131.
On heating with concentrated $$NaOH$$ solution in an inert atmosphere of $$C{O_2},$$ white phosphorus gives a gas. Which of the following statements is incorrect about the gas?
A.
It is highly poisonous and has smell like rotten fish.
B.
Its solution in water decomposes in the presence of light.
In the laboratory, phosphine $$\left( {P{H_3}} \right)$$ is prepared by heating white phosphorus with concentrated $$NaOH$$ solution in an inert atmosphere of $$C{O_2}.$$
$${P_4} + 3NaOH + 3{H_2}O \to $$ $$P{H_3} + 3Na{H_2}P{O_2}$$
$${P{H_3}}$$ is less basic than $$N{H_3}.$$ Due to larger size of $$P,$$ the electron density is diffused over a larger region and thus, the ability to donate electron pair is less.
132.
Which of the following has the minimum heat of dissociation :
A.
$${\left( {C{H_3}} \right)_3}N: \to B{F_3}$$
B.
$${\left( {C{H_3}} \right)_3}N: \to B{\left( {C{H_3}} \right)_2}F$$
C.
$${\left( {C{H_3}} \right)_3}N: \to B{\left( {C{H_3}} \right)_3}$$
D.
$${\left( {C{H_3}} \right)_3}N: \to B\left( {C{H_3}} \right){F_2}$$
Boron form $$BF_4^ - $$ in which boron atom extends its coordination number to 4 by utilising empty $$p$$ - orbital. It cannot extend its coordination number beyond 4 due to non - availability of $$d$$ - orbitals.
136.
Sulphur trioxide is not directly dissolved in water to form sulphuric acid because
A.
$$S{O_3}$$ does not react with water to form acid
B.
$$S{O_3}$$ gets oxidised to $${H_2}S{O_3}$$ when dissolved in water
C.
it results in the formation of dense fog of sulphuric acid which is difficult to condense
D.
sulphur trioxide is insoluble in water due to its covalent nature
Answer :
it results in the formation of dense fog of sulphuric acid which is difficult to condense
$${B_2}{O_3}$$ is acidic and reacts with basic (metallic) oxides forming metal borates. Aluminium and gallium oxides are amphoteric and those of indium and thallium are basic in their properties.
138.
Oxidation number of $$Cl\,\,{\text{in}}\,CaOC{l_2}$$ (bleaching powder) is:
A.
zero, since it contains $$C{l_2}$$
B.
$$ - 1,$$ since it contains $$Cl$$
C.
$$ + 1$$, since it contains $$Cl{O^ - }$$
D.
$$ + 1\,{\text{and}}\, - 1$$ since it contains $$Cl{O^ - }\,{\text{and}}\,C{l^ - }$$
Answer :
$$ + 1\,{\text{and}}\, - 1$$ since it contains $$Cl{O^ - }\,{\text{and}}\,C{l^ - }$$
$$CaOC{l_2} - $$ or it can also be written as
$$Ca\mathop {(OCl)}\limits_{{x_1}} \mathop {Cl}\limits_{{x_2}} $$
hence oxidation no of $$Cl\,\,{\text{in}}\,\,OC{l^ - }\,{\text{is}}$$
$$\eqalign{
& {{ - 2\, + }}\,{x_2} = - 1 \cr
& {x_2} = 2 - 1 = + 1 \cr} $$
now oxidation no. of another $$Cl$$ is $$ - 1$$ as it is present as $$C{l^ - }$$.
139.
What are the products formed in the reaction of xenon hexafluoride with silicon dioxide ?