Compounds of Alkali Metals

Q: Why does sodium form a peroxide but potassium a superoxide?

The larger the oxygen anion, the larger the cation it needs for a stable lattice. The superoxide ion is the largest, so it is stabilised only by the large potassium, rubidium and caesium ions. Sodium is of medium size and holds the peroxide ion, while the small lithium ion keeps the simple oxide ion.

Q: Why is potassium superoxide used in breathing apparatus?

Potassium superoxide absorbs exhaled carbon dioxide and releases oxygen in the same reaction: four moles of KO2 take up two moles of carbon dioxide and give three moles of oxygen. Moisture in the breath helps the reaction. This makes it useful in submarines, spacecraft and mine rescue equipment.

Q: Why are alkali metal superoxides coloured and paramagnetic?

The superoxide ion has thirteen valence electrons, with three in the two antibonding pi star orbitals, so one electron is unpaired. The unpaired electron makes the compound paramagnetic, and the easy excitation of the pi star electrons absorbs visible light, giving yellow to orange colours. Peroxides have all electrons paired and are colourless.

Q: Why is lithium fluoride almost insoluble in water?

Both lithium and fluoride ions are very small, so they pack closely and the lattice enthalpy is very high, about 1036 kJ per mole. The hydration of the two ions cannot release enough energy to break this lattice, so only about 0.13 g of lithium fluoride dissolves in 100 g of water.

Q: Why does lithium carbonate decompose on heating while sodium carbonate does not?

The small lithium ion strongly polarises the large carbonate ion, weakening one carbon oxygen bond. Lithium oxide, made of two small ions, has a very high lattice enthalpy, so the decomposition to lithium oxide and carbon dioxide is favourable. The larger sodium ion barely polarises carbonate, so sodium carbonate only melts.

Q: Why are lithium halides soluble in organic solvents?

The lithium ion is so small that it distorts the electron cloud of a halide ion, giving the lithium halogen bond some covalent character. Covalent compounds dissolve in solvents such as ethanol, acetone and ethyl acetate, and lithium chloride dissolves even in pyridine. The effect is strongest for lithium iodide, the most covalent halide.

Q: What kind of questions on alkali metal compounds appear in JEE Advanced?

JEE Advanced favours matching and multiple correct questions: the main product of burning each metal, the products with water or carbon dioxide, the bond order and magnetism of peroxide and superoxide ions, and counting paramagnetic species in a list. Solubility and thermal stability orders with their lattice energy reasons are also common.

Q: Which reactions of Na2O2 and KO2 should I memorise for JEE Advanced?

For JEE Advanced learn five equations: sodium peroxide with water giving sodium hydroxide and hydrogen peroxide, with carbon dioxide giving sodium carbonate and oxygen, and with cold dilute sulphuric acid giving hydrogen peroxide; potassium superoxide with water giving hydroxide, hydrogen peroxide and oxygen, and with carbon dioxide giving carbonate and three moles of oxygen.

ChemistryThe s-Block ElementsFor JEE aspirants

The compounds of alkali metals are almost all ionic, and two ideas explain nearly all of their chemistry: how well the cation and anion match in size, and how strongly the tiny $L i^{+}$ ion polarises its neighbours. This page covers the oxides, peroxides and superoxides (with their molecular orbital picture), hydroxides, halides and salts of oxoacids of the alkali metals, with the stability and solubility trends and the lithium exceptions. JEE Advanced asks for the reaction products, the magnetic behaviour and the reasons behind each trend.

On this page1Three oxides2MO and magnetism3With water and

C O_{2}

4Hydroxides5Halides6Solubility7Oxo-salts8Heating rules9Solved examples

Key Formulas - Quick Reference

★ Must learnMain product in excess air: Li → $L i_{2} O$ , Na → $N a_{2} O_{2}$ , K/Rb/Cs → $M O_{2}$ (larger cation stabilises larger anion).
★ Must learnHydrolysis: $M_{2} O + H_{2} O \to 2MOH$ ; $M_{2} O_{2} + 2 H_{2} O \to 2MOH + H_{2} O_{2}$ ; $2M O_{2} + 2 H_{2} O \to 2MOH + H_{2} O_{2} + O_{2}$ .
★ Must learnBond order and magnetism: $O_{2}$ 2 (para), $O_{2}^{-}$ 1.5 (para, 1 unpaired e $^{-}$ ), $O_{2}^{2 -}$ 1 (dia).
$C O_{2}$ scrubbing: $4K O_{2} + 2C O_{2} \to 2 K_{2} C O_{3} + 3 O_{2}$ ; $2N a_{2} O_{2} + 2C O_{2} \to 2N a_{2} C O_{3} + O_{2}$ .
Basic strength and solubility of hydroxides: $LiOH < NaOH < KOH < RbOH < CsOH$ .
★ Must learnHalides: m.p. $MF > MCl > MBr > MI$ ; $Δ_{f} H^{\circ}$ of fluorides less negative down the group, of Cl, Br, I more negative.
Solubility: LiF low (high lattice enthalpy), CsI low (small hydration enthalpies); lithium halides dissolve in organic solvents.
★ Must learnHeat: $L i_{2} C O_{3} \to L i_{2} O + C O_{2}$ ; $4LiN O_{3} \to 2L i_{2} O + 4N O_{2} + O_{2}$ ; $2NaN O_{3} \to 2NaN O_{2} + O_{2}$ ; stability of $M_{2} C O_{3}$ and $MHC O_{3}$ rises down the group.

1. Nature of Alkali Metal Compounds

Alkali metals lose their single $n s^{1}$ electron so easily that nearly all their compounds are ionic, with the metal as $M^{+}$ . What changes from compound to compound is the anion, and the stability of any ionic solid depends on its lattice enthalpy, which is largest when cation and anion are small and similar in size. The only real exceptions come from lithium: $L i^{+}$ is so small that it distorts large anions and gives its compounds some covalent character.

Compound class	Formula	Key point	Page
Oxide / peroxide / superoxide	$M_{2} O$ , $M_{2} O_{2}$ , $M O_{2}$	type depends on cation size	this page
Hydroxide	MOH	strongest bases known	this page
Halide	MX	ionic, high melting; LiI most covalent	this page
Carbonate, hydrogencarbonate, nitrate	$M_{2} C O_{3}$ , $MHC O_{3}$ , $MN O_{3}$	thermally stable except Li	this page
Hydride, nitride	$M^{+} H^{-}$ , $L i_{3} N$ only	stability LiH > NaH > … > CsH	Alkali Metals

2. Oxides, Peroxides and Superoxides

2.1 What forms on burning

On combustion in excess air lithium forms mainly the oxide $L i_{2} O$ (with some $L i_{2} O_{2}$ ), sodium forms the peroxide $N a_{2} O_{2}$ (with some $Na O_{2}$ ), and potassium, rubidium and caesium form superoxides $M O_{2}$ . Under controlled conditions all three types can be made for every metal; the list above is what forms naturally.

Figure 1: Main combustion products in excess air. Minor products:

L i_{2} O_{2}

with Li,

Na O_{2}

with Na. In all three the metal is +1; only the oxygen species changes.

The reason is lattice enthalpy. A large anion ( $O_{2}^{-}$ ) packs well only around a large cation; next to a tiny $L i^{+}$ it would leave the anions touching each other. So the stability of peroxides and superoxides increases as the metal ion becomes larger. A small cation can also polarise a peroxide or superoxide ion until it breaks into the simple oxide ion, which is another reason lithium stops at $L i_{2} O$ .

Exam Trick Bigger cation, bigger oxygen. Read the oxygen species from the metal: Li holds

O^{2 -}

, Na holds

O_{2}^{2 -}

, K, Rb, Cs hold

O_{2}^{-}

. The metal is always +1, so the formula follows:

M_{2} O

M_{2} O_{2}

M O_{2}

2.2 Colour and magnetism

Oxides and peroxides are colourless when pure (commercial $N a_{2} O_{2}$ looks pale yellow because of a little $Na O_{2}$ ). Superoxides are yellow or orange and paramagnetic. Both properties come from the unpaired electron of $O_{2}^{-}$ .

Figure 2: Each added electron enters

π^{*} 2 p

. Superoxide has one unpaired electron (KO

_{2}

is paramagnetic and coloured); peroxide has none (Na

_{2}

_{2}

is diamagnetic and colourless).

JEE Advanced

Build the oxygen species from $O_{2}$ and read everything from $π^{*} 2 p$ . $O_{2}$ has 12 valence electrons: $(σ 2 s)^{2} (σ^{*} 2 s)^{2} (σ 2 p)^{2} (π 2 p)^{4} (π^{*} 2 p)^{2}$ . Each extra electron enters $π^{*} 2 p$ , which weakens and lengthens the O-O bond.

Species	Found in	$π^{*}$ electrons	Bond order	O-O / pm	Unpaired e $^{-}$
$O_{2}^{+}$	$O_{2}^{+} [Pt F_{6}]^{-}$	1	2.5	112	1
$O_{2}$	air	2	2	121	2
$O_{2}^{-}$	$K O_{2}$	3	1.5	128	1
$O_{2}^{2 -}$	$N a_{2} O_{2}$ , $Ba O_{2}$	4	1	149	0

The ozonide ion $O_{3}^{-}$ (in red $K O_{3}$ , made from $KOH$ and ozone) has 19 valence electrons, so it too is paramagnetic with one unpaired electron. Questions that list ' $K O_{2}$ , $K O_{3}$ , $N a_{2} O_{2}$ , $Ba O_{2}$ ' and ask how many are paramagnetic expect the answer 2.

2.3 Reaction with water

All three oxide types are easily hydrolysed and give a strong alkali. The peroxide also gives hydrogen peroxide, and the superoxide gives hydrogen peroxide and oxygen:

M_{2} O + H_{2} O 2 M^{+} + 2O H^{-}

M_{2} O_{2} + 2 H_{2} O 2 M^{+} + 2O H^{-} + H_{2} O_{2}

2M O_{2} + 2 H_{2} O 2 M^{+} + 2O H^{-} + H_{2} O_{2} + O_{2}

Figure 3: Count what comes off with water. Oxide: base only; peroxide: base +

H_{2} O_{2}

; superoxide: base +

H_{2} O_{2}

O_{2}

. With

C O_{2}

the superoxide gives out more

O_{2}

than it takes in

C O_{2}

Peroxide,

N a_{2} O_{2}

O_{2}^{2 -}

, bond order 1, diamagnetic, colourless. With water: NaOH +

H_{2} O_{2}

(cold, dilute acid gives

H_{2} O_{2}

cleanly). Strong oxidant and bleach.

Superoxide,

K O_{2}

O_{2}^{-}

, bond order 1.5, paramagnetic, yellow-orange. With water: KOH +

H_{2} O_{2}

O_{2}

. Supplies

O_{2}

while absorbing

C O_{2}

2.4 Reaction with carbon dioxide and uses

The oxides are basic, so they absorb $C O_{2}$ to give carbonates. Peroxides and superoxides release oxygen at the same time, which makes $K O_{2}$ ideal for breathing apparatus in submarines, spacecraft and mine-rescue sets:

N a_{2} O + C O_{2} N a_{2} C O_{3}

4K O_{2} + 2C O_{2} 2 K_{2} C O_{3} + 3 O_{2}

Figure 4:

K O_{2}

removes

C O_{2}

and supplies

O_{2}

at the same time (3 mol

O_{2}

per 2 mol

C O_{2}

), which is why it is used in submarines, space capsules and mine-rescue sets.

Sodium peroxide is widely used as an oxidising agent in inorganic chemistry (for example, fusing chromite ore to give chromate) and as a bleach for wood pulp, paper and fabrics, since with water or dilute acid it gives $H_{2} O_{2}$ :

N a_{2} O_{2} + H_{2} S O_{4} cold, dilute N a_{2} S O_{4} + H_{2} O_{2}

Key idea

Oxygen species grow with the cation:

O^{2 -}

(Li),

O_{2}^{2 -}

(Na),

O_{2}^{-}

(K to Cs). Only the superoxide has an unpaired electron, and only it gives

O_{2}

with water.

2.5 Predicting the oxide in one pass

Figure 5: Two questions fix the product: is it Li, is it Na? Everything heavier gives the superoxide. Then count the products with water.

Quick Recall: tap to check

Why is

K O_{2}

paramagnetic?

The superoxide ion

O_{2}^{-}

has one unpaired electron in a

π^{*} 2 p

orbital.

Which gas is given off when

K O_{2}

reacts with water?

Oxygen:

2K O_{2} + 2 H_{2} O \to 2KOH + H_{2} O_{2} + O_{2}

Which has the longer O-O bond,

N a_{2} O_{2}

K O_{2}

N a_{2} O_{2}

: peroxide bond order 1 (149 pm) against superoxide 1.5 (128 pm).

3. Hydroxides

The hydroxides are made by the reaction of the oxides (or the metals) with water. They are white crystalline solids and the strongest of all bases. They dissolve freely in water, with the evolution of much heat because the ions are intensely hydrated. In the solid, each hydroxide is an ionic lattice of $M^{+}$ and $O H^{-}$ ; the larger and less polarising the cation, the more completely the M-O bond is ionic, so basic strength and solubility increase from LiOH to CsOH.

Figure 6: All five hydroxides are strong bases, the strongest of all bases. Basic strength rises from LiOH to CsOH as the M-O bond becomes more ionic (bars are schematic).

Hydroxide solutions absorb $C O_{2}$ from air: $2NaOH + C O_{2} \to N a_{2} C O_{3} + H_{2} O$ . This is why NaOH pellets become coated with carbonate and KOH is used as a $C O_{2}$ absorbent.
LiOH is the least soluble and the weakest base of the group, and on strong heating it loses water: $2LiOH \to L i_{2} O + H_{2} O$ . The other alkali metal hydroxides melt and even boil without decomposing (another Li-Mg similarity: Mg(OH) $_{2}$ also gives the oxide).

4. Halides

The alkali metal halides MX (X = F, Cl, Br, I) are high-melting, colourless crystalline solids. They are made by treating the oxide, hydroxide or carbonate with the aqueous hydrohalic acid, for example $N a_{2} C O_{3} + 2HCl \to 2NaCl + H_{2} O + C O_{2}$ . All have large negative enthalpies of formation.

4.1 Enthalpy of formation

For the fluorides, $Δ_{f} H^{\circ}$ becomes less negative down the group; for the chlorides, bromides and iodides the reverse is true (overall). For any one metal, $Δ_{f} H^{\circ}$ always becomes less negative from fluoride to iodide.

Figure 7: For fluorides

Δ_{f} H^{\circ}

becomes less negative down the group (LiF

- 616

to CsF

- 554

kJ mol

^{- 1}

); for chlorides, bromides and iodides the reverse holds overall (LiI

- 270

to CsI

- 347

). For any one metal, F is always the most negative.

Why the fluorides run the other way:

Δ_{f} H^{\circ}

is dominated by the lattice enthalpy. From Li to Cs the cost of making the gaseous ion (sublimation + ionisation) falls by about 226 kJ mol

^{- 1}

(679 to 453). With the tiny

F^{-}

ion the lattice enthalpy falls even more, about 296 kJ mol

^{- 1}

(LiF 1036, CsF 740), so the fluorides become less stable. With the large

I^{-}

ion the lattice enthalpy falls only about 157 kJ mol

^{- 1}

(LiI 757, CsI 600), the cheaper ion wins, and CsI is the most stable iodide.

4.2 Melting and boiling points

Melting and boiling points always follow fluoride > chloride > bromide > iodide. A larger anion sits farther from the cation, so the lattice enthalpy and the melting point fall together.

Figure 8: Melting point follows lattice enthalpy: fluoride > chloride > bromide > iodide for every alkali metal, because a larger anion sits farther from the cation.

4.3 Solubility

All the alkali metal halides are soluble in water except that LiF is only sparingly soluble (about 0.13 g in 100 g water). Dissolving costs the lattice enthalpy and pays back the hydration enthalpies of both ions. LiF is insoluble because its lattice enthalpy is very high; CsI is the least soluble iodide because both of its large ions have small hydration enthalpies. Ions of very different size dissolve best.

Figure 9: LiF is insoluble because of its high lattice enthalpy; CsI is the least soluble iodide because both of its ions have small hydration enthalpies (NCERT). Mismatched pairs dissolve best.

Lithium halides other than LiF are also soluble in ethanol, acetone and ethyl acetate, and LiCl dissolves even in pyridine. The small $L i^{+}$ polarises the large halide ions, giving the bonds partial covalent character, and covalent compounds dissolve in organic solvents. The effect is greatest with the largest anion, so LiI is the most covalent lithium halide.

Exam Trick Matched sizes are hard to dissolve. Small with small (LiF) has too much lattice enthalpy; large with large (CsI) has too little hydration. Small with large (LiI, CsF) dissolves best. The same rule explains the Group 2 solubility trends.

Key idea

For halides, the anion size sets the melting point (F highest) and, with the cation size, the solubility. LiF and CsI are the two awkward members.

Quick Recall: tap to check

Which alkali metal fluoride has the most negative

Δ_{f} H^{\circ}

LiF (

- 616

kJ mol

^{- 1}

): fluorides become less negative down the group.

Why is LiCl soluble in acetone but NaCl is not?

L i^{+}

polarises

C l^{-}

, giving LiCl partial covalent character; NaCl is fully ionic.

Arrange NaF, NaCl, NaBr, NaI by melting point.

NaF (1269 K) > NaCl (1074 K) > NaBr (1020 K) > NaI (934 K).

5. Salts of Oxoacids

An oxoacid has its acidic proton on a hydroxyl group attached to an atom that also carries an oxo group: carbonic acid $H_{2} C O_{3}$ is $OC (OH)_{2}$ and sulphuric acid is $O_{2} S (OH)_{2}$ . The alkali metals form salts with all the oxoacids. These salts are generally soluble in water and thermally stable. The carbonates $M_{2} C O_{3}$ and, in most cases, the hydrogencarbonates $MHC O_{3}$ are highly stable to heat, and their stability increases down the group as the electropositive character grows.

5.1 The lithium exceptions

Lithium carbonate is not so stable to heat. The very small $L i^{+}$ polarises the large $C O_{3}^{2 -}$ ion, weakening a C-O bond, and the salt breaks down to the more stable $L i_{2} O$ (small ions, very high lattice enthalpy) and $C O_{2}$ . Lithium hydrogencarbonate does not exist as a solid at all.

Figure 10: A small, highly charged cation distorts a large anion.

L i^{+}

polarises

C O_{3}^{2 -}

enough to split off

C O_{2}

, so

L i_{2} C O_{3}

is the only alkali metal carbonate that decomposes easily.

L i_{2} C O_{3} heat L i_{2} O + C O_{2}

Nitrates show the same split. Lithium nitrate gives the oxide, while the other alkali metal nitrates give the nitrite and oxygen:

4LiN O_{3} heat 2L i_{2} O + 4N O_{2} + O_{2}

2NaN O_{3} heat 2NaN O_{2} + O_{2}

Figure 11: Heat tests that separate lithium from the rest. Lithium salts end as the oxide (like Mg); the others keep the metal as carbonate or nitrite.

Why a small cation makes a carbonate unstable, in energy terms. For $M_{2} C O_{3} (s) \to M_{2} O (s) + C O_{2} (g)$ , the step that changes most with the cation is lattice enthalpy. The oxide ion ( $r \approx 140$ pm) is much smaller than the carbonate ion, so the gain in lattice enthalpy on going from carbonate to oxide is large only when the cation is small too. For $L i^{+}$ the gain pays for breaking the C-O bond; for $N a^{+}$ to $C s^{+}$ it does not. The same argument, with $M^{2 +}$ , gives the Group 2 order $BeC O_{3} < MgC O_{3} < CaC O_{3} < SrC O_{3} < BaC O_{3}$ .

Key idea

Salts of large oxoanions are stable with large cations.

L i^{+}

polarises them, so lithium carbonate and nitrate end as

L i_{2} O

, just as the magnesium salts do.

Figure 12: Compounds of alkali metals on one page. Two ideas explain nearly every branch: size match between cation and anion, and the polarising power of

L i^{+}

Quick Recall: tap to check

Which is more stable to heat,

L i_{2} C O_{3}

C s_{2} C O_{3}

C s_{2} C O_{3}

: stability of carbonates increases with cation size.

What does

NaN O_{3}

give on heating?

2NaN O_{3} \to 2NaN O_{2} + O_{2}

(the nitrite, unlike lithium nitrate).

6. Solved Examples

Solved Example 1

Why is

K O_{2}

paramagnetic?

Solution:

$K O_{2}$ contains the superoxide ion $O_{2}^{-}$ . It has 13 valence electrons, with three in the two $π^{*} 2 p$ orbitals, so one electron is unpaired. Any species with an unpaired electron is paramagnetic.

Solved Example 2

Which compound gives both hydrogen peroxide and oxygen when added to water?
(A)

L i_{2} O

(B)

N a_{2} O_{2}

(C)

K O_{2}

(D)

N a_{2} O

Solution:

Answer: (C). $2K O_{2} + 2 H_{2} O \to 2KOH + H_{2} O_{2} + O_{2}$ . The oxides give only the hydroxide, and the peroxide gives the hydroxide and $H_{2} O_{2}$ but no $O_{2}$ .

Solved Example 3

Which statements about the superoxide ion are correct? (One or more options.)
(A) Its bond order is 1.5.
(B) It is diamagnetic.
(C) Its O-O bond is longer than in

O_{2}

.
(D) It is stabilised by large cations such as

K^{+}

Solution:

Answer: (A), (C), (D). $O_{2}^{-}$ has 3 electrons in $π^{*} 2 p$ : bond order $\frac{8 - 5}{2} = 1.5$ (8 bonding and 5 antibonding electrons, counting the 2s orbitals). One unpaired electron makes it paramagnetic, so (B) is wrong. Lower bond order means a longer bond (128 pm against 121 pm). Large cations give it a high lattice enthalpy.

Solved Example 4

A breathing set contains 1.00 kg of

K O_{2}

. How much

O_{2}

can it release, and how much

C O_{2}

can it absorb, by the reaction

4K O_{2} + 2C O_{2} \to 2 K_{2} C O_{3} + 3 O_{2}

? (K = 39.10, O = 16.00, C = 12.01)

Solution:

$M (K O_{2}) = 71.10$ g mol $^{- 1}$ , so $n = \frac{1000}{71.10} = 14.07$ mol.

$O_{2}$ : $14.07 \times \frac{3}{4} = 10.55$ mol $= 10.55 \times 32.00 =$ 338 g.

$C O_{2}$ : $14.07 \times \frac{2}{4} = 7.03$ mol $= 7.03 \times 44.01 =$ 309 g.

The canister gives out more mass of $O_{2}$ than the $C O_{2}$ it takes in.

Solved Example 5

Among

K O_{2}

N a_{2} O_{2}

L i_{2} O

K O_{3}

Ba O_{2}

and

Rb O_{2}

, how many are paramagnetic? (Integer answer.)

Solution:

Paramagnetic species need an unpaired electron: superoxides ( $O_{2}^{-}$ ) and ozonides ( $O_{3}^{-}$ ). Peroxides ( $O_{2}^{2 -}$ ) and oxides ( $O^{2 -}$ ) are diamagnetic.

$K O_{2}$ , $K O_{3}$ and $Rb O_{2}$ : Answer: 3.

Solved Example 6

LiF is almost insoluble in water, and CsI is less soluble than NaI. Give one reason for each.

Solution:

LiF: both ions are small, so the lattice enthalpy (1036 kJ mol $^{- 1}$ ) is too large for the hydration enthalpies to pay back.

CsI: both ions are large, so their hydration enthalpies are small; the lattice enthalpy is low too, but hydration falls more. NaI, with a small cation and large anion, dissolves more easily.

Solved Example 7

The correct order of thermal stability is
(A)

L i_{2} C O_{3} > N a_{2} C O_{3} > K_{2} C O_{3}

(B)

K_{2} C O_{3} > N a_{2} C O_{3} > L i_{2} C O_{3}

(C)

N a_{2} C O_{3} > L i_{2} C O_{3} > K_{2} C O_{3}

(D) all equal

Solution:

Answer: (B). A larger cation polarises the carbonate ion less, so stability increases down the group; $L i_{2} C O_{3}$ decomposes to $L i_{2} O$ and $C O_{2}$ .

Practice Questions

Find the oxidation state of sodium in $N a_{2} O_{2}$ . (NCERT 10.4)Answer: +1. The peroxide ion $O_{2}^{2 -}$ carries the charge $- 2$ ; each O is $- 1$ .
The hydroxides and carbonates of sodium and potassium are easily soluble in water while those of magnesium and calcium are sparingly soluble. Explain. (NCERT 10.20)Answer: Singly charged, large $N a^{+}$ and $K^{+}$ give low lattice enthalpies that hydration easily overcomes; $M g^{2 +}$ and $C a^{2 +}$ are doubly charged and smaller, so their lattice enthalpies are much higher.
What happens when (i) sodium is dropped in water, (ii) sodium is heated in a free supply of air, (iii) sodium peroxide dissolves in water? (NCERT 10.25)Answer: $2Na + 2 H_{2} O \to 2NaOH + H_{2}$ (the metal melts and darts about); $2Na + O_{2} \to N a_{2} O_{2}$ ; $N a_{2} O_{2} + 2 H_{2} O \to 2NaOH + H_{2} O_{2}$ .
Write balanced equations for the reactions of (a) $N a_{2} O_{2}$ with water, (b) $K O_{2}$ with water, (c) $N a_{2} O$ with $C O_{2}$ . (NCERT 10.28)Answer: (a) $N a_{2} O_{2} + 2 H_{2} O \to 2NaOH + H_{2} O_{2}$ ; (b) $2K O_{2} + 2 H_{2} O \to 2KOH + H_{2} O_{2} + O_{2}$ ; (c) $N a_{2} O + C O_{2} \to N a_{2} C O_{3}$ .
How many litres of $O_{2}$ at STP (22.7 L mol $^{- 1}$ ) are released when 1 mol of $K O_{2}$ reacts completely with $C O_{2}$ ?Answer: $\frac{3}{4}$ mol $O_{2}$ = 17.0 L.
Which of $L i_{2} O$ , $N a_{2} O_{2}$ and $K O_{2}$ is coloured and paramagnetic?Answer: $K O_{2}$ (one unpaired electron in $O_{2}^{-}$ ).
Why is LiI more covalent than LiF?Answer: The large $I^{-}$ ion is polarised much more easily by $L i^{+}$ than the small, hard $F^{-}$ ion.

Common Mistakes to Avoid

Watch out

Writing $Na O_{2}$ or $N a_{2} O$ as the main product of burning sodium. The main product is the peroxide $N a_{2} O_{2}$ .
Giving K an oxidation state of +4 in $K O_{2}$ or Na +2 in $N a_{2} O_{2}$ : the metal is always +1; the oxygen is $- \frac{1}{2}$ or $- 1$ .
Calling $N a_{2} O_{2}$ paramagnetic. Peroxide has all electrons paired; only superoxides and ozonides are paramagnetic.
Forgetting the $O_{2}$ in the hydrolysis of a superoxide, or adding $O_{2}$ to the hydrolysis of a peroxide.
Reversing the fluoride trend: $Δ_{f} H^{\circ}$ of MF becomes less negative from LiF to CsF.
Saying all alkali metal halides are very soluble: LiF is sparingly soluble and CsI is the least soluble iodide.
Writing $LiN O_{3}$ → $LiN O_{2}$ + $O_{2}$ . Lithium nitrate gives $L i_{2} O$ , $N O_{2}$ and $O_{2}$ .
Assuming $LiHC O_{3}$ can be isolated like $NaHC O_{3}$ . It exists only in solution.

Frequently Asked Questions

Why does sodium form a peroxide but potassium a superoxide?

The larger the oxygen anion, the larger the cation it needs for a stable lattice. The superoxide ion is the largest, so it is stabilised only by the large potassium, rubidium and caesium ions. Sodium is of medium size and holds the peroxide ion, while the small lithium ion keeps the simple oxide ion.

Why is potassium superoxide used in breathing apparatus?

Potassium superoxide absorbs exhaled carbon dioxide and releases oxygen in the same reaction: four moles of KO2 take up two moles of carbon dioxide and give three moles of oxygen. Moisture in the breath helps the reaction. This makes it useful in submarines, spacecraft and mine rescue equipment.

Why are alkali metal superoxides coloured and paramagnetic?

The superoxide ion has thirteen valence electrons, with three in the two antibonding pi star orbitals, so one electron is unpaired. The unpaired electron makes the compound paramagnetic, and the easy excitation of the pi star electrons absorbs visible light, giving yellow to orange colours. Peroxides have all electrons paired and are colourless.

Why is lithium fluoride almost insoluble in water?

Both lithium and fluoride ions are very small, so they pack closely and the lattice enthalpy is very high, about 1036 kJ per mole. The hydration of the two ions cannot release enough energy to break this lattice, so only about 0.13 g of lithium fluoride dissolves in 100 g of water.

Why does lithium carbonate decompose on heating while sodium carbonate does not?

The small lithium ion strongly polarises the large carbonate ion, weakening one carbon oxygen bond. Lithium oxide, made of two small ions, has a very high lattice enthalpy, so the decomposition to lithium oxide and carbon dioxide is favourable. The larger sodium ion barely polarises carbonate, so sodium carbonate only melts.

Why are lithium halides soluble in organic solvents?

The lithium ion is so small that it distorts the electron cloud of a halide ion, giving the lithium halogen bond some covalent character. Covalent compounds dissolve in solvents such as ethanol, acetone and ethyl acetate, and lithium chloride dissolves even in pyridine. The effect is strongest for lithium iodide, the most covalent halide.

What kind of questions on alkali metal compounds appear in JEE Advanced?

JEE Advanced favours matching and multiple correct questions: the main product of burning each metal, the products with water or carbon dioxide, the bond order and magnetism of peroxide and superoxide ions, and counting paramagnetic species in a list. Solubility and thermal stability orders with their lattice energy reasons are also common.

Which reactions of Na2O2 and KO2 should I memorise for JEE Advanced?

For JEE Advanced learn five equations: sodium peroxide with water giving sodium hydroxide and hydrogen peroxide, with carbon dioxide giving sodium carbonate and oxygen, and with cold dilute sulphuric acid giving hydrogen peroxide; potassium superoxide with water giving hydroxide, hydrogen peroxide and oxygen, and with carbon dioxide giving carbonate and three moles of oxygen.

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