Due to strong $$H—F$$ bond, $${H^ + }$$ ions are not easily removed due to higher electronegativity $$(EN)$$ of $$F.$$
Hence more bond dissociation energy required.
$${\text{Acidic}}\,{\text{nature}} \propto \frac{1}{{{\text{Bond}}\,{\text{dissociation}}\,{\text{energy}}}}$$
So, $$HF$$ is not a stronger acid than $$HCl.$$
43.
Which of the following is incorrect for white and red phosphorus ?
Both white and red phosphorus are not soluble in $$C{S_2}$$ only white phosphorus is soluble in $$C{S_2}.$$
44.
The formation of the oxide ion $${O^{2 - }}\left( g \right),$$ from oxygen atom requires first an exothermic and then an endothermic step as shown below,
$$O\left( g \right) + {e^ - } \to {O^ - }\left( g \right);$$ $${\vartriangle _f}{H^ \circ } = - 141\,kJmo{l^{ - 1}}$$
$${O^ - }\left( g \right) + {e^ - } \to {O^{2 - }}\left( g \right);$$ $${\vartriangle _f}{H^ \circ } = \, + 780\,kJmo{l^{ - 1}}$$
Thus, process of formation of $${O^{2 - }}$$ in gas phase is unfavourable even though $${O^{2 - }}$$ is isoelectronic with neon. It is due to the fact that
A.
electron repulsion outweighs the stability gained by achieving noble gas configuration
B.
$${O^ - }$$ ion has comparatively srnaller size than
oxygen atom
C.
Oxygen is more electronegative
D.
addition of electron in oxygen result in large size of the ion
Answer :
electron repulsion outweighs the stability gained by achieving noble gas configuration
Since, electron repulsion predominate over the stability gained by achieving noble gas configuration. Hence, formation of $${O^{2 - }}$$ in gas phase is unfavourable.
45.
Which of the following statements is not correct about the structure of $$PC{l_5}?$$
A.
$$PC{l_5}$$ has a square pyramidal structure.
B.
Three equatorial $$P - Cl$$ bonds are equivalent.
C.
The two axial bonds are different and longer than equatorial bonds.
D.
In solid state it exists as $${\left[ {PC{l_4}} \right]^ + }{\left[ {PC{l_6}} \right]^ - }.$$
Answer :
$$PC{l_5}$$ has a square pyramidal structure.
When excess of carbon dioxide is passed in lime water, calcium carbonate is converted to calcium bicarbonate which is soluble, hence the milkiness due to calcium carbonate disappears.
Boric acid can be considered as an acid because its molecule accepts $$O{H^ - }$$ from water, releasing proton.
\[\underset{\text{Acid}}{\mathop{{{H}_{3}}B{{O}_{3}}}}\,+\underset{\text{Base}}{\mathop{{{H}_{2}}O}}\,\rightleftharpoons \underset{\begin{smallmatrix}
\text{Conjugate} \\
\text{base}
\end{smallmatrix}}{\mathop{B\left( OH \right)_{4}^{-}}}\,+\underset{\begin{smallmatrix}
\text{Conjugate} \\
\text{acid}
\end{smallmatrix}}{\mathop{{{H}^{+}}}}\,\] Remember In the given options to the question, (A), (B) and (C) are correct as all of these sentences have more or less similar meaning but here (C) option is the most appropriate one as it gives complete explanation of the fact that how boric acid can be combined with an acid.