URGENT HELP PLEASE balance equation !20 points!

Compare which element would have larger first ionization energy: an alkali metal in Period 2 or an alkali metal in Period 4?

maria [59]

Answer:

An alkali metal present in period 2 have larger first ionization energy.

Explanation:

Ionization energy:

The amount of energy required to remove the electron from the atom is called ionization energy.

Trend along period:

As we move from left to right across the periodic table the number of valance electrons in an atom increase. The atomic size tend to decrease in same period of periodic table because the electrons are added with in the same shell. When the electron are added, at the same time protons are also added in the nucleus. The positive charge is going to increase and this charge is greater in effect than the charge of electrons. This effect lead to the greater nuclear attraction. The electrons are pull towards the nucleus and valance shell get closer to the nucleus. As a result of this greater nuclear attraction atomic radius decreases and ionization energy increases because it is very difficult to remove the electron from atom and more energy is required.

Trend along group:

As we move down the group atomic radii increased with increase of atomic number. The addition of electron in next level cause the atomic radii to increased. The hold of nucleus on valance shell become weaker because of shielding of electrons thus size of atom increased.

As the size of atom increases the ionization energy from top to bottom also decreases because it becomes easier to remove the electron because of less nuclear attraction and as more electrons are added the outer electrons becomes more shielded and away from nucleus. Thus alkali metal present in period 2 have larger ionization energy because of more nuclear attraction as compared to the alkali metal present in period 4.

6 0

2 years ago

A decay series starts with the synthetic isotope ²³⁹₉₂U. The first four steps are emissions of a β⁻ particle, another β⁻, an a p

VladimirAG [237]

<u>Thorium series</u> could start by this sequence.

<h3>Brief explanation</h3>

To write balanced equations for nuclear decay processes. It's important to remember that the mass number and the atomic numbers must be balanced. And so what that means is that if we look at an elements nuclear symbol, the atomic number is the bottom number and the top number, the superscript, is the mass number, and so when we add them up on both sides, they have to be equal. There are two different ways in which decay can occur.

In this, series one is through beta decay, which means that the following particle is produced. The other is Alpha Decay, which produces this particle. Both are products. So if we start off with uranium to 39 you read it in nuclear notation, which means we have to find the atomic number just 92 and it undergoes beta decay.

So that means that it produces this particle find the second particle we used the atomic number, so 92 equals minus one plus x, where X equals 93 which is Neptune IAM. The mass number of our new isotope is zero plus X equals to 39 where X equals to 39. This product becomes the reactant in my next decay, which is also a beta decay. And to find the unknown element we do the same here.

Except for that it's 93 equals minus one plus x, where X is 94 which is P u plutonium, and the mass number is zero plus X equals to 39 or to 39. The next decay starts with the isotope that we just form to 39 p. U. This time it's an Alpha decay. So we produce this particle to find the unknown. Element 94 equals two plus x, where X equals 92 which takes us back to uranium.

Find the mass number of this isotope 2 39 equals four plus X, where X equals to 35. Finally, for the last decay, you have another Alpha decay starting with uranium to 35 making an alpha particle. The atomic number will be 90 which is T H and the top is 2 31 For the mass number. This begins the natural decay, series of thorium .