P - Block Elements MCQ Questions & Answers in Inorganic Chemistry | Chemistry
Learn P - Block Elements MCQ questions & answers in Inorganic Chemistry are available for students perparing for IIT-JEE, NEET, Engineering and Medical Enternace exam.
531.
Which of the following acids cannot be stored in glass?
A
$$HF$$
B
$$HCl$$
C
$${H_2}S{O_4}$$
D
$$HI$$
Answer :
$$HF$$
$$HF$$ attacks glass and gets dissolved in it.
$$N{a_2}Si{O_3} + 6HF \to N{a_2}Si{F_6} + 3{H_2}O$$
532.
Ionisation enthalpy $$\left( {{\Delta _i}{H_1}\,kJ\,mo{l^{ - 1}}} \right)$$ for the elements of Group 13 follows the order
A
$$B > Al > Ga > In > Tl$$
B
$$B < Al < Ga < In < Tl$$
C
$$B < Al > Ga < In > Tl$$
D
$$B > Al < Ga > In < Tl$$
Answer :
$$B > Al < Ga > In < Tl$$
The decrease in ionisation enthalpy from $$B$$ to $$Al$$ is associated with increase in size. The observed discontinuity in the ionisation enthalpy values between $$Al$$ and $$Ga,$$ and between $$In$$ and $$Tl$$ are due to inability of $$d$$ - and $$f$$ - electrons, which have low screening effect, to compensate the increase in nuclear charge.
$$B{\left( {OH} \right)_3}$$ orthoboric acid is acidic in nature and can accept $$O{H^ - }$$ ions to release $${H^ + }$$ in solution.
$${H_3}B{O_3} + 2{H_2}O \to $$ $${H_3}{O^ + } + {\left[ {B{{\left( {OH} \right)}_4}} \right]^ - }$$
535.
Boron nitride can be represented by the given structure.
The structure of $$BN$$ is similar to
A
graphite
B
diamond
C
benzene
D
pyridine
Answer :
graphite
No explanation is given for this question. Let's discuss the answer together.
536.
In $${P_4}{O_{10}}$$ each $$P$$ atom is linked with .......... $$O$$ atoms
A
$$2$$
B
$$3\,$$
C
$$4$$
D
$$5$$
Answer :
$$4$$
In $${P_4}{O_{10}}$$ each $$P$$ atom is linked to 4 oxygen atoms as can be confirmed by its structure. It is linked to three oxygen atoms by single bond and with one oxygen atom by double bond.
537.
\[N{{a}_{2}}{{B}_{4}}{{O}_{7}}\cdot 10{{H}_{2}}O\xrightarrow{\Delta }X\] \[\xrightarrow{\Delta }Y+Z\]
$$X, Y$$ and $$Z$$ in the reaction are
A
$$X = N{a_2}{B_4}{O_7},Y = NaB{O_2},$$ $$Z = {B_2}{O_3}$$
B
$$X = N{a_2}{B_4}{O_7},Y = {B_2}{O_3},$$ $$Z = {H_3}B{O_3}$$
\[N{{a}_{2}}{{B}_{4}}{{O}_{7}}\cdot 10{{H}_{2}}O\xrightarrow{\Delta }\underset{\left( X \right)}{\mathop{N{{a}_{2}}{{B}_{4}}{{O}_{7}}}}\,\] \[\xrightarrow{\Delta }\underset{\left( Y \right)}{\mathop{2NaB{{O}_{2}}}}\,+\underset{\left( Z \right)}{\mathop{{{B}_{2}}{{O}_{3}}}}\,\]
538.
A black powder when heated with conc. $$HCl$$ gives a greenish yellow gas. The gas acts as an oxidising and a bleaching agent. When it is passed over slaked lime, a white powder is formed which is a ready source of the greenish yellow gas. The black powder and white powder respectively are
A
$$KCl{O_3}\,{\text{and}}\,NaCl{O_3}$$
B
$$Mn{O_2}\,{\text{and}}\,Ca{\left( {OCl} \right)_2}$$
539.
Aluminium is usually found in $$ + 3$$ oxidation state. In
contrast, thallium exists in $$ + 1\,\,{\text{and}} + 3$$ oxidation states. This is due to:
A
inert pair effect
B
diagonal relationship
C
lattice effect
D
lanthanoid contraction
Answer :
inert pair effect
Due to the inert pair effect, thallium exists in more than one oxidation state. Also, for thallium + 1 oxidation state is more stable than + 3 oxidation state.
540.
In the commercial electrochemical process for aluminium extraction the electrolyte used is
A
$$Al{\left( {OH} \right)_3}$$ in $$NaOH$$ solution
B
an aqueous solution of $$A{l_2}{\left( {S{O_4}} \right)_3}.$$
C
a molten mixture of $$A{l_2}{O_3}$$ and $$N{a_3}Al{F_6}$$
D
a molten mixture of $$AlO\left( {OH} \right)$$ and $$Al{\left( {OH} \right)_3}$$
Answer :
a molten mixture of $$A{l_2}{O_3}$$ and $$N{a_3}Al{F_6}$$
$$A{l_2}{O_3}$$ is electrolyte, while $$N{a_3}Al{F_6}$$ is used to decrease the melting point of $$A{l_2}{O_3}$$ and to increase the conductivity.