Use LEFT and RIGHT arrow keys to navigate between flashcards;
Use UP and DOWN arrow keys to flip the card;
H to show hint;
A reads text to speech;
14 Cards in this Set
- Front
- Back
|
Why do d block elements have high melting point? |
High melting point of d block elements is high due to involvement of electrons from (n-1)d in addition to ns electrons in the interatomic bonding |
|
|
Why do Mn and Tc have low melting point |
M.pt and b.pt depend on intermolecular forces. The stronger the forces, the higher is the m.pt and b.pt. Due to half filled configuration of both Mn and Tc, the enthalpy of atomization is low, hence, the intermolecular forces are low and, therefore, their low melting point. |
|
|
Why do the transition elements exhibit higher enthalpies of atomisation? |
Because of involvement of large no. Of unpaired electrons[(n-1)d and ns electrons] in interatomic bonding. |
|
|
Why do 3d elements have almost the same size even when the nuclear charge increases fron Sc to Zn? |
This is because the 3d electrons shield the 4s electrons more effectively than the outer shell electrons shield one another. Due to this they also have almost same 1st ionization enthalpy. |
|
|
The three terms responsible for ionization enthalpy are-? |
(1) attraction of each electron towards nucleus (2) repulsion between electrons (3) exchange energy |
|
|
What is the role of exchange energy in determining ionisation enthalpy? |
Loss in exchange energy results in stabilty. Stability leads to higher ionisation enthalpy. |
|
|
Which of the 3d transition metals exhibit the highest no. of oxidation states and why? |
Mangenese(Mn) shows highest no. Of oxidation states due to availability of max. Unpaired electrons |
|
|
What does the standard electrode potential[ e.g.- E°(M^3+/M^2+)] reflect about the transition elements? |
Their stability in a particular state E.g- the high value of E°(M^3+/M^2+) of Zn2+ indicates its high stability in that form. |
|
|
How is flourine able to stabilize high oxidation states? |
The ability of flourine to stabilise higher oxidation states is due to 1) higher lattice enthalpy/energy. E.g- CoF3 2)higher bond enthalpy terms for higher covalant compounds. E.g- VF5 and CrF6 |
|
|
What happens when Cu2+ reacts with iodine |
It oxidises I- to I2 |
|
|
Why is Cu(II) more stable than Cu(I)? |
It is due to more negative hydration energy of Cu2+(aq) than Cu+, which more than compensates for second ionization enthalpy of Cu. |
|
|
Which is the only iron compound which has higher oxidation state above Fe2O3? |
Ferrates(VI)- (FeO4)2- Note: they also decompose into Fe2O3 and O2 |
|
|
How is oxygen able to stablizise higher oxidation states? |
Due to its ability to form multiple bonds |
|
|
Why do transition elements form a large number of complex compounds? |
Due to 1) comparitively smaller size of metal ions 2) their higher ionic charges 3) availabilty of d orbitals for bond formation |
