2 Diamond is \(\mathrm{sp}^{3}\) hybridized and graphite is \(\mathrm{sp}^{2}\) hybridized.
3 Both diamond and graphite are used as dry lubricants.
4 Diamond and graphite have two dimensional network.
Explanation:
In diamond each carbon is bonded with four other carbon atoms. So, hybridisation of carbon atom is \(\mathrm{sp}^{3}\). In graphite each carbon is bonded with three other carbon atoms. So, hybridisation of carbon atom is \(\mathrm{sp}^{2}\).
NEET - 2022
CHXI11:THE P-BLOCK ELEMENTS
316151
Thermodynamically, the most stable form of carbon is
1 diamond
2 graphite
3 peat
4 coal
Explanation:
Graphite is thermodynamically more stable than diamond because its free energy of formation is \(1.9 \mathrm{kJmol}^{-}\)lower at room temperature and ordinary \(\mathrm{P}\).
CHXI11:THE P-BLOCK ELEMENTS
316152
Carbon- 60 contains
1 20 pentagons and 12 hexagons
2 12 pentagons and 20 hexagons
3 30 pentagons and 30 hexagons
4 24 pentagons and 36 hexagons
Explanation:
Conceptual Questions
CHXI11:THE P-BLOCK ELEMENTS
316153
\(\mathrm{C}-\mathrm{C}\) bond length is maximum in
NEET Test Series from KOTA - 10 Papers In MS WORD
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CHXI11:THE P-BLOCK ELEMENTS
316150
Choose the correct statement :
1 Diamond is covalent and graphite is ionic.
2 Diamond is \(\mathrm{sp}^{3}\) hybridized and graphite is \(\mathrm{sp}^{2}\) hybridized.
3 Both diamond and graphite are used as dry lubricants.
4 Diamond and graphite have two dimensional network.
Explanation:
In diamond each carbon is bonded with four other carbon atoms. So, hybridisation of carbon atom is \(\mathrm{sp}^{3}\). In graphite each carbon is bonded with three other carbon atoms. So, hybridisation of carbon atom is \(\mathrm{sp}^{2}\).
NEET - 2022
CHXI11:THE P-BLOCK ELEMENTS
316151
Thermodynamically, the most stable form of carbon is
1 diamond
2 graphite
3 peat
4 coal
Explanation:
Graphite is thermodynamically more stable than diamond because its free energy of formation is \(1.9 \mathrm{kJmol}^{-}\)lower at room temperature and ordinary \(\mathrm{P}\).
CHXI11:THE P-BLOCK ELEMENTS
316152
Carbon- 60 contains
1 20 pentagons and 12 hexagons
2 12 pentagons and 20 hexagons
3 30 pentagons and 30 hexagons
4 24 pentagons and 36 hexagons
Explanation:
Conceptual Questions
CHXI11:THE P-BLOCK ELEMENTS
316153
\(\mathrm{C}-\mathrm{C}\) bond length is maximum in
2 Diamond is \(\mathrm{sp}^{3}\) hybridized and graphite is \(\mathrm{sp}^{2}\) hybridized.
3 Both diamond and graphite are used as dry lubricants.
4 Diamond and graphite have two dimensional network.
Explanation:
In diamond each carbon is bonded with four other carbon atoms. So, hybridisation of carbon atom is \(\mathrm{sp}^{3}\). In graphite each carbon is bonded with three other carbon atoms. So, hybridisation of carbon atom is \(\mathrm{sp}^{2}\).
NEET - 2022
CHXI11:THE P-BLOCK ELEMENTS
316151
Thermodynamically, the most stable form of carbon is
1 diamond
2 graphite
3 peat
4 coal
Explanation:
Graphite is thermodynamically more stable than diamond because its free energy of formation is \(1.9 \mathrm{kJmol}^{-}\)lower at room temperature and ordinary \(\mathrm{P}\).
CHXI11:THE P-BLOCK ELEMENTS
316152
Carbon- 60 contains
1 20 pentagons and 12 hexagons
2 12 pentagons and 20 hexagons
3 30 pentagons and 30 hexagons
4 24 pentagons and 36 hexagons
Explanation:
Conceptual Questions
CHXI11:THE P-BLOCK ELEMENTS
316153
\(\mathrm{C}-\mathrm{C}\) bond length is maximum in
2 Diamond is \(\mathrm{sp}^{3}\) hybridized and graphite is \(\mathrm{sp}^{2}\) hybridized.
3 Both diamond and graphite are used as dry lubricants.
4 Diamond and graphite have two dimensional network.
Explanation:
In diamond each carbon is bonded with four other carbon atoms. So, hybridisation of carbon atom is \(\mathrm{sp}^{3}\). In graphite each carbon is bonded with three other carbon atoms. So, hybridisation of carbon atom is \(\mathrm{sp}^{2}\).
NEET - 2022
CHXI11:THE P-BLOCK ELEMENTS
316151
Thermodynamically, the most stable form of carbon is
1 diamond
2 graphite
3 peat
4 coal
Explanation:
Graphite is thermodynamically more stable than diamond because its free energy of formation is \(1.9 \mathrm{kJmol}^{-}\)lower at room temperature and ordinary \(\mathrm{P}\).
CHXI11:THE P-BLOCK ELEMENTS
316152
Carbon- 60 contains
1 20 pentagons and 12 hexagons
2 12 pentagons and 20 hexagons
3 30 pentagons and 30 hexagons
4 24 pentagons and 36 hexagons
Explanation:
Conceptual Questions
CHXI11:THE P-BLOCK ELEMENTS
316153
\(\mathrm{C}-\mathrm{C}\) bond length is maximum in
