Chemistry 12 Chapter 8 Exemplar Solutions the P and F Block Elements Part 4

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III. Short Answer Type

Question 32:

Why does copper not replace hydrogen from acids?


Cu shows positive value.

Question 33:

Why values for and are more negative than expected?


Hint: Negative values for and are related to stabilities of half filled and fully filled configuration respectively. But for value is related to the highest negative enthalpy of hydration.

Question 34:

Why first ionisation enthalpy of is lower than that of ?


Ionisation enthalpy of is lower due to stability of and the value for is higher because its electron comes out from orbital.

Question 35:

Transition elements show high melting points. Why?


The high melting points of transition metals are attributed to the involvement of greater number of electrons in the interatomic metallic bonding from (n-1) d-orbitals in addition to ns electrons

Question 36:

When is treated with, a white precipitate is formed. Explain the reaction with the help of chemical equation.



Question 37:

Out of and , which is more stable and why?



is more stable than . The stability of rather than is due to the much more negative than .

Question 38:

When a brown compound of manganese (A) is treated with it gives a gas (B). The gas taken in excess, reacts with NH3 to give an explosive compound (C). Identify compounds A, B and C.


Question 39:

Although fluorine is more electronegative than oxygen, but the ability of oxygen to stabilise higher oxidation states exceeds that of fluorine. Why?


Hint: It is due to the ability of oxygen to form multiple bonds to metals

Question 40:

Although and ions have same number of unpaired electrons but the magnetic moment of is . and that of Why?


Hint: Due to symmetrical electronic configuration there is no orbital contribution in ion. However appreciable orbital contribution takes place in ion.

Question 41:

Ionisation enthalpies of , Pr and are higher than and . Why?


Hint: It is because in the beginning, when orbitals begin to be occupied, they will penetrate less into the inner core of electrons. The electrons will therefore, be more effectively shielded from the nuclear charge than electrons of the corresponding lanthanoids. Therefore outer electrons are less firmly held and they are available for bonding in the actinoids.

Question 42:

Although belongs to and belongs to transition series but it is quite difficult to separate them. Why?


Hint: Due to lanthanoid contraction, they have almost same size and .

Question 43:

Although oxidation states is the characteristic oxidation state of lanthanoids but cerium shows oxidation state also. Why?


It is because after losing one more electron acquires stable electronic configuration.

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