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11.31     Write balanced equations for:

           (vi)\; B_{2}H_{6}+NH_{3}\rightarrow

When diborane reacts with the ammonia it forms borazine on strong heating, which is also called inorganic benzene.

11.31     Write balanced equations for:

           (v)\; Al+NaOH\rightarrow

Aluminium on reacting with a strong alkali like sodium hydroxide with the presence of moisture gives sodium tetrahydroxoaluminate complex and liberates hydrogen gas.

11.31     Write balanced equations for:

         (iv)\; H_{3}BO_{3}\overset{\Delta }{\rightarrow}

Orthoboric acid on heating gives metaboric acid with the removal of 4 molecules of water and on further heating, metaboric acid converts to tetraboric acid and later on Boron trioxide (boric anhydride)

11.38.    If the starting material for the manufacture of silicones is  RSiCl_{3} , write the structure of the product formed.

 undergoes condensation polymerisation yields straight chain polymer and water molecule as a by-product. The final product is called silicone.  

11.31     Write balanced equations for:

         (iii)\; NaH+B_{2}H_{6}\rightarrow

When diborane reacts with the sodium hydride in the presence of ether, it gives sodium borohydride.

11.37.    Elements of group 14
                (a) exhibit oxidation state of +4 only

                (b) exhibit oxidation state of +2 and +4

                (c)  form  M^{2-}  and  M^{4+}  ions

                (d) form M^{2+}  and  M^{4+}  ions

Elements of group 14 have 4 valance electrons. so the oxidation state of the group is +4 and +2. Due to the inert pair effect, the stability of +2 is more on moving down the group. So, the correct option is B

11.31     Write balanced equations for:

       (ii)\; B_{2}H_{6}+H_{2}O\rightarrow

On treating diborane with water it gives orthoboric as the main product and liberates hydrogen gas.

11.31     Write balanced equations for:

                 (i)\; BF_{3}+LiH\rightarrow

 

                

                 

When boron trifluoride reacts with lithium hydride it gives diborane and lithium fluoride

11.36.    Thermodynamically the most stable form of carbon is
                   (a) diamond          (b) graphite
                   (c) fullerenes        (d) coal

Graphite is a thermodynamically most stable form of carbon. So, the correct option is B

11.35.    The type of hybridisation of boron in diborane is

                (a)\: \: \, \! \! sp\textbf          (b) \, sp^{2}     (c)\, sp^{3}     (d)\, dsp^{2}

In diborane, boron is  hybridised. So, the correct option is C

11.34.    Boric acid is polymeric due to
                (a) its acidic nature              (b)  the presence of hydrogen bonds
                (c) its monobasic nature       (d)  its geometry

Option B is the correct answer, Because of the inter-molecular hydrogen bonding in the boric acid, it forms the polymeric chain.

11.33.    An aqueous solution of borax is

                (a) neutral        (b) amphoteric

                (c) basic          (d) acidic

An aqueous solution of borax is basic in nature. It is salt of a strong base () and weak acid () Option C is the correct answer.

11.32.    Give one method for industrial preparation and one for laboratory preparation of C\! \, O  and C\! \, O_{2}  each.

Laboratory method of preparation of   and - Carbon dioxide can be prepared by treating calcium carbonate with dil. hydrochloric acid. Carbon monoxide can be prepared by the dehydration of methanoic acid(formic acid) with conc. sulphuric acid at 373K.  The commercial method of preparation of    and -  can be prepared by heating limestone.  can be prepared by passing steam over hot...

11.30     A certain salt X, gives the following results.
              (i) Its aqueous solution is alkaline to litmus.
              (ii) It swells up to a glassy material Y on strong heating.
              (iii) When conc. H_{2}SO_{4} is added to a hot solution of X,white crystal of an acid Z separates out.
              Write equations for all the above reactions and identify X, Y and Z.

As per the given information, the compound X is a salt of a strong base and a weak acid because salt is alkaline to litmus. And when X is heated it swells to become Y. Thus, this must be borax because borax on heating, loses water molecule and swells to form sodium metaborate if we continue to heat it becomes glassy material Y. Hence Y should be a mixture of two compound boric anhydrides and...

11.29     What do you understand by

         (c) catenation

Catenation- The same atoms of element link with another same atom via a strong covalent bond to form long chains or branches. This type of property is known as catenation. Carbon and silicon show this property.

11.29     What do you understand by

   (b) allotropy 

Allotropy- The existence of an element in more than one form. The various forms of the same element are known as allotropes. They have the same chemical properties but different physical properties. For example- Carbon exists in three allotropic forms: diamond, fullerene and the graphite.

11.29     What do you understand by

   (a) inert pair effect 

Allotropy is the existence of one element in more than one form. These have the same chemical properties but the different physical properties. The various forms of elements are called allotropes. As we know that carbon has three allotropes are named as:- Diamond, graphite and fullerene

11.26     Classify the given oxide as neutral, acidic, basic or amphoteric. Write suitable chemical equations to show their nature.

                   CO,B_{2}O_{3},SiO_{2},CO_{2},Al_{2}O_{3},PbO_{2},Tl_2O_{3}

 is a neutral Acidic oxides- Being acidic, the oxides react with bases to form a salt. So take base = sodium hydroxide ()  these compounds react with a base to form salts. ex.- sodium metaborate, sodium silicate and sodium carbonate respectively. Basic Oxides-  It reacts with an acid to form salts. let acid = hydrochloric acid (). for example thallium chloride. Amphoteric Oxides- These...

11.28    When metal X is treated with sodium hydroxide, a white precipitate (A) is obtained, which is soluble in excess of NaOH to give soluble complex (B). Compound (A) is soluble in dilute HCl to form compound (C). The compound (A) when heated strongly gives (D), which is used to extract metal. Identify (X), (A), (B), (C) and (D). Write suitable equations to support their identities.

In given information, when X is treated with the sodium hydroxide it gives A(white ppt) and it is soluble in excess of . Thus X must be aluminium. So, the white ppt of A is aluminium hydroxide.  The complex B is soluble in excess of sodium hydroxide So it means it is a sodium tetrahydroxy aluminate(II) complex. Now, when we add dilute HCl to aluminium hydroxide it gives the compound C...

11.27  In some of the reactions thallium resembles aluminium, whereas in others it resembles with group I metals. Support this statement by giving some evidences.

Thallium belongs to group 13 of the periodic table. It shows +3 and +1 oxidation state. Due to the inert pair effect, the stability of the +1 oxidation state is more, on moving down the group. We know that aluminium and alkali metals +3 and +1 oxidation states.  Thallium displays both oxidation states. Therefore it resembles both aluminium and alkali metals. Like , it also forms compounds such...
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