When ammonia dissolve in water, it form ammonium hydroxide which is ionise as given below
$$N{H_3} + {H_2}O \to N{H_4}OH \rightleftharpoons NH_4^ + + O{H^ - }$$
614.
\[\begin{align}
& M{{g}_{3}}{{B}_{2}}\xrightarrow{HC{{l}_{\left( aq \right)}}}\left[ X \right]+MgC{{l}_{2}} \\
& \left[ X \right]+{{H}_{2}}O\xrightarrow{HC{{l}_{\left( aq \right)}}}\left[ Y \right]+{{H}_{2}} \\
\end{align}\]
For $$\left[ X \right]$$ and $$\left[ Y \right]$$ the incorrect choice is
A
$$\left[ X \right]$$ is $${B_2}{H_6}$$ and $$\left[ Y \right]$$ is $${H_3}B{O_3}$$
B
$$\left[ X \right]$$ with air and $$\left[ Y \right]$$ on strong heating ( red heat ) give same compound
C
In $$\left[ Y \right],$$ boron completes its octet by removing $${H^ + }$$ from water molecule
\[\begin{align}
& M{{g}_{3}}{{B}_{2}}+6HC{{l}_{\left( aq \right)}}\to \underset{\left( X \right)}{\mathop{{{B}_{2}}{{H}_{6}}}}\,+3MgC{{l}_{2}} \\
& \underset{\left( X \right)}{\mathop{{{B}_{2}}{{H}_{6}}}}\,+6{{H}_{2}}O\to \underset{\left( Y \right)}{\mathop{2{{H}_{3}}B{{O}_{3}}}}\,+6{{H}_{2}} \\
& \underset{\left( X \right)}{\mathop{{{B}_{2}}{{H}_{6}}}}\,+3{{O}_{2}}\to {{B}_{2}}{{O}_{3}}+3{{H}_{2}}O \\
& \underset{\left( Y \right)}{\mathop{{{H}_{3}}B{{O}_{3}}}}\,\xrightarrow{\Delta }{{B}_{2}}{{O}_{3}} \\
\end{align}\]
In $$Y,B$$ completes it octet by removing \[{{H}^{+}}\] from water molecule.
$$\because \,C{l_2}$$ is more reactive than bromine
619.
Bond angle in $${H_2}O\left( {{{104.5}^ \circ }} \right)$$ is higher than the bond angle of $${H_2}S\left( {{{92.1}^ \circ }} \right).$$ The difference is due to
A
$$O$$ is diatomic and $$S$$ is tetra-atomic
B
difference in electronegativity of $$S$$ and $$O$$
C
difference in oxidation states of $$S$$ and $$O$$
D
difference in shapes of hybrid orbitals of $$S$$ and $$O.$$
Answer :
difference in electronegativity of $$S$$ and $$O$$