P - Block Elements MCQ Questions & Answers in Inorganic Chemistry | Chemistry
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211.
Which of the following metals does not show inert pair effect?
A
$$Tl$$
B
$$Ga$$
C
$$In$$
D
$$Al$$
Answer :
$$Al$$
No explanation is given for this question. Let's discuss the answer together.
212.
$$N{H_4}Cl{O_4} + HN{O_3}\left( {dil.} \right) \to $$ $$HCl{O_4} + \left[ X \right]$$
\[\left[ X \right]\xrightarrow{\Delta }Y\left( g \right)\]
$$\left[ X \right]$$ and $$\left[ Y \right]$$ are respectively -
A
$$N{H_4}N{O_3}\,\,\& \,\,{N_2}O$$
B
$$N{H_4}N{O_2}\,\,\& \,\,{N_2}$$
C
$$HN{O_4}\,\,\& \,\,{O_2}$$
D
$${\text{None of these}}$$
Answer :
$$N{H_4}N{O_3}\,\,\& \,\,{N_2}O$$
\[\begin{align}
& N{{H}_{4}}Cl{{O}_{4}}+HN{{O}_{3}}\to HCl{{O}_{4}}+\underset{\left( X \right)}{\mathop{N{{H}_{4}}N{{O}_{3}}}}\, \\
& N{{H}_{4}}N{{O}_{3}}\xrightarrow{\Delta }\underset{\left( Y \right)}{\mathop{{{N}_{2}}O}}\,+2{{H}_{2}}O \\
\end{align}\]
213.
Mark the correct statements about halogens.
A
Electron affinity of halogens is in the order $$F > Cl > Br > I.$$
B
$$HF$$ is the strongest hydrohalic acid.
C
$${F_2}$$ has lower bond dissociation energy than $$C{l_2}.$$
D
All halogens show variable oxidation states.
Answer :
$${F_2}$$ has lower bond dissociation energy than $$C{l_2}.$$
$${F_2}$$ has lower bond dissociation energy than $$C{l_2}$$ due to its small size which results in interelectronic repulsion.
214.
An example of a double salt is
A
bleaching powder
B
$${K_4}\left[ {Fe{{\left( {CN} \right)}_6}} \right]$$
C
hypo
D
potash alum
Answer :
potash alum
Double salts are additon or molecular compounds which are formed by two apparently saturated compounds but they lose their identity
when dissolved in water. The most common example of double salt is potash alum $${K_2}S{O_4} \cdot A{l_2}{\left( {S{O_4}} \right)_3} \cdot 24{H_2}O.$$
215.
Which of the following is a cyclic phosphate ?
A
$${H_3}{P_3}{O_{10}}$$
B
$${H_6}{P_4}{O_{13}}$$
C
$${H_5}{P_5}{O_{15}}$$
D
$${H_7}{P_5}{O_{16}}$$
Answer :
$${H_5}{P_5}{O_{15}}$$
$${H_5}{P_5}{O_{15}}$$ or $${\left( {HP{O_3}} \right)_5}.$$ It is metaphosphoric acid which is a cyclic phosphate.
216.
Boric acid is a weak monobasic acid and acts as Lewis acid
A
all hydrogen atoms lie in one plane and boron atoms lie in a plane perpendicular to this plane
B
2 boron atoms and 4 terminal hydrogen atoms lie in the same plane and 2 bridging hydrogen atoms lie in the perpendicular plane
C
4 bridging hydrogen atoms and boron atoms lie in one plane and two terminal hydrogen atoms lie in a plane perpendicular to this plane
D
all the atoms are in the same plane
Answer :
2 boron atoms and 4 terminal hydrogen atoms lie in the same plane and 2 bridging hydrogen atoms lie in the perpendicular plane
No explanation is given for this question. Let's discuss the answer together.
218.
Ammonia is a Lewis base. It forms complexes with cations. Which one of the following cations does not form complex with ammonia?
A
$$A{g^ + }$$
B
$$C{u^{2 + }}$$
C
$$C{d^{2 + }}$$
D
$$P{b^{2 + }}$$
Answer :
$$P{b^{2 + }}$$
No explanation is given for this question. Let's discuss the answer together.
219.
$$Ge\left( {{\text{II}}} \right)$$ compounds are powerful reducing agents whereas $$Pb\left( {{\text{IV}}} \right)$$ compounds are strong oxidants. It is because
A
$$Pb$$ is more electropositive than $$Ge$$
B
ionization potential of lead is less than that of $$Ge$$
C
ionic radii of $$P{b^{2 + }}$$ and $$P{b^{4 + }}$$ are larger than
those of $$G{e^{2 + }}$$ and $$G{e^{4 + }}$$
D
of more pronounced inert pair effect in lead than in $$Ge$$
Answer :
of more pronounced inert pair effect in lead than in $$Ge$$
$$Ge\left( {{\text{II}}} \right)$$ tends to acquire $$Ge\left( {{\text{IV}}} \right)$$ state by loss of electrons. Hence it is reducing in nature. $$Pb\left( {{\text{IV}}} \right)$$ tends to acquire $$Pb\left( {{\text{II}}} \right)O.S.$$ by gain of electrons. Hence it is oxidising in nature. This is due to inert pair effect.
220.
On controlled hydrolysis and condensation, $${R_3}SiCl$$ yields
A
B
C
D
Answer :
$${R_3}SiCl$$ on hydrolysis forms only a dimer.
$${R_3}SiOH + HOSi{R_3} \to {R_3}Si - O - Si{R_3}.$$