561.
Which of the following is not correct ?
A
$$Ge{\left( {OH} \right)_2}$$ is amphoteric
B
$$GeC{l_2}$$ is more stable than $$GeC{l_4}$$
C
$$Ge{O_2}$$ is weakly acidic
D
$$GeC{l_4}$$ in $$HCl$$ forms $${\left[ {GeC{l_2}} \right]^{2 - }}ion$$
Answer :
$$GeC{l_2}$$ is more stable than $$GeC{l_4}$$
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$$G{e^{4 + }}$$ is more stable than $$G{e^{2 + }}.$$ Hence $$GeC{l_4}$$ is more stable than $$GeC{l_2}$$
562.
The least number of oxyacids are formed by :
A
Chlorine
B
Nitrogen
C
Fluorine
D
Sulphur
Answer :
Fluorine
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Flourine is the most electronegative element & has least tendency to form double bonds.
563.
Which of the following is used to prepare $$C{l_2}$$ gas at room temperature from concentrated $$HCl?$$
A
$$NaOH$$
B
$${H_2}S$$
C
$$KMn{O_4}$$
D
$$C{r_2}{O_3}$$
Answer :
$$KMn{O_4}$$
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$$KMn{O_4}$$ oxidises halogen acids to halogen.
$$2KMn{O_4} + 16HCl \to $$ $$2KCl + 2MnC{l_2} + 8{H_2}O + 5C{l_2}$$
$$\therefore KMn{O_4}$$ is used to prepare $$C{l_2}$$ from concentrated $$HCl.$$
564.
$$\left( {S{i_2}{O_5}} \right)_n^{2n - }$$ anion is obtained when :
A
no oxygen of a $$SiO_4^{4 - }$$ tetrahedron is shared with another $$SiO_4^{4 - }$$ tetrahedron
B
one oxygen of a $$SiO_4^{4 - }$$ tetrahedron is shared with another $$SiO_4^{4 - }$$ tetrahedron
C
two oxygen of a $$SiO_4^{4 - }$$ tetrahedron are shared with another $$SiO_4^{4 - }$$ tetrahedron
D
three oxygen of a $$SiO_4^{4 - }$$ tetrahedron are shared with another $$SiO_4^{4 - }$$ tetrahedron
Answer :
three oxygen of a $$SiO_4^{4 - }$$ tetrahedron are shared with another $$SiO_4^{4 - }$$ tetrahedron
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Total No. oxygen atoms per silicon atom
$$ = \frac{1}{2} + \frac{1}{2} + \frac{1}{2} + 1 = 2.5$$
∴ Formula $$S{i_2}O_5^{2 - }.$$
565.
Ammonia can be dried by
A
$$\,{\text{Conc}}{\text{.}}\,{H_2}S{O_4}$$
B
$${P_2}{O_5}$$
C
$${\text{Anhydrous}}\,\,CuS{O_4}\,\,$$
D
none
Answer :
none
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None; it reacts with all given compounds. It forms addition compounds with them.
It can be dried over any metal oxide.
566.
Oxygen will directly react with each of the following elements except?
A
$$P$$
B
$$Cl$$
C
$$Na$$
D
$$S$$
Answer :
$$Cl$$
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Out of $$P, Na, S$$ and $$Cl,$$ chlorine does not react directly but $$Na, P$$ and $$S$$ react with oxygen directly.
$$\,{P_4} + 5{O_2} \to {P_4}{O_{10}}$$
$$\,\,\,\,\,S + {O_2} \to S{O_2}$$
$$4Na + {O_2} \to 2N{a_2}O$$
567.
The reason behind the lower atomic radius of $$Ga$$ as compared to $$Al$$ is
A
poor screening effect of $$d$$ - electrons for the outer electrons from increased nuclear charge
B
increased force of attraction of increased nuclear charge on electrons
C
increased ionisation enthalpy of $$Ga$$ as compared to $$Al$$
D
anomalous behaviour of $$Ga.$$
Answer :
poor screening effect of $$d$$ - electrons for the outer electrons from increased nuclear charge
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Presence of 10 $$d$$ - electrons in gallium which have poor screening effect on the outer electrons increases the force of attraction between the outermost electrons and nuclear charge resulting in decrease in atomic radius.
568.
A translucent white waxy solid $$(A)$$ reacts with excess of chlorine to give a yellowish white powder $$(B).$$ $$(B)$$ reacts with organic compounds containing $$-OH$$ group and converts them into chloro derivatives. $$(B)$$ on hydrolysis gives $$(C)$$ and is finally converted to phosphoric acid. $$(A), (B)$$ and $$(C)$$ are
A
$${P_4},PC{l_3},{H_3}P{O_4}$$
B
$${P_4},PC{l_5},{H_3}P{O_3}$$
C
$${P_4},PC{l_5},POC{l_3}$$
D
$${P_4},PC{l_3},POC{l_3}$$
Answer :
$${P_4},PC{l_5},POC{l_3}$$
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$$\mathop {{P_4}}\limits_{\left( A \right)} + 10C{l_2} \to \mathop {4PC{l_5}}\limits_{\left( B \right)} $$
$$PC{l_5} + {C_2}{H_5}OH \to $$ $${C_2}{H_5}Cl + POC{l_3} + HCl$$
$$PC{l_5} + {H_2}O \to \mathop {POC{l_3} + }\limits_{\left( C \right)} 2HCl$$
$$POC{l_3} + 3{H_2}O \to {H_3}P{O_4} + 3HCl$$
569.
There is a large number of carbon compounds due to
A
tetravalency of carbon
B
strong catenation property of carbon
C
allotropic property of carbon
D
non-metallic character of carbon
Answer :
strong catenation property of carbon
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No explanation is given for this question. Let's discuss the answer together.
570.
Sulphur trioxide can be obtained by which of the following reaction?
A
\[CaS{{O}_{4}}+C\xrightarrow{\Delta }\]
B
\[F{{e}_{2}}{{\left( S{{O}_{4}} \right)}_{3}}\xrightarrow{\Delta }\]
C
\[S+{{H}_{2}}S{{O}_{4}}\xrightarrow{\Delta }\]
D
\[{{H}_{2}}S{{O}_{4}}+PC{{l}_{5}}\xrightarrow{\Delta }\]
Answer :
\[F{{e}_{2}}{{\left( S{{O}_{4}} \right)}_{3}}\xrightarrow{\Delta }\]
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\[\begin{align}
& \left( \text{A} \right)\,CaS{{O}_{4}}+C\xrightarrow{\Delta }CaO+S{{O}_{2}}+CO \\
& \left( \text{B} \right)\,F{{e}_{2}}{{\left( S{{O}_{4}} \right)}_{3}}\xrightarrow{\Delta }F{{e}_{2}}{{O}_{3}}+3S{{O}_{3}} \\
& \left( \text{C} \right)\,S+2{{H}_{2}}S{{O}_{4}}\xrightarrow{\Delta }3S{{O}_{2}}+2{{H}_{2}}O \\
\end{align}\]
\[\left( \text{D} \right)\,{{H}_{2}}S{{O}_{4}}+PC{{l}_{5}}\xrightarrow{\Delta }\] \[\underset{\text{Chloro sulphonic acid}}{\mathop{S{{O}_{3}}HCl}}\,+POC{{l}_{3}}+HCl\]