391.
Which of the following hydrides is least stable to hydrolysis?
A
$$C{H_4}$$
B
$$Si{H_4}$$
C
$$Sn{H_4}$$
D
$$Pb{H_4}$$
Answer :
$$Si{H_4}$$
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Silicon hydrides are most easily hydrolysed.
$$Si{H_4} + 3{H_2}O \to {H_2}Si{O_3} + 4{H_2}$$
$$Si{H_4} + 2NaOH + {H_2}O \to $$ $$N{a_2}Si{O_3} + 4{H_2}$$
392.
One mole of calcium phosphide on reaction with excess water gives
A
one mole of phosphine
B
two moles of phosphoric acid
C
two moles of phosphine
D
one mole of phosphorus pentoxide
Answer :
two moles of phosphine
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$$C{a_3}{P_2} + 6{H_2}O \to 3Ca{\left( {OH} \right)_2} + 2P{H_3}\,;\,{\text{i}}{\text{.e}}\,\,2$$ moles of phosphine are produced.
393.
Alum helps in purifying water by
A
forming $$Si$$ complex with clay partiles
B
sulphate part which combines with the dirt and removes
it
C
coagulaing the mud particles
D
making mud water soluble.
Answer :
coagulaing the mud particles
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Alum furnishes $$A{I^{3 + }}$$ ions which bring about coagulation of negatively charged clay particles, bacteria etc.
394.
Which one of the following ionic species has the greatest proton affinity to form stable compound?
A
$$H{S^ - }$$
B
$$NH_2^ - $$
C
$${F^ - }$$
D
$${I^ - }$$
Answer :
$${F^ - }$$
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Fluorine is the most electronegative element in the periodic table. So, it has the greatest proton affinity to form stable compounds.
395.
Select the correct option regarding the properties of dioxygen.
A
Dioxygen never reacts with metals.
B
Dioxygen is diamagnetic in nature.
C
Combination of dioxygen with other elements is highly exothermic process.
D
Dioxygen liquefies at 55 $$K$$ and freezes at 90 $$K.$$
Answer :
Combination of dioxygen with other elements is highly exothermic process.
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No explanation is given for this question. Let's discuss the answer together.
396.
Which of the following xenon-oxo compounds may not be obtained by hydrolysis of xenon fluorides ?
A
$$Xe{O_2}{F_2}$$
B
$$XeO{F_4}$$
C
$$Xe{O_3}$$
D
$$Xe{O_4}$$
Answer :
$$Xe{O_4}$$
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\[\begin{align}
& Xe{{F}_{6}}+{{H}_{2}}O\xrightarrow{\text{Partial}}XeO{{F}_{4}}+2HF \\
& Xe{{F}_{6}}+2{{H}_{2}}O\xrightarrow{\text{Partial}}Xe{{O}_{2}}{{F}_{2}}+4HF \\
& Xe{{F}_{6}}+3{{H}_{2}}O\xrightarrow{\text{Complete}}Xe{{O}_{3}}+6HF \\
\end{align}\]
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
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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
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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
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$$Zn$$ 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 $$HN{O_3}$$ )
$$\eqalign{
& [Zn + \mathop {2HN{O_3}}\limits_{\left( {{\text{Nearly}}\,\,6\% } \right)} \,\, \to Zn{\left( {N{O_3}} \right)_2} + 2H] \times 4 \cr
& HN{O_3} + 8H\,\,\,\,\,\,\,\, \to N{H_3} + 3{H_2}O \cr
& N{H_3} + HN{O_3}\,\,\,\, \to N{H_4}N{O_3} \cr
& \overline {4Zn + 10HN{O_3} \to 4Zn{{\left( {N{O_3}} \right)}_2} + N{H_4}N{O_3} + 3{H_2}O} \cr} $$