The p-Block Elements — JEE Mains Chemistry Formulas and Traps

Group 13: Periodic Trends and the Inert Pair Effect

Group 13 trends in radius, ionisation enthalpy and electronegativity

Element	Atomic radius (pm)	M³⁺ radius (pm)	First ionisation enthalpy (kJ/mol)	Electronegativity
B	85	27	801	2.0
Al	143	53.5	577	1.5
Ga	135	62.0	579	1.6 Smaller than Al and with a slightly higher ionisation enthalpy: poor shielding by 3d electrons.
In	167	80.0	558	1.7
Tl	170	88.5	589	1.8

NCERT values. Read each column on its own: the atomic radius dips at Ga, the M³⁺ radius does not, and the ionisation enthalpy is lowest at In, not Tl.

Group 13 melting points, boron's lattice and gallium's liquid range

Element	Melting point (K)	Boiling point (K)	Density (g/cm³)	What to remember
B	2453	3923	2.35	Giant covalent $\mathrm{B_{12}}$ network: very hard, highest melting point
Al	933	2740	2.70	Light metal; made passive by concentrated $\mathrm{HNO_3}$ , which coats it with oxide
Ga	303	2676	5.90	Liquid from 303 K to 2676 K, the widest liquid range; used in high-temperature thermometers The lowest melting point in the group, and still a liquid in boiling water.
In	430	2353	7.31	Soft metal that melts above gallium
Tl	576	1730	11.85	The densest member of the group

Melting points fall from boron to gallium and then rise a little: B > Al > Tl > In > Ga.

The inert pair effect in group 13: Tl⁺ is more stable than Tl³⁺

Element	More stable oxidation state	E° for M³⁺ reduction (V)	How M³⁺ behaves
Al	+3 only	$-1.66$ ( $\mathrm{Al^{3+}/Al}$ )	Very stable; hard to reduce
Ga	+3	$-0.56$ ( $\mathrm{Ga^{3+}/Ga}$ )	Stable; +1 appears only in salts such as $\mathrm{GaAlCl_4}$
In	+3	$-0.34$ ( $\mathrm{In^{3+}/In}$ )	Stable; $\mathrm{In^{+}}$ is easily oxidised back to +3
Tl	+1	$+1.26$ ( $\mathrm{Tl^{3+}}$ reduced to $\mathrm{Tl^{+}}$ )	Strong oxidising agent The positive potential is the inert pair effect in numbers: Tl³⁺ is eager to become Tl⁺.

A more positive reduction potential means the ion is more easily reduced, so a stronger oxidising agent.

Common traps

The atomic radius is not a smooth rise

The order

\mathrm{B < Al < Ga < In < Tl}

looks natural but is wrong. Gallium is smaller than aluminium, so the correct order is

\mathrm{B < Ga < Al < In < Tl}

. The

\mathrm{M^{3+}}

radius, by contrast, does rise steadily.

Thallium does not have the lowest ionisation enthalpy

Indium has the lowest first ionisation enthalpy in group 13, 558 kJ/mol. Thallium's is higher, 589 kJ/mol, because its 4f and 5d electrons shield poorly.

Electronegativity does not simply fall down group 13

It falls from boron (2.0) to aluminium (1.5), then rises a little through gallium, indium and thallium. A statement that it decreases down the whole group is false.

Gallium thermometers are for HIGH temperatures

Gallium is useful in thermometers because it stays liquid up to about 2676 K. It freezes at 303 K, so a gallium thermometer cannot measure a low temperature such as the freezing point of brine.

Boron's hardness is not metallic bonding

Boron is a non-metal. Its high melting point and hardness come from a giant covalent network of

\mathrm{B_{12}}

icosahedra, not from metallic bonds.

Not every group 13 element has a stable +1 state

The +1 state becomes important only for the heavier elements and is truly stable only for thallium. Boron and aluminium show +3; a statement that all group 13 elements show a highly stable +1 state is false.

TlI₃ is not thallium(III) iodide

Thallium in

\mathrm{TlI_3}

is +1. The compound is

\mathrm{Tl^{+}}

with the tri-iodide ion

\mathrm{I_3^{-}}

, because

\mathrm{Tl^{3+}}

is a strong enough oxidant to turn iodide into iodine.

In GaAlCl₄, gallium is +1

The salt is

\mathrm{Ga^{+}[AlCl_4]^{-}}

. All four chlorines surround aluminium; gallium is the separate cation. Giving gallium +3 would leave the formula unbalanced.

Boron and Aluminium Compounds

Learn this subtopic in the notes

Borax, the borax bead test and boric acid

Borax bead and boric acid

\mathrm{Na_2B_4O_7 \xrightarrow{\Delta} 2NaBO_2 + B_2O_3} \qquad \mathrm{B(OH)_3 + 2H_2O \rightarrow [B(OH)_4]^{-} + H_3O^{+}}

Structure and preparation of diborane and borazine

Feature	Diborane, B₂H₆	Borazine, B₃N₃H₆
Shape	Non-planar: the two $\mathrm{BH_2}$ ends lie in one plane, the two bridging H above and below it	Planar six-membered ring of alternating B and N
Bonds	Four terminal 2-centre-2-electron B–H bonds and two bridging 3-centre-2-electron B–H–B bonds	Only ordinary 2-centre-2-electron bonds, with π electrons delocalised round the ring Banana bonds belong to diborane, never to borazine.
Hybridisation of boron	About $sp^3$	$sp^2$
Bond angles and lengths	Terminal H–B–H 122°, bridge H–B–H 97°; terminal B–H 119 pm, bridging B–H 134 pm	All six B–N bonds equal in length
With water	$\mathrm{B_2H_6 + 6H_2O \rightarrow 2B(OH)_3 + 6H_2}$	$\mathrm{B_3N_3H_6 + 9H_2O \rightarrow 3B(OH)_3 + 3NH_3 + 3H_2}$
Acid-base nature	Lewis acid; split by bases such as $\mathrm{NMe_3}$	Polar B–N bonds make it more reactive than benzene

The terminal H–B–H angle is wider than the bridge angle, so the terminal bonds have more s character and less p character.

Boron and aluminium halides as Lewis acids: back-bonding and maximum covalency

Species	Covalency of the central atom	Shape	Why
$\mathrm{BF_3}$	3	Trigonal planar	Electron deficient; back-bonding from F partly fills boron's empty p orbital
$\mathrm{[BF_4]^{-}}$	4 (oxidation state still +3)	Tetrahedral	Fluoride donates a pair into boron's empty orbital
$\mathrm{BF_6^{3-}}$	Would need 6	Does not exist	Boron has no d orbitals, so four bonds is its limit The reason NCERT gives is the missing d orbitals.
$\mathrm{[AlF_6]^{3-}}$	6	Octahedral	Aluminium uses its 3d orbitals
$\mathrm{[Al(H_2O)_6]^{3+}}$	6	Octahedral, $sp^3d^2$	Formed when aluminium chloride dissolves in acidified water
$\mathrm{Al_2Cl_6}$	4	Two tetrahedra sharing an edge of two bridging Cl	Each aluminium completes its octet through a chlorine lone pair

Covalency counts the bonds round the atom; it is not the oxidation state. Boron is +3 in both BF₃ and [BF₄]⁻.

Common traps

Cupric metaborate is blue-green, not colourless

In the oxidising flame copper gives blue-green copper(II) metaborate,

\mathrm{Cu(BO_2)_2}

. The colourless one is copper(I) metaborate,

\mathrm{CuBO_2}

, formed in the reducing (luminous) flame.

Boric acid is monobasic, not tribasic

The three OH groups in

\mathrm{B(OH)_3}

do not ionise. Boric acid accepts one hydroxide ion from water, so it releases one

\mathrm{H^{+}}

per molecule and is a weak acid.

Diborane has two 3-centre bonds, not four

Of the eight B–H links in

\mathrm{B_2H_6}

, four are ordinary terminal bonds. The four bridging links make just two 3-centre-2-electron bonds, one for each bridging hydrogen.

Diborane is not planar and its boron is not sp²

The two bridging hydrogens sit above and below the plane of the four terminal hydrogens, and each boron is roughly

sp^3

. The flat,

sp^2

molecule is borazine.

BH₃ is a Lewis acid, not a Lewis base

Boron in

\mathrm{BH_3}

has only six electrons and an empty p orbital, so it accepts an electron pair, for example from

\mathrm{NMe_3}

. It has no lone pair to donate.

Strongest back-bonding means weakest Lewis acid

Back-bonding is strongest in

\mathrm{BF_3}

because boron's 2p and fluorine's 2p orbitals are the same size. That filling of boron's empty orbital makes

\mathrm{BF_3}

the WEAKEST Lewis acid of the boron trihalides, not the strongest.

Covalency 4 does not mean oxidation state +4

In

\mathrm{[BF_4]^{-}}

boron forms four bonds, but its oxidation state is

x + 4(-1) = -1

, so

x = +3

.

Group 14: Carbon, Silicon, Tin and Lead

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Group 14 trends, carbon's allotropes and silicones

Element	Covalent radius (pm)	First ionisation enthalpy (kJ/mol)	Electronegativity	What sets it apart
C	77	1086	2.5	Catenation and $p\pi$ – $p\pi$ bonds; maximum covalency 4; allotropes
Si	118	786	1.8	Uses d orbitals: $\mathrm{[SiF_6]^{2-}}$ exists; $\mathrm{SiO_2}$ is acidic; forms silicones
Ge	122	761	1.8	$\mathrm{GeO_2}$ acidic; $\mathrm{[GeCl_6]^{2-}}$ exists
Sn	140	708	1.8	+4 more stable than +2; oxides amphoteric
Pb	146	715	1.9	+2 more stable than +4; oxides amphoteric Lead's ionisation enthalpy is a little HIGHER than tin's: poor shielding by 4f and 5d electrons.

The ionisation enthalpy falls from C to Sn, then rises slightly at Pb.

Inert pair effect in tin and lead: which ions oxidise and which reduce

Ion	Preferred state of the element	Behaves as	Evidence
$\mathrm{Sn^{2+}}$	+4	Reducing agent	$\mathrm{E^\circ(Sn^{4+}/Sn^{2+}) = +0.15\ V}$ : $\mathrm{Sn^{2+}}$ is easily oxidised
$\mathrm{Sn^{4+}}$	+4	Stable; a very weak oxidant at most	Same small potential, $+0.15\ V$
$\mathrm{Pb^{2+}}$	+2	Stable	The 6s pair stays out of bonding
$\mathrm{Pb^{4+}}$	+2	Strong oxidising agent	$\mathrm{E^\circ(Pb^{4+}/Pb^{2+}) = +1.67\ V}$ , the most positive here The strongest oxidant among these p-block ions.
$\mathrm{Tl^{3+}}$	+1	Strong oxidising agent	$\mathrm{Tl^{3+}}$ reduced to $\mathrm{Tl^{+}}$ : $+1.26\ V$
$\mathrm{Tl^{+}}$	+1	Stable	The 6s pair stays out of bonding

The more positive the reduction potential, the stronger the oxidising agent.

Tests for the lead ion in salt analysis

Reagent added to Pb²⁺	Product	Colour	What happens next
Dilute HCl	$\mathrm{PbCl_2}$	White	Dissolves on heating the water
$\mathrm{H_2S}$	$\mathrm{PbS}$	Black	Dissolves in hot dilute $\mathrm{HNO_3}$ to give $\mathrm{Pb(NO_3)_2}$
$\mathrm{K_2CrO_4}$	$\mathrm{PbCrO_4}$	Yellow	Dissolves in NaOH as $\mathrm{Na_2[Pb(OH)_4]}$ Charge 2−, four OH groups: coordination number 4.
KI	$\mathrm{PbI_2}$	Yellow	Dissolves in hot water and returns as golden spangles on cooling
Dilute $\mathrm{H_2SO_4}$	$\mathrm{PbSO_4}$	White	Dissolves in ammonium acetate solution

Chloride, sulphate and nitrate of lead are the white or colourless ones; chromate and iodide are yellow; sulphide is black.

Common traps

Carbon's allotropy comes from pπ–pπ bonds, not pπ–dπ

Carbon has no d orbitals. Its allotropes arise from catenation and its ability to form

p\pi

–

p\pi

multiple bonds with itself.

C₆₀ has 20 six-membered and 12 five-membered rings

It is easy to swap the numbers. Buckminsterfullerene has twelve pentagons, each surrounded only by hexagons, and twenty hexagons.

Covalent radius increases down group 14

From C to Pb each element adds a shell, so the covalent radius increases. A statement that it decreases in a regular manner is false.

The lower state is not always the reducing one

\mathrm{Sn^{2+}}

is a reducing agent but

\mathrm{Pb^{2+}}

is not. What matters is each element's preferred state: +4 for tin, +2 for lead.

Pb⁴⁺ is not stable like Sn⁴⁺

Both are +4, but lead's 6s pair resists bonding, so

\mathrm{Pb^{4+}}

is a strong oxidising agent while

\mathrm{Sn^{4+}}

is stable.

Lead chromate in NaOH gives a 2− complex with four OH

The product is

\mathrm{Na_2[Pb(OH)_4]}

: lead(II) with four hydroxide groups, coordination number 4, overall charge 2−. It is not neutral and not six-coordinate.

Lead nitrate is not a confirmatory test

Every confirmatory test for

\mathrm{Pb^{2+}}

makes an insoluble salt: chromate, iodide or sulphate. Lead nitrate is soluble, so its formation confirms nothing.

Group 15: Periodic Trends and Hydrides

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Group 15 trends and the anomalous behaviour of nitrogen

Element	Covalent radius (pm)	First ionisation enthalpy (kJ/mol)	Electronegativity	Character
N	70	1402	3.0	Non-metal, diatomic gas $\mathrm{N_2}$ No d orbitals: maximum covalency 4.
P	110	1012	2.1	Non-metal, $\mathrm{P_4}$ molecules
As	121	947	2.0	Metalloid
Sb	141	834	1.9	Metalloid
Bi	148	703	1.9	Metal, the only one in the group

The biggest steps are between N and P in every column; below arsenic the changes are small.

Hydrides of group 15 from NH₃ to BiH₃

Hydride	H–E–H angle (°)	Boiling point (K)	E–H bond enthalpy (kJ/mol)	Character
$\mathrm{NH_3}$	107.8	238.5	389	Most stable and most basic; weakest reducing agent; hydrogen bonded
$\mathrm{PH_3}$	93.6	185.5	322	Lowest boiling point in the group: no hydrogen bonding and a small molar mass The lowest boiling point is PH₃, not NH₃.
$\mathrm{AsH_3}$	91.8	210.6	297	Less basic and more reducing than $\mathrm{PH_3}$
$\mathrm{SbH_3}$	91.3	254.6	255	Highest boiling point of the four: the largest dispersion forces

BiH₃, not listed because it is too unstable to measure well, continues every trend: least stable, least basic, strongest reducing agent.

Common traps

The N–N single bond is weaker AND shorter than P–P

Nitrogen's small size makes the N–N bond short, but the lone pairs on the two nitrogens are then close and repel, so the bond is weaker than P–P. A statement that it is 'weaker and longer' or 'stronger' is false.

The +5 state becomes LESS stable down group 15

Because of the inert pair effect, +3 grows more stable and +5 less stable from N to Bi. Bismuth(V) is a strong oxidising agent.

Nitrogen's multiple bonds are pπ–pπ

Nitrogen has no d orbitals, so it cannot form

d\pi

–

p\pi

bonds. Its maximum covalency is 4, never 5 or 6.

Boiling point does not rise steadily down group 15

Phosphine boils lowest. Ammonia is raised by hydrogen bonding, so the order is

\mathrm{PH_3 < AsH_3 < NH_3 < SbH_3}

, not a steady rise from NH₃.

Ammonia is the weakest reducing agent, not the strongest

The N–H bond is the strongest E–H bond in the group, so ammonia gives up hydrogen least easily. Reducing power rises down the group to

\mathrm{BiH_3}

.

Basicity decreases down group 15

The lone pair on a large atom such as Sb or Bi is spread out and held loosely in a large orbital, so it binds a proton poorly. Basicity falls from

\mathrm{NH_3}

to

\mathrm{BiH_3}

.

Nitrogen and Its Compounds

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Preparing dinitrogen and nitric acid, and the nitrogen tests

Dinitrogen, the Ostwald process and the brown ring

\mathrm{NH_4Cl + NaNO_2 \rightarrow N_2 + 2H_2O + NaCl} \qquad \mathrm{3NO_2 + H_2O \rightarrow 2HNO_3 + NO} \qquad \mathrm{[Fe(H_2O)_6]^{2+} + NO \rightarrow [Fe(H_2O)_5(NO)]^{2+} + H_2O}

Oxides of nitrogen: oxidation states, structures and nature

Oxide	Oxidation state of N	Structure	Nature
$\mathrm{N_2O}$	+1	Linear N≡N–O; one N–N bond	Neutral; colourless gas
$\mathrm{NO}$	+2	N=O with one unpaired electron	Neutral; colourless gas
$\mathrm{N_2O_3}$	+3	O=N–NO₂; one N–N bond	Acidic; blue solid
$\mathrm{NO_2}$	+4	Bent, odd electron on N; one N=O and one N–O	Acidic; brown gas The odd-electron oxide that dimerises to N₂O₄.
$\mathrm{N_2O_4}$	+4	O₂N–NO₂; one N–N bond, no bridging O	Acidic; colourless
$\mathrm{N_2O_5}$	+5	O₂N–O–NO₂; one N–O–N bridge, no N–N bond	Acidic; colourless solid, the anhydride of $\mathrm{HNO_3}$

Only the two lowest oxides are neutral; only N₂O₅ bridges its nitrogens through oxygen.

Common traps

N₂O₄ has no bridging oxygen

When two

\mathrm{NO_2}

molecules combine, the unpaired electrons on nitrogen form an N–N bond:

\mathrm{O_2N{-}NO_2}

. The number of bridging oxygen atoms is zero.

N₂O and NO are neutral, not acidic

Of the oxides of nitrogen, only

\mathrm{N_2O}

and

\mathrm{NO}

are neutral; they form no acid with water.

\mathrm{N_2O_3}

,

\mathrm{NO_2}

,

\mathrm{N_2O_4}

and

\mathrm{N_2O_5}

are acidic.

Air does not form NO because the reaction is endothermic

Nitrogen and oxygen need about 2000 K, as in lightning, to form NO, because

\mathrm{N_2 + O_2 \rightarrow 2NO}

absorbs heat. The reason is not that nitrogen oxides are unstable.

The brown ring holds NO, not NO₂

Iron(II) reduces nitrate to nitric oxide, and NO bonds to iron as

\mathrm{[Fe(H_2O)_5(NO)]^{2+}}

. No complex of

\mathrm{NO_2}

or

\mathrm{N_2O}

forms.

Dilute nitric acid on lead sulphide gives NO, not N₂O

\mathrm{3PbS + 8HNO_3 \rightarrow 3Pb(NO_3)_2 + 2NO + 3S + 4H_2O}

. The products are lead nitrate, sulphur, nitric oxide and water; nitrous oxide is not formed.

Phosphorus and Its Oxoacids

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Allotropes of phosphorus and reactions of white phosphorus and its chlorides

Key reactions of phosphorus

\mathrm{P_4 + 3NaOH + 3H_2O \rightarrow PH_3 + 3NaH_2PO_2} \qquad \mathrm{P_4 + 8SOCl_2 \rightarrow 4PCl_3 + 4SO_2 + 2S_2Cl_2} \qquad \mathrm{PCl_3 + 3H_2O \rightarrow H_3PO_3 + 3HCl} \qquad \mathrm{PCl_5 + 4H_2O \rightarrow H_3PO_4 + 5HCl}

Basicity and reducing power of phosphorus oxoacids

Counting rule for phosphorus oxoacids

\text{basicity} = \text{number of } \mathrm{P{-}OH} \text{ groups} \qquad \text{non-ionisable H} = \text{number of } \mathrm{P{-}H} \text{ bonds} \qquad \mathrm{P{-}H} \text{ present} \Rightarrow \text{reducing}

Oxoacids of phosphorus: formulas, oxidation states and bonds

Acid	Formula	Oxidation state of P	Bonds in the structure
Hypophosphorous (phosphinic)	$\mathrm{H_3PO_2}$	+1	Two P–H, one P–OH, one P=O
Orthophosphorous (phosphonic)	$\mathrm{H_3PO_3}$	+3	One P–H, two P–OH, one P=O
Pyrophosphorous	$\mathrm{H_4P_2O_5}$	+3	Two P–H, two P–OH, two P=O, one P–O–P
Hypophosphoric	$\mathrm{H_4P_2O_6}$	+4	One P–P, four P–OH, two P=O Hypophosphoric (+4, P–P bond) is not hypophosphorous (+1, two P–H).
Orthophosphoric	$\mathrm{H_3PO_4}$	+5	Three P–OH, one P=O
Pyrophosphoric	$\mathrm{H_4P_2O_7}$	+5	Four P–OH, two P=O, one P–O–P
Cyclotrimetaphosphoric	$\mathrm{(HPO_3)_3}$	+5	A ring with three P–O–P, three P–OH, three P=O
Phosphorus(V) oxide	$\mathrm{P_4O_{10}}$	+5	Six P–O–P bridges and four P=O (the anhydride, not an acid)

Pyrophosphorous is +3 with P–H bonds; pyrophosphoric is +5 with none.

Common traps

Thionyl chloride gives PCl₃, not PCl₅ or POCl₃

\mathrm{P_4 + 8SOCl_2 \rightarrow 4PCl_3 + 4SO_2 + 2S_2Cl_2}

. The by-products are sulphur dioxide and disulphur dichloride, not chlorine.

Heating red phosphorus gives α-black, not β-black

α-black phosphorus comes from red phosphorus in a sealed tube at 803 K. β-black comes from white phosphorus at 473 K under high pressure.

Hypophosphorous is +1; hypophosphoric is +4

The '-ous' acid

\mathrm{H_3PO_2}

has phosphorus at +1 with two P–H bonds. The '-ic' acid

\mathrm{H_4P_2O_6}

has phosphorus at +4 with a P–P bond.

Pyrophosphoric acid has only one P–O–P bridge

\mathrm{H_4P_2O_7}

is two

\mathrm{H_3PO_4}

units joined by losing one water, so there is exactly one P–O–P. It is the cyclic trimer

\mathrm{(HPO_3)_3}

that has three.

H₃PO₃ is dibasic, not tribasic

One of its three hydrogens is bonded to phosphorus and never ionises. Phosphorous acid gives

\mathrm{NaH_2PO_3}

and

\mathrm{Na_2HPO_3}

, never

\mathrm{Na_3PO_3}

.

H₃PO₂ with NaOH gives NaH₂PO₂

Hypophosphorous acid has one P–OH, so it takes one NaOH and the salt keeps both P–H hydrogens:

\mathrm{NaH_2PO_2}

. Writing

\mathrm{NaH_2PO_3}

changes the phosphorus compound.

Complete hydrolysis of PCl₃ gives H₃PO₃

Every P–Cl becomes P–OH, but one of the three ends up as P–H after rearrangement to

\mathrm{HP(O)(OH)_2}

. The product has two ionisable hydrogens and one non-ionisable.

Group 16: Oxygen and Sulphur

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Redox reactions of sulphur compounds and the tests for sulphide and sulphite

Sulphur redox reactions

\mathrm{Cr_2O_7^{2-} + 3SO_2 + 2H^{+} \rightarrow 2Cr^{3+} + 3SO_4^{2-} + H_2O} \qquad \mathrm{S_8 + 12OH^{-} \rightarrow 4S^{2-} + 2S_2O_3^{2-} + 6H_2O} \qquad \mathrm{2S_2O_3^{2-} + I_2 \rightarrow S_4O_6^{2-} + 2I^{-}}

Group 16 trends: oxygen's anomalies, hydrides and oxides

Hydride	Melting point (K)	H–E bond enthalpy (kJ/mol)	H–E–H angle (°)	Acid strength (Ka)
$\mathrm{H_2O}$	273	463	104	$1.8 \times 10^{-16}$ Hydrogen bonding makes water melt highest, though it is the lightest.
$\mathrm{H_2S}$	188	347	92	$1.3 \times 10^{-7}$
$\mathrm{H_2Se}$	208	276	91	$1.3 \times 10^{-4}$
$\mathrm{H_2Te}$	222	238	90	$2.3 \times 10^{-3}$

The bond enthalpy falls down the group, so acid strength and reducing power rise: H₂Te is the strongest acid and strongest reducing agent of the four.

Oxoacids of sulphur: structures, S=O bonds and oxidation states

Acid	Formula	Oxidation state of S	S=O bonds	Link between units
Sulphurous	$\mathrm{H_2SO_3}$	+4	1	One unit; a lone pair on S
Sulphuric	$\mathrm{H_2SO_4}$	+6	2	One unit, two S–OH
Thiosulphuric	$\mathrm{H_2S_2O_3}$	Average +2; the two S differ	1	A terminal S doubly bonded to the central S, in place of one O
Dithionic	$\mathrm{H_2S_2O_6}$	+5, both S alike	4	A direct S–S bond
Pyrosulphuric (oleum)	$\mathrm{H_2S_2O_7}$	+6	4	One S–O–S bridge
Peroxodisulphuric (Marshall's)	$\mathrm{H_2S_2O_8}$	+6	4	One O–O peroxo bridge Still +6: the two peroxo oxygens are −1 each.
Polythionic	$\mathrm{H_2S_xO_6}$	Ends +5, chain 0	4	A chain of S atoms between two $\mathrm{SO_3H}$ groups

Each S=O bond carries one π bond, so counting S=O counts the π bonds of every acid here except thiosulphuric, which also has an S=S.

Common traps

Oxygen does not show only −2

Oxygen is −1 in peroxides, +1 in

\mathrm{O_2F_2}

and +2 in

\mathrm{OF_2}

. A statement that it shows only −2 is false.

Down group 16, +4 becomes MORE stable than +6

The inert pair effect makes the lower state more stable for the heavier elements. So the stability of +6 falls and that of +4 rises from S to Po.

Ozone has six lone pairs, not five

In

\mathrm{O_3}

the central oxygen has one lone pair and the two end oxygens have five between them (two on one, three on the other), six in all.

Rhombic sulphur is the room-temperature form

Monoclinic sulphur is stable only above 369 K. At room temperature the stable crystalline form is rhombic sulphur.

Pyrosulphuric acid has an S–O–S bridge, not a peroxo bond

\mathrm{H_2S_2O_7}

joins its two sulphurs through one oxygen. The O–O peroxo bond belongs to

\mathrm{H_2S_2O_8}

.

Marshall's acid needs concentrated sulphuric acid

Electrolysing dilute sulphuric acid or dilute sodium sulphate just splits water. Peroxodisulphuric acid forms at the anode only from a concentrated solution at high current density.

Lead acetate paper turns black from lead sulphide

Hydrogen sulphide gives black PbS. Lead sulphite is white, so a statement that the black colour is lead sulphite is false.

The green colour is chromium(III) sulphate, not Cr₂O₃

In acidified solution, dichromate is reduced to

\mathrm{Cr^{3+}}

, which stays in solution as

\mathrm{Cr_2(SO_4)_3}

.

\mathrm{Cr_2O_3}

is the green solid from heating ammonium dichromate.

Bromine takes thiosulphate further than iodine

Both oxidise thiosulphate. Iodine, the weaker oxidant, stops at

\mathrm{S_4O_6^{2-}}

; bromine, the stronger one, goes on to

\mathrm{SO_4^{2-}}

. Thiosulphate is oxidised in both cases, never reduced.

Groups 17 and 18: Halogens and Noble Gases

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Oxidising power and disproportionation of the halogens

Chlorine with alkali

\mathrm{Cl_2 + 2OH^- \xrightarrow{\text{cold, dilute}} Cl^- + ClO^- + H_2O} \qquad \mathrm{3Cl_2 + 6OH^- \xrightarrow{\text{hot, conc.}} 5Cl^- + ClO_3^- + 3H_2O}

Interhalogen shapes, halogen oxoacids and xenon fluorides

Lone pairs on the central atom

\mathrm{XX'_n}:\ \text{lone pairs on X} = \dfrac{7-n}{2},\ n = 1, 3, 5, 7 \qquad \mathrm{XeF_n}:\ \text{lone pairs on Xe} = \dfrac{8-n}{2}

Halogen properties: bond enthalpy, electron gain enthalpy and hydrogen halides

Halogen (hydride)	X–X bond enthalpy (kJ/mol)	Electron gain enthalpy (kJ/mol)	HX boiling point (K)	HX melting point (K)
F (HF)	158.8	$-333$	293	190 Weak F–F bond and a less negative electron gain enthalpy than Cl: both from fluorine's small size.
Cl (HCl)	242.6	$-349$	189	159
Br (HBr)	192.8	$-325$	206	185
I (HI)	151.1	$-296$	238	222

Chlorine leads in both bond enthalpy and electron gain enthalpy. HF boils highest; HI melts highest.

Common traps

F₂ does not have the highest bond enthalpy

Lone-pair repulsion between the two small fluorine atoms weakens the F–F bond. The order is

\mathrm{Cl_2 > Br_2 > F_2 > I_2}

, so chlorine is highest.

HF boils highest but does not melt highest

Hydrogen bonding lifts HF's boiling point above HI's. For melting points the larger dispersion forces in HI win, so HI melts highest:

\mathrm{HCl < HBr < HF < HI}

.

Chlorine, not fluorine, has the most negative electron gain enthalpy

The order of the magnitudes is

\mathrm{Cl > F > Br > I}

. A statement that it is F > Cl > Br > I is false.

Cold dilute alkali gives hypochlorite, not chlorate

Chlorine with cold, dilute alkali gives

\mathrm{Cl^-}

and

\mathrm{ClO^-}

in a 1 : 1 ratio. Chlorate,

\mathrm{ClO_3^-}

, forms only with hot, concentrated alkali.

A +7 oxoanion cannot disproportionate

Disproportionation needs an intermediate state, so the halogen can be both oxidised and reduced. In

\mathrm{ClO_4^-}

or

\mathrm{BrO_4^-}

the halogen is already at +7, its highest state.

FeI₃ does not exist

Iron(III) is a strong enough oxidant to turn iodide into iodine, so it cannot sit beside three iodides.

\mathrm{FeI_2}

is known, but

\mathrm{FeX_3}

exists only for F, Cl and Br.

XX′₅ is square pyramidal, not trigonal bipyramidal

Five bonds and one lone pair make six electron pairs. The pairs point to the corners of an octahedron, and with one corner held by the lone pair the atoms form a square pyramid.

An interhalogen is not a halate

\mathrm{BrF_5}

has bromine at +5, the same oxidation state as bromate,

\mathrm{BrO_3^-}

, but it is an interhalogen compound, not an oxoanion.

Noble gases have very low boiling points

Their atoms attract each other only by weak dispersion forces. That is why they liquefy only at very low temperatures, so any statement that they have high boiling points is false.

The p-Block Elements formulas

Group 13: Periodic Trends and the Inert Pair Effect

Group 13 trends in radius, ionisation enthalpy and electronegativity

Group 13 melting points, boron's lattice and gallium's liquid range

The inert pair effect in group 13: Tl⁺ is more stable than Tl³⁺

Common traps

Boron and Aluminium Compounds

Borax, the borax bead test and boric acid

Structure and preparation of diborane and borazine

Boron and aluminium halides as Lewis acids: back-bonding and maximum covalency

Common traps

Group 14: Carbon, Silicon, Tin and Lead

Group 14 trends, carbon's allotropes and silicones

Inert pair effect in tin and lead: which ions oxidise and which reduce

Tests for the lead ion in salt analysis

Common traps

Group 15: Periodic Trends and Hydrides

Group 15 trends and the anomalous behaviour of nitrogen

Hydrides of group 15 from NH₃ to BiH₃

Common traps

Nitrogen and Its Compounds

Preparing dinitrogen and nitric acid, and the nitrogen tests

Oxides of nitrogen: oxidation states, structures and nature

Common traps

Phosphorus and Its Oxoacids

Allotropes of phosphorus and reactions of white phosphorus and its chlorides

Basicity and reducing power of phosphorus oxoacids

Oxoacids of phosphorus: formulas, oxidation states and bonds

Common traps

Group 16: Oxygen and Sulphur

Redox reactions of sulphur compounds and the tests for sulphide and sulphite

Group 16 trends: oxygen's anomalies, hydrides and oxides

Oxoacids of sulphur: structures, S=O bonds and oxidation states

Common traps

Groups 17 and 18: Halogens and Noble Gases

Oxidising power and disproportionation of the halogens

Interhalogen shapes, halogen oxoacids and xenon fluorides

Halogen properties: bond enthalpy, electron gain enthalpy and hydrogen halides

Common traps

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