Periodic Table Trends 1. What is the trend in ionic radius across a period? Down a group? 2. What is the trend in electronegativity across a period? Down a group? Using your knowledge of atomic structure, what the explanation for this trend?

2 Answers
  1. Ionic radii decreases across a period.
    Ionic radii increases down a group.
  2. Electronegativity increases across a period.
    Electronegativity decreases down a group.

Explanation:

Ionic radii decreases across a period.
This is due to the fact that metal cations lose electrons, causing the overall radius of an ion to decrease. Non-metal cations gain electrons, causing the overall radius of an ion to decrease, but this happens in reverse (compare fluorine to oxygen and nitrogen, which one gains the most electrons).

Ionic radii increases down a group.
In a group, all the ions have the same charge as they have the same valency (that is, the same number of valence electrons on the highest energy level sub-orbital). Therefore, ionic radii increase down a group as more shells are added (per period).

2.

Electronegativity increases across a period.
This is because the number of protons in the nucleus increases across the period. That causes attraction to bonding pairs of electrons more strongly. (Shielding effect or other factors aside, this is the simplest answer.)

Electronegativity decreases down a group.
Similarly (but opposite) to ionic radii, the electronegativity decreases due to the longer distance between the nucleus and the valence electron shell, hence decreasing the attraction, making the atom have less of an attraction for electrons or protons.

Mar 25, 2018

Ionic radii: Decreases then increases as you go along the period
Electronegativity: Increases as you along and decreases as you go down period.

Explanation:

This is more complicated in regard to ionic radius we have to be careful to recognize if it is an anion (negative) or a cation (positive)

If it is a anion we can see that it has one more electron than its atom. Take Carbon it has 6 electrons and 6 protons, if we add an electron then there is 7 electrons and 6 protons the additional electron increases the repulsive forces between the electrons making the radius increase.

Whilst with a cation it has one less electron than its atom. So now the carbon cation has 5 electrons and 6 protons. The loss of an electron decreases the repulsive forces decreasing the radius size.

Now what we have to look at is what kind of ions the elements in the periodic table become to look at how ionic radius changes along a period. If we take row three we know that a stable state is either 2,8 or 2,8,8 for it's energy levels. So an element will gain electrons/lose electrons to be in those states.
So Na(sodium) Mg (magnesium) and Al (aluminium) have less than 4 electrons in outer shell.

This means that they will be more likely to lose as getting to 2,8 is easier to get to than 2,8,8 so they will all become Cations. In addition each successive one will lose more electrons to get to the 2,8 stage i.e. Na will lose 1, Mg 2, Al 3. So as you go along the ionic radius will decrease.

The opposite will happen to P (phosphorous) S (sulfur) and Cl (Chlorine) as it is easier to go to 2,8,8 they will gain electrons, so they are anions. So as each one gains less electrons to get to the stage as you go along each one ionic radius will be smaller than the one before.

Ar (argon) won't gain or lose so there will be no change and Si (silicon) can do either but usually we say it becomes a cation and loses all 4 electrons, so has the smallest radius of all elements in the third row.

Going down the general rule is that the ionic radius will increase as the electrons are in further away valence shell (outer shell).

In regard to electro-negativity as you go along a period it increases as the atomic radii across a period are getting smaller so the electron is closer to the nucleus making it harder to remove.

As you go down it is easier to remove as it is further away as it is in a further energy level and their is additional shielding reducing the attractive forces from the in-between electron shells.