Oxidation state properties and inert pair effect of group (IV) A , group 14 element.

General group Trend.

The general electronic configuration of group 14 of P-block element follows ns^₂ np^₂which is as below.

Electronic configuration.

General Valance shell electronic configuration = ns² np_²

Physical properties.

1.Covalent radii.

The covalent radii increase down the group.The difference in size between Si and Ge is less than might be otherwise expected because Ge has a full 3d shell which shields the nuclear charge rather ineffectively.In a similar way, the small difference in size between Sn and Pb is because of the filling of the 4f shell.

2.Ionisation energy.

The ionisation energy decreased from the C to Si but then change in an irregular way this is because the effect of filling the d and f shells.The amount of energy required to form M⁴⁺ ions is extremely large and hence simple ionic compounds are rare.The only elements which will give a large enough electronegativity difference to giver ionic character are F and O .The compound SnF₂, PbF_2, SnF₄, PbF₄, PbO₂ are significantly ionic, but the only significant metal ion is Pb²⁺.

3.Melting point.

C has the extremely high melting point.Si melts appreciably lower than C but the value of Si and Ge are still high.They all have the very stable diamond type of lattice, so require a diamond type of lattice.Melting involving the breaking the strong covalent bond of a lattice, so require a lot of energy.The melting point decreases on descending a group because the M-M bonds become weaker as the atom increase in size.Sn and Pb are3 metallic and have many melting points.They don't use all four all four outer electrons for metallic bonding.

4. Metallic and non-metallic character.

The change from non-metal to metal with the increasing atomic number is well understood in group 14, where silicon and carbon are non-metal.The increasing in metallic character shows itself in the structures and appearance of the element, in physical properties such as malleability and electrical conductivity, and in chemical properties such as increased tendency to form M2+ ions and the acidic or basic properties of the oxides and hydroxides.

Oxidation state on the basis if group Iva elements.

These element exhibit oxidation state +2 and and+4.If all the valance electron participates in bonding then oxidation state +4 results.However, if only np electron participate in bonding then oxidation state +2 results.

Since, inert pair effect on increasing on defending the group from carbon to lead.Stability of oxidation state +2 increase from carbon-lead. and that of +4 decrease.

Stability of oxidation state +4 is as below.

$$C^{+4}>Si^{+4}>Ge^{+4}>Sn^{+4}>Pb^{+4}$$

$$C^{+4}>Si^{+4}>Ge^{+4>}Sn^{+4}=stable$$

$$Pb^{+4}=Unstable$$

Stability of Oxidation state +2 is as below.

$$C^{+2}<Si^{+2}<Ge^{2+}<Sn^{+2}<Pb^{+2}$$

$$C^{+2}=unstable$$

$$Pb^{+2}=most \,stable$$

$$Si^{+2}<Ge^{+2}<Sn^{+2}=less\, stable\, but\, exist$$

In general, except lead, all elements in this group exhibit oxidation state +4, However, lead exhibit oxidation state +2.

Causes of di valency and tri valency.

$$Oxidation\,state +4=ΔHfE>E_exct$$

$$Oxidation\, State +2=ΔHfE>E_exct$$

Question.

Explain why Pb(IV) act as oxidisiging agent but not Pb(II).

$$Pb^{+4}+2e^-\xrightarrow[]{reduction}Pb^{+2}$$

$$Pb^{+4}=Unstable$$

$$Pb^{+2}=stable$$

$$Pb^{+4}=itself \,\, reduced$$

$$Pb^{+4}=oxidising\, agent$$

$$Pd^{+2}→No \,reduction$$

$$Pd^{+2}→no\, oxidizing \,property$$

Pb being heavier P-block element exhibit inert pair effect.Oxidation state +4 is unstable for Pb but oxidation state +2 is stable.So, Pb⁺⁴ has strong tendency to gain two electron and to be reduced into Pb^+2.

Since Pb⁺⁴ itself reduced into Pb⁺² it acts as oxidising agent.

However, Pb⁺² is stable and has no tendency to gain any electron at all.Since Pb⁺² does not undergo further reduction it has any oxidising property.

An inert pair on the basis of the group Iva element.

Carbon and silicon show +4 oxidation state while Ge, Sn and Pb show+2 as well as +4 oxidation state.Group 14 (IVA) elements have outer electronic configuration ns²np².When all four valence electron is lost, we get the elements in +4 oxidation state i,e M⁺⁴ cation is formed.The stability of M+4 cation decreased from Ge⁺⁴ to Pb^+4.

$$i,e\, Ge^{4+}>Sn^{4+}>Pb^{4+}$$

When only two np electrons from the ns² np² are lost, we get the element in +2 oxidation state i,e.M²⁺ cation is formed.In this case the s- an electron in the outermost shell remaining paired and not participate in bonding.This is called inert pair effect.The stability of +2 oxidation state increases on descending down the group.Lead is the lowermost element of group IVA hence inert pair effect is more marked in Pb and therefore Pb⁺² compounds are more stable than Pb⁺⁴ compounds.

Compounds of Ge⁺² and Sn⁺² are less stable than those of Ge⁴⁺ and Sn⁴⁺ respectively and hence compounds of Ge²⁺ and Sn²⁺ are readily changed into those of Ge⁴⁺ and Sn⁴⁺. In other words, compounds of a Ge²⁺ act as reducing agent.

$$Ge^{2+}Compounds\xrightarrow{Oxidation\,(-2e^-)}Ge^{4+}Compound$$

$$Ge^{2+}Compounds=less\, stable$$

$$Ge^{4+}Compound=more\, stable$$

Pb²⁺ compounds are more stable than Pb⁴⁺ compounds and hence the compounds of Pb⁴⁺ are readily changed (reduced) into those of Pb²⁺.In other words compounds of Pb⁴⁺ act as strong oxidising agent.

$$Pb^{4+}Compounds\xrightarrow{reduction\,(+2e^-)}Pb^{2+}$$

$$Pb^{4+}Compounds=less\, stable$$

$$Pb^{2+}=more\, stable$$

Reference.

F.A.Cotton and Wilkinson G. Basic inorganic Chemistry. John,Wiley and Sons (Asia), 2007.

Lee., J.D. Concise Inorganic Chemistry. fifth edition. New Delhi: Oxford University Press., 2008.

Sharma, M.L and P.N Chaudhary. A textbook of B.S.C chemistry. Kathmandu Nepal: Ekta Books Thapathali Kathmandu, 2011.

$$SiCl_4+h-NH_2\overset{liqNH_3}{\rightarrow}Si(NH_2)_4+4HCl$$

$$[NH_3+HCl \rightarrow NH_4Cl]×4$$

$$\overline{SiCl_4+8NH_3\rightarrowSi(NH_2)_4 +4 NH_4Cl }$$