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D-Block elements/ Transition elements: Properties and explanation of How and Why?

Jul 24, 2018 18:03 IST
    d-Block elements
    d-Block elements

    Every year questions are being asked in nearly all the competitive examinations like NEET, JEE, UPTU, WBJE, etc. from d-block element.

    Let’s understand the definition first

    d- Block Elements:

    D-block elements occupy the space between the s-block (most electro positive) and p-block (most electro negative) in the periodic table. Since they bridge the two blocks and show a transition in the properties from the metals to the non-metals, they are also called Transition elements.

    The last electron of this block elements enter into the penultimate shell (The outermost shell of an atom is called valence shell while second last shell is called penultimate shell).

    Any element which has incompletely filled d-orbital in its ground state or in one of the most common oxidation state.

    The configuration must be d1 to d9

    if the configuration is d to d10 , it cannot be a transition element

    eg: Zn, Cd, Hg: the electronic configuration of all the three elements are: 3 d10 3S2, 4 d10 4S2, 5 d10 5S2

    The most common oxidation state of all the three states is: Zn+2, Cd+2, Hg+2 and here the configuration of all the 3 elements are 3 d10, 4 d10 and 5 d10

    So, from above explanation, these all 3 elements are not transition elements

    These all three elements also do not show the typical properties of transition elements except their ability to form the complex compounds

    Let’s see the general electronic configuration of d- block elements: (n-1) d1-10 ns 1-2

    Here d- orbital belongs to penultimate shell (one shell before the last shell)

    Now let’s talk about the transition series:  The set / collection of transition metals is called transition series.

    There are 4 transition series in the transition elements that is 1st, 2nd, 3rd and 4th transition series.

    Properties of transition elements:

    1. Atomic radius:

    As we move from left to right, atomic radius decrease first because of increment of effective nuclear charges and after that it becomes constant (due to the poor shielding effect of d-electrons) and then after it starts increasing towards end because of the e-- e- repulsion which are higher as compare to the increase in the effective nuclear charge.

    So, the trend of change of the atomic radius within the transition series is: decrease, constant and then increase

    2. They can form alloy:

    Transition elements can form alloy because of almost same size, so one atom can take the place of other atom in the crystal lattice.

    3. Transition elements can show variable oxidation states:

    The energy gap between : (n-1) d and ns orbital is very small so, electron from both the orbital can participate in bonding and therefore they can show variable oxidation states

    4. Metallic character:

    Metallic character depends on the presence of unpaired electron which provide then a strong metallic bond and because of strong metallic bond, the melting and boiling points of transition elements are high

    5. Density:

    Density is also high because of presence of unpaired electrons

    6. Enthalpy of atomization:

    The energy required to separate two elements from their atomic state is known as Enthalpy of atomization

    Enthalpy of atomization is high in transition elements and the reason is again the presence of unpaired electrons.

    Eg: when we will move from left to right means from Sc to Zn, the enthalpy of atomization will increase till Mn and then decrease towards the end that is Zn.

    SC                          Mn                   Zn

    3d1 4S2               5d5 4S2               3d10 4S2   

    Number of unpaired electrons:

    In Zn: 1

    In Mn: 5

    In Zn: 0

    So moving from left to right, the number of unpaired electrons will increase till Mn and then decrease till Zn.

    7. Complex formation:

    Because of their small size, high charge and presence of vacant d-orbital

    8. Formation of non- Non-stoichiometric compounds:

    The compound in which the ratio of the transition elements is not fixed is known as Non-stoichiometric compounds

    Eg: Fe0.98 O1

    Reason:  due to the defects in the crystal lattice

                     Show variable valances

    9. Transition compounds can show catalytic property:

    Eg: In hydrogenation reaction, Ni-Pt and Pd are being used as catalyst

         In contact process, V2O5 is used as catalyst which is transition compound

    Reason:

    Transition elements shows variable oxidation states and it is possible when they will form the intermediate during the reaction.

    Large surface area of the transition elements so that the reactants can easily get adsorbed

    10. Thermodynamic stability of transition compounds:

    Let’s talk about the sum of ionization energies of NI(II) and Pt(II)

     

    IE1+ IE2

    IE1+ IE2+IE3+IE4

    Ni

     

    2.49X103

    11.29X103

    Pt

     

    2.66X103

    9.36X103

    Concept: lower the ionization energy of the transition element, more will be the stability of transition compounds

    In the above table if we see, IE1+ IE2 for Ni (II) is lesser than IE1+ IE2 of Pt, therefore, the Ni(ii) compound is more stable than Pt(II)

    Let’s compare the +4 oxidation state for Ni and Pt.

    From the above table we can see that IE1+ IE2+IE3+IE4 for Ni (IV) is higher than IE1+ IE2+IE3+IE4 of Pt (IV), so we can say that

    Pt(IV) compounds are more stable than Ni(IV)compounds

    Eg:

    Q. K2NiCl6   - does not exist

                       K2Pt Cl6   - Exist,  Why??

    Reason:

    In both the compounds, Pt and Ni elements are in +4 oxidation state. As we know that the +4 oxidation state of Pt is more stable than +4 oxidation state of Ni.

    Let us discuss the electrode potential and oxidation state:

    Concept: Lower the electrode potential; grater will be the stability of higher oxidation state of the metal in aqueous solution.

    Eg: let us see the Electrode potential of few transition elements given below

    E Mn+3/Mn+2

    + 1.57 V

    E Fe+3/ Fe+2

    + 0.77 V

    Arrange the given metal ions according to their stability?

    Explanation:

    In case of Mn+3/ Mn+2, Mn+2 is having d5 configuration which is half filled and because of the same, this configuration is more satiable and hiving highest E value

    In case of Fe+3/ Fe+2, the E value is lower as compare to the Mn+3/Mn+2. This means the tendency of Fe+3 to get reduced into Fe+2 is very less.

    Fe+3 is having the d5 configuration and so it is already stable and will not like to reduce into Fe+2

    Let’s have a look on the few questions based on the oxidation states.

    Q. Why Cu+ is unstable in aqueous solution?

    Ans: it undergoes disproportionation reaction (same species gets oxidised as well as reduced)

    2Cu+  ....................................> Cu+2    +    Cu

    Then why Cu+2 is more stable than Cu+ , although the ionization energy of Cu+2 is much higher than Cu+

    More negative hydration enthalpy of Cu+2 compensates for the high ionization enthalpy that is why Cu+2 is more stable.

    Now in low oxidation states, ionic compounds are formed where as in high oxidation state covalent compounds are formed

    Conclusion: here we have covered all the important features of d-block elements and also discussed the few important questions. It is very important to get the clear concept about the topic of this chapter.

    For such informative notes and other study material in English and Hindi language, keep visiting jagranjosh.com

    Also Read:

    UP Board: Class 10 and Class 12th

    All about NIOS

    Bihar Board: Class 10 and Class 12th

    Rajasthan Board: Class 10 and Class 12th

    Maharashtra Board: Class 10 and Class 12th

     

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