Assertion (A) Components of a mixture of red and blue inks can be separated by distributing the components between stationary and mobile phases in paper chromatography.
Reason (R) The coloured components of inks migrate at different rates because paper selectively retains different components according to the difference in their partition between the two phases.
1 What is meant by hybridisation? Compound $\mathrm{CH}_2=\mathrm{C}=\mathrm{CH}_2$ contains $s p$ or $s p^2$-hybridised carbon atoms. Will it be a planar molecule?
Hybridisation is mixing of atomic orbitals to form new hybrid orbitals. The new orbital have the same total electron capacity as the old ones. The properties and energies of the new hybridised orbitals are an average of the unhybridised orbitals.
Hybridisation can be found by counting number of $\sigma-$ bonds around the carbon atom.
$$\begin{aligned} 3 \sigma & =s p^2 \text {-hybridisation } \\ 2 \sigma & =s p \text {-hybridisation } \end{aligned}$$
In allene, carbon atoms 1 and 3 are $s p^2$-hybridised as each one of them is joined by a double bond. And, carbon atom 2 is $s p$-hybridised as it has two double bonds at each of its side. Therefore, the two $\pi$-bonds are perpendicular to each other, in allene, as shown below.
$\mathrm{H}_a$ and $\mathrm{H}_b$ lie in the plane of paper while $\mathrm{H}_c$ and $\mathrm{H}_d$ lie in a plane perpendicular to the plane of the paper. Hence, the allene molecule as a whole is non-planar.
Benzoic acid is an organic compound. Its crude sample can be purified by crystallisation from hot water. What characteristic differences in the properties of benzoic acid and the impurity make this process of purification suitable?
Benzoic acid can be purified by hot water because of following characteristics
(i) Benzoic acid is more soluble in hot water and less soluble in cold water.
(ii) Impurities present in benzoic acid are either insoluble in water or more soluble in water to such an extent that they remain in solution as the mother liquor upon crystallisation.
Two liquids $(A)$ and $(B)$ can be separated by the method of fractional distillation. The boiling point of liquid $(A)$ is less than boiling point of liquid $(B)$. Which of the liquids do you expect to come out first in the distillate ? Explain.
If the difference in boiling points of two liquids is not much, fractional distillation is used to separate them. In this technique, fractionating column is fitted over the mouth of the round bottom flask.
When vapours of a liquid mixture are passed through a fractionating column, the vapours of the low boiling liquid $(A)$ will move up while those of the high boiling liquid will condense and fall back into the flask. Therefore, liquid $(A)$ with low boiling point will distill first.
You have a mixture of three liquids $A, B$ and $C$. There is a large difference in the boiling points of $A$ and rest of the two liquids i.e., $B$ and $C$. Boiling point of liquids $B$ and $C$ are quite close. Liquid $A$ boils at a higher temperature than $B$ and $C$ and boiling point $B$ is lower than $C$. How will you separate the components of the mixture. Draw a diagram showing set up of the apparatus for the process.
The boiling points are in the order of $A>C>B$.
Liquid $A$ can be separated from rest of the mixture of liquid $B$ and $C$ by simple distillation $B$ and $C$ can be separated by fractional distillation.
Due to the fact that boiling point of $A$ is much higher than those of liquids $B$ and $C$. This can be done by using apparatus as shown in figure (I). As the boiling points of liquid $(B)$ and $(C)$ are quite close but much lower than that of $A$, hence, mixture of liquids $(B)$ and $(C)$ will distill together leaving behind liquid $(A)$.
On further heating liquid $(A)$ will distill over. Now, place the mixtures of liquid $(B)$ and $(C)$ in a flask fitted with fractionating column as illustrated in figure (II). On fractional distillation, liquid $(B)$ will distill over first and then liquid $(C)$ as former possess lower boiling point than that of later.