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{:("List-I","List-II"),((P)C(n)H(2n+2),(...

`{:("List-I","List-II"),((P)C_(n)H_(2n+2),(1)"Alkyne"),((Q) C_(n)H_(2n),(2)"Alkene"),((R ) C_(n)H_(2n-2),(3)"Alkane"),((S)C_(n)H_(2n-1),(4)"Alkyl group"):}`

A

P-3, Q-2, R-1, S-4

B

P-1, Q-2, R-3, S-4

C

P-2, Q-3, R-1, S-4

D

P-3, Q-1, R-2, S-4

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem of matching the general formulas of hydrocarbons with their respective names, we can follow these steps: ### Step 1: Identify the General Formula for Alkanes - The general formula for alkanes is \( C_nH_{2n+2} \). - This means for every n carbon atoms, there are \( 2n + 2 \) hydrogen atoms. **Hint:** Remember that alkanes are saturated hydrocarbons with single bonds. ### Step 2: Identify the General Formula for Alkenes - The general formula for alkenes is \( C_nH_{2n} \). - This indicates that for every n carbon atoms, there are \( 2n \) hydrogen atoms. **Hint:** Alkenes contain at least one double bond between carbon atoms. ### Step 3: Identify the General Formula for Alkynes - The general formula for alkynes is \( C_nH_{2n-2} \). - This shows that for every n carbon atoms, there are \( 2n - 2 \) hydrogen atoms. **Hint:** Alkynes contain at least one triple bond between carbon atoms. ### Step 4: Identify the General Formula for Alkyl Groups - The general formula for alkyl groups is \( C_nH_{2n-1} \). - This means that for every n carbon atoms, there are \( 2n - 1 \) hydrogen atoms. **Hint:** Alkyl groups are derived from alkanes by removing one hydrogen atom. ### Step 5: Match Each Formula with Its Name - **P: \( C_nH_{2n+2} \)** matches with **(3) Alkane**. - **Q: \( C_nH_{2n} \)** matches with **(2) Alkene**. - **R: \( C_nH_{2n-2} \)** matches with **(1) Alkyne**. - **S: \( C_nH_{2n-1} \)** matches with **(4) Alkyl group**. ### Final Matching: - P → 3 (Alkane) - Q → 2 (Alkene) - R → 1 (Alkyne) - S → 4 (Alkyl group) ### Summary of Matches: - **P (C_nH_{2n+2})** → **(3) Alkane** - **Q (C_nH_{2n})** → **(2) Alkene** - **R (C_nH_{2n-2})** → **(1) Alkyne** - **S (C_nH_{2n-1})** → **(4) Alkyl group**
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Similar Questions

Explore conceptually related problems

Match the following the compounds of column I with column II. {:(,"Column-I",,"Column-II",),((A),C_(n)H_(2n+2),(p),"Alkynes",),((B),C_(n)H_(2n),(q),"Alkenes",),((C),C_(n)H_(2n-2),(r),"Cyclohexane",),((D),C_(6)H_(12),(s),"Paraffins or alkanes",):}

Alkynes : C_(n)H_(2n-2) :: Alkenes : _______________

Knowledge Check

  • The formula C_(n)H_(2n-2) shows

    A
    Alkene & Alkyne
    B
    Alkyne & Alkadiyne
    C
    Alkane & Alkadiene
    D
    Alkyne & Alkadiene
  • C_(n)H_(2n)O_(2) is the formula of

    A
    Carbohydrate
    B
    Fatty acid
    C
    Fat
    D
    Nucleic acid.
  • C_(n)H_(2n)O_(n) is the formula of

    A
    Fatty acid
    B
    Fat
    C
    Glycerol
    D
    Carbohydrate.
  • Similar Questions

    Explore conceptually related problems

    Prove that ^nC_(0)^(2n)C_(n)-^(n)C_(1)^(2n-2)C_(n)+^(n)C_(2)^(2n-4)C_(n)-...=2^(n)

    Prove that ^nC_(0)^(2n)C_(n)-^(n)C_(1)^(2n-1)C_(n)+^(n)C_(2)xx^(2n-2)C_(n)++(-1)^(n)sim nC_(n)^(n)C_(n)=1

    Prove that (""^(2n)C_(0))^(2)-(""^(2n)C_(1))^(2)+(""^(2n)C_(2))^(2)-…+(""^(2n)C_(2n))^(2)=(-1)^(n)*""^(2n)C_(n) .

    {:(List-I,List-II),((A)"Energy",(1)(2pize^(2))/(nh)),((B)"Velocity",(2)(-2pi^(2)mz^(2)e^(4))/(n^(2)h^(2))),((C)"Rydberg constant",(3)(2pi^(2)mz^(2)e^(4))/(h^(3)c)),((D)"Radius",(4)(n^(2)h^(2))/(4pi^(2)mze^(2))),(,(5)(-4pi^(2)mz^(2)e^(4))/(n^(2)h^(2))):} The correct match is

    General formula C_(n) H_(2n-2) represents