Periodic Trends & Exceptions

Generally, size increases down a group. However, Gallium (135 pm) is smaller than Aluminium (143 pm) due to the poor shielding of 3d electrons (d-block contraction) which pulls the outer shell closer.

This is the most "jumbled" group. $Tl$ has a very high Ionization Enthalpy due to poor shielding of 4f electrons (Lanthanoid contraction), making it harder to remove electrons than $In$, $Al$, or $Ga$.

Due to the Inert Pair Effect, the heavier elements prefer the +1 state over the group oxidation state (+3). $Tl$ is stable as $Tl^+$, while $Al$ is stable as $Al^{3+}$.

Since $Tl$ prefers the +1 state, $Tl^{3+}$ is a strong oxidizing agent (wants to gain electrons). Conversely, $Al$ prefers +3, so $Al$ metal is a strong reducing agent.

This is counter-intuitive. Fluorine is most electronegative, but $BF_3$ is the weakest acid. This is due to strong $p\pi - p\pi$ back bonding between the small B (2p) and F (2p) orbitals, which satisfies Boron's electron deficiency. Large Iodine cannot back-bond effectively with small Boron.

C-C bonds are extremely strong ($348 kJmol^{-1}$). As size increases down the group, M-M bond strength decreases, reducing catenation. $Pb$ does not show catenation.

Note the flip at the end. $Pb$ has a higher IE than $Sn$ due to the Lanthanoid contraction (poor shielding of f-electrons), increasing effective nuclear charge on $Pb$.

Boiling point generally increases with mass (Van der Waals forces). However, $NH_3$ has an anomalously high BP due to Hydrogen Bonding. It sits between $As$ and $Sb$.

As the central atom becomes larger and less electronegative, the bond pairs differ further apart, reducing repulsion. Pure p-orbitals are used for bonding in heavier elements (Drago's rule suggestion).

As the size of the central atom increases (N to Bi), the M-H bond length increases and bond dissociation enthalpy decreases ($NH_3$: 389 kJ/mol vs $SbH_3$: 255 kJ/mol), making them less stable.

Directly inverse to thermal stability. Since $Bi-H$ bonds break easily, $BiH_3$ is the strongest reducing agent (gives H easily).

High electron density on the small Nitrogen atom makes the lone pair available for donation. Large atoms disperse the charge, reducing basicity.

Oxygen has the least negative value in the group (except Polonium) due to its small size and high inter-electronic repulsion. Sulphur has the most negative value.

$H_2O$ has the highest BP due to strong hydrogen bonding. Among the rest, BP increases with molecular mass (Van der Waals forces).

Same logic as Group 15. Repulsion decreases as the central atom size increases.

Bond dissociation energy decreases down the group ($H-Te$ is weak). A weaker bond releases $H^+$ more easily, increasing acidity.

The O-O single bond is weaker than S-S due to large electron-electron repulsion between the lone pairs on the small Oxygen atoms. This makes catenation easier for Sulphur ($S_8$) than Oxygen.

Chlorine has the most negative $\Delta_{eg}H$ in the periodic table. Fluorine is less negative than Chlorine because its tiny 2p orbitals cause inter-electronic repulsion, resisting the addition of an electron.

$F_2$ has an anomalously low bond energy (lower than $Cl$ and $Br$) because of large lone pair-lone pair repulsion between the two close Fluorine atoms.

$H-I$ bond is the longest and weakest (lowest dissociation enthalpy), so it dissociates most easily to release $H^+$.

$HF$ is a liquid with the highest BP due to Hydrogen Bonding. The others follow the mass trend (Van der Waals).

Fluorine is the strongest oxidant due to high hydration enthalpy of $F^-$ and low bond dissociation enthalpy of $F_2$.

$Cl$ is more electronegative, pulling electron density away from the O-H bond, making the release of $H^+$ easier.

As the oxidation state of the central atom increases (+1 to +7), the negative charge on the conjugate base is more delocalized (resonance stabilization), making the acid stronger.

The 4d and 5d series have virtually the same radii (e.g., $Zr \approx Hf$) due to the Lanthanoid Contraction. The filling of 4f orbitals (poor shielding) pulls the 5d shells in.

Maxima is at $d^5$ due to maximum unpaired electrons participating in metallic bonding. $Mn$ has a surprisingly low MP because its stable half-filled $d^5$ structure leads to weaker delocalization/metallic bonding.

Radius decreases while mass increases, leading to a sharp increase in density.

Magnetic moment ($\mu = \sqrt{n(n+2)}$) depends on the number of unpaired electrons ($n$). $Mn^{2+}$ ($d^5$) has the maximum (5) unpaired electrons.

These elements have a fully filled $d^{10}s^2$ configuration, making them very stable and resistant to electron removal. This explains why they are often not considered "typical" transition metals.

Fluorine is small and highly electronegative; it can stabilize the highest oxidation states of transition metals. Iodine usually stabilizes lower states ($CuI$).

Due to Lanthanoid Contraction, the size of the cation decreases ($La^{3+} \to Lu^{3+}$). The M-OH bond becomes more covalent and harder to break (less $OH^-$ release).

Removing the 3rd electron from $Mn$ ($d^5 \to d^4$) breaks the stable half-filled configuration, requiring huge energy. Removing it from $Fe$ ($d^6 \to d^5$) achieves stability, so it is easier.

Smaller ions have higher charge density and attract more water molecules. This is why hydrated radius order is inverse: $Li^+_{(aq)} > Cs^+_{(aq)}$.

Fajans' Rule. Larger anions ($I^-$) are more easily polarized by the cation, leading to higher covalent character.

In $Cl_2O$, the large Cl atoms repel each other (steric hindrance), opening up the angle. In $OF_2$, the bonding electrons are closer to F, reducing lp-bp repulsion, but the steric crowding in $Cl_2O$ dominates.

Dispersion forces (Van der Waals) increase with atomic size/mass. Helium has the lowest boiling point of any substance.

Hydration energy decreases faster than lattice energy down the group. $Be^{2+}$ is small and heavily hydrated, making it soluble.

Large cations stabilize large anions ($CO_3^{2-}$). Small $Be^{2+}$ polarizes the carbonate ion strongly, causing it to decompose easily.

$TlI_3$ is actually $Tl^+(I_3)^-$, not Thallium(III) iodide. Tl prefers +1 due to inert pair effect.

Single N-N bonds are weak due to repulsion between lone pairs on the small N atoms. P-P bonds are stronger, so P shows better catenation.

In $NH_3$, the orbital dipole and bond dipoles add up. In $NF_3$, the Fluorine pulls electrons resulting in bond dipoles opposing the orbital dipole.

Regular increase with size and Van der Waals forces. $I_2$ is a solid, $Br_2$ liquid, others gases.

For same electron count, higher nuclear charge (protons) pulls the shell in tighter. $Al^{3+}$ has most protons, smallest size.

$NO_2^+$ is linear (sp, 180). $NO_2$ has one odd electron (bent). $NO_2^-$ has a full lone pair (bent, more repulsion than odd electron).

They have stable octets. You must add energy to force an electron into a higher principal quantum number shell.

$NCl_3$ has vacant d-orbitals on Cl to accept water's lone pair. $NF_3$ has no vacant d-orbitals on N or F. $CCl_4$ doesn't hydrolyze (no d-orbitals), but $SiCl_4$ does.

(Often asked in context of preparation). Strong $\pi$-acceptors have high trans-effect.

Bond length increases, bond strength decreases.

You must memorize which are amphoteric. $CO, NO, N_2O$ are Neutral. $GeO$ is acidic.

They have an odd number of valence electrons (11 and 17 valence electrons respectively). They dimerize ($N_2O_2, N_2O_4$) to become diamagnetic.