P - Block Elements MCQ Questions & Answers in Inorganic Chemistry | Chemistry
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1.
Which one of the following is the correct statement?
A
Boric acid is a protonic acid
B
Beryllium exhibits coordination number of six
C
Chlorides of both beryllium and aluminium have bridged
chloride structures in solid phase
D
$${B_2}{H_6}2N{H_3}$$ is known as ‘inorganic benzene’
Answer :
Chlorides of both beryllium and aluminium have bridged
chloride structures in solid phase
The correct formula of inorganic benzene is $${B_3}{N_3}{H_6}$$ so (D) is incorrect statement
incorrect statement.
The coordination number exhibited by beryllium is $$4$$ and not $$6$$ so statement (B) is incorrect.
Both $$BeC{l_2}\,{\text{and}}\,AlC{l_3}$$ exhibit bridged structures in solid so (C) is correct statement.
2.
Which of the following are peroxoacids of sulphur?
A
$${H_2}S{O_5}\,\,{\text{and}}\,\,{H_2}{S_2}{O_8}$$
B
$${H_2}S{O_5}\,\,{\text{and}}\,\,{H_2}{S_2}{O_7}$$
C
$${H_2}{S_2}{O_7}\,\,{\text{and}}\,\,{H_2}{S_2}{O_8}$$
D
$${H_2}{S_2}{O_6}\,\,{\text{and}}\,\,{H_2}{S_2}{O_7}$$
$${H_2}S{O_5}\,\,{\text{and}}\,\,{H_2}{S_2}{O_8}$$
Both $${H_2}S{O_5}$$ and $${H_2}{S_2}{O_8}$$ have peroxide linkages.
3.
Which of the following bonds will be most polar?
A
$$N—Cl$$
B
$$O—F$$
C
$$N—F$$
D
$$N—N$$
Answer :
$$N—F$$
The polar character arises due to the difference in electronegativity. The electronegativity difference of $$N—F$$ bond is maximum, so it is more polar bond.
4.
Bauxite ore is generally contaminated with impurity of oxides of two elements $$X$$ and $$Y.$$ Which of the following statement is correct ?
A
$$X$$ is a non-metal and belongs to the third period while $$Y$$ is a metal and belongs to the fourth period.
B
One of two oxides has three-dimensional polymeric structure.
C
Both (A) and (B) are correct.
D
None of the above.
Answer :
Both (A) and (B) are correct.
Two oxides present in bauxite as an impurity are $$Si{O_2}$$ and $$F{e_2}{O_3}.$$ $$Si$$ belongs to the third period and $$Fe$$ to the fourth period. $$Si{O_2}$$ has a three dimensional structure.
5.
$$AlC{l_3}$$ achieves stability by forming a dimer. In trivalent state the compound is hydrolysed in water. $$AlC{l_3}$$ in acidified aqueous solution forms
8.
From the following information
$$X + {H_2}S{O_4} \to Y$$ ( a colourless and irritating gas )
$$Y + {K_2}C{r_2}{O_7} + {H_2}S{O_4} \to $$ ( green coloured solution )
Identify the pair $$X$$ and $$Y.$$
A
$$\bar Cr,HCl$$
B
$$SO_3^{2 - },S{O_2}$$
C
$${S^{2 - }},{H_2}S$$
D
$$CO_3^{2 - },C{O_2}$$
Answer :
$$SO_3^{2 - },S{O_2}$$
\[\underset{\left( X \right)}{\mathop{SO_{3}^{2-}}}\,+2{{H}^{\oplus }}\to S{{O}_{2}}+{{H}_{2}}O\]
\[\underset{\left( Y \right)}{\mathop{3S{{O}_{2}}}}\,+{{K}_{2}}C{{r}_{2}}{{O}_{7}}+{{H}_{2}}S{{O}_{4}}\to \] \[{{K}_{2}}S{{O}_{4}}+\underset{\begin{smallmatrix}
\text{Green coloured} \\
\text{solution}
\end{smallmatrix}}{\mathop{C{{r}_{2}}{{\left( S{{O}_{4}} \right)}_{2}}+{{H}_{2}}O}}\,\]
9.
The structure of diborane $$\left( {{B_2}{H_6}} \right)$$ contains
A
four $$2c - 2e$$ bonds and four $$3c - 2e$$ bonds
B
two $$2c - 2e$$ bonds and two $$3c - 3e$$ bonds
C
two $$2c - 2e$$ bonds and four $$3c - 2e$$ bonds
D
four $$2c - 2e$$ bonds and two $$3c - 2e$$ bonds
Answer :
four $$2c - 2e$$ bonds and two $$3c - 2e$$ bonds
Indiborane structure $${{B_2}{H_6}}$$ there are two $$2c - 2e$$ bonds and two $$3c - 2e$$ bonds (see structure of diborane).
Structure of $${{B_2}{H_6}}$$ :
10.
What is the correct observation when $$B{r_2}$$ is treated with $$NaF, NaCl$$ and $$NaI$$ taken in three test-tubes labelled as $$(X), (Y)$$ and $$(Z)?$$
A
$${F_2}$$ is liberated in $$(X)$$ and $$C{l_2}$$ in $$(Y).$$
B
Only $${I_2}$$ is liberated in $$(Z).$$
C
Only $$C{l_2}$$ is liberated in $$(Y).$$
D
Only $${F_2}$$ is liberated in $$(X).$$
Answer :
Only $${I_2}$$ is liberated in $$(Z).$$
More reactive halogen can displace less reactive halogen from its salt solution. $${F_2}$$ can displace $$C{l_2},B{r_2},{I_2}$$ from their salt solutions while $$C{l_2}$$ can displace $$B{r_2}$$ and $${I_2}$$ from $$NaBr$$ and $$NaI.B{r_2}$$ can displace only $${I_2}$$ from $$NaI.$$
$$2NaI + B{r_2} \to 2NaBr + {I_2} \uparrow $$