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.  

Trends in oxidation state of d-block elements Part-3

 Oxidation States (oxidation number)-describes loss/gain of electron by an atom

·      Stability of oxidation state of an element is defined by –

i)      Electronic configuration

ii)      Sum of the ionisation enthalpies

iii)    SEP value

iv)    Hydration enthalpy

·      Sc(+3) and Zn(+2) exhibit only one oxidation state

·      Except scandium, (which has +3 oxidation state) for the elements of first transition series +2 oxidation state is the most common. This state arises due to loss of 4s-electron

·      In 3d series highest oxidation state is +7 (Mn)

·      In d-block series highest oxidation state is +8 (Os, Ru)

·      All transition elements, except first and last member of the series, exhibit a number of oxidation state

·      Generally, within the transition series, the highest oxidation state increases with increase of atomic number, reaching to a maximum in the middle and then starts decreasing.

·      The variable oxidation states of a transition metal is due to the involvement of (n-1)d and, outer ns electrons in bonding as the energies of ns and (n-1)d subshells are nearly equal.

·      In a group of d-block elements, the higher states are more stable for heavier elements.

For example, in group 6, Mo(vi) and W(vi) are more stable than Cr(vi). For example, dichromate having Cr(vi) is a strong. Oxidising agent in acidic medium while MoO3and W03 are stable. oxides.

·      Higher oxidation states are exhibited when ns and (n-1)d-electrons take part in bonding.

·      higher oxidation states are found in compound with fluorine and oxygen because fluorine and oxygen are most electronegative in nature

       Higher oxidation states in oxides are normally more stable than fluorides due to capability of oxygen to form multiple bonds.

·      In p-block lower oxidation states of heavier elements are more stable while in d-block heavier element, higher oxidation state are more stable. Due to inert pair effect found in p-block elements.

            For example, in group 6, Mo(vi) and W(vi) are more stable than Cr(vi). For example, dichromate having Cr(vi) is a strong. Oxidising agent in acidic medium while MoO3and W03 are stable. oxides.

·      Some of the transition metals form compounds in zero oxidation state or lower oxidation state In transition element when a complex compound has ligands capable of π-acceptor character in addition to the σ-bonding or ligand having ability of forming synergic bond/back bonding

Example – Ni(CO)4, Fe(CO)₅

Problems-1. (on the basis of electronic configuration)

a). Ti⁴⁺ (3d^o4s⁰) is more stable than Ti^{3 +} (3d¹4s^o) 

Ans - due to completely filled orbitals in Ti4+ which is most stable than partially filled orbital

b).  Mn²⁺⁽3d⁵4s⁰) is more stable than Mn^{3 +} (3d⁴ 4s⁰)

Ans - due to half-filled orbitals which   is most stable than partially filled orbital

 

 c). Fe³⁺(3d⁵4s⁰) is more stable than Fe^{2 +} (3d⁶ 4s⁰) 

Ans- due to half-filled orbitals which   is most stable than partially filled orbital

Problem-2 (on the basis of sum of ionisation enthalpy)

a). Ni²⁺is more stable than Pt²⁺ while Pt⁴⁺ is more stable than Ni⁴⁺

  Ans- due to sum of IE₁ + IE₂ for Ni²⁺ is lesser than IE₁ + IE₂ for Pt²⁺ while IE₁ + IE₂+ IE₃ + IE₄ for Pt⁴⁺ is smaller than Ni⁴⁺ .

Problem -3 ( on the basis of hydration enthalpy)

a).Cu⁺ is not stable in aqueous solution than Cu²⁺ why?

Ans – since energy is required to remove one electron from Cu⁺.large hydration enery evolve during hydration of Cu⁺ that compensates it .

Problem-4 Name the element which do not show variable oxidation state – Sc

·      Problem -5 which of the 3d- series of transition metal exhibit variable oxidation state and why ?  Ans--Mn due to the involvement of (n-1)d and, outer ns electrons in bonding as the energies of ns and (n-1)d subshells are nearly equal.

Click here to see next The Inner Transition f-Block Elements

Click here to see d & f block elements slide presentation

Bond parameter- Bond length, bond angle and bond energy

 Bond parameter-

 Bond Length-

 The average distance between the nucleus of two atoms is known as bond length, normally it is represented in Å or pm  

Factors Affecting Bond Length

(a) electronegativity: - Bond length proportional to EN

                                             Therefore               H—F < H—Cl < H—Br < H—I

(b) Bond order or number of bonds: - Bond length α 1/Number of bonds or bond order

                                                                     ex.        C—C,      >      C = C  >    C = C

                                                 bond length          154pm             134pm      120pm

Bond Angle- The minimum angle between any two adjacent bonds is known as bond angle. It is represented in degree (°), min (') and second (")

Factors affecting the bond angle –

a). Hybridization— bond angle is proportional to  % 's' character

                                   BeCl2     > BCl3           > CCl4

          bond Angle      180°          120°          109.28'

Hybridizations-           sp               sp2              sp3

% s- character            50%           33.33%       25%

b). Lone pair —When hybridization is same, lone pair are different

      Bond angle is inversely proportional to No. of lone pair Example-

                               CH₄                        NH₃                     H₂O

Hybridization     sp³                      sp³                       sp³

Bond angle        109.28'                107°                     104.5°         

No of lone pair    0                       1                            2

c) nature of Central Atom

                 Bond angle is proportional to the Electronegativity of central atom

Example -

                       NH3         >       PH3       > AsH3

 Bond angle  107°                   93°            91°  

 - Electronegativity decreases from N to As , so,  Bond angle will decrease

d). Nature of Side atom bonded to a central atom

       Bond angle is proportional to size of the side atom  but is inversely proportional to electronegativity of bonded atom

  Example-

    PF₃ < PCl₃ < PBr₃ < PI₃   ---->   Reason - (Electronegativity of side atom decrease)

    OF₂ < Cl₂O < Br₂O ---->   Reason - (Electronegativity of side atom decrease)

     SF₂ < SCl₂ < SBr₂ ---->   Reason - (Electronegativity of side atom decrease

HYBRIDISATION

 

HYBRIDISATION

   ·      It is introduced by Pauling and Slater, to explain equivalent nature of covalent bonds in a molecule.

   ·      Mixing of different shape and approximate equal energy atomic orbitals, and redistribution of energy to form new orbitals, of same shape & same energy. These new orbitals are called hybrid orbitals and the phenomenon is called hybridization.

 Characteristic of Hybridization

(1) Hybridization is a mixing of orbitals and not electrons. Therefore, in hybridization full filled, half-filled and empty orbitals may take part.

(2) Number of the hybrid orbitals formed is always be equal  to number of atomic orbitals which  participates in the process of hybridization.

(3) The number of hybrid orbitals on central atom of a molecule or ion

 (4) One element can represent many hybridization state depending on experimental conditions for example, C showing sp, sp² and sp³ hybridization in its compounds.

(5) Hybrid orbitals are designated as sp, sp², sp³ etc.

 (6) The order of repulsion between lp – lp > lp – bp > bp – bp

(7) The directional properties in hybrid orbital is more than atomic orbitals. Therefore, hybrid orbitals form stronger sigma bond. The directional property of different hybrid orbitals will be in following order. sp < sp2 < sp3 < sp3d < sp3d 2 < sp3d3 since it depends upon the directional nature of orbitals.

Types of Hybridization

(A) sp hybridization

 (a) In this hybridization one s  & one p – orbital of an atom are mixed to give two new hybrid orbitals which are equivalent in shape & energy known as sp hybrid orbitals.

 (b) These two sp hybrid orbitals are arrange in straight line & at bond angle 180°.

(B) sp² Hybridization:

(a) In this hybridization one s & two p orbitals are mixed to give three new sp2 hybrid orbitals which are in same shape & equivalent energies.

 (b) These three sp2 hybrid orbitals are at angle of 120° & giving trigonal planar shape.

(C) sp³ Hybridization:

 (I) In this hybridization one  ‘ s’ orbital and three ‘p’ orbitals of an atom of a molecule or ion, are mixed to give four new hybrid orbitals called as sp3 hybrid orbitals.

 (II) The angle between hybrid orbitals will be 109° 28'

(D) sp³d Hybridization:

(I) In this hybridization one s orbital, three p orbitals and one d orbital are mixed to give five new hybrid orbitals which are equivalent in shape and energy called as sp3d hybrid orbitals.

(II) Out of these five hybrid orbitals, three hybrid orbitals are at 120° angle and two hybrid orbitals are perpendicular to the plane of three hybrid orbitals that is trigonal planar, the shape of molecule becomes trigonal bipyramidal. For example, PF5 showing sp3d hybridisation

(C) sp³d²Hybridisation:

 (I) In this hybridization, one s-orbital, three p-orbitals & two d-orbitals (dz2 , dx2–y2 ) are mixed to give six new hybrid orbitals known as sp3d 2 hybrid orbitals.

(II) The geometry of molecule obtained from above six hybrid orbitals will be symmetrical octahedral or square bipyramidal.

 (III) The angle between all hybrid orbitals will be 90°. Example: SF₆, AlF6 ^–3, PF₆ ^–, ICl5, XeF4, XeOF_4, ICl₄ ^–

(F) sp³d³ Hybridization:

(I) In this hybridization, one s-orbital, three p-orbitals & three d-orbitals are mixed to give seven new hybrid orbitals known as sp³d ³ hybrid orbitals.

(II) These seven sp³d ³ orbitals are configurated in pentagonal bipyramidal shape.

(III) Five bond angles are of 72° and 10 bond angles of 90°. (IV) The following examples showing sp3d 3 hybridization –IF₇ & XeF₆

_{explaination is coming soon}