NEET Test Series from KOTA - 10 Papers In MS WORD
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CHXI12:ORGANIC CHEMISTRY SOME BASIC PRINCIPLES AND TECHNIQUES
317091
The ratio of the number of \(\mathrm{sp}, \mathrm{sp}^{2}\) and \(\mathrm{sp}^{3}\) orbitals, respectively in the compound is \(\mathrm{CH}_{3}-\mathrm{CH}=\mathrm{C}=\mathrm{CH}-\mathrm{C} \equiv \mathrm{C}-\mathrm{CH}_{3}\)
CHXI12:ORGANIC CHEMISTRY SOME BASIC PRINCIPLES AND TECHNIQUES
317092
Assertion : The \({\text{sp}}\) hybrid orbital contains more \({\text{s}}\) 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 \(50 \% \mathrm{~s}\) character is more electronegative than \(\mathrm{sp}^{2}\) or \(\mathrm{sp}^{3}\) 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 \(\mathrm{H}_{2} \mathrm{C}=\mathrm{O}\)
CHXI12:ORGANIC CHEMISTRY SOME BASIC PRINCIPLES AND TECHNIQUES
317091
The ratio of the number of \(\mathrm{sp}, \mathrm{sp}^{2}\) and \(\mathrm{sp}^{3}\) orbitals, respectively in the compound is \(\mathrm{CH}_{3}-\mathrm{CH}=\mathrm{C}=\mathrm{CH}-\mathrm{C} \equiv \mathrm{C}-\mathrm{CH}_{3}\)
CHXI12:ORGANIC CHEMISTRY SOME BASIC PRINCIPLES AND TECHNIQUES
317092
Assertion : The \({\text{sp}}\) hybrid orbital contains more \({\text{s}}\) 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 \(50 \% \mathrm{~s}\) character is more electronegative than \(\mathrm{sp}^{2}\) or \(\mathrm{sp}^{3}\) 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 \(\mathrm{H}_{2} \mathrm{C}=\mathrm{O}\)
CHXI12:ORGANIC CHEMISTRY SOME BASIC PRINCIPLES AND TECHNIQUES
317091
The ratio of the number of \(\mathrm{sp}, \mathrm{sp}^{2}\) and \(\mathrm{sp}^{3}\) orbitals, respectively in the compound is \(\mathrm{CH}_{3}-\mathrm{CH}=\mathrm{C}=\mathrm{CH}-\mathrm{C} \equiv \mathrm{C}-\mathrm{CH}_{3}\)
CHXI12:ORGANIC CHEMISTRY SOME BASIC PRINCIPLES AND TECHNIQUES
317092
Assertion : The \({\text{sp}}\) hybrid orbital contains more \({\text{s}}\) 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 \(50 \% \mathrm{~s}\) character is more electronegative than \(\mathrm{sp}^{2}\) or \(\mathrm{sp}^{3}\) 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 \(\mathrm{H}_{2} \mathrm{C}=\mathrm{O}\)
CHXI12:ORGANIC CHEMISTRY SOME BASIC PRINCIPLES AND TECHNIQUES
317091
The ratio of the number of \(\mathrm{sp}, \mathrm{sp}^{2}\) and \(\mathrm{sp}^{3}\) orbitals, respectively in the compound is \(\mathrm{CH}_{3}-\mathrm{CH}=\mathrm{C}=\mathrm{CH}-\mathrm{C} \equiv \mathrm{C}-\mathrm{CH}_{3}\)
CHXI12:ORGANIC CHEMISTRY SOME BASIC PRINCIPLES AND TECHNIQUES
317092
Assertion : The \({\text{sp}}\) hybrid orbital contains more \({\text{s}}\) 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 \(50 \% \mathrm{~s}\) character is more electronegative than \(\mathrm{sp}^{2}\) or \(\mathrm{sp}^{3}\) 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 \(\mathrm{H}_{2} \mathrm{C}=\mathrm{O}\)
