuration of Rn is: [Xe] 6s2 4f14 5d10 6p6. This means that you are suming 2 + 14 + 10 + 6 = 32 electrons with respect to the element Xe.
You can verity that the atomic number of Xe is 54, so when you add 32 you get 54 + 32 = 86, which is the atomic number of Rn.
Again, as per Aufbau's rules, the next element of the same group or period is when the 6 electrons of the 7p orbital are filled. For that, they have to pass 32 elements whose orbitals are:
7s2 5f14 6d10 7p6: count the electrons added: 2 + 14 + 10 + 6 = 32, and that is why the next element wil have atomic number 86 + 32 = 118.
Now, when you go for a new series, you find a new type of orbital, the g orbital, for which the model predict there are 18 electrons to fill.
So the next element of the group will have ; 2 + 18 + 14 + 10 + 6 = 50 electrons, which means that the atomic number of this, not yet discovered element, has atomic number 118 + 50 = 168.
By the way the element with atomic number 118 was already discovdered at its symbol is Og. You can search that information in internet.
Answers: 118 and 168
The correct answer from the choices given is the third option. Covalent compounds have low boiling points. Also, their melting points are low. Covalent bonds have relatively low attractions which results to these properties. The bonds are easily broken by taking energy or adding energy.
Answer is: a. Rubidium (Rb) is more reactive than strontium (Sr) because strontium atoms must lose more electrons.
The ionization energy (Ei) is the minimum amount of energy required to remove the valence electron, when element lose electrons, oxidation number of element grows (oxidation process).
Alkaline metals (group 1), in this example rubidium, have lowest ionizations energy and easy remove valence electrons (one electron), they are most reactive metals.
Earth alkaline metals (group 2), in this example strontium, have higher ionization energy than alkaline metals, because they have two valence electrons, they are less reactive.
Rubidium electron configuration: ₃₇Rb 1s²2s²2p⁶3s²3p⁶3d¹⁰4s²4p⁶5s¹; one valence electron is 5s¹ orbital.
Strontium electron configuration: ₃₈Sr 1s²2s²2p⁶3s²3p⁶3d¹⁰4s²4p⁶5s²; two valence electrons is 5s² orbital.
Answer:
The solution to this question can be defined as follows:
Explanation:
Please find the attached file for the solution:
The question is incomplete, the complete question is;
Choose the aqueous solution that has the highest boiling point. These are all solutions of nonvolatile solutes and you should assume ideal van't Hoff factors where applicable. 0.100 m C6H12O6 0.100 m AlCl3 0.100 m NaCl 0.100 m MgCl2 They all have the same boiling point.
Answer:
AlCl3 0.100 m
Explanation:
Let us remember that the boiling point elevation is given by;
ΔTb = Kb m i
Where;
ΔTb = boiling point elevation
Kb = boiling point constant
m = molality of the solution
i = Van't Hoff factor
We can see from the question that all the solutions possess the same molality, ΔTb now depends on the value of the Van't Hoff factor which in turn depends on the number of particles in solution.
AlCl3 yields four particles in solution, hence ΔTb is highest for AlCl3 . The solution having the highest value of ΔTb also has the highest boiling point.
