P - Block Elements MCQ Questions & Answers in Inorganic Chemistry | Chemistry
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551.
It is because of inability of $$n{s^2}$$ electrons of the valence shell to participate in bonding that
A
$$S{n^{2 + }}$$ is reducing while $$P{b^{4 + }}$$ is oxidising
B
$$S{n^{2 + }}$$ is oxidising while $$P{b^{4 + }}$$ is reducing
C
$$S{n^{2 + }}$$ and $$P{b^{2 + }}$$ are both oxidising and reducing
D
$$S{n^{4 + }}$$ is reducing while $$P{b^{4 + }}$$ is oxidising
Answer :
$$S{n^{4 + }}$$ is reducing while $$P{b^{4 + }}$$ is oxidising
The inability of $$n{s^2}$$ electrons of the valence shell to participate in bonding is called as inert pair effect. Due to this effect, the lower oxidation state becomes more stable on descending the group. Thus, $$S{n^{2 + }}$$ is a reducing agent while $$P{b^{4 + }}$$ act as an oxidising agent.
552.
The correct order of the oxidation states of nitrogen in $$NO,{N_2}O,N{O_2}\,{\text{and}}\,{N_2}{O_3}$$ is:
A
$$N{O_2} < NO < {N_2}{O_3} < {N_2}O$$
B
$$N{O_2} < {N_2}{O_3} < NO < {N_2}O$$
C
$${N_2}O < \,{N_2}{O_3} < NO < N{O_2}$$
D
$${N_2}O < \,NO < {N_2}{O_3} < N{O_2}$$
Answer :
$${N_2}O < \,NO < {N_2}{O_3} < N{O_2}$$
(oxide)
(oxidation state)
$${N_2}O$$
$$ + 1$$
$$NO$$
$$ + 2$$
$${N_2}{O_3}$$
$$ + 3$$
$$N{O_2}$$
$$ + 4$$
So, $${N_2}O < NO < {N_2}{O_3} < N{O_2}$$
553.
Fill in the blanks by choosing an appropriate option. $$\underline {\left( {\text{i}} \right)} $$ is a synthetic radioactive element of group 15 having electronic configuration $$\underline {\left( {{\text{ii}}} \right)} .$$
No explanation is given for this question. Let's discuss the answer together.
554.
Why do boron and aluminium halides behave as Lewis acids?
A
Both halides $$\left( {M{X_3}} \right)$$ can accept electrons from a donor to complete their octet.
B
Both halides $$\left( {M{X_3}} \right)$$ can donate a pair of electrons.
C
Both halides $$\left( {M{X_3}} \right)$$ are covalent polymeric structures.
D
Both halides $$\left( {M{X_3}} \right)$$ react with water to give hydroxides and $$HCl.$$
Answer :
Both halides $$\left( {M{X_3}} \right)$$ can accept electrons from a donor to complete their octet.
Both boron and aluminium in their trihalides $$\left( {M{X_3}} \right)$$ possess six electrons in their valence shell. To complete the octet they can accept a lone pair of electrons acting as Lewis acids.
555.
Which of the following is coloured
A
$$NO$$
B
$${N_2}O$$
C
$$S{O_3}$$
D
$${\text{None }}$$
Answer :
$${\text{None }}$$
All are colourless gases.
556.
The number of $$S - S$$ bonds in $$S{O_3},{S_2}O_3^{2 - },{S_2}O_6^{2 - }$$ and $${S_2}O_8^{2 - }$$ respectively are
A
1, 0, 0, 1
B
1, 0, 1, 0
C
0, 1, 1, 0
D
0, 1, 0, 1
Answer :
0, 1, 1, 0
Hence (C) is the correct option.
557.
What happens when silicon is heated with methyl chloride in presence of copper as a catalyst at $$573\,K?$$
A
Methyl substituted chlorosilanes are formed.
B
Only $$M{e_4}Si$$ is formed.
C
Polymerised chains of $${\left( {C{H_3}} \right)_3}SiCl$$ are formed.
D
Silicones are formed.
Answer :
Methyl substituted chlorosilanes are formed.
Methyl substituted chlorosilanes are formed hydrolysis of dimethyl dichlorosilane give $$M{e_2}Si{\left( {OH} \right)_2}$$ which polymerise to give silicones.
\[2C{{H}_{3}}Cl+Si\xrightarrow[570\,K]{Cu\,\text{powder}}\] \[{{\left( C{{H}_{3}} \right)}_{2}}SiC{{l}_{2}}+C{{H}_{3}}SiC{{l}_{3}}+\] \[{{\left( C{{H}_{3}} \right)}_{3}}SiCl+{{\left( C{{H}_{3}} \right)}_{4}}Si\]
558.
Number of sigma bonds in $${P_4}{O_{10}}$$ is
A
$$6$$
B
$$7$$
C
$$17$$
D
$$16$$
Answer :
$$16$$
559.
Match the column I with column II and mark the appropriate choice.
Column I
Column II
a.
$${H_2}S{O_3}$$
1.
+6, dibasic
b.
$${H_2}S{O_5}$$
2.
+5, dibasic
c.
$${H_2}{S_2}{O_6}$$
3.
+6, monobasic
d.
$${H_2}S{O_4}$$
4.
+4, dibasic
A
a - 1, b - 2, c - 3, d - 4
B
a - 2, b - 3, c - 1, d - 4
C
a - 3, b - 4, c - 2, d - 1
D
a - 4, b - 3, c - 2, d - 1
Answer :
a - 4, b - 3, c - 2, d - 1
No explanation is given for this question. Let's discuss the answer together.
560.
Which of the following fluorides does not exist?
A
$$N{F_5}$$
B
$$P{F_5}$$
C
$$As{F_5}$$
D
$$Sb{F_5}$$
Answer :
$$N{F_5}$$
Nitrogen does not form pentahalide because it does not have vacant $$d$$-orbital.