Intermolecular forces vs Thermal Energy

 

Thermal energy

·      It is the energy of a body arising by motion of its particles/atoms/molecules

·       Is proportional to the temperature of substance.

·      It is the measure of average kinetic energy of molecules of matter

·      Is responsible for movement of particles

·      This movement of particle is called thermal motion

Intermolecular forces vs Thermal Energy

    ·      Intermolecular forces tend to keep the molecules together but thermal energy of molecules tends to keep them apart.

     ·      Three states of matter are the result of the balance between intermolecular forces and the thermal energy of the molecules.

 

·      Gas ----------> Liquid ----------->Solid

Predominance of intermolecular interactions

·      Gas <--------- Liquid <----------Solid

Predominance of thermal energy

p-block d-and f-block elements Questions

 

Catalytic properties of transition metals

 

Catalytic properties of transition metals

    ·      Good catalysts due to

   a.  the presence of free valencies and  variable oxidation states

   b.  these provide surface for reaction 

  

some Examples are as

    ·      Iron (III) catalyses the reaction between iodide and persuphate ions-

2I^- +   S₂O₈^2- -------> I₂ + 2SO₄^2-

Explanation of catalytic action of Fe³⁺ in above reaction

Step-1 

2Fe³⁺ + 2I^-    -------> 2Fe²⁺ + 2I₂

Step-2

 2 Fe²⁺ +   S₂O₈^2- -------> 2Fe³⁺   + 2SO₄^2-

in this catalytic action of Fe, Fe shows variable oxidation state i.e. Fe³⁺ changes into Fe²⁺ Again into Fe³⁺

    ·      Pt-used as a catalyst in the manufacture of H₂S0₄.       

    ·      Fe-used as a catalyst in the manufacture of NH₃ by Haber process. A small amount of molybdenum is added as a promoter.

    ·      Ni.-used as a catalyst in the hydrogenation of oils.

    ·      V₂0₅-used as a catalyst for the oxidation of S0₂ into S0₃ for the manufacture of H₂S0₄ in the contact process.

    ·       Mn02-used as a catalyst in the decomposition of KCI0₃ for preparation of oxygen.

Formation of Interstitial Compounds and Alloy

 

·      

 Formation of Interstitial Compounds-

     ·      interstitial compounds are those which are formed when small atoms like H, C, N, B etc are trapped inside the crystal lattice of metals

      ·       The general characteristic physical and chemical properties of these compounds are:

        a).  High melting points which are higher than those of pure metals.

        b) Retain metallic conductivity i.e. of pure metals.

        c). Very hard and some borides have hardness as that of diamond.

        d). Chemically inert.

Alloy Formation –

     ·     transition elements form alloy because of similar in atomic size and other characteristics of transition metal

     ·      Alloys are hard and having high melting point.

             e.g., Brass- (Cu + Zn) 

                     Bronze -(Cu + Sn) etc.

       Hg when mix with other metals form semisolid amalgam except Fe, Co, Ni, Li.

Colours of Transition Metal Ions

 

Colours of Transition Metal Ions

·      Colour in transition metal ions depends upon presence of unpaired electron which show d-d transition of unpaired electron from t_2g to e_g set of energies when electron absorbs energy to jump from t_2g to e_g and come back to t_2g from e_g by emission of energy that appear with colour

Ex.:

 Ti⁺⁴, ------ [Ar] 3d⁰4s⁰ ----- no of unpaired electron ----->Zero---> colourless

 V⁺⁵ -----[Ar] 3d⁰4s⁰ -----no of unpaired electron ----->Zero---> colourless

Ans - Because of absence due to presence of paired electrons which do not show d-d transition

Ex-

 Sc^2+, ---------[Ar] 3d¹4s⁰ ------no of unpaired electron ----->1e^- ---> coloured

Cr³⁺ ---------[Ar] 3d³4s⁰ ------no of unpaired electron ----->3e^- ---> coloured

Ans- due to presence of unpaired electrons which show d-d transition.

Magnetic Property of Transition Metals –

 

Magnetic Property of Transition Metals –

    ·      Diamagnetic substances contain electron pairs with opposite spins and are repelled by applied magnetics filled.

 Ex.: Ti⁺⁴,------ [Ar] 3d⁰4s⁰  ----- no of unpaired electron ----->Zero---> Diamagnetic 

         V⁺⁵,-----[Ar] 3d⁰4s⁰  -----no of unpaired electron ----->Zero---> Diamagnetic 

        Sc³⁺ ------[Ar] 3d⁰4s⁰  ------no of unpaired electron ----->Zero---> Diamagnetic 

        Zn  --------[Ar] 3d¹⁰4s² ------no of unpaired electron ----->Zero---> Diamagnetic 

         Hg---------[Ar] 3d¹⁰4s² ------no of unpaired electron ----->Zero---> Diamagnetic 

        Cd  ------- [Ar] 3d¹⁰4s²  ------ no of unpaired electron ----->Zero---> Diamagnetic

 

     ·      Paramagnetic substances contain unpaired electron spins or unpaired electrons and are attracted strongly in applied magnetic field.

 Sc^2+, ---------[Ar] 3d¹4s⁰ ------no of unpaired electron ----->1e^- ---> paramagnetic

Cr³⁺ ---------[Ar] 3d³4s⁰ ------no of unpaired electron ----->3e^- ---> paramagnetic

    ·      Paramagnetic character increase with increase in no of unpaired electron.

·      Each unpaired electron having magnetic moment associated with its spin angular momentum and orbital angular momentum.so magnetic momentum can be calculated based on spin only formula

    µ=  √n(n+2)  B.M. (Bohr Magnetons)

 

Standard electrode potential -SEP/SRP

 

Standard electrode potential 

·      Stability of the compounds depends upon electrode potentials. Electrode potential value depends upon

 a). enthalpy of sublimation(atomisation enthalpy) of the metal

b).  ionisation enthalpy and

c).  hydration enthalpy

M(s) ---------> M⁺(aq) + e    

Δ_total H (Total energy change)

Total energy change (Δ_total H) 

= Δ_a H + Δ_i H + Δ_Hyd H

·      The smaller the value of total energy change for a particular oxidation state in aqueous solution, greater will be the stability of that oxidation state.

·      The electrode potentials are the measure the total energy change.

·      The lower the electrode potential, i.e., more negative the standard reduction potential of the electrode, more stable is the oxidation state of the transition metal in the aqueous medium.

Trends in M²⁺/M Standard Electrode Potentials

·      It is evident that there is no regular trend in the E⁰ (M²⁺/M) values. This is due to irregular variation of ionisation energies and sublimation energies (atomisation energy) of the atoms of the members of the transition series.

·      The lower (Less negative) of E⁰ (M²⁺/M) values along the series is due to increase in the first and second ionisation energies.

·      The lower  the SEP value, the more is the stability of O.S. of a metal in aqueous.

·      Mn, Ni and Zn have more negative SRP/SEP (E⁰ (M²⁺/M) values than expected because.

Reason: Mn²⁺& Zn²⁺ have 3d⁵ & 3d¹⁰ stable electronic configuration

  Since  Ni  has highest hydration energy. (Δ_hydH Ni⁺² = –2121 kJ/mole)

·      Copper having positive E⁰ value because the sum of the first and second ionisation enthalpies for copper is very high due to exceptionally high second ionisation enthalpy. This is not compensated by the hydration enthalpy(Δ_hydH).

Therefore, it does not liberate hydrogen from acids. It reacts only with oxidising acids such as HN0₃ and cone. H₂S0₄.

Trends in M³⁺/M²⁺ Standard Electrode Potentials-

·      E⁰ value for Sc³⁺/Sc²⁺ is very low. Hence, Sc³⁺ is stable. This is due to its noble gas configuration [Ar]3d⁰.

·      E⁰value for Mn³⁺/Mn²⁺ is high. This reflects that Mn²⁺ state is stable due to d⁵ configuration.

·      For Fe³⁺/Fe²⁺ couple, the value of E⁰  is comparatively low because Fe³⁺ is extra  stable due to 3d⁵.

·      The comparatively low value for vanadium(V) is due to stability of V²⁺having half-filled t_2g level.

 

·      Problem- 1. Mn²⁺ is most stable than Fe³⁺ while both having same electronic configuration Ans due to low value (less negative) of E⁰ for Mn

     Problem -2 Why is Cr²⁺ reducing and Mn³⁺is  oxidising when both have d⁴ configuration?

Ans- Cr²⁺ ----->Cr³⁺ + e^-

      3d⁴               3d³ having half-filled t_2g level

     Mn³⁺ + e^- --------> Mn

     d⁴                           d⁵

less stable                most stable due to half-filled orbital

Problem-3 Why is Cr²⁺ reducing and Mn³⁺is  oxidising when both have d⁴ configuration

Cr²⁺ is stronger reducing reagent due to negative value of SEP(E⁰value)

Whereas

Cr²⁺----------> Cr³⁺

 d⁴                        d³

Fe²⁺---------> Fe³⁺

d⁶                        d⁵

Problem-4 why the Mn³⁺/Mn²⁺ couple is much more positive than that for Cr³⁺/Cr²⁺ or Fe³⁺/Fe²⁺

Because Mn²⁺ has most stable electronic configuration as 3d⁵ so it requires lager third ionisation enthalpy than others so it has much more positive SEP.

Problem – 5. Cu¹⁺ is less table than Cu²⁺

Cu⁺ (aq) + e^_   -----------------> Cu(s)                            E^o = 0.52V

Cu²⁺ (aq) + e^_   -----------------> Cu(s)                           E^o = 0.34V

Due to large (+)ve standard electrode potential of Cu⁺ is more reduced easily than Cu²⁺

In other word Cu⁺  compound give disproportionate reaction in aqueous

  Cu⁺ (aq) -----------------> Cu(s)   +  Cu⁺²    

Due to large (-)ve  hydration enthalpy of   Cu⁺²     which  more than compensate for second ionisation enthalpy of Cu

                                         

Chemical reactivity and E^o values-

·      The E^o values for M²⁺/M across a series increase toward positive value that indicates a decreasing tendency to form divalent cation across the series.

·      Mn, Ni and Zn have more negative SRP/SEP (E⁰ (M²⁺/M) values than expected because.

  Reason: Mn²⁺& Zn²⁺ have 3d⁵ & 3d¹⁰ stable electronic configuration

  Since Ni has highest hydration energy. (Δ_hydH Ni⁺² = –2121 kJ/mole)

·      Mn³⁺ and Co³⁺ ions are strongest oxidising agents in aqueous solution because their E^o values for M³⁺/M²⁺ are large (+) ve .

While Ti²⁺ V²⁺ and Cr²⁺ are strong reducing agent will liberate hydrogen from a dilute acid, since they have negative SEP value.

Note- Lower (large negative) value of E^o -------> element behave as a reducing reagent

            higher(large positive) value of E^o --------> element behave as a oxidising reagent.