Foundations — Tetravalency of carbon; hybridization recap (sp, sp², sp³)

Before you touch the main recap, we build every symbol it quietly assumes. Nothing below is skipped: if the parent note wrote it, we earn it here first.

1. The atom picture: nucleus + electron rooms

The picture: think of the nucleus as a lamp and each electron as a moth. The moth doesn't crash into the lamp and doesn't fly away — it lives in a region around it. That region is what we will call an orbital.

Why the topic needs this: every bond carbon makes is really two electrons shared between two nuclei. To understand bonding we must first know where electrons live.

2. Atomic number $Z$ and counting electrons

Read the symbol out loud: "$Z$ equals six" means "carbon has six protons, and a neutral carbon has six electrons to place."

Why the topic needs it: the parent note opens with "$Z=6$". We must place those exact 6 electrons into rooms to see why carbon looks two-handed at first.

3. Orbitals: the rooms, and their shapes ($s$ and $p$)

There are two shapes we need:

• $s$ orbital — a ball centred on the nucleus. Same in every direction (no preferred direction). We write the ball with a leading number for its shell: $1s$, $2s$, ...
• $p$ orbital — a dumbbell (two lobes) pointing along one axis. There are three of them, one per axis, called $p_x$, $p_y$, $p_z$. The little subscript just says which way the dumbbell points.

Why the topic needs it: the parent writes $2s$, $2p_x$, $2p_y$, $2p_z$ and later "one $s$ + three $p$". Those are these rooms.

4. Shells and the superscript count: reading $1s^{2}2s^{2}2p^2$

Decode carbon's ground state, symbol by symbol: $1s^{2}2s^{2}2p_x^{1}2p_y^{1}2p_z^0$

• $1s^2$ → inner ball, 2 electrons (full).
• $2s^2$ → second-shell ball, 2 electrons (full).
• $2p_x^1$ → x-dumbbell, 1 electron.
• $2p_y^1$ → y-dumbbell, 1 electron.
• $2p_z^0$ → z-dumbbell, 0 electrons (empty).

Add the superscripts: $2+2+1+1+0=6=Z$. Good — all six placed.

Why one-per-room? Two electrons are both negative, so they repel. Parking them in separate dumbbells keeps them farther apart, which is lower energy — cheaper than cramming both into $p_x$. That is exactly why carbon's last two electrons read $2p_x^{1}2p_y^1$ (one each) and not $2p_x^2$.

The picture: three empty seats in a row on an empty bus — strangers sit one per seat before anyone doubles up.

The picture: two rooms ($p_x$, $p_y$) each hold one lonely electron; the $p_z$ room is empty; the $2s$ room is packed with a pair.

Why the topic needs it: the parent's whole "carbon looks divalent" puzzle is: only two unpaired electrons ($2p_x^1,2p_y^1$) are available to pair up with other atoms. Counting superscripts (and knowing Hund's rule put them in separate rooms) is how you see that.

5. Covalent bond and valency

The picture: each unpaired electron is a free hand; a bond is a handshake between two atoms' hands. Two lonely electrons ⇒ only two handshakes ⇒ the atom seems divalent.

Why the topic needs it: this is the exact paradox the parent resolves — carbon has two lonely hands but makes four handshakes in ${\text{CH}}_4$.

6. Promotion: moving one electron up a room

$\underset{\text{2 lonely electrons}}{\underbrace{1s^{2}2s^{2}2p_x^{1}2p_y^{1}2p_z^0}}⟶\underset{\text{4 lonely electrons}}{\underbrace{1s^{2}2s^{1}2p_x^{1}2p_y^{1}2p_z^1}}$

Notice the $2s$ superscript dropped $2\to 1$ and $2p_z$ rose $0\to 1$. Now four rooms each hold one lonely electron ⇒ four free hands.

Why the topic needs it: this single jump is why carbon can be tetravalent at all.

7. Angles and the degree symbol: $109.5^{\circ }$, $120^{\circ }$, $180^{\circ }$

The picture: stand at the central atom, look down two bonds like two roads leaving a roundabout — the bond angle is how wide apart those roads are.

Why the topic needs it: the parent's whole geometry table ($180^{\circ }$ linear, $120^{\circ }$ planar, $109.5^{\circ }$ tetrahedral) is nothing but bond angles. Degrees are the ruler.

8. $\sigma$ (sigma) and $\pi$ (pi): two kinds of bond

• Single bond = 1 $\sigma$.
• Double bond = 1 $\sigma$ + 1 $\pi$.
• Triple bond = 1 $\sigma$ + 2 $\pi$.

Why the topic needs it: the parent's key rule — "$\pi$ bonds don't count in the steric number" — only makes sense once you know $\pi$ is the sideways extra layer that uses leftover dumbbells, not a fresh bonding hand. See the deeper page Sigma and Pi bonds — orbital overlap.

9. Lone pair

The picture: a pair of electrons parked at home instead of out shaking hands. In water's oxygen there are two such parked pairs.

Why the topic needs it: the steric-number recipe adds "lone pairs on the central atom". You must be able to spot parked pairs.

10. Hybridization and the symbols $sp$, $s{p}^2$, $s{p}^3$

Now the superscripts on $sp$, $s{p}^2$, $s{p}^3$ are just how many $p$ dumbbells joined one $s$ ball:

Why the topic needs it: these three symbols are the topic. Everything else (angle, bond length, acidity) is read off from them — see Bond length and bond strength trends and Acidity of terminal alkynes.

11. Steric number: the counting recipe

Every symbol in this line is now defined: $\sigma$ (§8), lone pair (§9), "central atom" = the one whose shape you want. And crucially $\pi$ (§8) is absent — it never enters, because it uses leftover dumbbells, not hybrid hands.

$\text{SN}=2\Rightarrow sp,\text{SN}=3\Rightarrow s{p}^2,\text{SN}=4\Rightarrow s{p}^3$

Why the topic needs it: SN is the single machine that turns a Lewis structure into a geometry. Feeds straight into VSEPR theory and molecular geometry.

Prerequisite map

Equipment checklist

Test yourself — cover the right side. If any answer is fuzzy, re-read that section before the main note.