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Mistake Bank: P-Block (Gr 13 & 14) | Chemca

Mistake Bank: P-Block (Gr 13 & 14) | Chemca

The Mistake Bank

Chapter 11: P-Block Elements (Group 13 & 14)

Where metal meets non-metal, and logic meets exceptions.

Acidity of Boron Halides

Back Bonding

Scenario: Arrange $BF_3$, $BCl_3$, and $BBr_3$ in increasing order of Lewis Acidic strength.

What Students Do

Student thinks: "Fluorine is most electronegative, so it pulls electrons from Boron, making Boron most electron-deficient."

$$ BF_3 > BCl_3 > BBr_3 $$

(Ignores back-bonding!)

The Correct Way

Size Matching Matters!

In $BF_3$, both B and F are $2p-2p$. Effective back-bonding ($p\pi-p\pi$) satisfies Boron's hunger.

In $BBr_3$, the overlap is $2p-4p$ (very poor). Boron stays hungry.

$$ BBr_3 > BCl_3 > BF_3 $$

Hydrolysis of Chlorides

Reaction Mechanism

Scenario: Does $CCl_4$ hydrolyze in water like $SiCl_4$?

What Students Do

Student assumes same group members behave similarly.

Writes: $CCl_4 + H_2O \rightarrow C(OH)_4 + HCl$

(Carbon has no vacancy!)

The Correct Way

No d-orbitals in Carbon!

For hydrolysis, water's oxygen needs to attack the central atom. Carbon ($n=2$) has no empty d-orbitals to accept the lone pair.

$SiCl_4$ has empty d-orbitals, so it hydrolyzes.

Answer: $CCl_4$ does NOT hydrolyze.

The Inert Pair Effect

Oxidation States

Scenario: Compare stability of $Pb^{4+}$ and $Pb^{2+}$. Which is a better Oxidizing Agent?

What Students Do

Student assumes group oxidation state (+4) is always most stable.

Thinks $Pb^{4+}$ is stable and inert.

The Correct Way

Heavy metals prefer lower states!

Due to poor shielding by f-electrons, the s-electrons of the outer shell ($6s^2$) refuse to participate (Inert Pair).

$Pb^{2+}$ is stable. $Pb^{4+}$ is unstable.

Thus, $Pb^{4+}$ desperately wants to gain electrons to become $Pb^{2+}$, making it a Strong Oxidizing Agent.

Structure of Silica (SiO2)

Bonding

Scenario: Draw the structure of $SiO_2$.

What Students Do

Student thinks it's like $CO_2$.

Draws linear molecule: $O=Si=O$

(Si cannot form stable $\pi$-bonds with Oxygen!)

The Correct Way

It's a 3D Network Solid!

Silicon forms single bonds with 4 Oxygens, and each Oxygen bonds with 2 Silicons.

This forms a giant covalent lattice (Quartz), which is why sand is a hard solid, while $CO_2$ is a gas.

Banana Bonds in Diborane

Chemical Bonding

Scenario: In $B_2H_6$, are all B-H bonds equal?

What Students Do

Student assumes symmetry like Ethane ($C_2H_6$).

Answer: "Yes, all bonds are identical single bonds."

The Correct Way

Terminal vs Bridge Bonds!

1. 4 Terminal B-H bonds: Normal 2c-2e bonds.

2. 2 Bridge B-H-B bonds: 3c-2e bonds (Banana bonds).

The bridge bonds are longer and weaker than terminal bonds.

AlCl3 in Water

Compounds

Scenario: Does $AlCl_3$ exist as a dimer ($Al_2Cl_6$) in water?

What Students Do

Student memorizes "$AlCl_3$ is a dimer" and applies it everywhere.

Answer: "Yes."

The Correct Way

High Hydration Energy!

In non-polar solvents (like benzene), it is a dimer.

In water, the high hydration energy breaks the dimer bridges.

It exists as octahedral ions: $[Al(H_2O)_6]^{3+} + 3Cl^-$.

Confess Your Sins!

"P-Block is the wild west of the periodic table. Did you get shot down by an exception?"

Did one of these catch you? Or do you have a different horror story from your last exam?

Scroll down to the comments section below and tell us:

"Which mistake were you making?"

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