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Question 14 Marks

Two allotropic forms of carbon which are crystalline in nature, are diamond and graphite. They differ physically but
chemically they are similar. Diamond is the hardest crystalline form of carbon. In diamond, each carbon atom islinked to four other carbon atoms by covalent bonds. In graphite, each carbon atom is linked to three other carbonatoms by covalent bond. Graphite is relatively soft and greasy. It is also a good conductor of electricity. The C-Cbond length in graphite is 141.5 pm while in diamond it is 154 pm .
i. Which is a good conductor of heat and electricity- graphite or diamond? Explain.
ii. Which binding force is present in the structure of diamond?
iii. Why Diamond is not a good conductor of electricity and heat?
OR
Draw the structure of diamond showing bond between carbons.

Answer

i. Graphite is the good conductor of heat and electricity because In graphite one carbon atom is attached to three other carbonatoms. One electron of carbon remains free. Due to this free valence electron graphite is an thermal and electrical conductor.
ii. Covalent bonding is present between carbon atoms in diamond i.e. they share their electrons with each other.
iii. In diamond, one carbon is attached to four other carbon atoms hence it has no free electron. So it do not show heating and electrical conductivity.
OR

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Question 24 Marks

More than three million carbon compounds have been discovered in the field of chemistry. The diversity of these compounds is due to the capacity of carbon atoms for bonding with one another as well as with other atoms.
Most of the carbon compounds are poor conductors of electricity and have low melting and boiling points.
a. Write the molecular formula of first two members of homologous series having functional group
b. Given below are the formulae of some functional groups:

Write the name of these functional groups.
c. What would be observed on adding a $5 \%$ alkaline potassium permanganate drop by drop to some warm ethanol taken in a test tube? State the role of $\mathrm{KMnO}_{4}$ in the reaction and write the chemical equation for the reaction involved.
OR
Write the name of the compound formed when ethanol is heated at 443 K temperature with excess of conc.$\mathrm{H}_{2} \mathrm{SO}_{4}$. What is the role of conc. $\mathrm{H}_{2} \mathrm{SO}_{4}$ in the reaction? Write the chemical equation for the reactioninvolved.

Answer

a. $\mathrm{CH}_{3} \mathrm{Br}$
$\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Br}$
b. Aldehyde
Ketone
c. The colour of $\mathrm{KMnO}_{4}$ disappears;
$\mathrm{KMnO}_{4}$ acts as an oxidizing agent.
$CH _3 CH _2 OH \xrightarrow{\text { Alkaline } KMnO _4+\text { Heat }} CH _3 COOH$
OR
Ethene
Conc. $\mathrm{H}_{2} \mathrm{SO}_{4}$ acts as a dehydrating agent.
$ \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH} \xrightarrow[443 \mathrm{~K}]{\text { Conc. } \mathrm{H}_{2} \mathrm{SO}_{4}+\text { Heat }} \mathrm{CH}_{2}=\mathrm{CH}_{2}+\mathrm{H}_{2} \mathrm{O}$

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