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.
$${H_2}{S_2}{O_6}\,\,{\text{and}}\,\,{H_2}{S_2}{O_7}$$
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.
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 $$Y.$$
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 $$(Y).$$
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 $$NaI.$$
$$2NaI + B{r_2} \to 2NaBr + {I_2} \uparrow $$