Class 11th
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New answer posted
11 months agoContributor-Level 10
4.28. (a) Acetylene C2? H2? contains 3? bonds and 2? bonds.
(b) Ethylene C2? H4? contains 5? bonds and 1? bond.
New answer posted
11 months agoContributor-Level 10
(a) H-COOH
CH3—COOH
CH3CH2—COOH
CH3CH2CH2—COOH
CH3CH2CH2CH2—COOH
(b) CH3COCH3
CH3COCH2CH3
CH3COCH2CH2CH3
CH3COCH2CH2CH2CH3
CH3CO (CH3)4CH3
(c) H—CH=CH2
CH3CH=CH2
CH3CH2CH=CH2
CH3CH2CH2CH=CH2
CH3CH2CH2CH2CH=CH2
New answer posted
11 months agoContributor-Level 10
(a) 2, 2-Demethylpentane
(b) 2, 4, 7-Trimethyloctane. For two alkyl groups on the same carbon its locant is repeated twice, 2, 4, 7-locant set is lower than 2, 5, 7.
(c) 2- Chloro-4-methylpentane. Alphabetical order of substituents, (d) But-3-yn-1-ol. Lower locant for the principal functional group, i.e., alcohol.
New answer posted
11 months ago
Contributor-Level 10
(a) Propylbenzene (b) 3-Methylpentanenitrite (c) 2, 5-Dimethylheptane
(d) 3-Bromo- 3-chloroheptane (e) 3-Chloropropanal (f) 2, 2-Dichloroethanol
New question posted
11 months agoNew question posted
11 months agoNew answer posted
11 months agoContributor-Level 10
Single bond contains only one sigma bond and double bond contains one sigma and one pi bond.
- In this structure, nine single bonds and three double bonds are present. So, there are 12σ and 3π bonds present.
- In this structure, eighteen single bonds are present. So, there are only 18σ bonds present.
- In this structure, four single bonds are present. So, there are only 4σ bonds present.
- In this structure, four single bonds and two double bonds are present. So, there are 6σ and 2π bonds present.
- In this structure, five single bonds and one double bond are present. So, there are 6σ and 1π bonds present.
- In this structure, seven single bonds
New answer posted
11 months agoContributor-Level 10
4.26. In BF3, B atom is sp2 hybridised. In NH3, N is sp3 hybridised.
After the reaction, hybridisation of B changes from sp2 to sp3, while the hybridisation of N remains the same.
New question posted
11 months agoNew answer posted
11 months agoContributor-Level 10
4.25. Electronic configuration of 13Al in ground state= 1s2 2s2 2p6 3s2 3p1
Electronic configuration of 13Al in excited state = 1s2 2s2 2p6 3s1 3px13py1
Hence, hybridisation will be sp2 and this makes the geometry to be Trigonal planar (in case of AlCl3).
In AlCl–4, the empty 3pz orbital is also involved. So, the hybridisation is sp3 and the shape is tetrahedral
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