Home
Class 12
CHEMISTRY
(i) Account for the fact that Eu^(2+) is...

(i) Account for the fact that `Eu^(2+)` is a strong reducing agent.
(ii) Why has Zn lowest enthalpy of atomization in 3d series ?

Text Solution

Verified by Experts

(i) `Eu^(2+)` is a strong reducing agent because it changes `Eu^(3+)` by losing as electron, which is comparatively more stable.
(ii) In Zn, metallic bond is weak due to completely filled electronic configuration i.e., `[Ar]4s^(2)3d^(10)`. Therefore,its enthalpy of atomization is the lowest in teh 3d series of elements.
Doubtnut Promotions Banner Mobile Dark
|

Topper's Solved these Questions

  • D-AND -F BLOCK ELEMENTS

    DINESH PUBLICATION|Exercise Higher order Thinking skill (HOTS)|11 Videos
  • D-AND -F BLOCK ELEMENTS

    DINESH PUBLICATION|Exercise NCERT Exampler , Multiple Choice Questions|21 Videos
  • D-AND -F BLOCK ELEMENTS

    DINESH PUBLICATION|Exercise Additional Important Question|28 Videos
  • CYANIDES, ISOCYANIDES, NITROCOMPOUNDS AND AMINES

    DINESH PUBLICATION|Exercise Unit test - 8|20 Videos
  • ELECTROCHEMISTRY

    DINESH PUBLICATION|Exercise ADDITIONAL NUMERICAL PROBLEMS FOR PRACTICE|12 Videos

Similar Questions

Explore conceptually related problems

Give reasons for the following: Eu^(2+) is a strong reducing agent.

Why Zinc has lowest melting point in 3d series?

Account for the fact that SO_(2) can act as an oxidising and reducing agent while SO_(3) is only oxidising agent.

Account for the following : (a) CuCl_(2) is more stable than Cu_(2)Cl_(2) . (b) Atomic radii of 4d and 5d series elements are nearly same. (c ) Hydrochloric acid is not used in permanganate titrations. or Account for the following : (a) Eu^(2+) is a strong reducing agent. (b) Orange colour of dichromate ion changes to yellow in alkaline medium. (c ) E^(@) ( M^(2+) // M ) values for transiton metals show irregular variation.

In 3d series (Sc to Zn), the enthalpy of atomization of Zn is low. Why ?

In 3d series, the enthalpy of atomisation of zinc is the lowest Why?

(a) How will you prepare : (i) K_(2)MnO_(4) from MnO_(2) ? (ii) Na_(2)Cr_(2)O_(7) from Na_(2)CrO_(4) ? (b) Account for the following: (i) Mn^(2+) is more stable than Fe^(2+) towards oxidation to +3 state. (ii) The enthalpy of atomisation is lowest for Zn is 3d series of the transition elements. (iii) Actinoid elements show wide range of oxidation states.

(a) How do you prepare : (i) K_(2)MnO_(4) from MnO_(2) (ii) Na_(2)Cr_(2)O_(7) from Na_(2)CrO_(4) ? (b) Account for the following : (i) Mn^(2+) is more stable than Fe^(2+) towards oxidation to +3 state. (ii) The enthalpy of atomisation is lowest for Zn in 3d series of transition elements. (iii) Actinoid elements show wide range of oxidation states. or (i) Name the elements of 3d transition series which shows maximum number of oxidation states. Why dows it show so ? (ii)Which transition element of 3d series has positive E_(M^(+)//M)^(@) value and why ? (iii) Out of Cr^(3+) and Mn^(3+) which is stronger oxidizing agent and why ? (iv) Name a member of the lanthanoid series which is well known to exhibit +2 oxidation state. (v) Complete the equation : MnO_(4)^(-) +8 H^(+) + 5e^(-) rarr

Why is H_(3)PO_(2) a stronger reducing agent than H_(3PO_(3) ?

In the 3d series (Sc = 21 to Zn = 30) : (i) Which elements shows maximum number of oxidation states ? (ii) Which elements shows only +3 oxidation state ? (iii) Which elements has the lowest enthalpy of atomization?