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Question: Consider the following statements : \({{S}_{1}}:\,{{N}_{2}}{{H}_{4}}\,is\,pyramidal\,about\,each\,...

Consider the following statements :
S1:N2H4ispyramidalabouteachNatom{{S}_{1}}:\,{{N}_{2}}{{H}_{4}}\,is\,pyramidal\,about\,each\,N\,atom
S2:NH2OHispyramidalabouteachNatomandbentabouttheOatom{{S}_{2}}:\,N{{H}_{2}}OH\,is\,pyramidal\,about\,each\,N\,atom\,and\,bent\,about\,the\,O\,atom
S3:CH3COClistrigonalaboutthecarbonatom(attachedtoOandCl){{S}_{3}}:\,C{{H}_{3}}COCl\,is\,trigonal\,about\,the\,carbon\,atom\,(attached\,to\,O\,and\,Cl)
Select the correct statement.
A.S1andS2only{{S}_{1}}\,and\,{{S}_{2}}\,only
B.S1andS3only{{S}_{1}}\,and\,{{S}_{3}}\,only
C.S2andS3only{{S}_{2}}\,and\,{{S}_{3}}\,only
D.S1,S2andS3{{S}_{1}},\,{{S}_{2\,}}\,and\,{{S}_{3}}

Explanation

Solution

Before solving this question, we should first know about the structures of N2H4{{N}_{2}}{{H}_{4}}, NH2OHN{{H}_{2}}OHand CH3COClC{{H}_{3}}COCl then find out which of the above statements are true. N2H4{{N}_{2}}{{H}_{4}}is hydrazine, NH2OHN{{H}_{2}}OHis hydroxylamine, and CH3COClC{{H}_{3}}COCl is acetyl chloride.

Complete answer:
N2H4{{N}_{2}}{{H}_{4}} is an inorganic compound known as Hydrazine is a simple pnictogen hydride that is very toxic but controlled by a solution. It is a colorless liquid that is flammable and has an odor like ammonia. Polymer foams are prepared with the help of it. They are used as a precursor in pharmaceuticals, in numerous rocket fuels.

As we can see in this structure, N2H4{{N}_{2}}{{H}_{4}}is pyramidal about each N atom.
NH2OHN{{H}_{2}}OH is an inorganic compound known as hydroxylamine. It is a pure material that is unstable crystalline and is white. It is used as an aqueous solution. It is also used in the preparation of oximes.

As we can see in this structure, NH2OHN{{H}_{2}}OH Pyramidal about each N atom and bent about the O atom.
CH3COClC{{H}_{3}}COClis an organic compound known as acetyl chloride, a derivative of acetic acid. It comes under the category of acyl halides. It is a volatile liquid that is colorless and corrosive.

As we can see in this structure, CH3COClC{{H}_{3}}COCl is Trigonal about the carbon atom.

So, Option (D) S1,S2andS3{{S}_{1}},\,{{S}_{2\,}}\,and\,{{S}_{3}} is correct.

Note:
Uses of –
N2H4{{N}_{2}}{{H}_{4}}– Cancer research, medicines, water boilers, inhibitors.
NH2OHN{{H}_{2}}OH- Purification of aldehydes and ketones, antioxidant for fatty acids.
CH3COClC{{H}_{3}}COCl- Dyes, Organic synthesis, quantitative analysis.