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Match the catalysts (Column I) with products (Column II).

             ColumnI                                          Column II

            Catalyst                                               Product

      (A)   V_{2}O_{5}                                          (i) Polyethylene     

      (B)  TiCl_{4}/Al(Me)_{3}                      (ii) ethanal

      (C)   PdCl_{2}                                       (iii)   H_{2}SO_{4}

      (D) Iron\; Oxide                               (iv)   NH_{3}

 

 

  • Option 1)

     (A)-(iii); (B)-(iv);(C)-(i);(D)-(ii)

  • Option 2)

      (A)-(ii);(B)-(iii);(C)-(i);(D)-(iv)

  • Option 3)

     (A)-(iii);(B)-(i);(C)-(ii);(D)-(iv)

  • Option 4)

     (A)-(iv);(B)-(iii);(C)-(ii);(D)-(i)

 
      Contact Process - In this process SO2 obtained by burning of S or iron pyrites which is catalytically oxidised to  in presence of timely divided Pt or  as catalyst -       High density Polythene (HDP) - -  Polymerization of ethene in a hydrocarbon solvent under low pressure in presence of triethylaluminium and titanium tetrachloride (Ziegier. natta catalyst) -  Chain growth,...

Diborane (B_{2}H_{6}) reacts independenttly with O_{2}  and H_{2}O to produce, respectively:

 

  • Option 1)

    H_{3}BO_{3}\; and\; B_{2}O_{3}

  • Option 2)

    B_{2}O_{3}\; and\; [BH_{4}]^{-}

  • Option 3)

    B_{2}O_{3}\; and\;H_{3}BO_{3}

     

  • Option 4)

    HBO_{2}\; and\;H_{3}BO_{3}

    Diborane - Simplest boron hydride, dimer of BH3, colourless and highly toxic gas - wherein Hybridisation of boron in diborane is sp3       Preparation of diborane in laboratory - By the oxidation of sodium borohydride with I2 - wherein       Preparation of diborane from boron halides - -       Oxides of boron family - Acidity decreases and basicity increases down the group General...

The correct statement among the following is : 

  • Option 1)

    \left ( SiH_{3} \right )_{3}N is pyramidal and less basic than \left (CH_{3} \right )_{3}N.
     

  • Option 2)

    \left ( SiH_{3} \right )_{3}Nnis planar and less basic than  \left (CH_{3} \right )_{3}N.

  • Option 3)

    \left ( SiH_{3} \right )_{3}N is planar and more basic than \left (CH_{3} \right )_{3}N.

     

  • Option 4)

    \left ( SiH_{3} \right )_{3}N is pyramidal and more basic than \left (CH_{3} \right )_{3}N.

  Electronegativity of Nitrogen family - Gradually decreases down the group. More electronegative than group 14 -       Positive oxidation state of Nitrogen family - Shows +3 and +5 oxidation states (except N2) Stability of +3O.S increases down the group due to inert pair effect - wherein +3 when only p electron are involved and +5 when all valence five (s and p) electrons are...

The basic structural unit of feldspar, zeolites, mica, and asbestos is :

 

  • Option 1)

    (SiO_{3})^{2-}

  • Option 2)

    SiO_{2}

  • Option 3)

    (SiO_{4})^{4-}

  • Option 4)

  Zeolites - Hydrated alumino silicate having negative charge which is balanced by cations such as Na+, K+ or Ca2+ - wherein Eg. ZSM-5       Silicates - Compounds with basic unit SiO44-, each Si atom is linked to four oxygen atom tetrahedrally, joined by corners - wherein       (i) chemical formula zeolite:  .It is also known as hydrated aluminium silicate. (ii) Mica : It is a potassium...

The noble gas that does NOT occur in the atmosphere is :

  • Option 1)

    He

  • Option 2)

    Kr

  • Option 3)

    Ne

  • Option 4)

    Rn

 are present in atmosphere  but  is not present in atmosphere  [Reason :- option should be  not  because  is not noble gas]Option 1)Option 2)Option 3)Option 4)Rn

An organic compound 'A' is oxidized with Na_{2}O_{2} followed by boiling with HNO_{3}. The resultant solution is then treated with ammonium molybdate to yield a yellow precipitate.

Based on above observation, the element present in the given compound is : 


 

  • Option 1)

    Sulphur

  • Option 2)

    Nitrogen

  • Option 3)

    Phosphorus 

  • Option 4)

    Fluorine

According to question :   is used for detection of phosphorus. This yellow coloured precipitate helps us to conclude that Phosphorus is present in the organic compound. Option 1)SulphurOption 2)NitrogenOption 3)Phosphorus Option 4)Fluorine

The number of pentagons in C_{60} and trigons (triangles) in white phosphorus, respectively, are :

  • Option 1)

    20\: and\: 3

  • Option 2)

     12\: and\: 4

  • Option 3)

     12\: and\: \: 3

  • Option 4)

     20 \: and \: 4

Number of pentagons in  Number of triangles in white Option 1)Option 2) Option 3) Option 4) 

The oxoacid of sulphur that does not contain bond between sulphur atoms is:

  • Option 1)

    H_{2}S_{4}O_{6}

  • Option 2)

    H_{2}S_{2}O_{3}

  • Option 3)

    H_{2}S_{2}O_{7}

  • Option 4)

    H_{2}S_{2}O_{4}

 
: This molecule does not have bond between sulphur atoms .  Option (3) is correct. Option 1) Option 2) Option 3) Option 4)

The amorphous form of silica is :

  • Option 1)

    Tridymite

  • Option 2)

    Kieselguhr

  • Option 3)

    Cristobalite

  • Option 4)

    Quartz

  Silicon Dioxide/ Silica/ Quartz - Covalent, three dimensional solid network in which each silicon is covalently bond to four oxygen atoms (sp3 hybridisation) forming a tetrahedral structure - wherein     Function of quartz - As piezoelectric material in clocks, radio, television broadcasting and mobile communication. -         Quartz, tridymite, cristobalite is crystalline form, and...

The correct statement among I to III regarding group 13 element oxide are , 

I. Boron trioxide is acidic.

II. Oxides of aluminium and gallium are amphoteric.

III. Oxides of indium and thallium are basic .

  • Option 1)

    I and II only

  • Option 2)

    I , II and III

  • Option 3)

    I and III only

  • Option 4)

    II and III only 

  Oxides of boron family - Acidity decreases and basicity increases down the group General formula: M2O3 - wherein Aluminium oxide is amphoteric           I.     is acidic oxide in nature . II.    are amphoteric oxides. III.    are basic oxides. All statements are correct.Option 1)I and II onlyOption 2)I , II and IIIOption 3)I and III onlyOption 4)II and III only 

The C - C bond length is maximum in:

 

  • Option 1)

    graphite

  • Option 2)

    C_{70}

  • Option 3)

    C_{60}

  • Option 4)

    diamond

Answer is Diamond  (1) Graphic     (partial double bind character, Bond length ) (2) Diamond  (single bond, Bond length  ) (3)                    ----  (partial double Bond character, Bond length  &  ) (4) Option 1)graphiteOption 2)Option 3)Option 4)diamond
orthoboric_acid.PNG The correct statement(s) for orthoboric acid is/are (A) It behaves as a weak acid in water due to self ionisation. (B) Acidity of its aqueous solution increases upon addition of ethylene glycol. (C) It has a three dimensional Structure due to bonding (D) It is a weak electrolyte in water.
The answer to the following question of Lewis acid for finding the correct statement is given below:  
The bond dissociation energy of B-F in BF_{3} is 646 kJ mol -1 whereas that of C-F in CF_{4} is 515 kJ mol -1 The correct reason for higher B-F bond dissociation energy as compared to that of C-F is
The bond dissociation energy of B-F in BF3 is 646 kJ mol -1 whereas that of C-F in CF4 is 515 kJ mol -1 The correct reason for higher B-F bond dissociation energy as compared to that of C-F is Bond length between two atoms. the atomic size of boron is smaller than carbon, so the strength of the bond of B-F will higher than C-F.

 The number of S=O and S−OH bonds present in peroxodisulphuric acid and pyrosulphuric acid respectively are :

  • Option 1)

     (2 and 2) and (2 and 2)

  • Option 2)

     (2 and 4) and (2 and 4)

  • Option 3)

     (4 and 2) and (2 and 4)

  • Option 4)

     (4 and 2) and (4 and 2)

     

 
Option 1)  (2 and 2) and (2 and 2) Option 2)  (2 and 4) and (2 and 4) Option 3)  (4 and 2) and (2 and 4) Option 4)  (4 and 2) and (4 and 2)  

A metal ‘M’ reacts with nitrogen gas to afford ‘M3N’. ‘M3N’ on heating at high temperature gives back ‘M’ and on reaction with water produces a gas ‘B’.  Gas ‘B’ reacts with an aqueous solution of CuSO4 to form a deep blue compound.  ‘M’ and ‘B’ respectively are :

  • Option 1)

     Li and NH3

  • Option 2)

     Ba and N2

  • Option 3)

    Na and NH3

  • Option 4)

    Al and N2

     

 
Option 1)  Li and NH3 Option 2)  Ba and N2 Option 3) Na and NH3 Option 4) Al and N2  

The chloride that CANNOT get hydrolysed is:

  • Option 1)

    SnCl_{4}

     

  • Option 2)

    PbCl_{4}

  • Option 3)

    CCl_{4}

  • Option 4)

    SiCl_{4}

  Hydrolysis of carbon tetrachloride - CCl4 does not undergo hydrolysis - wherein Due to unavailability of d-orbitals       Hydrolysis of silicon tetrachloride - SiCl4 undergo hydrolysis to give silicic acid -Awe know that wherein Due to the presence of vacant d-orbitals In chloride the carbon atom have not any vacant orbital for hydrolysis.    Option 1)  Option 2)Option 3)Option 4)

The relative stability of +1 oxidation state of group 13 elements follows the order :

 

  • Option 1)

    Ga < Al < In < TI

  • Option 2)

    Al < Ga < TI < In

  • Option 3)

    Al < Ga < In < TI

  • Option 4)

    TI < In < Ga < Al

  Oxidation state of boron family - Boron show +3 covalence, this tendency decreases down the group, other members have +3 oxidation state form ionic compounds. - wherein Lower elements also show +1 ionic state due to inert pair effect. E.g. Tl+, Ga+   Inert pair effect - The phenomenon in which outer shell s electrons (ns2) penetrate to (n-1) d-electrons and thus become closer to nucleus and...

The hydride that is NOT electron deficient is :

  • Option 1)

    B_{2}H_{6}

  • Option 2)

    GaH_{3}

     

  • Option 3)

    AlH_{3}

  • Option 4)

    SiH_{4}

     

  Thermal Stability of Hydride - HO>HS>HSe>HTe>HPo - wherein Thermal stability increases with increase in electronegativity difference       Thermal Stability of Hydride - HO>HS>HSe>HTe>HPo - wherein Thermal stability increases with increase in electronegativity difference  is not electron deficient.   Option 1)Option 2)  Option 3)Option 4)  

The number of 2-center-2-electron and 3-certer-2- electron bonds in B_{2}H_{6} respectively are :

  • Option 1)

    4 and 2

  • Option 2)

    2 and 4

  • Option 3)

    2and 2

  • Option 4)

    2and 1

  Structure of diborane - It has three centre-electron pair bonds, also called as banana-bond B2H6 has two types of B-H bond. Normal covalent bond (2c-2e bond) Bond between three atoms, (B-Hb-B), (3c-2e bond) - wherein po As we have learned in Boron Hydroxide the number of 2 centre 2 e-  and 3 centre 2 e- bonds are 4 and 2 respectively .    Option 1)4 and 2Option 2)2 and 4Option 3)2and...

The pair that contain two P-H bonds in each of the  oxoacids is

:

 

 

 

  • Option 1)

    H_{3}PO_{3}and H_{3}PO_{2}

  • Option 2)

    H_{4}P_{2}O_{5}and H_{4}P_{2}O_{6}

  • Option 3)

    H_{4}P_{2}O_{5}and H_{3}PO_{3}

  • Option 4)

    H_{3}PO_{2}and H_{4}P_{2}O_{5}

  Hypophosphorus Acid - H3PO2 (+1), monobasic - wherein   As we have learned is oxoacids . H3PO2 and H4P2O5 Both have to P-H bonds .  Option 1)Option 2)Option 3)Option 4)
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