P - Block Elements MCQ Questions & Answers in Inorganic Chemistry | Chemistry
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391.
Which of the following hydrides is least stable to hydrolysis?
397.
Hydrolysis of one mole of peroxodisulphuric acid produces
A
two moles of sulphuric acid
B
two moles of peroxomonosulphuric acid
C
one mole of sulphuric acid and one mole of peroxomonosulphuric acid
D
one mole of sulphuric acid, one mole of
peroxomonosulphuric acid and one mole of hydrogen peroxide.
Answer :
one mole of sulphuric acid, one mole of
peroxomonosulphuric acid and one mole of hydrogen peroxide.
398.
The type of hybridisation of boron in diborane is
A
$$sp$$ - hybridisation
B
$$s{p^2}$$ - hybridisation
C
$$s{p^3}$$ - hybridisation
D
$$s{p^3}{d^2}$$ - hybridisation
Answer :
$$s{p^3}$$ - hybridisation
No explanation is given for this question. Let's discuss the answer together.
399.
Which of the following is not matched correctly with its use?
A
Piezoelectric material - Quartz
B
Ion -exchangers - Graphite
C
Filtration plants - Silica
D
Electrical insulators - Silicones
Answer :
Ion -exchangers - Graphite
Zeolites are used as ion-exchangers in softening of hard water.
400.
$$Zn$$ gives $${H_2}$$ gas with $${H_2}S{O_4}$$ and $$HCl$$ but not with $$HN{O_3}$$ because
A
$$Zn$$ act as oxidising agent when react with $$HN{O_3}$$
B
$$HN{O_3}$$ is weaker acid than $${H_2}S{O_4}$$ and $$HCl$$
C
In electrochemical series $$Zn$$ is placed above hydrogen
D
$$NO_3^ - $$ is reduced in preference to hydronium ion
Answer :
$$NO_3^ - $$ is reduced in preference to hydronium ion
$$Zn$$ have lower value of $$E_{cell}^ \circ $$ and easily gives oxidation. $$Zn$$ is present above $${H_2}$$ in electrochemical series. So, it liberates hydrogen gas from $$dilu.$$ $$HCl/{H_2}S{O_4}.$$ But $$HN{O_3}$$ is an oxidising agent. The hydrogen obtained in this reaction is converted into $${H_2}O.$$ In $$HN{O_3},NO_3^ - $$ $$ion$$ is reduced and gives $$N{H_4}N{O_3},{N_2}O,NO$$ and $$N{O_2}$$ ( based upon the concentration of $$HN{O_3}$$ )
$$\eqalign{
& [Zn + \mathop {2HN{O_3}}\limits_{\left( {{\text{Nearly}}\,\,6\% } \right)} \,\, \to Zn{\left( {N{O_3}} \right)_2} + 2H] \times 4 \cr
& HN{O_3} + 8H\,\,\,\,\,\,\,\, \to N{H_3} + 3{H_2}O \cr
& N{H_3} + HN{O_3}\,\,\,\, \to N{H_4}N{O_3} \cr
& \overline {4Zn + 10HN{O_3} \to 4Zn{{\left( {N{O_3}} \right)}_2} + N{H_4}N{O_3} + 3{H_2}O} \cr} $$