4 Marks Questions

Question 14 Marks

In order to overcome the scarcity of drinking water in a remote village in Gujarat, Arnav and Aariv two young entrepreneurs still in their high school, have developed a unique water purifier that is capable of converting sea water into drinking water. It works on the principle of concentration difference between two solutions.
i. Name the phenomenon/process based on which this product is made? (1)
ii. How difference in concentration of solutions help in converting sea water into drinking water?
iii. What arrangement they must have created in their product to covert sea water into drinking water?
OR
iii. Equimolar solutions of NaCl and glucose are not isotonic. Why?

Answer

i. The product is based on the phenomenon of Reverse Osmosis between solutions of two different concentration.
ii. When solutions of two different concentration are separated by a semipermeable membrane and excess pressure is applied on the solution of higher concentration, solvent flow from higher concentration to lower concentration. This is called Reverse Osmosis and same can be used to treat seawater and convert into drinking water.
iii. Following arrangement must have been made:
i. Use of a semipermeable membrane.
ii. Separate Compartment having seawater and Drinking Water separated by semipermeable membrane.
iii. Excess pressure applied in compartment having sea water.

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Question 24 Marks

The unique behaviour of $Cu ,$ having a positive $E ^{\circ}$ accounts for its inability to liberate $H _2$ from acids. Only oxidising acids $($nitric and hot concentrated sulphuric$)$ react with $Cu ,$ the acids being reduced. The stability of the half $-$ filled $d$ sub $-$ shell in $Mn ^{2+}$ and the completely filled $d ^{10}$ configuration in $Zn ^{2+}$ are related to their $E ^{\circ}$ values, whereas $E ^{\circ}$ for $Ni$ is related to the highest negative $\Delta_{h y d} H ^{\circ}$. An examination of the $E{\left(M^{3+} / M^{2+}\right)}^o$ values the low value for $Sc$ reflects the stability of $Sc ^{3+}$ which has a noble gas configuration. The comparatively high value for $Mn$ shows that $Mn ^{2+}\left( d ^5\right)$ is particularly stable, whereas a comparatively low value for $Fe$ shows the extra stability of $Fe ^{3+}\left( d ^5\right)$. The comparatively low value for $V$ is related to the stability of $V ^{2+} \ ($half $-$ filled $t _{2 g}$ level$)$.

i. Why $Zn$ has high value for $M^{3+}/M^{2+}$ Standard Electrode Potentials?
ii. Transition metals, despite high $E^\circ$ oxidation, are poor reducing agents. Justify.
iii. Why is $Cr^2+$ reducing and Mn³+ oxidising when both $Cr$ and $Mn$ have $d^4$ configuration?
$OR$
iii. Why $Cu^{2+}$ is more stable than $Cu^+$?

Answer

i. Due to the removal of an electron from the stable $d ^{10}$ configuration of $Zn ^{2+}$.
ii. Transition metals despite having high $E ^{\circ}$ oxidation, are poor reducing agents because of their high heat of vaporization, high ionisation energies and low heats of hydration.
iii. $Cr ^{2+}$ is reducing as its configuration changes from $d _4$ to $d _3$, the having a half $-$ filled $t _{2 g}$ level.
On the other hand, the change from $Mn ^{3+}$ to $Mn ^{2+}$ results in the half $-$ filled $\left( d _5\right)$ configuration which has extra stability.

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Chemistry 4 Marks Questions

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