# Periodic Properties

Chemistry I · Atoms, Bonds and Reaction Rates · https://tryals.app/learn/chemistry-i/periodic-properties

## One Cause, Many Trends

Almost every periodic trend follows from the **effective nuclear charge** $Z_{\text{eff}}$, the net positive pull an outer electron actually feels, once inner electrons have screened part of the nuclear charge:

$$Z_{\text{eff}} = Z - S$$

where $S$ is the shielding constant. Core electrons shield well; electrons in the same shell shield each other only weakly.

**Across a period** (left to right), $Z$ rises by one per element while the new electrons enter the *same* shell and barely shield each other. $Z_{\text{eff}}$ therefore climbs steadily, pulling the outer shell inward. **Down a group**, each step adds a whole new shell and a full core of shielding electrons, so $Z_{\text{eff}}$ stays roughly constant while $n$ jumps.

| Property | Across a period | Down a group |
|---|---|---|
| Atomic radius | Decreases | Increases |
| Ionisation energy | Increases | Decreases |
| Electron affinity | More negative | Less negative |
| Metallic character | Decreases | Increases |

**Ionisation energy** is the energy to remove the outermost electron. Successive ionisations always cost more, and there is a huge jump the moment you break into a full core, that jump is how the group number can be read off experimental data. Two irregularities are worth knowing: boron dips below beryllium because its electron leaves a higher $2p$ rather than a filled $2s$, and oxygen dips below nitrogen because its fourth $2p$ electron must pair up and suffer extra repulsion.

**Cations** are always smaller than their parent atom, losing an electron often empties a whole shell and leaves the rest more tightly held. **Anions** are always larger, because the added electron increases repulsion without increasing $Z$.

> **Common pitfall:** explaining "atoms get smaller across a period" by saying electrons are added. Electrons are indeed added, but they enter the same shell and shield poorly, so the rising nuclear charge wins. It is the *proton* count, felt through weak shielding, that shrinks the atom.

## Practice questions

10 of this lesson's 13 practice questions, with answers. The full set is in the app.

### 1. For a valence electron of silicon ($Z = 14$), the shielding constant is $S = 9.85$ by Slater’s rules. Compute the effective nuclear charge, to two decimal places.

**Answer:** 4.15 (within ±0.03)

**Why:** $Z_{\text{eff}} = Z - S = 14 - 9.85 = 4.15$. The valence electron feels roughly a $+4$ pull rather than the full $+14$, which is why silicon is far larger than its nuclear charge alone would suggest.

Page: https://tryals.app/practice/chemistry-i/periodic-properties/for-a-valence-electron-of-silicon-z-14-the-shielding-constant-is

### 2. Why does atomic radius decrease from left to right across a period?

A. Nuclear charge rises while the added electrons shield each other poorly
B. Outer electrons are gradually lost, causing valence shells to collapse
C. Electron–electron repulsion falls steadily as subshells are completed
D. Additional inner shielding layers are inserted, pulling outer shells inward

**Answer:** A. Nuclear charge rises while the added electrons shield each other poorly

**Why:** Each step adds a proton and an electron, but the electron joins the same shell where it screens the nucleus only weakly. The net pull on the outer shell therefore grows and the atom contracts.

Page: https://tryals.app/practice/chemistry-i/periodic-properties/why-does-atomic-radius-decrease-from-left-to-right-across-a-period

### 3. A sodium cation is smaller than a neutral sodium atom.

**Answer:** True

**Why:** True, sodium’s single $3s$ electron is its whole outer shell, so $\mathrm{Na^+}$ is left with the much smaller neon core, and the unchanged $+11$ nucleus holds those 10 electrons more tightly still.

Page: https://tryals.app/practice/chemistry-i/periodic-properties/a-sodium-cation-is-smaller-than-a-neutral-sodium-atom

### 4. Boron has a LOWER first ionisation energy than beryllium, breaking the general trend. Why?

A. Boron’s added electron pairs up in the 2s orbital, causing mutual repulsion
B. Boron has a significantly larger radius, reducing the effective nuclear charge
C. Boron’s outermost electron is in a higher-energy 2p orbital, not the filled 2s
D. Boron achieves a stable full outer valence shell upon losing its electron

**Answer:** C. Boron’s outermost electron is in a higher-energy 2p orbital, not the filled 2s

**Why:** Beryllium ($2s^2$) loses an electron from a filled, lower-energy $2s$. Boron ($2s^2 2p^1$) loses its lone $2p$ electron, which sits higher and is more shielded, so it costs less despite the larger $Z$. (Pairing repulsion is the explanation for the *oxygen* dip, not this one.)

Page: https://tryals.app/practice/chemistry-i/periodic-properties/boron-has-a-lower-first-ionisation-energy-than-beryllium-breaking

### 5. Which properties increase going DOWN a group?

A. First ionisation energy
B. Atomic radius
C. Metallic character
D. Number of occupied shells

**Answer:** B. Atomic radius; C. Metallic character; D. Number of occupied shells

**Why:** Going down adds shells, so radius and shell count rise, and the loosely held outer electron makes the element more metallic. Ionisation energy *falls*, because that distant, well-shielded electron is easier to remove.

Page: https://tryals.app/practice/chemistry-i/periodic-properties/which-properties-increase-going-down-a-group

### 6. Successive ionisation energies of an element (kJ/mol) are 738, 1451, 7733, 10540. Which group number does the huge jump identify, that is, how many valence electrons does the atom have?

**Answer:** 2

**Why:** The first two electrons cost 738 and 1451 kJ/mol; the third costs over five times as much because it breaks into the noble-gas core. Two valence electrons means group 2, this is magnesium.

Page: https://tryals.app/practice/chemistry-i/periodic-properties/successive-ionisation-energies-of-an-element-kj-mol-are-738-1451

### 7. Match each species to its size relative to the neutral atom.

**Answer:**

- A cation formed by losing the whole outer shell → Much smaller than the atom
- An anion formed by adding an electron → Larger than the atom
- An atom one place further right in the period → Slightly smaller than the atom
- An atom one place further down the group → Considerably larger than the atom

**Why:** Emptying a shell contracts an ion sharply; an extra electron adds repulsion without extra nuclear charge and expands it; one step right shrinks slightly; one step down adds a whole shell and grows a lot.

Page: https://tryals.app/practice/chemistry-i/periodic-properties/match-each-species-to-its-size-relative-to-the-neutral-atom

### 8. Electron affinity generally becomes more negative going DOWN a group.

**Answer:** False

**Why:** False, the trend runs the other way. Down a group the added electron enters a shell further from the nucleus and better shielded, so less energy is released and the affinity becomes *less* negative. It is going ACROSS a period that makes it more negative, peaking at the halogens.

Page: https://tryals.app/practice/chemistry-i/periodic-properties/electron-affinity-generally-becomes-more-negative-going-down-a-group

### 9. Oxygen has a lower first ionisation energy than nitrogen despite possessing a higher nuclear charge. What accounts for this apparent contradiction, and what does it reveal about valence electrons?

A. The additional outer electron experiences superior shielding from the core
B. Paired electrons in a subshell repel each other, lowering removal energy
C. Inner shells expand slightly to counterbalance the increased nuclear charge
D. Higher nuclear charges invariably reduce electron binding in open subshells

**Answer:** B. Paired electrons in a subshell repel each other, lowering removal energy

**Why:** Added core shielding would require new inner shells, whereas both elements share identical cores. Inner shells do not expand outward with higher Z, and increased nuclear charge generally binds electrons tighter; here, intra-orbital repulsion uniquely destabilises the pair.

Page: https://tryals.app/practice/chemistry-i/periodic-properties/oxygen-has-a-lower-first-ionisation-energy-than-nitrogen-despite

### 10. Arrange these atoms from LARGEST to smallest atomic radius.

**Answer:**

1. Potassium
2. Sodium
3. Magnesium
4. Oxygen

**Why:** K lies below Na and so is largest. Mg is one step right of Na and therefore slightly smaller. O is in period 2 with a high effective nuclear charge and is smallest by a wide margin, moving down a group changes size far more than moving across a period.

Page: https://tryals.app/practice/chemistry-i/periodic-properties/arrange-these-atoms-from-largest-to-smallest-atomic-radius