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Engineering
254 Views   |  

Due to the presence of an unpaired electron, free radicals are

  • Option 1)

    chemically reactive

  • Option 2)

    chemically inactive

  • Option 3)

    anions

  • Option 4)

    cations

 

As we learnt in

Alkyl free Radical -

Chemical species in which carbon atom bearing one unpaired electron is called alkyl  free radical

- wherein

CH_{3}\dot{CH_{2}}

alkyl  free radical

 

 Due presence of unpaired electron force radicals are chemically highy reactive. 


Option 1)

chemically reactive

This option is correct

Option 2)

chemically inactive

This option is incorrect

Option 3)

anions

This option is incorrect

Option 4)

cations

This option is incorrect

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Engineering
454 Views   |  

Lattice energy of an ionic compound depends upon

  • Option 1)

    charge on the ion only

  • Option 2)

    size of the ion only

  • Option 3)

    packing of the ion only

  • Option 4)

    charge and size of the ion

 
As we learnt in Ionic Radii - The ionic radius is the distance between the nucleus of an ion and the point where the nucleus exerts its influence on the electron cloud.     Energy released while forming a lattice is directly proportional to the charges of the ions and inversely proportional to the distance between them. Option 1) charge on the ion only This option is incorrect. Option...
Engineering
166 Views   |  

Based on lattice energy and other considerations which one of the following alkali metal chlorides is expected to have the highest melting point?

  • Option 1)

    LiCl

  • Option 2)

    NaCl

  • Option 3)

    KCl

  • Option 4)

    RbCl

 
As we learnt in Alkali Metal - Each period starts with an alkali metal whose outermost electronic configuration is ns1 -    Despite the fact that  has higher lattice energy than ,   has lower melting point because of its covalent nature. Also, as we go down the group's lattice energy generally decreases as size of alkali metal atom increases. Option 1) This option is incorrect. Option...
Engineering
120 Views   |  

In which of the following arrangements the order is NOT according to the property indicated against it?

  • Option 1)

    Al^{3+}<Mg^{2+}<Na^{+}<F^{-}- increasing ionic size

  • Option 2)

    B<C<N<O   - increasing first ionisation enthalpy

  • Option 3)

    I<Br<F<Cl    increasing electron gain enthalpy    (with negative sign)

  • Option 4)

    Li<Na<K<Rb  - increasing metallic radius

 
As we learnt in Factors affecting I.E - Removal of an electron from half filled and full filled atom require more energy than expected. When we look into option 2, we find a discrepancy there that O is shown to have a greater I.E than N. Usually, I.E increases along a period but as N has half filled orbitals, removal of electrons requires more energy. Option 1) - increasing ionic size This...
Engineering
152 Views   |  

During the process of electrolytic refining of copper, some metals present as impurity settle as ' anode mud ' These are

  • Option 1)

    Sn \: and\: Ag

  • Option 2)

    Pb \: and\: Zn

  • Option 3)

    Ag \: and\: Au

  • Option 4)

    Fe \: and \: Ni

 
As we learnt in Electrolytic Refining - Many of the metals such as Cu,Au,Ag,Al,Hb etc are purified by this method. - wherein The impure metal is made anode while a thin sheet of pure metal acts as a cathode.     In electrolytic refining of copper, electropositive impurities like Fe, Zn, etc dissolve in the solution and less electropositive impurities like Ag, Au, etc collect below the...
Engineering
97 Views   |  

Tertiary alkyl halides are practically inert to substitution by S_N2 mechanism because of

  • Option 1)

    insolubility

  • Option 2)

    instability

  • Option 3)

    inductive effect

  • Option 4)

    steric hindrance

 
As we learnt in SN2 Substitution Nucleophilic bimolecular - These reactions proceed in only one step. - wherein       reaction is influenced due to steric hindrance.   Option 1) insolubility This option is incorrect. Option 2) instability This option is incorrect. Option 3) inductive effect This option is incorrect. Option 4) steric hindrance This option is correct.
Engineering
592 Views   |  

The best reagent to convert pent­-3-­en­-2-­ol into pent­-3-­en­-2-­one is

  • Option 1)

    acidic permanganate

  • Option 2)

    acidic dichromate

  • Option 3)

    chromic anhydride in glacial acetic acid

  • Option 4)

    pyridinium chloro­-chromate

 
As we learnt in Oxidation of alcohol - –  Also known as dehydrogenation reaction. –  Tertiary alcohol do not undergo oxidation due to lack of  -  hydrogen - wherein           Pyridinium chloro-chromate (PCC) is specific for the conversion. Option 1) acidic permanganate This option is incorrect. Option 2) acidic dichromate This option is incorrect. Option 3) chromic anhydride in glacial...
Engineering
610 Views   |  

Alkyl halides react with dialkyl copper reagents to give

  • Option 1)

    alkenes

  • Option 2)

    alkyl copper halides

  • Option 3)

    alkanes

  • Option 4)

    alkenyl halides

 
As we learnt in Reaction of alkyl halide with R2CuLi - Corey House alkane synthesis, alkane is  obtained as a product. - wherein        (alkane) Option 1) alkenes This option is incorrect. Option 2) alkyl copper halides This option is incorrect. Option 3) alkanes This option is correct. Option 4) alkenyl halides This option is incorrect.
Engineering
525 Views   |  

Elimination of bromine from 2­-bromobutane results in the formation of

  • Option 1)

    equimolar mixture of 1 and 2-­butene

  • Option 2)

    predominantly 2­-butene

  • Option 3)

    predominantly 1-­butene

  • Option 4)

    predominantly 2­-butyne

 
As we learnt in Reaction of alkyl halide with KOH (alc) - - elimination reaction take place and produces alkenes. - wherein    The reaction will give us Saytzeff's product predominantly which is 2 Butane.   Option 1) equimolar mixture of 1 and 2-­butene Incorrect option Option 2) predominantly 2­-butene Correct option Option 3) predominantly 1-­butene Incorrect option Option 4) predominantly...
Engineering
109 Views   |  

p- cresol reacts with chloroform in alkaline medium to give the compound A which adds hydrogen cyanide to form the compound B. The latter on acidic hydrolysis gives chiral carboxylic acid. The structure of the carboxylic acid is

  • Option 1)

  • Option 2)

  • Option 3)

  • Option 4)

 
As we learnt in Hydrolysis of Trihalogen derivatives - Three halogen atoms are attached to the same carbon atom yield carboxylic acid.  - wherein     Option 1) This option is incorrect. Option 2) This option is correct. Option 3) This option is incorrect. Option 4) This option is incorrect.
Engineering
422 Views   |  

Which of the following is a polyamide?

  • Option 1)

    Teflon

  • Option 2)

    Nylon­-66

  • Option 3)

    Terylene

  • Option 4)

    Bakelite

 
As we learnt in Nylon 6,6 - - Condensation polymerization of hexamethylenediamine with adipic acid under high pressure. - Copolymer, step growth - wherein - High tensile strength - Used in textile fabrics, bristles for brushes.    Nylon 6,6 is a polyamide.   Option 1) Teflon This option is incorrect. Option 2) Nylon­-66 This option is correct. Option 3) Terylene This option is...
Engineering
142 Views   |  

In both DNA and RNA, heterocyclic base and phosphate ester linkages are at

  • Option 1)

    C_{5}'\; and\: C_{2}' respectively of the sugar molecule

  • Option 2)

    C_{2}'\; and\: C_{5}'  respectively of the sugar molecule

  • Option 3)

    C_{1}'\; and\: C_{5}' respectively of the sugar molecule

  • Option 4)

    C_{5}'\; and\: C_{1}'  respectively of the sugar molecule

 
As we learnt in Nucleotide - Made up of three components : Pentose sugar Phosphoric acid Nitrogenous base - wherein    In DNA and RNA heterocylic base and phosphate ester are at respectively of the sugar molecule Option 1) respectively of the sugar molecule This option is incorrect Option 2)   respectively of the sugar molecule This option is incorrect Option 3) respectively of the...
Engineering
178 Views   |  

Which of the following is fully fluorinated polymer?

  • Option 1)

    Neoprene

  • Option 2)

    Teflon

  • Option 3)

    Thiokol

  • Option 4)

    PVC

 
As we learnt in Polytetrafluroethene / Teflon - -  Polymerization of tetrafluroethene (F2C = CF2) with free radical or persulphate at high pressure. -  Homopolymer. - wherein -  Chemically inert and resistant to attack by corrosive reagents. -  Used in making oil seals, gaskets, non-stick surface       Teflon is a homopolymer of tetrafluoroethene. Option 1) Neoprene This option is...
Engineering
129 Views   |  

The molecular shapes of SF_{4},\; CF_{4}\; \; and\; \; XeF_{4}  are

  • Option 1)

    the same with 2, 0 and 1 lone pairs of electrons on the central atom respectively.

  • Option 2)

    the same with 1, 1 and 1 lone pair of electrons on the central atoms respectively.

  • Option 3)

    different with 0, 1 and 2 lone pairs of electrons on the central atom respectively.

  • Option 4)

    different with 1, 0 and 2 lone pairs of electrons on the central atom respectively.

 
As we learnt in  Structure of Xenon tetrafluoride - Square planar and hybridisation is Sp3d2 - wherein    (lone pair electron = 2) XeF4 is square planar with sp3d2 hybridization. (lone pair electron=0) CF4 is tetrahedral with sp3 hybridization. (lone pair electron =1) SF4 is trigonal bipyramidal with sp3d hybridization. Option 1) the same with 2, 0 and 1 lone pairs of electrons on the...
Engineering
135 Views   |  

The number of hydrogen atom(s) attached to phosphorus atom in hypophosphorous acid is

  • Option 1)

    zero

  • Option 2)

    two

  • Option 3)

    one

  • Option 4)

    three

 
As we learnt in  Hypophosphorus Acid - H3PO2 (+1), monobasic - wherein    The number of hydrogen atoms attached to P is 2. Option 1) zero This option is incorrect. Option 2) two This option is correct. Option 3) one This option is incorrect. Option 4) three This option is incorrect.
Engineering
154 Views   |  

The correct order of the thermal stability of hydrogen halides (H-X)  is

  • Option 1)

    H\! I>H\! Br>H\! Cl>H\! F

  • Option 2)

    H\! F>H\! Cl>H\! Br>H\! I

  • Option 3)

    H\! Cl<H\! F>H\! Br<H\! I

  • Option 4)

    H\! I>H\! Cl<H\! F>H\! Br

 
As we learnt in  Thermal Stability of Hydride - HF>HCl>HBr>HI - wherein Thermal stability increases with increase in electronegativity difference     As the bond strength increases, stability decreases. Thermal Stability : Option 1) This option is incorrect. Option 2) This option is correct. Option 3) This option is incorrect. Option 4) This option is incorrect.
Engineering
109 Views   |  

In silicon dioxide

  • Option 1)

    each silicon atom is surrounded by four oxygen atoms and each oxygen atom is bonded to two silicon atoms

  • Option 2)

    each silicon atom is surrounded by two oxygen atoms and each oxygen atom is bonded to two silicon atoms

  • Option 3)

    silicon atom is bonded to two oxygen atoms

  • Option 4)

    there are double bonds between silicon and oxygen atoms

 
As we learnt in  Oxide of Silicon - SiO2 form three dimensional network - wherein Due to lack of formation of with oxygen    Structure of SiO2 form 3-dimensional network. Option 1) each silicon atom is surrounded by four oxygen atoms and each oxygen atom is bonded to two silicon atoms This option is correct. Option 2) each silicon atom is surrounded by two oxygen atoms and each oxygen atom is...
Engineering
251 Views   |  

Which of the following oxides is amphoteric in character?

  • Option 1)

    CaO\;

  • Option 2)

    \; CO_2\;

  • Option 3)

    \; SiO_2\;

  • Option 4)

    \; SnO_2

 
As we learnt in  Order of basic oxides of oxygen - Alaklimetal oxidesalkaline earth metal oxides transition metal oxides  -     Acidic oxides of oxygen - Non metals oxides are generally acidic  - wherein CO2,SO2,SO3,NO2,N2O5        is the amphoteric in nature as it acts as both acid and base   Option 1) This is incorrect option Option 2) This is incorrect option Option 3) This is...
Engineering
503 Views   |  

The lanthanide contraction is responsible for the fact that

  • Option 1)

     Zr  and Y have about the same radius

  • Option 2)

     Zr and Nb have similar oxidation state

  • Option 3)

     Zr  and Hf have about the same radius

  • Option 4)

    Zr and Zn have the same oxidation state

 
As we learnt in Lanthanoid contraction - The filling of if orbits before 5d orbits result regular decrease in atomic radii called Lanthanoid contraction or can say the size of radii not much increase. - wherein      Due to Lanthanoid contraction size of 5d-series elements is smaller than 4d-series element, that's why Zr and Hf have almost same atomic radii. Option 1)    and  have about the...
Engineering
175 Views   |  

The number and type of bonds between two carbon atoms in calcium carbide are

  • Option 1)

    one sigma, one pi

  • Option 2)

    one sigma, two pi

  • Option 3)

    two sigma, one pi

  • Option 4)

    two sigma, two pi

 
As we lernt in Carbides of Be and Al - Liberate methane on reaction with water - wherein    one sigma and two  bond.    Option 1) one sigma, one pi This option is incorrect. Option 2) one sigma, two pi This option is correct. Option 3) two sigma, one pi This option is incorrect. Option 4) two sigma, two pi This option is incorrect.
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