NCERT Exercise Solution Of Chemistry (CH1-16) Class-12

A Hydrocarbon with the molecular formula , C5H10 belongs to the group with a General molecular formula CnH2n . Therefore , It May either be an Alkene Or a Cycloalkane . Since Hydrocarbon Does Not React With Chlorine In The Dark , it cannot be an Alkene . Thus it should Be a Cycloalkane . Further , The Hydrocarbon Gives a Single Monochloro Compound , C5H9Cl By Reacting With Chlorine In Bright Sunlight . Since a Single MonoChloro Compound Is Formed , The Hydrocarbon Must Contain H atoms That Are All Equivalent . Also , As All H Atoms Of A Cycloalkane Are Equivalent , The Hydrocarbon Must Be a Cycloalkane . Hence  , The Said Compound Is Cyclopentane .                     CYCLOPENTANE THE REACTION INVOLVED IN THE QUESTION ARE : AMAZON AD pTron Bassbuds Lite V2 in-Ear True Wireless Bluetooth Headphones (TWS) with Mic - (Black)

CCl4 is a Symmetrical Molecule . Therefore , The Dipole Moments Of All Four C-Cl Bonds Cancel Each Other . Hence , its resultant Dipole Moment Is Zero . As Shown In Figure , In CHCl3 , The resultant of Dipole Moments Of Two C-Cl Bonds Is Opposed By The Resultant Of Dipole Moments Of One C-H Bond and One C-Cl Bond . Since the resultant of one C-H Bond And One    C-Cl Bond Dipole moments is smaller than two C-Cl Bonds , the opposition is to a small extent . As a Result , CHCl3 Has A Small Dipole Moment Of 1.08 D. On The Other Hand , In Case Of CH2Cl2 , The Resultant Of The Dipole Moments Of Two  C-Cl Bonds Is Strengthened By The Resultant Of The Dipole Moments Of Two C-H Bonds . As a result , CH2Cl2 Has A Higher Dipole Moment Of 1.60 D Than ChCl3 i.e. , CH2Cl2  Has The Highest Dipole Moment . Hence , The Given Compound Can Be Arranged In The Increasing Order Of Dipole Moments As :-  CCl4<CHCl3<CH2Cl2 AMAZON ADVERTISEMENT