Answer

(i) B to TI

The electric configuration of group 13 elements is $n s^{2} n p^{1}$. Therefore, the most common oxidation state exhibited by them should be +3 . However, it is only boron and aluminium which practically show the +3 oxidation state. The remaining elements, i.e., $Ga$, In, $TI$, show both the +1 and +3 oxidation states. On moving down the group, the +1 state becomes more stable. For example, $TI(+1)$ is more stable than $TI(+3)$. This is because of the inert pair effect. The two electrons present in the s-shell are strongly attracted by the nucleus and do not participate in bonding. This inert pair effect becomes more and more prominent on moving down the group. Hence, Ga (+1) is unstable, In (+1) is fairly stable, and $TI(+1)$ is very stable.

The stability of the +3 oxidation state decreases on moving down the group.

(ii) $C$ to $Pb$

The electronic configuration of group 14 elements is $n s^{2} n p^{2}$. Therefore, the most common oxidation state exhibited by them should be +4 . However, the +2 oxidation state becomes more and more common on moving down the group. $C$ and Si mostly show the +4 state. On moving down the group, the higher oxidation state becomes less stable. This is because of the inert pair effect. Thus, although $Ge, Sn$, and $Pb$ show both the +2 and +4 states, the stability of the lower oxidation state increases and that of the higher oxidation state decreases on moving down the group.

C $\quad Si \quad Ge \quad Sn Pb$

stability of +4 state decreases

Answer

Boron and thallium belong to group 13 of the periodic table. In this group, the +1 oxidation state becomes more stable on moving down the group. $BCl_3$ is more stable than $TICl_3$ because the +3 oxidation state of $B$ is more stable than the +3 oxidation state of $TI$. In $TI$, the +3 state is highly oxidising and it reverts back to the more stable +1 state.

Answer

The electric configuration of boron is $n s^{2} n p^{1}$. It has three electrons in its valence shell. Thus, it can form only three covalent bonds. This means that there are only six electrons around boron and its octet remains incomplete. When one atom of boron combines with three fluorine atoms, its octet remains incomplete. Hence, boron trifluoride remains electron-deficient and acts as a Lewis acid.

Answer

Being a Lewis acid, $BCl_3$ readily undergoes hydrolysis. Boric acid is formed as a result.

$BCl_3+3 H_2 O \longrightarrow 3 HCl+B(OH)_3$

$CCl_4$ completely resists hydrolysis. Carbon does not have any vacant orbital. Hence, it cannot accept electrons from water to form an intermediate. When $CCl_4$ and water are mixed, they form separate layers.

$CCl_4+H_2 O \longrightarrow$ No reaction

Answer

Boric acid is not a protic acid. It is a weak monobasic acid, behaving as a Lewis acid.

$B(OH)_3+2 HOH \longrightarrow[B(OH)_4]^{-}+H_3 O^{+}$

It behaves as an acid by accepting a pair of electrons from ${ }^{\text{âe" }} OH$ ion.

Answer

On heating orthoboric acid $(H_3 BO_3)$ at $370 K$ or above, it changes to metaboric acid $(HBO_2)$. On further heating, this yields boric oxide $B_2 O_3$.

$ H_3 BO_3 \xrightarrow[370 K]{\Delta} HBO_2 \xrightarrow[\text{ red hot }]{\Delta} B_2 O_3 $

Metaboric acid Boric oxide

Answer

(i) $BF_3$

As a result of its small size and high electronegativity, boron tends to form monomeric covalent halides. These halides have a planar triangular geometry. This triangular shape is formed by the overlap of three $s p^{2}$ hybridised orbitals of boron with the sporbitals of three halogen atoms. Boron is $s p^{2}$ hybridised in $BF_3$.

(ii) $BH_4{ }^{-}$

Boron-hydride ion $(BH_4)$ is formed by the $s p^{3}$ hybridisation of boron orbitals. Therefore, it is tetrahedral in structure.

Answer

A substance is called amphoteric if it displays characteristics of both acids and bases. Aluminium dissolves in both acids and bases, showing amphoteric behaviour.

(i) $2 Al _{(s)}+6 HCl _{(o q)} \longrightarrow 2 Al _{(a q)}^{3+}+6 Cl _{(\alpha q)}^{-}+3 H _{2(g)}$

(ii) $2 Al _{(s)}+2 NaOH _{(x q)}+6 H_2 O _{(l)} \longrightarrow 2 Na^{+}[Al(OH)_4] _{(a q)}^{-}+3 H _{2(g)}$

Answer

In an electron-deficient compound, the octet of electrons is not complete, i.e., the central metal atom has an incomplete octet. Therefore, it needs electrons to complete its octet.

(i) $BCl_3$

$BCl_3$ is an appropriate example of an electron-deficient compound. $B$ has 3 valence electrons. After forming three covalent bonds with chlorine, the number of electrons around it increases to 6 . However, it is still short of two electrons to complete its octet. (ii) $SiCl_4$

The electronic configuration of silicon is $n s^{2} n p^{2}$. This indicates that it has four valence electrons. After it forms four covalent bonds with four chlorine atoms, its electron count increases to eight. Thus, $SiCl_4$ is not an electron-deficient compound.

Answer

(a) $CO_3^{2-}$

(b) $HCO_3^{-}$ion

(A)

(B)

There are only two resonating structures for the bicarbonate ion.

Answer

The state of hybridisation of carbon in: (a) $CO_3^{2-}$

$C$ in $CO_3^{2-}$ is $s p^{2}$ hybridised and is bonded to three oxygen atoms.

(b) Diamond

Each carbon in diamond is $s p^{3}$ hybridised and is bound to four other carbon atoms.

(c) Graphite

Each carbon atom in graphite is $sp^2 $ hybridised and is bound to three other carbon atoms

Answer

Answer

(a) Lead belongs to group 14 of the periodic table. The two oxidation states displayed by this group is +2 and +4 . On moving down the group, the +2 oxidation state becomes more stable and the +4 oxidation state becomes less stable.

This is because of the inert pair effect. Hence, $PbCl_4$ is much less stable than $PbCl_2$. However, the formation of $PbCl_4$ takes place when chlorine gas is bubbled through a saturated solution of $PICl_2$.

$ PbCl _{2(s)}+Cl _{2(g)} \longrightarrow PbCl _{4(l)} $

(b) On moving down group IV, the higher oxidation state becomes unstable because of the inert pair effect. $Pb(IV)$ is highly unstable and when heated, it reduces to $Pb(II)$.

$ PbCl _{4(l)} \xrightarrow{\Delta} PbCl _{2(s)}+Cl _{2(g)} $

(c) Lead is known not to form $Pbl_4$. $Pb(+4)$ is oxidising in nature and lis reducing in nature. A combination of $Pb(IV)$ and iodide ion is not stable. lodide ion is strongly reducing in nature. $Pb(IV)$ oxidises $I^{a}$ to $I^{2}$ and itself gets reduced to $Pb(II)$.

$ PbI_4 \longrightarrow PbI_2+I_2 $

Answer

The B-F bond length in $BF_3$ is shorter than the B- $F$ bond length in $BF_4^{-} . BF_3$ is an electron-deficient species. With a vacant $p$-orbital on boron, the fluorine and boron atoms undergo $p$ т- $p \pi$ mack-bonding to remove this deficiency. This imparts a double-bond character to the B-F bond.

This double-bond character causes the bond length to shorten in $BF_3(130 pm)$. However, when $BF_3$ coordinates with

the fluoride ion, a change in hybridisation from $ sp^2( in BF_3) $ to $ sp^3(in BF^- _4 ) $ occur. Boron now forms 4 - bond and

the double-bond character is lost. This accounts for a B- $F$ bond length of 143 pm in $BF_4^{-}$ion.

Tetrahedral structure

Answer

As a result of the difference in the electronegativities of $B$ and $Cl$, the $B-Cl$ bond is polar in nature. However, the $BCl_3$ molecule is non-polar. This is because $BCl_3$ is trigonal planar in shape. It is a symmetrical molecule. Hence, the respective dipole-moments of the $B-Cl$ bond cancel each other, thereby causing a zero-dipole moment.

Answer

Hydrogen fluoride (HF) is a covalent compound and has a very strong intermolecular hydrogen-bonding. Thus, it does not provide ions and aluminium fluoride (AIF) does not dissolve in it. Sodium fluoride (NaF) is an ionic compound and when it is added to the mixture, AIF dissolves. This is because of the availability of free $F^{\hat{\epsilon} \epsilon^{*}}$. The reaction involved in the process is: $AlF_3+3 NaF$ $Na_3[AlF_6]$

Sodium hexafluoroaluminate(III)

When boron trifluoride $(BF_3)$ is added to the solution, aluminium fluoride precipitates out of the solution. This happens because the tendency of boron to form complexes is much more than that of aluminium. Therefore, when $BF_3$ is added to the solution, $B$ replaces $Al$ from the complexes according to the following reaction:

$ Na_3[AlF_6]+3 BF_3 \longrightarrow 3 Na[BF_4]+AlF_3 $

Answer

Carbon monoxide is highly-poisonous because of its ability to form a complex with haemoglobin. The CO-Hb complex is more stable than the $O_2-Hb$ complex. The former prevents $Hb$ from binding with oxygen. Thus, a person dies because of suffocation on not receiving oxygen. It is found that the $CO-Hb$ complex is about 300 times more stable than the $O_2-Hb$ complex.

Answer

Carbon dioxide is a very essential gas for our survival. However, an increased content of $CO_2$ in the atmosphere poses a serious threat. An increment in the combustion of fossil fuels, decomposition of limestone, and a decrease in the number of trees has led to greater levels of carbon dioxide. Carbon dioxide has the property of trapping the heat provided by sunrays. Higher the level of carbon dioxide, higher is the amount of heat trapped. This results in an increase in the atmospheric temperature, thereby causing global warming.

Answer

(a) Diborane

$B_2 H_6$ is an electron-deficient compound. $B_2 H_6$ has only 12 electrons - 6 efrom $6 H$ atoms and 3 eeach from $2 B$ atoms. Thus, after combining with $3 H$ atoms, none of the boron atoms has any electrons left. X-ray diffraction studies have shown the structure of diborane as:

2 boron and 4 terminal hydrogen atoms $(H_t)$ lie in one plane, while the other two bridging hydrogen atoms $(H_b)$ lie in a plane perpendicular to the plane of boron atoms. Again, of the two bridging hydrogen atoms, one $H$ atom lies above the plane and the other lies below the plane. The terminal bonds are regular two-centre two-electron $(2 c-2 e)$ bonds, while the two bridging ( $B-H-B)$ bonds are three-centre two-electron $(3 c-2 e)$ bonds.

(b) Boric acid

Boric acid has a layered structure. Each planar $BO_3$ unit is linked to one another through $H$ atoms. The $H$ atoms form a covalent bond with $BO_3$ unit, while a hydrogen bond is formed with another $BO_3$ unit. In the given figure, the dotted lines represent hydrogen bonds.

Answer

(a) When heated, borax undergoes various transitions. It first loses water molecules and swells. Then, it turns into a transparent liquid, solidifying to form a glass-like material called borax bead.

$ \underset{\text{ Borax }}{Na_2 B_4 O_7 \cdot 10 H_2 O} \xrightarrow{\text{ Sodium }} \underset{\text{ metaborate }}{Na_2 B_4 O_7} \xrightarrow{\Delta} 2 NaBO_2+\underset{\text{ Boric }}{B_2 O_3} $

(b) When boric acid is added to water, it accepts electrons from ${ }^{at} OH$ ion.

$ B(OH)_3+2 HOH \longrightarrow[B(OH)_4]^{-}+H_3 O^{+} $

(c) Al reacts with dilute $NaOH$ to form sodium tetrahydroxoaluminate(III). Hydrogen gas is liberated in the process.

$ 2 Al _{(s)}+2 NaOH _{(a q)}+6 H_2 O _{(l)} \longrightarrow 2 Na^{+}[Al(OH)_4] _{(o q)}^{-}+3 H _{2(g)} $

(d) $BF_3$ (a Lewis acid) reacts with $NH_3$ (a Lewis base) to form an adduct. This results in a complete octet around $B$ in $BF_3$.

$ F_3 B+: NH_3 \longrightarrow F_3 B \to: NH_3 $

Answer

(a) When silicon reacts with methyl chloride in the presence of copper (catalyst) and at a temperature of about $537 K$, a class of organosilicon polymers called methyl-substituted chlorosilanes ( $MeSiCl_3, Me_2 SiCl_2, Me_3 SiCl$, and $Me_4 Si$ ) are formed.

(b) When silicon dioxide $(SiO_2)$ is heated with hydrogen fluoride $(HF)$, it forms silicon tetrafluoride $(SiF_4)$. Usually, the Si-O bond is a strong bond and it resists any attack by halogens and most acids, even at a high temperature. However, it is attacked by HF.

$SiO_2+4 HF \longrightarrow SiF_4+2 H_2 O$

The $SiF_4$ formed in this reaction can further react with $HF$ to form hydrofluorosilicic acid.

$SiF_4+2 HF \longrightarrow H_2 SiF_6$

(c) When $CO$ reacts with $ZnO$, it reduces $ZnO$ to $Zn$. $CO$ acts as a reducing agent.

$ ZnO _{(s)}+CO _{(g)} \xrightarrow{\Delta} Zn _{(s)}+CO _{2(g)} $

(d) When hydrated alumina is added to sodium hydroxide, the former dissolves in the latter because of the formation of sodium meta-aluminate.

$ Al_2 O_3 \cdot 2 H_2 O+2 NaOH \longrightarrow 2 NaAlO_2+3 H_2 O $

Answer

(i) Concentrated $HNO_3$ can be stored and transported in aluminium containers as it reacts with aluminium to form a thin protective oxide layer on the aluminium surface. This oxide layer renders aluminium passive.

(ii) Sodium hydroxide and aluminium react to form sodium tetrahydroxoaluminate(III) and hydrogen gas. The pressure of the produced hydrogen gas is used to open blocked drains.

$ 2 Al+2 NaOH+6 H_2 O \longrightarrow 2 Na^{+}[Al(OH)_4]^{-}+3 H_2 $

(iii) Graphite has a layered structure and different layers of graphite are bonded to each other by weak van der Waals’ forces. These layers can slide over each other. Graphite is soft and slippery. Therefore, graphite can be used as a lubricant.

(iv) In diamond, carbon is $s p^{3}$ hybridised. Each carbon atom is bonded to four other carbon atoms with the help of strong covalent bonds. These covalent bonds are present throughout the surface, giving it a very rigid 3-D structure. It is very difficult to break this extended covalent bonding and for this reason, diamond is the hardest substance known. Thus, it is used as an abrasive and for cutting tools.

(v) Aluminium has a high tensile strength and is very light in weight. It can also be alloyed with various metals such as $Cu, Mn, Mg, Si$, and $Zn$. It is very malleable and ductile. Therefore, it is used in making aircraft bodies.

(vi) The oxygen present in water reacts with aluminium to form a thin layer of aluminium oxide. This layer prevents aluminium from further reaction. However, when water is kept in an aluminium vessel for long periods of time, some amount of aluminium oxide may dissolve in water. As aluminium ions are harmful, water should not be stored in aluminium vessels overnight.

(vii) Silver, copper, and aluminium are among the best conductors of electricity. Silver is an expensive metal and silver wires are very expensive. Copper is quite expensive and is also very heavy. Aluminium is a very ductile metal. Thus, aluminium is used in making wires for electrical conduction.

Answer

Ionisation enthalpy of carbon (the first element of group 14) is very high (1086 kJ/mol). This is expected owing to its small size. However, on moving down the group to silicon, there is a sharp decrease in the enthalpy ( $786 kJ)$. This is because of an appreciable increase in the atomic sizes of elements on moving down the group.

Answer

Although Ga has one shell more than AI, its size is lesser than Al. This is because of the poor shielding effect of the $3 d$-electrons. The shielding effect of $d$-electrons is very poor and the effective nuclear charge experienced by the valence electrons in gallium is much more than it is in the case of $Al$.

Answer

Allotropy is the existence of an element in more than one form, having the same chemical properties but different physical properties. The various forms of an element are called allotropes.

Diamond:

The rigid 3-D structure of diamond makes it a very hard substance. In fact, diamond is one of the hardest naturallyoccurring substances. It is used as an abrasive and for cutting tools.

Graphite:

It has $s p^{2}$ hybridised carbon, arranged in the form of layers. These layers are held together by weak van der Walls’ forces. These layers can slide over each other, making graphite soft and slippery. Therefore, it is used as a lubricant.

Answer

(1) $CO=$ Neutral

(2) $B_2 O_3=$ Acidic

Being acidic, it reacts with bases to form salts. It reacts with $NaOH$ to form sodium metaborate.

$B_2 O_3+2 NaOH \longrightarrow 2 NaBO_2+H_2 O$

(3) $SiO_2=$ Acidic

Being acidic, it reacts with bases to form salts. It reacts with $NaOH$ to form sodium silicate.

$SiO_2+2 NaOH \longrightarrow 2 Na_2 SiO_3+H_2 O$

(4) $CO_2=$ Acidic

Being acidic, it reacts with bases to form salts. It reacts with $NaOH$ to form sodium carbonate.

$CO_2+2 NaOH \longrightarrow Na_2 CO_3+H_2 O$

(5) $Al_2 O_3=$ Amphoteric

Amphoteric substances react with both acids and bases. $Al_2 O_3$ reacts with both $NaOH$ and $H_2 SO_4$.

$Al_2 O_3+2 NaOH \longrightarrow NaAlO_2$

$Al_2 O_3+3 H_2 SO_4 \longrightarrow Al_2(SO_4)_3+3 H_2 O$

(6) $PbO_2=$ Amphoteric

Amphoteric substances react with both acids and bases. $PbO_2$ reacts with both $NaOH$ and $H_2 SO_4$.

$PbO_2+2 NaOH \longrightarrow Na_2 PbO_3+H_2 O$

$2 PbO_2+2 H_2 SO_4 \longrightarrow 2 PbSO_4+2 H_2 O+O_2$

(7) $Tl_2 O_3=$ Basic

Being basic, it reacts with acids to form salts. It reacts with $HCl$ to form thallium chloride.

$Tl_2 O_3+6 HCl \longrightarrow 2 TlCl_3+3 H_2 O$

Answer

Thallium belongs to group 13 of the periodic table. The most common oxidation state for this group is +3 . However, heavier members of this group also display the +1 oxidation state. This happens because of the inert pair effect. Aluminium displays the +3 oxidation state and alkali metals display the +1 oxidation state. Thallium displays both the oxidation states. Therefore, it resembles both aluminium and alkali metals.

Thallium, like aluminium, forms compounds such as $TICl_3$ and $Tl_2 O_3$. It resembles alkali metals in compounds $Tl_2 O$ and $TICl$.

Answer

The given metal $X$ gives a white precipitate with sodium hydroxide and the precipitate dissolves in excess of sodium hydroxide. Hence, $X$ must be aluminium.

The white precipitate (compound A) obtained is aluminium hydroxide. The compound $B$ formed when an excess of the base is added is sodium tetrahydroxoaluminate(III).

$ 2 Al+3 NaOH \longrightarrow Al(OH)_3 \downarrow+3 Na^{+} $

$ \text{ Aluminium }(X) \text{ Sodium hydroxide } \quad \text{ White ppt.(A) } $

$$ \begin{equation*} Al(OH)_3+NaOH \longrightarrow \quad Na^{+}[Al(OH)_4]^{-} \tag{A} \end{equation*} $$

(Soluble complex B)

Now, when dilute hydrochloric acid is added to aluminium hydroxide, aluminium chloride (compound C) is obtained.

$$ \begin{equation*} Al(OH)_3+3 HCl \longrightarrow AlCl_3+3 H_2 O \tag{A} \end{equation*} $$

Also, when compound $A$ is heated strongly, it gives compound $D$. This compound is used to extract metal X. Aluminium metal is extracted from alumina. Hence, compound $D$ must be alumina.

$ 2 Al(OH)_3 \xrightarrow{\Delta} Al_2 O_3+3 H_2 O $

(A) (D)

Answer

(a) Inert pair effect

As one moves down the group, the tendency of s-block electrons to participate in chemical bonding decreases. This effect is known as inert pair effect. In case of group 13 elements, the electronic configuration is $n s^{2} n p^{1}$ and their group valency is +3 . However, on moving down the group, the +1 oxidation state becomes more stable. This happens because of the poor shielding of the $n s^{2}$ electrons by the $d$ - and $f$ - electrons. As a result of the poor shielding, the $n s^{2}$ electrons are held tightly by the nucleus and so, they cannot participate in chemical bonding.

(b) Allotropy

Allotropy is the existence of an element in more than one form, having the same chemical properties but different physical properties. The various forms of an element are called allotropes. For example, carbon exists in three allotropic forms: diamond, graphite, and fullerenes.

(c) Catenation

The atoms of some elements (such as carbon) can link with one another through strong covalent bonds to form long chains or branches. This property is known as catenation. It is most common in carbon and quite significant in $Si$ and $S$.

Answer

The given salt is alkaline to litmus. Therefore, $X$ is a salt of a strong base and a weak acid. Also, when Xis strongly heated, it swells to form substance $Y$. Therefore, Xmust be borax.

When borax is heated, it loses water and swells to form sodium metaborate. When heating is continued, it solidifies to form a glassy material $Y$. Hence, $Y$ must be a mixture of sodium metaborate and boric anhydride.

$ \begin{aligned} & Na_2 B_4 O_7+7 H_2 O \xrightarrow{\text{ water }} 2 NaOH+4 H_3 BO_3 \\ & \text{ Borax }(X) \quad \text{ Sodium hydroxide } \quad \text{ Orthoboric acid } \end{aligned} $

$ \underset{\text{ Borax }(X)}{Na_2 B_4 O_7 \cdot 10 H_2 O \xrightarrow{\Delta} \underset{\text{ Sodium metaborate }}{Na_2 B_4 O_7} \xrightarrow{\Delta} \underset{\begin{matrix} \text{ Boric anhydride } \\ \text{ (Glassy material) } \end{matrix} }{B_2 O_3}+2 NaBO_2} $

When concentrated acid is added to borax, white crystals of orthoboric acid $(Z)$ are formed.

$ Na_2 B_4 O_7 \cdot 10 H_2 O+H_2 SO_4 \xrightarrow{\Delta} Na_2 SO_4+4 H_3 BO_3+5 H_2 O $

Borax $(X)$

Orthoboric acid (Z)

Answer

(i)

$ 2 BF_3+6 LiH \longrightarrow B_2 H_6+6 LiF $

Boron trifluoride Lithium hydride Diborane Lithium fluoride

(ii) $B_2 H_6+6 H_2 O \longrightarrow 2 H_3 BO_3+6 H_2$

Diborane Water Orthoboric acid Hydrogen

(iii) $B_2 H_6+2 NaH \xrightarrow{\text{ ether }} 2 NaBH_4$

Diborane Sodium hydride Sodium borohydride

(iv)

Boron trioxide

(v) $2 Al+2 NaOH+6 H_2 O \longrightarrow 2 Na^{+}[Al(OH)_4] _{(a q)}^{-}+3 H_2$

Sodium tetrahydroxoaluminate(III)

(vi) $3 B_2 H_6+6 NH_3 \longrightarrow 3[BH_2(NH_3)_2]^{+}[BH_4]^{-} \longrightarrow 2 B_3 N_3 H_6+12 H_2$

Borazene

Answer

Caron dioxide

In the laboratory, $CO_2$ can be prepared by the action of dilute hydrochloric acid on calcium carbonate. The reaction involved is as follows:

$CaCO_3+2 HCl _{(a q)} \longrightarrow CaCl _{2(a q)}+CO _{2(g)}+H_2 O _{(j)}$

$CO_2$ is commercially prepared by heating limestone. The reaction involved is as follows: $CaCO_3 \xrightarrow{\Delta} CaO+CO_2 \uparrow$

Caron monoxide

In the laboratory, $CO$ is prepared by the dehydration of formic acid with conc. $H_2 SO_4$, at $373 K$. The reaction involved is as follows:

$HCOOH \xrightarrow[\text{ conc. } H_2 SO_4]{373 K} H_2 O+CO \uparrow$

$CO$ is commercially prepared by passing steam over hot coke. The reaction involved is as follows:

$C _{(s)}+H_2 O _{(g)} \xrightarrow{473-1273 K} \underbrace{CO _{(g)}+H _{2(g)}} _{\text{water gas }}$

Answer

(c) Borax is a salt of a strong base $(NaOH)$ and a weak acid $(H_3 BO_3)$. It is, therefore, basic in nature.

Answer

(b) Boric acid is polymeric because of the presence of hydrogen bonds. In the given figure, the dotted lines represent hydrogen bonds.

Answer

(c) Boron in diborane is $s p^{3}$ hybridised.

Answer

(b) Graphite is thermodynamically the most stable form of carbon.

Answer

(b)The elements of group 14 have 4 valence electrons. Therefore, the oxidation state of the group is +4 . However, as a result of the inert pair effect, the lower oxidation state becomes more and more stable and the higher oxidation state becomes less stable. Therefore, this group exhibits +4 and +2 oxidation states.