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CHXI12:ORGANIC CHEMISTRY SOME BASIC PRINCIPLES AND TECHNIQUES

317091 The ratio of the number of $sp, {sp}^{2}$ and ${sp}^{3}$ orbitals, respectively in the compound is ${CH}_{3} - CH = C = CH - C \equiv C - {CH}_{3}$

1

1 : 1 : 1

2

2 : 2 : 1

3

3 : 2 : 1

4

3 : 2 : 2

Explanation:

$H_{3} \underset{s p^{3}}{C} - H \underset{s p^{2}}{C} = \underset{s p}{C} = \underset{s p^{2}}{C} H - \underset{s p}{C} \equiv \underset{s p}{C} - \underset{s p^{3}}{C} H_{3}$
$s p h y b r i d o r b i t a l s = 3$
$s p^{2} h y b r i d o r b i t a l s = 2$
$s p^{3} h y b r i d o r b i t a l s = 2$ .

CHXI12:ORGANIC CHEMISTRY SOME BASIC PRINCIPLES AND TECHNIQUES

317092 Assertion :
The $sp$ hybrid orbital contains more $s$ character and hence it is closer to its nucleus.
Reason :
Hybridisation has no effect on the bond length and bond enthalpy in organic compounds.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but Reason is correct.

Explanation:

Hybridisation influences the bond length and bond enthalpy (strength) in organic compounds. Thus, sp hybridised carbon atom having hybrid orbitals with $50 % s$ character is more electronegative than ${sp}^{2}$ or ${sp}^{3}$ hybridised carbon atoms. So, the option (3) is correct.

CHXI12:ORGANIC CHEMISTRY SOME BASIC PRINCIPLES AND TECHNIQUES

317093 Write the state of hybridisation of carbon and shape in the following compound
$H_{2} C = O$

1

s p^{2}

hybridised carbon, trigonal planar

2

s p^{3}

hybridised carbon, tetrahedral

3

s p

hybridised carbon, linear

4 No hybridisation, trigonal planar

CHXI12:ORGANIC CHEMISTRY SOME BASIC PRINCIPLES AND TECHNIQUES

317094 Which has most electronegative carbon atom?

1

{CH}_{3} {CH}_{2} {CH}_{2} OH

2

{CH}_{2} = {CHCH}_{2} OH

3

CH \equiv {CCH}_{2} OH

4

{CH}_{2} = CHOH

Explanation:

$\underset{s p}{C} H \equiv \underset{s p}{C} \underset{s p^{3}}{C} H_{2} O H$
$E N : C_{s p} > C_{s p^{2}} > C_{s p^{3}}$ .

CHXI12:ORGANIC CHEMISTRY SOME BASIC PRINCIPLES AND TECHNIQUES

317091 The ratio of the number of $sp, {sp}^{2}$ and ${sp}^{3}$ orbitals, respectively in the compound is ${CH}_{3} - CH = C = CH - C \equiv C - {CH}_{3}$

1

1 : 1 : 1

2

2 : 2 : 1

3

3 : 2 : 1

4

3 : 2 : 2

Explanation:

$H_{3} \underset{s p^{3}}{C} - H \underset{s p^{2}}{C} = \underset{s p}{C} = \underset{s p^{2}}{C} H - \underset{s p}{C} \equiv \underset{s p}{C} - \underset{s p^{3}}{C} H_{3}$
$s p h y b r i d o r b i t a l s = 3$
$s p^{2} h y b r i d o r b i t a l s = 2$
$s p^{3} h y b r i d o r b i t a l s = 2$ .

CHXI12:ORGANIC CHEMISTRY SOME BASIC PRINCIPLES AND TECHNIQUES

317092 Assertion :
The $sp$ hybrid orbital contains more $s$ character and hence it is closer to its nucleus.
Reason :
Hybridisation has no effect on the bond length and bond enthalpy in organic compounds.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but Reason is correct.

Explanation:

Hybridisation influences the bond length and bond enthalpy (strength) in organic compounds. Thus, sp hybridised carbon atom having hybrid orbitals with $50 % s$ character is more electronegative than ${sp}^{2}$ or ${sp}^{3}$ hybridised carbon atoms. So, the option (3) is correct.

CHXI12:ORGANIC CHEMISTRY SOME BASIC PRINCIPLES AND TECHNIQUES

317093 Write the state of hybridisation of carbon and shape in the following compound
$H_{2} C = O$

1

s p^{2}

hybridised carbon, trigonal planar

2

s p^{3}

hybridised carbon, tetrahedral

3

s p

hybridised carbon, linear

4 No hybridisation, trigonal planar

CHXI12:ORGANIC CHEMISTRY SOME BASIC PRINCIPLES AND TECHNIQUES

317094 Which has most electronegative carbon atom?

1

{CH}_{3} {CH}_{2} {CH}_{2} OH

2

{CH}_{2} = {CHCH}_{2} OH

3

CH \equiv {CCH}_{2} OH

4

{CH}_{2} = CHOH

Explanation:

$\underset{s p}{C} H \equiv \underset{s p}{C} \underset{s p^{3}}{C} H_{2} O H$
$E N : C_{s p} > C_{s p^{2}} > C_{s p^{3}}$ .

CHXI12:ORGANIC CHEMISTRY SOME BASIC PRINCIPLES AND TECHNIQUES

317091 The ratio of the number of $sp, {sp}^{2}$ and ${sp}^{3}$ orbitals, respectively in the compound is ${CH}_{3} - CH = C = CH - C \equiv C - {CH}_{3}$

1

1 : 1 : 1

2

2 : 2 : 1

3

3 : 2 : 1

4

3 : 2 : 2

Explanation:

$H_{3} \underset{s p^{3}}{C} - H \underset{s p^{2}}{C} = \underset{s p}{C} = \underset{s p^{2}}{C} H - \underset{s p}{C} \equiv \underset{s p}{C} - \underset{s p^{3}}{C} H_{3}$
$s p h y b r i d o r b i t a l s = 3$
$s p^{2} h y b r i d o r b i t a l s = 2$
$s p^{3} h y b r i d o r b i t a l s = 2$ .

CHXI12:ORGANIC CHEMISTRY SOME BASIC PRINCIPLES AND TECHNIQUES

317092 Assertion :
The $sp$ hybrid orbital contains more $s$ character and hence it is closer to its nucleus.
Reason :
Hybridisation has no effect on the bond length and bond enthalpy in organic compounds.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but Reason is correct.

Explanation:

Hybridisation influences the bond length and bond enthalpy (strength) in organic compounds. Thus, sp hybridised carbon atom having hybrid orbitals with $50 % s$ character is more electronegative than ${sp}^{2}$ or ${sp}^{3}$ hybridised carbon atoms. So, the option (3) is correct.

CHXI12:ORGANIC CHEMISTRY SOME BASIC PRINCIPLES AND TECHNIQUES

317093 Write the state of hybridisation of carbon and shape in the following compound
$H_{2} C = O$

1

s p^{2}

hybridised carbon, trigonal planar

2

s p^{3}

hybridised carbon, tetrahedral

3

s p

hybridised carbon, linear

4 No hybridisation, trigonal planar

CHXI12:ORGANIC CHEMISTRY SOME BASIC PRINCIPLES AND TECHNIQUES

317094 Which has most electronegative carbon atom?

1

{CH}_{3} {CH}_{2} {CH}_{2} OH

2

{CH}_{2} = {CHCH}_{2} OH

3

CH \equiv {CCH}_{2} OH

4

{CH}_{2} = CHOH

Explanation:

$\underset{s p}{C} H \equiv \underset{s p}{C} \underset{s p^{3}}{C} H_{2} O H$
$E N : C_{s p} > C_{s p^{2}} > C_{s p^{3}}$ .

CHXI12:ORGANIC CHEMISTRY SOME BASIC PRINCIPLES AND TECHNIQUES

317091 The ratio of the number of $sp, {sp}^{2}$ and ${sp}^{3}$ orbitals, respectively in the compound is ${CH}_{3} - CH = C = CH - C \equiv C - {CH}_{3}$

1

1 : 1 : 1

2

2 : 2 : 1

3

3 : 2 : 1

4

3 : 2 : 2

Explanation:

$H_{3} \underset{s p^{3}}{C} - H \underset{s p^{2}}{C} = \underset{s p}{C} = \underset{s p^{2}}{C} H - \underset{s p}{C} \equiv \underset{s p}{C} - \underset{s p^{3}}{C} H_{3}$
$s p h y b r i d o r b i t a l s = 3$
$s p^{2} h y b r i d o r b i t a l s = 2$
$s p^{3} h y b r i d o r b i t a l s = 2$ .

CHXI12:ORGANIC CHEMISTRY SOME BASIC PRINCIPLES AND TECHNIQUES

317092 Assertion :
The $sp$ hybrid orbital contains more $s$ character and hence it is closer to its nucleus.
Reason :
Hybridisation has no effect on the bond length and bond enthalpy in organic compounds.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but Reason is correct.

Explanation:

Hybridisation influences the bond length and bond enthalpy (strength) in organic compounds. Thus, sp hybridised carbon atom having hybrid orbitals with $50 % s$ character is more electronegative than ${sp}^{2}$ or ${sp}^{3}$ hybridised carbon atoms. So, the option (3) is correct.

CHXI12:ORGANIC CHEMISTRY SOME BASIC PRINCIPLES AND TECHNIQUES

317093 Write the state of hybridisation of carbon and shape in the following compound
$H_{2} C = O$

1

s p^{2}

hybridised carbon, trigonal planar

2

s p^{3}

hybridised carbon, tetrahedral

3

s p

hybridised carbon, linear

4 No hybridisation, trigonal planar

CHXI12:ORGANIC CHEMISTRY SOME BASIC PRINCIPLES AND TECHNIQUES

317094 Which has most electronegative carbon atom?

1

{CH}_{3} {CH}_{2} {CH}_{2} OH

2

{CH}_{2} = {CHCH}_{2} OH

3

CH \equiv {CCH}_{2} OH

4

{CH}_{2} = CHOH

Explanation:

$\underset{s p}{C} H \equiv \underset{s p}{C} \underset{s p^{3}}{C} H_{2} O H$
$E N : C_{s p} > C_{s p^{2}} > C_{s p^{3}}$ .