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As we learnt in Mole Fraction - It is the ratio of moles of solute or moles of solvent to moles of the solution. - wherein If a substance ‘A’ dissolves in substance ‘B’ and their number of moles are  and  respectively; then the mole fractions of A and B are given as Mole fraction of A = (number of moles of A)/(number of moles of solution ) = /(+ )    At   ,                                      ...

 

 

a lone pair refers to a pair of valence electrons that are not shared with another atom and is sometimes called an unshared pair or non-bonding pair. Lone pairs are found in the outermost electron shell of atoms. They can be identified by using a Lewis structure.
Engineering
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Which of the following arrangements shows the schematic alignment of magnetic moments of antiferromagnetic substance?

  • Option 1)

  • Option 2)

  • Option 3)

  • Option 4)

 

As we learned

The substances which are expected to passes para-magnetism or ferromagnetism on the basis of unpaired electrons but actually, they possess zero net magnetic moments are called anti ferromagnetic substance.


Option 1)

Option 2)

Option 3)

Option 4)

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Engineering
127 Views   |  

The major product of the following reaction is :

  • Option 1)

  • Option 2)

  • Option 3)

  • Option 4)

 
As we learned    Friedel Crafts reaction - Anisole reacts with an alkyl halide and acyl halide introducing alkyl and acyl groups in ortho and para positions. - wherein       Option 1) Option 2) Option 3) Option 4)
Engineering
357 Views   |  

The increasing order of nitration of the following compounds is :

  • Option 1)

    (b) < (a) <  (c) < (d)

  • Option 2)

    (a) < (b) <  (c) < (d)

  • Option 3)

    (b) < (a) < (d) < (c) 

  • Option 4)

    (a) < (b) < (d) < (c) 

 
As we learnt in  Gabriel Phthalimide Synthesis - Further alkylation can be stopped and a pure primary amine can be obtained by alkylation of phthalimide ( Gabriel synthesis ) followed by hydrolysis. - wherein     Here the aniline is least reactive due to formation of anilinium ion in acidic medium.    Option 1) (b) < (a) <  (c) < (d) Option 2) (a) < (b) <  (c) < (d) Option 3) (b) < (a) <...
Engineering
140 Views   |  

The reagent(s) required for the following conversion are :

  • Option 1)

    (i) B_{2}H_{6}\; \; \; \; \; \; \; \; (ii)SnCl_{2}/HCl\; \; \; \; \; \; \; \; (iii)H_{3}O^{+}

     

     

     

  • Option 2)

    (i) LiAlH_{4}\; \; \; \; \; \; \; \; (ii)H_{3}O^{+}

  • Option 3)

    (i) B_{2}H_{6}\; \; \; \; \; \; \; \; (ii)DIBAL-H\; \; \; \; \; \; \; \; (iii)H_{3}O^{+}

  • Option 4)

    (i) NaBH_{4}\; \; \; \; \; \; \; \; (ii)Raney\: Ni/H_{2}\; \; \; \; \; \; \; \; (iii)H_{3}O^{+}

 
As we learned    Hydroboration oxidation reaction - Addition of water to an alkene is syn, anti-Markownikoff. - wherein       Option 1)       Option 2) Option 3) Option 4)
Engineering
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Which of the following will not exist in zwitter ionic form at pH =7 ?

  • Option 1)

  • Option 2)

  • Option 3)

  • Option 4)

 
As we learned    Zwitter ion - A molecule with dipolar ends ( an acidic end and a basic end ) - wherein In aqueous solution,  and  group undergo ionization  accepts  and becomes  and  losses  and becomes      The N atom of amide is not basic.      Option 1) Option 2) Option 3) Option 4)
Engineering
191 Views   |  

The copolymer formed by addition polymerization of styrene and acrylonitrile in the presence of peroxide is :

  • Option 1)

  • Option 2)

  • Option 3)

  • Option 4)

 
As we learned    Synthetic Polymers - Synthesized in laboratories from chemicals. - wherein Polythene, Nylon, Teflon, PVC       Option 1) Option 2) Option 3) Option 4)
Engineering
379 Views   |  

Which of the following will most readily give the dehydrohalogenation product ?

  • Option 1)

  • Option 2)

  • Option 3)

  • Option 4)

 
As we learned    Haloarene - Those organic compound in which halogen atom is directly attached to benzene is called haloarene. - wherein     Here dehydrohalogenation goes by E1cb and most stable carbonion formation is favoured in option (1)    Option 1) Option 2) Option 3) Option 4)
Engineering
138 Views   |  

The correct match between items of List-I and List-II is :

List - I                                                          List - II

(A) Coloured                                                (P) Steam impurity distillation

(B) Mixture of o-nitrophenol                         (Q) Fractional distillation

and p-nitrophenol                      

(C) Crude Naphtha                                      (R) Charcoal treatment

(D) Mixture of                                               (S) Distillation under reduced pressure

glycerol and sugars

  • Option 1)

    (A) - (R) , (B) - (S) , (C) - (P) , (D) - (Q)

  • Option 2)

    (A) - (R) , (B) - (P) , (C) - (S) , (D) - (Q)

  • Option 3)

    (A) - (R) , (B) - (P) , (C) - (Q) , (D) - (S)

  • Option 4)

    (A) - (P) , (B) - (S) , (C) - (R) , (D) - (Q)

 
As we learned  @2608 @2609 @2610   Option 1) (A) - (R) , (B) - (S) , (C) - (P) , (D) - (Q) Option 2) (A) - (R) , (B) - (P) , (C) - (S) , (D) - (Q) Option 3) (A) - (R) , (B) - (P) , (C) - (Q) , (D) - (S) Option 4) (A) - (P) , (B) - (S) , (C) - (R) , (D) - (Q)
Engineering
115 Views   |  

The main reduction product of the following compound with NaBH4 in methanol is :

  • Option 1)

  • Option 2)

  • Option 3)

  • Option 4)

 
As we learned    Alkoxy mercuration demercuration - Reaction between alkene and alcohol in the presence of trifluoroacetate. - wherein     NaBH4 can reduce carbonyl group but it cannot reduce amide.    Option 1) Option 2) Option 3) Option 4)
Engineering
108 Views   |  

Which of the following is the correct structure of Adenosine ?

  • Option 1)

  • Option 2)

  • Option 3)

  • Option 4)

 
As we learned    Nucleoside - Made up of two components : Pentose sugar Nitrogenous base -       Option 1) Option 2) Option 3) Option 4)
Engineering
190 Views   |  

The IUPAC name of the following compound is :

  • Option 1)

    4-methyl-3-ethylhex-4-ene

  • Option 2)

    3-ethyl-4-methylhex-4-ene

  • Option 3)

    4-ethyl-3-methylhex-2-ene

  • Option 4)

    4,4-diethyl-3-methylbut-2-ene

 

As we learned 

 

Nomenclature of organic compounds -

Longest chain is main chain, branched chain is substituent.

Locant value or position number should be minimum.

- wherein

IUPAC system of nomenclature

 

 

 

 


Option 1)

4-methyl-3-ethylhex-4-ene

Option 2)

3-ethyl-4-methylhex-4-ene

Option 3)

4-ethyl-3-methylhex-2-ene

Option 4)

4,4-diethyl-3-methylbut-2-ene

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Engineering
146 Views   |  

Xenon hexafluoride on partial hydrolysis produces compounds ‘X’ and ‘Y’. Compounds ‘X’ and ‘Y’ and the oxidation state of Xe are respectively :

  • Option 1)

    XeO_{2}\left ( +4 \right )\: and\: XeO_{3}\left ( +6 \right )

     

     

     

  • Option 2)

    XeOF_{4}\left ( +6 \right )\: and\: XeO_{3}\left ( +6 \right )

  • Option 3)

    XeO_{2}F_{2}\left ( +6 \right )\: and\: XeO_{2}\left ( +4 \right )

  • Option 4)

    XeOF_{4}\left ( +6 \right )\: and\: XeO_{2}F_{2}\left ( +4 \right )

 
As we learned    Oxy fluoride of Xenon - XeF4+H2O XeF6+H2O XeF6+2H2O - wherein Fluoride of Xe partially hydrolyse by water to give oxy  fluoride of Xe       Option 1)       Option 2) Option 3) Option 4)
Engineering
124 Views   |  

The decreasing order of bond angles in BF3, NH3, PF3 and I_{3}^{-}is :

  • Option 1)

    I_{3}^{-} > NH_{3}> PF_{3}> BF_{3}

     

     

     

  • Option 2)

    I_{3}^{-} >BF_{3}> NH_{3}> PF_{3}

  • Option 3)

    BF_{3} >I_{3}^{-}}>PF_{3}>NH_{3}

  • Option 4)

    BF_{3} >NH_{3}>PF_{3}>I_{3}^{-}}

 
As we learned    S - P Hybridisation - When one S - and one P - orbital belonging to the same main shell of an atom are mixed together to form two new SP - orbitals. - wherein shapes            Diagonal or Linear     sp2 Hybridisation - When one s - and two p - orbitals of the same shell of an atom mix to form three new equivalent orbitals.The hybridised orbital is called sp2 orbital. -...
Engineering
340 Views   |  

             (I)      (II)

H –– N - - - N - - - N

In hydrogen azide (above) the bond orders of bonds (I) and (II) are :

  • Option 1)

         (I)              (II)

    < 2\; \; \; \; \; \; \; > 2

     

     

     

  • Option 2)

    > 2\; \; \; \; \; \; \; < 2

  • Option 3)

    > 2\; \; \; \; \; \; \; > 2

  • Option 4)

    < 2\; \; \; \; \; \; \; < 2

 
As we learned    Bond Order - Bond order is defined as one half the difference between the number of electrons present in the bonding and the antibonding orbitals. - wherein       Option 1)      (I)              (II)       Option 2) Option 3) Option 4)
Engineering
130 Views   |  

The correct combination is :

  • Option 1)

    \left [ Ni\left ( CN \right )_{4} \right ]^{2-}-tetrahedral ;

    \left [ Ni\left ( CO \right )_{4} \right ]-paramagnetic

     

     

     

     

     

     

     

     

     

  • Option 2)

    \left [ NiCl_{4} \right ]^{2-}-paramagnetic;

    \left [ Ni\left ( CO \right )_{4} \right ]-tetrahedral

  • Option 3)

    \left [ NiCl_{4} \right ]^{2-}-square-planar;

    \left [ Ni\left ( CN \right )_{4} \right ]^{2-}-paramagnetic

  • Option 4)

    \left [ NiCl_{4} \right ]^{2-}-diamagnetic;

    \left [ Ni\left ( CO \right )_{4} \right ]-square-planar

 

As we learned 

 

Weak field Ligand -

These are the ligands which do not pair the electron in inner d-orbital.

- wherein

For example: N_{2}O, F^{-}, Cl^{-} etc    

 

 

Strong Field Ligand -

These are the ligands which pair the electron in inner d-orbital and makes required d-orbitals empty/available. 

- wherein

For example: CO, CN etc

 

 

Ni^{+2}\rightarrow \left [ Ar \right ]3d^{8},4S^{0},4p^{0}                         Ni\left ( o \right )\left [ Ar \right ]3d^{8},4S^{2},4p^{0}

Cl- is weak field ligand                                             CO is strong field ligand

(no pairing of electron)                                    \left [ Ar \right ]3d^{10} 4S^{0}4p^{0}

                                                                                              Sp^{3}

                                                                                    tetrahedral 

 

 


Option 1)

\left [ Ni\left ( CN \right )_{4} \right ]^{2-}-tetrahedral ;

\left [ Ni\left ( CO \right )_{4} \right ]-paramagnetic

 

 

 

 

 

 

 

 

 

Option 2)

\left [ NiCl_{4} \right ]^{2-}-paramagnetic;

\left [ Ni\left ( CO \right )_{4} \right ]-tetrahedral

Option 3)

\left [ NiCl_{4} \right ]^{2-}-square-planar;

\left [ Ni\left ( CN \right )_{4} \right ]^{2-}-paramagnetic

Option 4)

\left [ NiCl_{4} \right ]^{2-}-diamagnetic;

\left [ Ni\left ( CO \right )_{4} \right ]-square-planar

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Engineering
76 Views   |  

Identify the pair in which the geometry of the species is T-shape and squarepyramidal, respectively :

  • Option 1)

    CIF_{3}\; \; and\; \; \; IO_{4}^{-}

     

     

     

  • Option 2)

    ICl_{2}^{-}\; \; and\; \; \; ICl_{5}^{}

  • Option 3)

    XeOF_{2} \; \; \; and\; \; \; XeOF_{4}

  • Option 4)

    IO_{3}^{-}\; \; \; and\; \; \; IO_{2}F_{2}^{-}

 
As we learned  @4740 @4749   Option 1)       Option 2) Option 3) Option 4)
Engineering
110 Views   |  

A white sodium salt dissolves readily in water to give a solution which is neutral to litmus. When silver nitrate solution is added to the aforementioned solution, a white precipitate is obtained which does not dissolve in dil. nitric acid. The anion is :

  • Option 1)

    CO_{3}^{2-}

     

     

     

  • Option 2)

    SO_{4}^{2-}

  • Option 3)

    CI^{-}

  • Option 4)

    S^{2-}

 
As we learned   Sodium Chloride (NaCl) - Generally obtained by evapouration of sea water by sunlight - wherein It is purified by passing HCl gas to impure solution of NaCl. Concentration of increases, thus pure NaCl gets precipitated due to common ion effect                                         (Neutral)                                          (White...
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