Explain the following:
$(1)$ $Tl (NO_3)_2$ acts as an oxidizing agent.
$(2)$ Carbon shows catenation property but lead does not.
Explain the following:
$(1)$ $Tl (NO_3)_2$ acts as an oxidizing agent.
$(2)$ Carbon shows catenation property but lead does not.
Due to inert pair effect, $\mathrm{Tl}$ is more stable in $+1$ oxidation state than that of $+3$ oxidation state. Therefore, $\mathrm{Tl}\left(\mathrm{NO}_{3}\right)_{3}$ acts as strong oxidizing agent.
Carbon atoms have the tendency to link with one another through covalent bonds to form chains and rings. This property is called catenation. This is because $\mathrm{C}-\mathrm{C}$ bonds are very strong. Down the group the size increases and electronegativity decreases and there by, tendency to show catenation decreases. This can be clearly seen from bond enthalpies values. The order of catenation is $\mathrm{C} \gg \mathrm{Si}>\mathrm{Ge} \approx \mathrm{Sn}$. Lead does not show catenation.
Similar Questions
Complete the following chemical equations :
$Z + 3LiAl{H_4} \to X + 3LiF + 3Al{F_3}$
$X + 6{H_2}O \to Y + 6{H_2}$
$3X + 3{O_2}\xrightarrow{\Delta }{B_2}{O_3} + 3{H_2}O$
Complete the following chemical equations :
$Z + 3LiAl{H_4} \to X + 3LiF + 3Al{F_3}$
$X + 6{H_2}O \to Y + 6{H_2}$
$3X + 3{O_2}\xrightarrow{\Delta }{B_2}{O_3} + 3{H_2}O$
Assertion : Both $Be$ and $Al$ can form complexes such as $BeF_4^{2-}$ and $AlF_6^{3-}$ respectively, $BeF_6^{3-}$ is not formed.
Reason : In case of $Be$, no vacant $d-$ orbitals are present in its outermost shell.
Assertion : Both $Be$ and $Al$ can form complexes such as $BeF_4^{2-}$ and $AlF_6^{3-}$ respectively, $BeF_6^{3-}$ is not formed.
Reason : In case of $Be$, no vacant $d-$ orbitals are present in its outermost shell.
- [AIIMS 2015]
A certain salt $X,$ gives the following results.
$(i)$ Its aqueous solution is alkaline to litmus.
$(ii) $ It swells up to a glassy material $Y$ on strong heating.
$(iii)$ When conc. $H _{2} SO _{4}$ is added to a hot solution of $X,$ white crystal of an acid $Z$ separates out.
Write equations for all the above reactions and identify $X, Y$ and $Z$
A certain salt $X,$ gives the following results.
$(i)$ Its aqueous solution is alkaline to litmus.
$(ii) $ It swells up to a glassy material $Y$ on strong heating.
$(iii)$ When conc. $H _{2} SO _{4}$ is added to a hot solution of $X,$ white crystal of an acid $Z$ separates out.
Write equations for all the above reactions and identify $X, Y$ and $Z$
Boric acid is an acid because its molecule
Boric acid is an acid because its molecule
- [NEET 2016]
$(i)\,Al\,\xrightarrow{{{N_2}}}\,A$ $(ii)\,Al\,\xrightarrow{c}\,B$
Here $A$ and $B$ on hydrolysis respectively gives
$(i)\,Al\,\xrightarrow{{{N_2}}}\,A$ $(ii)\,Al\,\xrightarrow{c}\,B$
Here $A$ and $B$ on hydrolysis respectively gives