The P Block Elements Class 11 Notes CBSE Chemistry Chapter 11 (Free PDF Download)

Section – A (1 Mark Questions)

1. How does metallic and non-metallic character vary in a group?

Ans. The non-metals and the metals exist only in the p-block of the periodic table. The non-metallic character of elements decreases down the group. In fact the heaviest element in each p-block group is the most metallic in nature.

2. Third-period elements can expand their covalence above four. Briefly explain.

Ans. The third-period elements of p-groups included d-orbital, which can be utilized to form bond and expand octet.

3. Give two examples of electron deficient molecules.

Ans. BF3, Al2Cl6 

4. Arrange the following halides of boron in the increasing order of acidic character: BF3, BCl3, BBr3, BI3.

Ans. BF3 < BCl3 < BBr3 < BI3 

Since in BF3 there is back -bonding due to which it is less acidic among all .

5. Why CCl4 behaves as an electron precise molecule?

Ans. Carbon in CCl4, the number of electrons around the central atom in a molecule is eight and thus is electron precise molecule.

6. Lead unaffected by water. Why?

Ans. Lead is unaffected by water, probable because of a protective oxide film formation.

7. Why Diamond is the hardest substance known?

Ans. Diamond is the hardest substance on the earth because it is very difficult to break extended covalent bonding.

8. What is water gas composed of ?

Ans. The mixture of CO and H2 is known as water gas or synthesis gas.

9. What is the maximum covalence shown by N?

Ans. Nitrogen shows a maximum covalence of +4 because only four orbitals, one S and three P-orbitals are available for bonding in Nitrogen.

10. Bi(V) is a stronger oxidizing agent than Bi(III). Why?

Ans. Bi is more stable in +3 oxidation state in comparison to +5 due to inert pair effect. Therefore, Bi(V) has a strong tendency to act as oxidizing agent.

Section – B (2 Marks Questions)

11. SiCl4 forms [SiCl6]2– while CCl4 does not form [CCl6]2– Explain.

Ans. Carbon does not have d-orbitals and hence CCl4 does not combine with Cl– ions to give [CCl6]2–. On the other hand, silicon has vacant 3d-orbitals and thus can expand its covalency from 4 to 6. Therefore SiCl4 combines with Cl ions to form [SiCl6]2–.

12. Why does carbon form covalent compounds whereas lead forms ionic compounds?

Ans. Carbon cannot lose electrons to form C4 because the sum of four ionization enthalpies is very high. It cannot gain four electrons to form C4 because energetically it is not favourable. Hence, C forms only covalent compounds. Down the group 14, ionization enthalpies decrease, Pb being the last element has so low I.E. that it can lose electrons to form ionic compounds.

13. What are halides of carbon? Give few examples.

Ans. Carbon combines with halogens to form both simple and mixed tetrahalides. In the case of simple halides, all the four expected tetrahalides (e.g. CF4, CCl4, CBr4 and CCl4) are known to exist. The stability of the simple tetrahalides decreases with the increasing atomic mass of the halogen.

(CF4 > CCl4 > CBr4 > Cl4)

Amongst the mixed halides the better-known compounds are (CFCl3, CF2Cl2 and CCl3Br).

14. Why are boron halides & diborane referred to as “Electron deficient compounds”?

Ans. Boron in its halides has only six electrons in its valence shell, therefore it is short by two electrons to complete its octet. As a result, a molecule of boron halide can accept a pair of electrons from any electron-rich compound that is why boron halides are called electron-deficient compounds. In diborane, the total no. of valence electrons is not sufficient to completely fill the available orbitals. This gives an electron-deficient character to diborane.

15. All the elements of group 13 except thallium show a +3 oxidation state while it, shows a +1 oxidation state. Give reasons.

Ans. The valence shells of group 13-elements have two electrons in s-subshell & 1-electron in p-subshell. Therefore, they are expected to show a +3-oxidation state. In thallium, the electrons of the s-subshell do not take part in bond formation due to the inert pair effect & only one electron of the p-orbital participates in bond formation, thus it shows +1 oxidation state only.

16. Carbon different from other member of the group. Why?

Ans. Carbon differs from rest of the members of its group due to its smaller size, higher electro negativity, higher ionization enthalpy and unavailability of d-orbitals.

17. Covalence of carbon not expand beyond four. Explain briefly.

Ans. In carbon, only s and p orbitals are available for bonding and therefore, it can accommodate only four pairs of electrons around it. This limit the maximum covalence to four whereas other members can expand their covalence due to the presence of d-orbitals.

18. Boron forms no compounds in a unipositive state but thallium in a unipositive state is quite unstable. Why?

Ans. Boron has electronic configuration 2s22p1 & therefore forms compounds in a trivalent state. However, thallium prefers to form compounds in the +1 oxidation state rather than in the +3 oxidation state as suggested by its group number. This is due to the inert pair effect. According to this effect, the 6s2 electrons in the case of heavy metals preferably do not take part in bonding.

19. Why Borazine is more reactive than benzene?

Ans. Both Borazine & Benzene are isoelectronic. In benzene C = C bonds are non-polar while N = B bonds in borazine are polar in nature due to the presence of a co-ordinate bond between N & B atoms. As a result, addition is quite frequent in borazine while it is less in benzene because of delocalisation of pi-electron charge.

20. Why does CO2 have a linear shape with no dipole moment?

Ans. In CO2 molecule carbon atom undergoes sp hybridization. Two sp hybridized orbital of carbon atom overlap with two p-orbital’s of oxygen atoms to make two sigma bonds while other two electrons of carbon atom are involved in $p\pi -p\pi$ bonding with oxygen atom. This results in its linear shape [with both C-O bond of equal length (115 pm)] with no dipole moment.

PDF Summary - Class 11 Chemistry The p-Block Elements Notes (Chapter 13)

Introduction:

The p-block is made up of elements in groups 13 through 18 of the periodic table. Metals, metalloids, and non-metals are all found in the p-block. The electrical configuration of the p-block components is that of a generic valence shell configuration of $\text{n}{\text{s}}^2\text{n}{\text{p}}^{1-6}$. Because of their tiny size, strong electronegativity, and lack of d-orbitals, the initial members of each group from $13-17$ of the p-block elements differ in many ways from the other members of their respective groups.

When compared to the other members of the same group, the first member of a group has a stronger capacity to establish $$p\pi -p\pi$$ multiple connections between itself (e.g. $\text{C}=\text{C},\text{C}\equiv$ $\text{C},\text{N}\equiv \text{N}$ ) and to elements in the second row$(\text{e}.\text{g C}=\text{O},\text{C}=\text{N},\text{C}\equiv \text{N},\text{N}=\text{O})$. The group number minus 10 is the maximum oxidation of a p-block element. Due to the inert pair effect, the oxidation state two less than the maximum group oxidation state becomes more stable in groups 13 to 16. (s-subshell electrons' unwillingness to engage in chemical bonding)

Trends in Properties of P-Block Elements: 

(A) Group 13 Elements: The Boron Family:

Boron is a non-metal, while aluminium is a metal with numerous chemical similarities to boron, while gallium, indium, and thallium are virtually entirely metallic.

1. Electronic Configuration:

The valence shell electronic configuration of elements of boron family is $\text{n}{\text{s}}^2\text{n}{\text{p}}^1$.

2. Atomic Radii: 

As one moves along the group, an additional shell of electrons is added to each succeeding member, resulting in a rise in atomic radius. $\text{Ga}$ has a smaller atomic radius than $\text{Al}$. Because of the higher nuclear charge in gallium, the presence of an extra 10 d-electrons only provides a weak screening effect for the outer electrons. As a result, Gallium’s atomic radius( $135\text{pm}$ ) is less than that of Aluminium( $143\text{pm}$ ).

$\text{B}<\text{Ga}<\text{Al}<\text{In}<\text{Tl}$$\text{B}<\text{Ga}<\text{Al}<\text{In}<\text{Tl}$

3. Ionization Enthalpy:

The ionisation enthalpy values do not drop smoothly along the group as predicted from the overall patterns. The drop from $\text{B}$ to $\text{Al}$ is related with increases in size. The inability of $\text{d and f }$ electrons, which have a low screening effect, to compensate for the increase in nuclear charge is responsible for the observed discontinuity in ionisation enthalpy values between $\text{Al}$ and $\text{Ga}$, as well as between $\text{In}$ and $\text{Tl}$. The sum of each element's first three ionisation enthalpies is extremely high.

$$\text{B}>\text{Al}>\text{Ga}>\text{In}>\text{Tl}$$$$\text{B}>\text{Al}>\text{Ga}>\text{In}>\text{Tl}$$

4. Electronegativity:

Electronegativity falls from $\text{B}$ to $\text{Al}$ and then increases somewhat as you move down the group. This is due to differences in atomic size across the elements.

$\text{B}>\text{Tl}>\text{In}>\text{Ga}>\text{Al}$$\text{Tl}>\text{In}>\text{Ga}>\text{Al}$

5. Physical Properties:

Boron is a non-metallic element. It's a solid black colour that's incredibly hard. It comes in a variety of allotropic forms. Boron has an extraordinarily high melting point due to its extremely strong crystalline lattice. Soft metals having a low melting point and good electrical conductivity make up the rest of the group. Gallium, which has a low melting point( $303\text{K}$ ), might be liquid during the summer. Because of its high boiling point $(2676\text{K})$, it may be used to measure high temperatures. From boron through thallium, the density of the elements rises.

6. Chemical Properties:

Oxidation State and Trends in Chemical Reactivity:

The sum of boron's first three ionisation enthalpies is quite high due to its tiny size. This prohibits it from producing $+3$ ions, forcing it to exclusively produce covalent compounds. However, as we progress from $\text{B}$ to $\text{Al}$, the sum of the first three ionisation enthalpies of $\text{Al}$ falls significantly, allowing $\text{Al}$ to form $\text{A}{\text{l}}^{+3}$ ions. Effective nuclear charge, on the other hand, binds ns electrons closely down the group due to inadequate shielding, limiting their participation in bonding. As a result, bonding may include solely the p-orbital electron. In reality, both $+1$ and $+3$ oxidation states have been reported in $\text{Ga, In and Tl}$. For heavier elements, the relative stability of the $+1$oxidation state increases: $\text{Al}<\text{Ga}<\ln <\text{TI}$. The $+1$ oxidation state is the most common in thallium, with the $+3$oxidation state being the most oxidising. Energy considerations predict that compounds in the $+1$ oxidation state are more ionic than those in the $+3$oxidation state. The number of electrons surrounding the central atom in a molecule of these elements' compounds (for example, boron in $\text{B}{\text{F}}_{\text{3}}$) will be just six in the trivalent state.

These electron-deficient molecules have a proclivity to take a pair of electrons in order to reach a stable electronic state, and so act as Lewis acids. The inclination to act as Lewis acid reduces as the size of the group grows smaller. To create $\text{BC}{\text{l}}_{\text{3}}\text{.N}{\text{H}}_{\text{3}}$, $\text{BC}{\text{l}}_{\text{3}}$simply obtains a lone pair of electrons from ammonia. Most covalent compounds are hydrolysed in water when they are in a trivalent state. The trichloride forms a tetrahedral species ${\left[\text{M}{(\text{OH})}_4\right]}^{-}$when it is hydrolysed in water; aluminium chloride forms an octahedral ion ${\left[\text{Al}{\left({\text{H}}_2\text{O}\right)}_6\right]}^{3+}$ in an acidified aqueous solution.

1. Reactivity Towards Air:

Boron, in its crystalline state, is non-reactive. On the surface of aluminium, a very thin oxide layer develops, protecting the metal from further assault. ${\text{B}}_2{\text{O}}_3$ and $\text{A}{\text{l}}_2{\text{O}}_3$ are formed when amorphous boron and aluminium metal are heated in the air. Nitrides are created when dinitrogen is heated to a high temperature.

$2\text{E}(\text{s})+3{\text{O}}_2(\text{g})\stackrel{\mathrm{\Delta }}{\to }2{\text{E}}_2{\text{O}}_3(\text{s});2\text{E}(\text{s})+{\text{N}}_2(\text{g})\stackrel{\mathrm{\Delta }}{\to }2\text{EN}(\text{s})$

The character of these oxides changes as you move through the group. Boron trioxide is an acidic compound that forms metal borates when it combines with basic (metallic) oxides. The oxides of aluminium and gallium are amphoteric, while those of indium and thallium are basic.

2. Reactivity Towards Acids and Alkalies: 

Even at moderate temperatures, boron does not react with acids and alkalies; nevertheless, aluminium dissolves in mineral acids and aqueous alkalies, giving it an amphoteric property. Dihydrogen is generated when aluminium is dissolved in dilute $\text{HCl}$. Concentrated nitric acid, on the other hand, makes aluminium inactive by creating a protective oxide coating on the surface. When aluminium interacts with aqueous alkali, it produces dihydrogen.

$\text{2Al(s)+6HCl(aq)}\stackrel{}{\to }\text{2A}{\text{l}}^{\text{3 + }}\text{(aq) + 6C}{\text{l}}^{\text{ - }}\text{ + 3}{\text{H}}_{\text{2}}$

$\text{2Al(s) + 2NaOH(aq) + 6}{\text{H}}_{\text{2}}\text{O(l) }\stackrel{}{\to }\text{2N}{\text{a}}^{\text{ + }}{\left[\text{Al(OH}{\text{)}}_{\text{4}}\right]}^{\text{ - }}(\text{aq})\text{+}\text{3}{\text{H}}_{\text{2}}\text{(g)}$

$\text{sodium tetrahydroxoaluminate(III)}$

3. Reactivity Towards Halogens: 

These elements react with halogen to form trihalides (except $\text{Tl}{\text{I}}_3$ ). $2\text{E}(\text{s})+3{\text{X}}_2(\text{g})\to 2\text{E}{\text{X}}_3(\text{s})(\text{X}=\text{F},\text{ClBr},\text{l})$

Important Trends and Anomalous Properties of Boron: 

All of these elements' tri-chlorides, bromides, and iodides, which are covalent in nature, are hydrolysed in water. Except for boron, species like tetrahedral ${\left[\text{M}{(\text{OH})}_4\right]}^{-}$ and octahedral ${\left[\text{M}{\left({\text{H}}_2\text{O}\right)}_6\right]}^{3+}$ occur in aqueous medium. The highest covalence of boron is 4 owing to the absence of d orbitals. The highest covalence may be predicted beyond 4 since d-orbitals are accessible with Al and other elements.

Boron (B):

Occurrence: 

Boron may be found in the following minerals in nature:

1. Borax ${\left(\text{N}{\text{a}}^{+}\right)}_2{\text{B}}_4{\text{O}}_7^{2-}.10{\text{H}}_2\text{O}.$ (Boron is part of an anionic complex), 

2. Boric acid ${\text{H}}_3\text{B}{\text{O}}_3$,

3. Kernite $\text{N}{\text{a}}_2{\text{B}}_4{\text{O}}_7\cdot 4{\text{H}}_2\text{O}$ 

4. Colemanite $\text{C}{\text{a}}_2{\text{B}}_6{\text{O}}_{11}\cdot 5{\text{H}}_2\text{O}$

Extraction of Boron:

1. In the absence of oxygen, by reducing ${\text{B}}_2{\text{O}}_3$ with magnesium, sodium, or potassium:

$\text{N}{\text{a}}_2{\text{B}}_4{\text{O}}_7+2\text{HCl}+5{\text{H}}_2\text{O}\to 4{\text{H}}_3\text{B}{\text{O}}_3+2\text{NaCl}$

$2{\text{H}}_3\text{B}{\text{O}}_3\stackrel{\mathrm{\Delta }}{\to }{\text{B}}_2{\text{O}}_3+3{\text{H}}_2\text{O };{\text{B}}_2{\text{O}}_3+3\text{Mg}\stackrel{}{\to }2\text{B}+3\text{MgO}$

The product is heated with $\text{HCl}$ and filtered after ${\text{K}}_2\text{O}$ or $\text{MgO}$ dissolves, leaving elemental boron behind. Before being dried, it is thoroughly washed to remove the $\text{HCl}$This technique produces a dark amorphous boron powder $\text{B}$.

2. By heating potassium fluoroborate$\left(\text{KB}{\text{F}}_4\right)$ with potassium metal, you can make boron.

$\text{KB}{\text{F}}_4+3\text{K}\stackrel{}{\to }4\text{KF}+\text{B}$

It is then treated with dilute $\text{HCl}$ to remove $\text{KF}$ and $\text{B}$ is then washed and dried.

$\text{B}$ is then washed and dried after being treated with dilute$\text{HCl}$ to eliminate $\text{KF}$.

3. In tiny amounts in pure form (crystalline boron) as a result of 

1. Reduction of $\text{BB}{\text{r}}_3$with ${\text{H}}_2$on a heated titanium metal filament at $1275-1475\text{K}$. The vapours of $\text{B}{\text{r}}_2$are absorbed in$\text{Cu}$, and the vapours of boron that remain are condensed.

2. Decomposition of $\text{B}{\text{I}}_3$vapours using tungsten electrodes and a high-tension arc            ($80\text{kV}$ ). 

$2\text{B}{\text{I}}_3\stackrel{}{\to }2\text{B}\uparrow +3{\text{I}}_2\uparrow$ (VanArkel method).

Properties:

1. It comes in five different varieties, four of which are crystalline and one of which is amorphous. All crystalline forms are made up of clusters of  ${\text{B}}_{12}$units and are extremely hard. All crystalline forms are chemically inert and appear black. Melting points are in the range of ${2300}^{\circ }\text{C}$. The amorphous form, on the other hand, is brown and chemically active.

2. Reaction With Air:

$$2\text{E}(\text{s})+3{\text{O}}_2(\text{g})\to 2{\text{E}}_2{\text{O}}_3(\text{s});2\text{E}(\text{s})+{\text{N}}_2(\text{g})\to 2\text{EN}(\text{s})$$

3. Action of Alkalies and Acids:

$2\text{B}+2\text{NaOH}+2{\text{H}}_2\text{O}\stackrel{}{\to }2\text{NaB}{\text{O}}_2+3{\text{H}}_2$

$\text{2 B + 3}{\text{H}}_{\text{2}}\text{S}{\text{O}}_{\text{4}}\stackrel{\text{oxidation}}{\to }\text{2}{\text{H}}_{\text{3}}\text{B}{\text{O}}_{\text{3}}\text{ + 3S}{\text{O}}_{\text{2}}$

$\text{2B + 6HN}{\text{O}}_{\text{3}}\stackrel{\text{oxidation}}{\to }\text{2}{\text{H}}_{\text{3}}\text{B}{\text{O}}_{\text{3}}\text{ + 6N}{\text{O}}_{\text{2}}$

4. Reaction with $\text{Mg}$ and $\text{Ca}$ :

$3\text{Mg}+2\text{B}\stackrel{}{\to }\text{M}{\text{g}}_3{\text{B}}_2$

$\text{3Ca + 2B}\stackrel{}{\to }\text{C}{\text{a}}_{\text{3}}{\text{B}}_{\text{2}}$

$\text{M}{\text{g}}_3{\text{B}}_2$ on repetitive hydrolysis gives diborane. 

$\text{M}{\text{g}}_{\text{3}}{\text{B}}_{\text{2}}\text{ + 6HCl}\stackrel{\text{hydrolysis}}{\to }\text{ 3MgC}{\text{l}}_{\text{2}}\text{ + }{\text{B}}_{\text{2}}{\text{H}}_{\text{6}}\text{ ; }{\text{B}}_{\text{2}}{\text{H}}_{\text{6}}\text{ + 6}{\text{H}}_{\text{2}}\text{O}\stackrel{}{\to }\text{2}{\text{H}}_{\text{3}}\text{B}{\text{O}}_{\text{3}}\text{ + 6}{\text{H}}_{\text{2}}$ 

5. Reducing Properties:

$\text{3Si}{\text{O}}_{\text{2}}\text{ + 4 B}\stackrel{}{\to }2{\text{B}}_{\text{2}}{\text{O}}_{\text{3}}\text{ + 3Si}$

$\text{3C}{\text{O}}_{\text{2}}\text{ + 4 B}\stackrel{}{\to }2{\text{B}}_{\text{2}}{\text{O}}_{\text{3}}\text{ + 3C}$

It decays in steam liberating hydrogen gas. 

$\text{2B + 3}{\text{H}}_{\text{2}}\text{O(steam)}\stackrel{}{\to }{\text{B}}_{\text{2}}{\text{O}}_{\text{3}}\text{ + 3}{\text{H}}_{\text{2}}$

Uses:

Boron is utilised in the manufacture of high-impact steel and, since it absorbs neutrons, in reactor rods for atomic reaction control.

Compounds Of Boron:

• Boron Trioxide $\left({\text{B}}_2{\text{O}}_3\right)$ :

Preparation:

${\text{H}}_{\text{3}}\text{B}{\text{O}}_{\text{3}}\stackrel{}{\stackrel{\text{10}{\text{0}}^{\circ }\text{C}}{\to }}\text{ HB}{\text{O}}_{\text{2}}\stackrel{}{\stackrel{\text{16}{\text{0}}^{\text{o }}\text{C}}{\to }}{\text{H}}_{\text{2}}{\text{B}}_{\text{4}}{\text{O}}_{\text{7}}\stackrel{\text{red heat}}{\to }{\text{B}}_{\text{2}}{\text{O}}_{\text{3}}$

$\text{boric acid}$

Properties:

1. It is a slightly acidic oxide that forms borates when it interacts with alkalies or bases.$3\text{N}{\text{a}}_2\text{O}+{\text{B}}_2{\text{O}}_3\to 2\text{N}{\text{a}}_3\text{B}{\text{O}}_3$ (sodium orthoborate).

2. It forms orthoboric acid after a gradual reaction with water. It generates coloured compounds when heated with transition metal salts.${\text{H}}_2\text{O}+{\text{B}}_2{\text{O}}_3\to 2\text{HB}{\text{O}}_2;\text{HB}{\text{O}}_2+{\text{H}}_2\text{O}\to {\text{H}}_3\text{B}{\text{O}}_3$

${\text{B}}_2{\text{O}}_3+{\text{P}}_2{\text{O}}_5\rightleftharpoons 2\text{BP}{\text{O}}_4$

Orthoboric Acid $H_{3}B{O}_3$ :

Among the oxyacids of boron are

Preparation:

1. A concentrated solution of borax is treated with sulphuric acid to precipitate it.$\text{N}{\text{a}}_2{\text{B}}_4{\text{O}}_7+{\text{H}}_2\text{S}{\text{O}}_4+5{\text{H}}_2\text{O}\to \text{N}{\text{a}}_2\text{S}{\text{O}}_4+4{\text{H}}_3\text{B}{\text{O}}_3\downarrow$

2. ${\text{H}}_3\text{B}{\text{O}}_3$ is made by suspending powdered colemanite in water and filtering the surplus $\text{S}{\text{O}}_2$. White crystals of ${\text{H}}_3\text{B}{\text{O}}_3$are produced after filtering and chilling the filtrate.

$\text{C}{\text{a}}_2{\text{B}}_6{\text{O}}_{11}+2\text{S}{\text{O}}_2+11{\text{H}}_2\text{O}\stackrel{}{\to }2\text{Ca}{\left(\text{HS}{\text{O}}_3\right)}_2+6{\text{H}}_3\text{B}{\text{O}}_3$

Properties:

1. It is a weak monobasic acid, and the boron atom completes its octet in aqueous solution by eliminating $\text{O}{\text{H}}^{-}$ from water molecules:

$\text{B}(\text{OH}{)}_3(\text{aq})+2{\text{H}}_2\text{O}(\ell )\to \text{B}(\text{OH}{)}_4^{-}(\text{aq})+{\text{H}}_3{\text{O}}^{+}(\text{aq})$

As a result, it acts as a Lewis acid rather than a proton donor. When a polyhydroxy molecule like glycol or glycerol is added to its aqueous solution, it behaves as a strong acid. The great stability of the conjugate bone chelate complex accounts for the acidity.

Catechol and salicylic acids create comparable complexes, but ethanol does not.

When heated, it produces metaboric acid $\left(\text{HB}{\text{O}}_2\right)$ first, followed by boron trioxide.

${\text{H}}_{\text{3}}\text{B}{\text{O}}_{\text{3}}\stackrel{}{\stackrel{\text{10}{\text{0}}^{\circ }\text{C}}{\to }}\text{ HB}{\text{O}}_{\text{2}}\stackrel{}{\stackrel{\text{16}{\text{0}}^{\circ }\text{C}}{\to }}{\text{H}}_{\text{2}}{\text{B}}_{\text{4}}{\text{O}}_{\text{7}}\stackrel{\text{redheat}}{\to }{\text{B}}_{\text{2}}{\text{O}}_{\text{3}}$

$\text{boric acid}$

Orthoboric acid is oily to the touch and is less soluble in cold water than hot water. It has a multilayer structure with hydrogen bonds connecting planar units of $\text{B}{\text{O}}_3$.

Test For Borate Radical:

The evolved gas is burnt when boric acid is heated with ethyl alcohol, creating a green edged flame.

${\text{H}}_{\text{3}}\text{B}{\text{O}}_{\text{3}}\text{ + 3}{\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}\stackrel{}{\to }\text{B}{\left(\text{O}{\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\right)}_{\text{3}}\text{ + 3}{\text{H}}_{\text{2}}\text{O}$

$\text{ ethyl borate }\left(\text{volatile}\right)$

Uses:

It's an antiseptic, and the water solution may be used to cleanse your eyes. In addition, it is utilised in the glass, enamel, and pottery industries.

• Borax $\left(\mathbf{N}{\mathbf{a}}_{\mathbf{2}}{\text{B}}_4{\text{O}}_7\cdot 10{\text{H}}_2\text{O}\right):$

Preparation: 

Borax can be found in nature, but it can also be produced using the methods listed below.

1. From Colemanite mineral. When colemanite powder is heated in a solution of $\text{N}{\text{a}}_2\text{C}{\text{O}}_3$, the precipitation of $\text{CaC}{\text{O}}_3$ takes place.

When white crystals of borax precipitate, the filterate is cooled. On treatment with $\text{C}{\text{O}}_2$, the mother liquor changes $\text{NaB}{\text{O}}_2$ to $\text{N}{\text{a}}_2{\text{B}}_4{\text{O}}_7$, which precipitates out after crystallisation.

$\text{4NaB}{\text{O}}_{\text{2}}\text{ + C}{\text{O}}_{\text{2}}\stackrel{}{\to }\text{N}{\text{a}}_{\text{2}}\text{;}{\text{B}}_{\text{4}}{\text{O}}_{\text{7}}\text{ + N}{\text{a}}_{\text{2}}\text{C}{\text{O}}_{\text{3}}$

2. From orthoboric acid. The action of $$\text{N}{\text{a}}_{\text{2}}\text{C}{\text{O}}_{\text{3}}$$on orthoboric acid  produces borax.

$\text{4}{\text{H}}_{\text{3}}\text{B}{\text{O}}_{\text{3}}\text{ + N}{\text{a}}_{\text{2}}\text{C}{\text{O}}_{\text{3}}\stackrel{}{\to }\text{N}{\text{a}}_{\text{2}}{\text{B}}_{\text{4}}{\text{O}}_{\text{7}}\text{ + 6}{\text{H}}_{\text{2}}\text{O + C}{\text{O}}_{\text{2}}\uparrow$

Properties:

1. Borax is a white powder that is more soluble in hot water than cold water.

2. Because of its hydrolysis to weak acid ${\text{H}}_{\text{3}}\text{B}{\text{O}}_{\text{3}}$and strong alkali $\text{NaOH}$, its aqueous solution is alkaline.

$\text{N}{\text{a}}_{\text{2}}{\text{B}}_{\text{4}}{\text{O}}_{\text{7}}\text{ + 7}{\text{H}}_{\text{2}}\text{O}\stackrel{}{\to }\text{4}{\text{H}}_{\text{3}}\text{B}{\text{O}}_{\text{3}}\text{ + 2NaOH}$

3. When borax powder is heated, it expands at first owing to water loss in the form of steam, but then it transforms into a colourless clear borax bead at ${740}^{\circ }\text{C}$.

$\text{N}{\text{a}}_{\text{2}}{\text{B}}_{\text{4}}{\text{O}}_{\text{7}}\times \text{10}{\text{H}}_{\text{2}}\text{O}\stackrel{\mathrm{\Delta }}{\to }\text{N}{\text{a}}_{\text{2}}{\text{B}}_{\text{4}}{\text{O}}_{\text{7}}\text{ + 10}{\text{H}}_{\text{2}}\text{O}\uparrow$ $\text{N}{\text{a}}_{\text{2}}{\text{B}}_{\text{4}}{\text{O}}_{\text{7}}\stackrel{\text{74}{\text{0}}^{\text{o}}\text{C}}{\to }\text{2NaB}{\text{O}}_{\text{2}}\text{ + }{\text{B}}_{\text{2}}{\text{O}}_{\text{3}}\text{ (borax bead) }$

4. Action of Acids:

$\text{N}{\text{a}}_2{\text{B}}_4{\text{O}}_7+2\text{HCl}+5{\text{H}}_2\text{O}\stackrel{}{\to }2\text{NaCl}+4{\text{H}}_3\text{B}{\text{O}}_3\text{ (boric acid) }$

The white flakes of boric acid are produced by heating.$$\text{N}{\text{a}}_{\text{2}}{\text{B}}_{\text{4}}{\text{O}}_{\text{7}}\stackrel{\text{NaOH}}{\to }\text{ NaB}{\text{O}}_{\text{2}}\stackrel{{\text{H}}_{\text{2}}{\text{O}}_{\text{2}}}{\to }\text{ N}{\text{a}}_{\text{2}}\left[{\text{(OH)}}_{\text{2}}\text{ B(O - O}{\text{)}}_{\text{2}}\text{ B(OH}{\text{)}}_{\text{2}}\right]\text{ 6}{\text{H}}_{\text{2}}\text{O}$$

Correct formula of borax is $\text{N}{\text{a}}_{\text{2}}\left[{\text{B}}_{\text{4}}{\text{O}}_{\text{5}}{\text{(OH)}}_{\text{4}}\right]\text{.8}{\text{H}}_{\text{2}}\text{O}$

Bead Test: Boric anhydride reacts with certain metal salts such as, $\text{N}{\text{i}}^{\text{2 + }}\text{, C}{\text{u}}^{\text{2 + }}\text{, C}{\text{o}}^{\text{2 + }}\text{, M}{\text{n}}^{\text{2 + }}\text{, C}{\text{r}}^{\text{3 + }}$

To form coloured metaborates. The colour of the metaborates can be used to identify the metallic ions (cations) in salts.

$\text{N}{\text{a}}_{\text{2}}{\text{B}}_{\text{4}}{\text{O}}_{\text{7}}\text{.10}{\text{H}}_{\text{2}}\text{O}\underset{}{\stackrel{}{\underset{\text{ - 10}{\text{H}}_{\text{2}}\text{O}}{\overset{\mathrm{\Delta }}{\to }}}}\text{N}{\text{a}}_{\text{2}}{\text{B}}_{\text{4}}{\text{O}}_{\text{7}}\stackrel{\text{74}{\text{0}}^{\text{o}}\text{C}}{\to }\underset{\text{glassy mass }}{\underbrace{\text{NaB}{\text{O}}_{\text{2}}\text{ + }{\text{B}}_{\text{2}}{\text{O}}_{\text{3}}}}\text{ ; CuO + }{\text{B}}_{\text{2}}{\text{O}}_{\text{3}}\stackrel{}{\to }\text{Cu}{\left(\text{B}{\text{O}}_{\text{2}}\right)}_{\text{2}}\text{(blue bead)}$ 

Uses: It is utilised in borax bead test, in gold purification, as flux during welding of metals and in glass production process.

• Diborane $(B_2{H}_6)$: Binary compounds of $\text{B}$ with $\text{H}$ are called boron hydrides or boranes. These compounds form following two types of series:

${\text{B}}_{\text{n}}{\text{H}}_{\text{n}+4}-{\text{B}}_2{\text{H}}_6,{\text{B}}_5{\text{H}}_9,{\text{B}}_6{\text{H}}_{10},{\text{B}}_{10}{\text{H}}_{14}$

${\text{B}}_{\text{n}}{\text{H}}_{\text{n}+6}-{\text{B}}_4{\text{H}}_{10},{\text{B}}_5{\text{H}}_{11},{\text{B}}_6{\text{H}}_{12},{\text{B}}_9{\text{H}}_{15}$

The chemistry of diborane has piqued attention due to its use in a variety of synthetic processes, as well as the fact that the structure's elucidation aided in the clarification of fundamental ideas regarding the structure of electron deficient compounds.

Preparation:

1. $\text{4B}{\text{F}}_{\text{3}}\text{ + 3LiAl}{\text{H}}_{\text{4}}\stackrel{\text{ether}}{\to }\text{2}{\text{B}}_{\text{2}}{\text{H}}_{\text{6}}\text{ + 3LiF + 3Al}{\text{F}}_{\text{3}}$

2. $\text{2BC}{\text{l}}_{\text{3}}\text{ + 6}{\text{H}}_{\text{2}}\text{(excess)}\stackrel{\text{silent electric discharge}}{\to }{\text{B}}_{\text{2}}{\text{H}}_{\text{6}}\text{ + 6HCl}$ 

3. $\text{8B}{\text{F}}_{\text{3}}\text{ + 6LiH}\stackrel{\text{ether}}{\to }{\text{B}}_{\text{2}}{\text{H}}_{\text{6}}\text{ + 6LiB}{\text{F}}_{\text{4}}$ 

4. $\text{2NaB}{\text{H}}_{\text{4}}\text{ + }{\text{I}}_{\text{2}}\stackrel{\text{ether}}{\to }{\text{B}}_{\text{2}}{\text{H}}_{\text{6}}\text{ + 2NaI + }{\text{H}}_{\text{2}}$

5. $\text{3NaB}{\text{H}}_{\text{4}}\text{ + 4B}{\text{F}}_{\text{3}}\underset{\text{450K}}{\overset{\text{ether}}{\to }}\text{3NaB}{\text{F}}_{\text{4}}\text{ + 2}{\text{B}}_{\text{2}}{\text{H}}_{\text{6}}$

6. It can also be obtained by addition of $\text{NaB}{\text{H}}_4$ to concentrated ${\text{H}}_2\text{S}{\text{O}}_4$ or ${\text{H}}_3\text{P}{\text{O}}_4$. $\text{2NaB}{\text{H}}_{\text{4}}\text{ + }{\text{H}}_{\text{2}}\text{S}{\text{O}}_{\text{4}}\stackrel{}{\to }{\text{B}}_{\text{2}}{\text{H}}_{\text{6}}\text{ + 2}{\text{H}}_{\text{2}}\text{ + N}{\text{a}}_{\text{2}}\text{S}{\text{O}}_{\text{4}}$

$2\text{NaB}{\text{H}}_4+2{\text{H}}_3\text{P}{\text{O}}_4\stackrel{}{\to }{\text{B}}_2{\text{H}}_6+2{\text{H}}_2+2\text{Na}{\text{H}}_2\text{P}{\text{O}}_4$

7. $\text{2B}{\text{F}}_{\text{3}}\text{ + 6NaH}\stackrel{\text{450K}}{\to }{\text{B}}_{\text{2}}{\text{H}}_{\text{6}}\text{ + 6NaF}$ (Industrial method)

Properties:

1. Diborane is a colourless gas having boiling point $183\text{K}$ which is briskly decomposed by water with the formation of ${\text{H}}_{\text{3}}\text{B}{\text{O}}_{\text{3}}\mathrm{\&}{\text{H}}_{\text{2}}$ :

${\text{B}}_2{\text{H}}_6+6{\text{H}}_2\text{O}\stackrel{}{\to }2{\text{H}}_3\text{B}{\text{O}}_3+6{\text{H}}_2$

2. Diborane mixtures with air or oxygen spontaneously combust, releasing a substantial quantity of heat. Most other fuels have a lower heat of combustion per unit weight of fuel than diborane. As a result, it is utilised as a rocket fuel. ${\text{B}}_2{\text{H}}_6+3{\text{O}}_2\stackrel{}{\to }{\text{B}}_2{\text{O}}_3+3{\text{H}}_2\text{O }\mathrm{\Delta }\text{H}=-1976\text{kJ}\text{mo}{\text{l}}^{-1}$

3. Pyrolysis of diborane in sealed containers at temperatures above $375\text{K}$ is a complicated process that results in a mixture of boranes. eg, ${\text{B}}_4{\text{H}}_{10},{\text{B}}_5{\text{H}}_9,{\text{B}}_6{\text{H}}_{12}$, and ${\text{B}}_{10}{\text{H}}_{14}$ .

4. Organoboranes are formed when diborane reacts easily with alkenes and alkynes in ether solvents at normal temperature. The hydroboration reaction is the name given to this process.

${\text{B}}_2{\text{H}}_6+\text{HCl}\stackrel{}{\to }{\text{B}}_2{\text{H}}_5\text{Cl}+{\text{H}}_2$

${\text{B}}_2{\text{H}}_6+6\text{MeOH}\stackrel{}{\to }2\text{B}(\text{OMe}{)}_3+6{\text{H}}_2$

5. Cleavage reactions

1. ${\text{B}}_2{\text{H}}_6+2\text{M}{\text{e}}_3\text{N}\stackrel{}{\to }2\text{M}{\text{e}}_3\text{NB}{\text{H}}_3$

2. ${\text{B}}_2{\text{H}}_6+2\text{M}{\text{e}}_3\text{P}\stackrel{}{\to }2\text{M}{\text{e}}_3\text{PB}{\text{H}}_3$

3. ${\text{B}}_{\text{2}}{\text{H}}_{\text{6}}\text{ + 2CO}\stackrel{\text{20}{\text{0}}^{\text{o}}\text{C, 20atm}}{\to }\text{2B}{\text{H}}_{\text{3}}\text{CO}$ (borane carbonyl)

4. ${\text{B}}_2{\text{H}}_6+2\text{E}{\text{t}}_2\text{S}\stackrel{}{\to }2\text{E}{\text{t}}_2\text{SB}{\text{H}}_3$

5. $3{\text{B}}_2{\text{H}}_6+6\text{N}{\text{H}}_3\stackrel{\text{low temprature}}{\to }{\text{B}}_{\text{2}}{\text{H}}_{\text{6}}\text{.2N}{\text{H}}_{\text{3}}$  or ${\left[\text{B}{\text{H}}_{\text{2}}{\left(\text{N}{\text{H}}_{\text{3}}\right)}_{\text{2}}\right]}^{\text{ + }}{\text{BH}}_{\text{4}}^{\text{ - }}\stackrel{\text{20}{\text{0}}^{\text{o}}\text{C}}{\to }{\text{B}}_{\text{3}}{\text{N}}_{\text{3}}{\text{H}}_{\text{6}}\text{(borazole) + 12}{\text{H}}_{\text{2}}$

6. ${\text{B}}_2{\text{H}}_6+2\text{KOH}+2{\text{H}}_2\text{O}\stackrel{}{\to }2\text{KB}{\text{O}}_2+6{\text{H}}_2(\text{xv}){\text{B}}_2{\text{H}}_6+6\text{C}{\text{l}}_2\stackrel{}{\to }2\text{BC}{\text{l}}_3+6\text{HCl}$

7. ${\text{B}}_2{\text{H}}_6+2\text{LiH}\stackrel{}{\to }2\text{LiB}{\text{H}}_4$

Aluminium

Extraction (Hall-Heroult Process):

Bauxite $\left(\text{A}{\text{l}}_2{\text{O}}_3\cdot 2{\text{H}}_2\text{O}\right)$ is the mineral ore from which aluminium is derived. Bayere's method is used to purify the ore initially. The anhydrous $\text{A}{\text{l}}_2{\text{O}}_3$ is combined with and fused. When aluminium is acquired at the cathode, the fused mixture of $\text{N}{\text{a}}_{\text{3}}\text{Al}{\text{F}}_{\text{6}}\text{ \& Ca}{\text{F}}_{\text{2}}$is treated to electrolytic reduction. Hoope's method purifies aluminium.

Properties:

Aluminium is a silvery metal with a density of$2.7\text{g}/\text{cc}$ and a melting point of ${660}^{\circ }\text{C}$ , as well as being an excellent heat and electrical conductor. It is ductile and malleable.

1. Action of Air: Dry air has no effect on aluminium. However, wet air creates a thin coating of $\text{A}{\text{l}}_2{\text{O}}_3$ on its surface, which dulls its brilliance. It burns to create $\text{A}{\text{l}}_2{\text{O}}_3$and $\text{AlN}$ at extremely high temperatures.

2. Reaction with Halogens: 

When gaseous halogens travel through aluminum, the halide forms in an anhydrous state. $2\text{Al}+3\text{C}{\text{l}}_2\stackrel{}{\to }2\text{AlC}{\text{l}}_3$

3. Action of Alkalies: When heated with concentrated $\text{NaOH}$, it releases ${\text{H}}_2$gas and forms a colorless sodium aluminate solution.

$2\text{Al}+2\text{NaOH}+2{\text{H}}_2\text{O}\stackrel{}{\to }2\text{NaAl}{\text{O}}_2+3{\text{H}}_2\uparrow$

4. Action of Acids: Aluminium interacts with both dilute ${\text{H}}_2\text{S}{\text{O}}_4$and $\text{HCl}$but not with concentrated $\text{HN}{\text{O}}_3$, since concentrated $\text{HN}{\text{O}}_3$ causes aluminium to become passive, producing a protective oxide coating on the surface.

$2\text{Al}+3{\text{H}}_2\text{S}{\text{O}}_4\stackrel{}{\to }\text{A}{\text{l}}_2{\left(\text{S}{\text{O}}_4\right)}_3+3{\text{H}}_2\uparrow ;2\text{Al}+6\text{HCl}\stackrel{}{\to }2\text{AlC}{\text{l}}_3+3{\text{H}}_2\uparrow$

5. Reaction with ${\text{N}}_2$ : Aluminum nitride is produced when ${\text{N}}_2$gas is passed over heated aluminum. As a result, hot aluminum functions as a ${\text{N}}_2$absorber. 

$2\text{Al}+{\text{N}}_2\stackrel{}{\to }2\text{AlN}$

$\text{AlN}$ reacts with hot water to form $\text{Al}(\text{OH}{)}_3$ and $\text{N}{\text{H}}_3$.

6. Reaction with Water: Cold water has no effect on aluminium. Boiling water or steam attacks it extremely slowly. 

$2\text{Al}+3{\text{H}}_2\text{O}\stackrel{}{\to }\text{Al}(\text{OH}{)}_3+3{\text{H}}_2\uparrow$

7. Action of $\text{HgC}{\text{l}}_2$ Solution: Mercury is released when aluminium is added to a solution of $\text{HgC}{\text{l}}_2$.

$3\text{HgC}{\text{l}}_2+2\text{Al}\stackrel{}{\to }2\text{AlC}{\text{l}}_3+3\text{Hg}\downarrow$

8. Reduction of Oxides of Metals: When less reactive metal oxides are heated with aluminium; the less reactive metal is released. $3\text{Mn}{\text{O}}_2+4\text{Al}+\stackrel{\mathrm{\Delta }}{\to }2\text{A}{\text{l}}_2{\text{O}}_3+3\text{Mn};\text{ C}{\text{r}}_2{\text{O}}_3+2\text{Al}+\stackrel{\mathrm{\Delta }}{\to }\text{A}{\text{l}}_2{\text{O}}_3+2\text{Cr}$

Uses:

Aluminium is utilized for plating tanks, pipes, iron bars, and other steel items to prevent corrosion, manufacturing of aluminium cables, precise instruments, surgical apparatus, aircraft bodies, train coaches, motorboats, and automobiles.

Compounds of Aluminium:

• Aluminium Oxide $\left(\text{A}{\text{l}}_2{\text{O}}_3\right)$ :

It's also known as alumina. It may be found in the form of bauxite and corundum in nature. It can also be found in gemstones. Topaz yellow, sapphire blue, ruby red, amethyst violet, and emerald green are some of the most important aluminium oxide stones.

Preparation:

Pure $\text{A}{\text{l}}_2{\text{O}}_3$ is obtained by igniting $\text{A}{\text{l}}_2{\left(\text{S}{\text{O}}_4\right)}_3,\text{Al}(\text{OH}{)}_3$ or ammonium alum.

$\text{A}{\text{l}}_2{\left(\text{S}{\text{O}}_4\right)}_3+\stackrel{\mathrm{\Delta }}{\to }\text{A}{\text{l}}_2{\text{O}}_3+3\text{S}{\text{O}}_3\uparrow ;2\text{Al}(\text{OH}{)}_3+\stackrel{\mathrm{\Delta }}{\to }\text{A}{\text{l}}_2{\text{O}}_3+3{\text{H}}_2\text{O}\uparrow$

${\left(\text{N}{\text{H}}_4\right)}_2\text{S}{\text{O}}_4\cdot \text{A}{\text{l}}_2{\left(\text{S}{\text{O}}_4\right)}_3\cdot 24{\text{H}}_2\text{O}\stackrel{\mathrm{\Delta }}{\to }2\text{N}{\text{H}}_3\uparrow +\text{A}{\text{l}}_2{\text{O}}_3+4\text{S}{\text{O}}_2\uparrow +25{\text{H}}_2\text{O}\uparrow$

Properties:

It's a white amorphous powder that's insoluble in water but soluble in acids (forming, for example$\text{AlC}{\text{l}}_3$) and alkalies (forming$\text{NaAl}{\text{O}}_2$), making it amphoteric. It's a polar covalent molecule with a positive charge.

Uses:

It is employed in the extraction of aluminium, the creation of fake gems, the production of aluminium compounds, and the fabrication of furnace linings. As a catalyst in organic processes, it is a refractory substance.

• Aluminium Chloride $\left(\text{AlC}{\text{l}}_3,6{\text{H}}_2\text{O}\right)$ :

It is a colourless crystalline solid that is water soluble. It has a covalent bond. Anhydrous $\text{AlC}{\text{l}}_3$is a white solid that is deliquescent.

Preparation:

1. By mixing aluminium, $\text{A}{\text{l}}_2{\text{O}}_3$, or $\text{Al}(\text{OH}{)}_3$ in dilute $\text{HCl}$ :

$2\text{Al}+6\text{HCl}\stackrel{}{\to }2\text{AlC}{\text{l}}_3+3{\text{H}}_2\uparrow ;\text{ A}{\text{l}}_2{\text{O}}_3+6\text{HCl}\stackrel{}{\to }2\text{AlC}{\text{l}}_3+3{\text{H}}_2\text{O};\text{Al}(\text{OH}{)}_3+3\text{HCl}\stackrel{}{\to }\text{AlC}{\text{l}}_3+3{\text{H}}_2\text{O}$

The solution gained is filtered and crystallized and the crystals of $\text{AlC}{\text{l}}_3\cdot 6{\text{H}}_2\text{O}$ are obtained.

2. Anhydrous $\text{AlC}{\text{l}}_3$ is obtained by the action of $\text{C}{\text{l}}_2$ on heated aluminium.

3. By heating a mixture of $\text{A}{\text{l}}_2{\text{O}}_3$ and coke and passing chlorine over it. $\text{A}{\text{l}}_{\text{2}}{\text{O}}_{\text{3}}\text{ + 3C + 3C}{\text{l}}_{\text{2}}\stackrel{}{\to }\text{2AlC}{\text{l}}_{\text{3}}\text{(anhy}\text{.) + 3CO}\uparrow$ 

Properties:

1. Action of Heat:

When heated at a high temperature, hydrated salt is transformed to $\text{A}{\text{l}}_2{\text{O}}_3$. $2\text{AlC}{\text{l}}_3\cdot 6{\text{H}}_2\text{O}\stackrel{\mathrm{\Delta }}{\to }\text{A}{\text{l}}_2{\text{O}}_3+6\text{HCl}\uparrow +9{\text{H}}_2\text{O}$

2. Action of Moisture on Anhydrous $\text{AlC}{\text{l}}_3$ :

When open to air, anhydrous $\text{AlC}{\text{l}}_3$ yields white fumes of $\text{HCl}$ $\text{AlC}{\text{l}}_3+3{\text{H}}_2\text{O}\iff \text{Al}(\text{OH}{)}_3+3\text{HCl}\uparrow$

3. Action of $\text{N}{\text{H}}_3$: Anhydrous $\text{AlC}{\text{l}}_3$ absorbs $\text{N}{\text{H}}_3$ since the latter is a Lewis acid. $\text{AlC}{\text{l}}_3+6\text{N}{\text{H}}_3\to \text{AlC}{\text{l}}_3\cdot 6\text{N}{\text{H}}_3$ (white solid)

4. Action of $\text{NaOH}$ Solution:

When solution of $\text{NaOH}$is dropped into an aqueous solution of $\text{AlC}{\text{l}}_3$, a gelatinous precipitate of $\text{Al}(\text{OH}{)}_3$forms first, which dissolves in excess of $\text{NaOH}$solution to provide a colorless sodium aluminate solution.

$\text{AlC}{\text{l}}_3+3\text{NaOH}\to \text{Al}(\text{OH}{)}_3\downarrow +3\text{NaCl};\text{Al}(\text{OH}{)}_3+\text{NaOH}\to \text{NaAl}{\text{O}}_2+2{\text{H}}_2\text{O}$

This reaction is useful for determining the difference between an aluminium salt and salts of$\text{Mg},\text{Ca},\text{Sr and Ba}$. (When $\text{NaOH}$solution is added to their salt solutions, it creates a white hydroxide precipitate that does not dissolve in excess of$\text{NaOH}$.)

5. Action of $\text{N}{\text{H}}_4\text{OH}$ Solution:

When $\text{N}{\text{H}}_4\text{OH}$ solution is combined with a solution of $\text{AlC}{\text{l}}_3$, a white precipitate of $\text{Al}(\text{OH}{)}_3$ is obtained which is insoluble  in excess of $\text{N}{\text{H}}_4\text{OH}$. 

$\text{AlC}{\text{l}}_3+3\text{N}{\text{H}}_4\text{OH}\to \text{Al}(\text{OH}{)}_3\downarrow$ (white gelatinous) $+3\text{N}{\text{H}}_4\text{Cl}$

To differentiate an $\text{Al}$salt from a $\text{Zn}$salt, this reaction is important. (When using a $\text{Zn}$salt, a white precipitate forms of $\text{Zn}(\text{OH}{)}_2$ that dissolves in excess of $\text{N}{\text{H}}_4\text{OH}$ solution.)

6. Hydrolysis with Water:

When $\text{AlC}{\text{l}}_3$is dissolved in water, it undergoes fast hydrolysis, yielding $\text{Al}(\text{OH}{)}_3$ which is a weak base and $\text{HCl}$ which is a strong acid. As a result, litmus finds the solution to be acidic.

${\left[\text{Al}{\left({\text{H}}_2\text{O}\right)}_6\right]}^{3+}\rightleftharpoons {\left[\text{Al}{\left({\text{H}}_2\text{O}\right)}_5\text{OH}\right]}^{+2}+{\text{H}}^{+}$

The complex cation has a great affinity to undergo dimerization.

7. $4\text{LiH}+\text{AlC}{\text{l}}_3\stackrel{}{\to }\text{LiAl}{\text{H}}_4+3\text{LiCl}$