631.
$$Al{F_3}$$ is soluble in $$HF$$ only in presence of $$KF.$$
It is due to the formation of
A
$${K_3}\left[ {Al{F_3}{H_3}} \right]$$
B
$${K_3}\left[ {Al{F_6}} \right]$$
C
$$Al{H_3}$$
D
$$K\left[ {Al{F_3}H} \right]$$
Answer :
$${K_3}\left[ {Al{F_6}} \right]$$
View Solution
Discuss Question
Key Idea $$A{l^{3 + }}$$ shows maximum coordination number 6, thus it will form $$AlF_6^{3 - }.$$
$${Al{F_3}}$$ forms $${K_3}\left[ {Al{F_6}} \right]$$ when dissolved in $$HF$$ in the presence of $$KF$$ as shown below :
\[Al{{F}_{3}}+3KF\xrightarrow{HF}{{K}_{3}}\left[ Al{{F}_{6}} \right]\]
632.
In borax bead test which compound is formed?
A
$$Ortho$$ borate
B
$$Meta$$ borate
C
Double oxide
D
Tetra borate
Answer :
$$Meta$$ borate
View Solution
Discuss Question
In borax bead test the coloured $$meta$$ borates are formed by transition metal salts.
\[N{{a}_{2}}{{B}_{4}}{{O}_{7}}.10{{H}_{2}}O\xrightarrow[-10{{H}_{2}}O]{\Delta }\] \[N{{a}_{2}}{{B}_{4}}{{O}_{7}}\xrightarrow{\Delta }2NaB{{O}_{2}}+{{B}_{2}}{{O}_{3}}\]
\[{{B}_{2}}{{O}_{3}}+CuO\to \underset{\begin{smallmatrix}
\text{Cupric}\,\,\text{metaborate} \\
\left( \text{Blue}\,\,\text{bead} \right)
\end{smallmatrix}}{\mathop{Cu{{\left( B{{O}_{2}} \right)}_{2}}}}\,\]
633.
Which of the following phosphorus is the most reactive?
A
Red phosphorus
B
White phosphorus
C
Scarlet phosphorus
D
Violet phosphorus
Answer :
White phosphorus
View Solution
Discuss Question
White phosphorus has low ignition temperature.
So, it is most reactive.
634.
Each of the following is true for white and red phosphorus except that they
A
both are soluble in $$C{S_2}$$
B
can be oxidised by heating in air
C
consist of the same kind of atoms
D
can be converted into one another
Answer :
both are soluble in $$C{S_2}$$
View Solution
Discuss Question
White and red phosphorus are the main allotropes of phosphorus. White phosphorus is soluble in carbon disulphide whereas red phosphorus is insoluble in carbon disulphide.
635.
Which would quickly absorb oxygen?
A
Alkaline solution of pyrogallol
B
Conc. $${H_2}S{O_4}$$
C
Lime water
D
Alkaline solution of $$CuS{O_4}$$
Answer :
Alkaline solution of pyrogallol
View Solution
Discuss Question
Alkaline solution of pyrogallol is used to absorb dioxygen gas $$\left( {{O_2}} \right).$$
636.
In group 13, electronegativity first decreases from $$B$$ to $$Al$$ and then increases marginally down the group. This is because of
A
non-metallic nature of $$B$$
B
discrepancies in atomic size of elements
C
ability of $$B$$ and $$Al$$ to form $$p\pi - p\pi $$ multiple bonds
D
irregular trend in electronegativity throughout the periodic table.
Answer :
discrepancies in atomic size of elements
View Solution
Discuss Question
There is a large size difference between $$B$$ and $$Al.$$
637.
The decreasing order of boiling points of the following hydrides is
A
$$Sb{H_3} > As{H_3} > P{H_3} > N{H_3}$$
B
$$N{H_3} > Sb{H_3} > As{H_3} > P{H_3}$$
C
$$Sb{H_3} > N{H_3} > As{H_3} > P{H_3}$$
D
$$P{H_3} > As{H_3} > Sb{H_3} > N{H_3}$$
Answer :
$$Sb{H_3} > N{H_3} > As{H_3} > P{H_3}$$
View Solution
Discuss Question
$$N{H_3}$$ has abnormally high boiling points because of their tendency to form hydrogen bonds. The order of boiling point will be $$Sb{H_3} > N{H_3} > As{H_3} > P{H_3}$$
638.
The compound of xenon with zero dipole moment is
A
$$Xe{O_3}$$
B
$$Xe{F_4}$$
C
$$XeO{F_4}$$
D
$$Xe{O_2}$$
Answer :
$$Xe{F_4}$$
View Solution
Discuss Question
$$Xe{F_4}$$ has zero dipole moment. It has square planar structure due to which the bond moments of $$Xe - F$$ cancel each other.
639.
Often a ground glass stopper gets stuck in the neck of a glass bottle containing $$NaOH$$ solution. This is due to :
A
The presence of dirt particles in between.
B
The formation of solid silicate in between by the reaction of $$Si{O_2}$$ of glass with $$NaOH.$$
C
The formation of $$N{a_2}C{O_3}$$ in between by the reaction of $$C{O_2}$$ of air and $$NaOH.$$
D
Glass contains a boron compound which forms a precipitate with the $$NaOH$$ solution.
Answer :
The formation of solid silicate in between by the reaction of $$Si{O_2}$$ of glass with $$NaOH.$$
View Solution
Discuss Question
\[Si{{O}_{2}}+2NaOH\to \underset{\begin{smallmatrix}
\text{Sodium} \\
\text{silicate}
\end{smallmatrix}}{\mathop{N{{a}_{2}}Si{{O}_{3}}}}\,+{{H}_{2}}O\]
640.
Chlorine water on standing loses its colour and forms :
A
$$HCl\,\,{\text{only}}$$
B
$$HCl\,\,{\text{and}}\,\,HCl{O_2}$$
C
$$HCl\,\,{\text{and}}\,\,HOCl$$
D
$$HOCl\,\,{\text{and}}\,\,HOC{l_2}$$
Answer :
$$HCl\,\,{\text{and}}\,\,HOCl$$
View Solution
Discuss Question
Chlorinated water is yellow in colour on standing reaction occurs $$C{l_2} + {H_2}O \to HCl + HOCl$$