Mix Examples-Hydrogen and Its compounds Questions in English

(C) Sodium oxalate $(Na_2C_2O_4)$ reacts with concentrated $H_2SO_4$ to produce carbon monoxide $(CO)$,carbon dioxide $(CO_2)$,and sodium hydrogen sulfate $(NaHSO_4)$.
The chemical reaction is: $Na_2C_2O_4 + H_2SO_4 \rightarrow Na_2SO_4 + CO + CO_2 + H_2O$ (or with excess acid to form $NaHSO_4$).
Thus,the products are $CO$ and $CO_2$.

(C) The reaction of oxalic acid with concentrated $H_2SO_4$ is as follows:
$(COOH)_2 \xrightarrow{H_2SO_4} CO + CO_2 + H_2O$
The mixture of gases produced is $CO$ and $CO_2$.
When this mixture is passed through caustic potash $(KOH)$,$CO_2$ is acidic in nature and gets absorbed,while $CO$ is neutral and passes through.
The reaction between $CO_2$ and $KOH$ is:
$2KOH + CO_2 \to K_2CO_3 + H_2O$
Thus,$K_2CO_3$ (potassium carbonate) is formed.

(D) $I$. Atomic hydrogen is produced by passing an electric arc through hydrogen gas at low pressure,which is a correct statement.
$II$. Hydrogen is a weaker reducing agent than aluminum,so it cannot reduce $Al_2O_3$ to $Al$. This is a correct statement.
$III$. Finely divided metals like $Pd$,$Pt$,and $Ni$ have the property of adsorbing large volumes of hydrogen gas,which is a correct statement.
$IV$. Nascent hydrogen is highly reactive and is produced in situ during chemical reactions. The reaction $Na + C_2H_5OH \rightarrow C_2H_5ONa + [H]$ is a standard method to generate nascent hydrogen. This is a correct statement.
Therefore,all statements $I, II, III,$ and $IV$ are correct.

(A) The reaction of oxalic acid with conc. $H_2SO_4$ is: $(COOH)_2 \xrightarrow{\text{conc. } H_2SO_4, \Delta} CO + CO_2 + H_2O$.
$CO$ is a neutral gas,while $CO_2$ is an acidic gas.
Potassium hydroxide $(KOH)$ is a base and it absorbs the acidic gas,$CO_2$.
The reaction is: $2KOH + CO_2 \rightarrow K_2CO_3 + H_2O$.
Therefore,the product formed is $K_2CO_3$ and the gas absorbed is $CO_2$.

(D) $1$. Boiling point: The boiling point of $H_2O$ is $100 \ ^\circ C$ $(373 \ K)$,while the boiling point of $H_2O_2$ is $150 \ ^\circ C$ $(423 \ K)$. Thus,$H_2O < H_2O_2$ is correct.
$2$. Acidic nature: $H_2O_2$ is a stronger acid than $H_2O$ because the $O-H$ bond in $H_2O_2$ is more polar due to the electron-withdrawing effect of the second oxygen atom,and the resulting hydroperoxide ion $(HO_2^-)$ is more stable than the hydroxide ion $(OH^-)$. Thus,$H_2O < H_2O_2$ is correct.
$3$. Dipole moment: The dipole moment of $H_2O$ is $1.85 \ D$,while the dipole moment of $H_2O_2$ is $2.13 \ D$. Thus,$H_2O < H_2O_2$ is correct.
Since all statements are correct,the correct option is $D$.

(A) $(i)$ Hydrogen peroxide $(H_2O_2)$ is widely used as an antiseptic and bleaching agent. Hence,$(i)-(c)$.
$(ii)$ Calcium sulphate hemihydrate $(CaSO_4 \cdot \frac{1}{2}H_2O)$ is known as Plaster of Paris. Hence,$(ii)-(d)$.
$(iii)$ Potassium superoxide $(KO_2)$ is used in breathing apparatus and as an air purifier in submarines because it reacts with $CO_2$ to release $O_2$. Hence,$(iii)-(a)$.
$(iv)$ Calgon $(Na_6P_6O_{18})$ is used for water softening by sequestering calcium and magnesium ions. Hence,$(iv)-(b)$.
Therefore,the correct matching is $(i)-(c), (ii)-(d), (iii)-(a), (iv)-(b)$.

(B) $(i)$ Hydrogen peroxide $(H_2O_2)$ is widely used as an antiseptic.
$(ii)$ Calcium sulphate hemihydrate $(CaSO_4 \cdot \frac{1}{2}H_2O)$ is known as Plaster of Paris.
$(iii)$ Potassium superoxide $(KO_2)$ is used as an air purifier in submarines and space crafts because it absorbs $CO_2$ and releases $O_2$.
$(iv)$ Calgon $(Na_2[Na_4(PO_3)_6])$ is used for water softening.
Therefore,the correct match is: $(i-c, ii-d, iii-a, iv-b)$.

(A) The boiling point $(BP)$ depends on the strength of intermolecular forces.
$H_2O_2$ has strong hydrogen bonding and high molecular weight,leading to the highest $BP$ $(423 \ K)$.
$D_2O$ has stronger hydrogen bonding and higher mass than $H_2O$,so $BP(D_2O) > BP(H_2O)$ $(374.4 \ K > 373 \ K)$.
$D_2$ and $H_2$ are non-polar molecules held by weak van der Waals forces. Since $D_2$ has a higher molecular mass than $H_2$,$BP(D_2) > BP(H_2)$.
Therefore,the correct order is $H_2O_2 > D_2O > H_2O > D_2 > H_2$.

(C) The thermal decomposition of oxalic acid is: $H_2C_2O_4 \xrightarrow{\Delta } CO + CO_2 + H_2O$.
Thus,$A = CO$,$B = CO_2$,and $C = H_2O$.
$CO$ burns with a blue flame and is oxidized to $CO_2$.
Further,$CO + Cl_2 \to COCl_2 (D)$ (phosgene).
Then,$COCl_2 + 4NH_3 \to NH_2CONH_2 (E) + 2NH_4Cl$.
Therefore,$A, B, C \text{ and } E \text{ are } CO, CO_2, H_2O \text{ and } NH_2CONH_2$.

(D) When copper sulphate $(CuSO_4)$ is mixed with lime $(Ca(OH)_2)$ suspension,it forms a mixture known as "Bouillie Bordelaise" or Bordeaux mixture.
This mixture is widely used as a fungicide in agriculture.

(A) $(i)$ $Cr$
$(ii)$ $D_2O$ (Heavy water)
$(iii)$ Hydrogen
$(iv)$ Auto-oxidation

(A) $H_2O_2$ is used as a disinfectant $(1-s)$.
$D_2O$ is used as a moderator in nuclear reactors $(2-p)$.
Liquid $H_2$ is used as a rocket fuel $(3-q)$.
Zeolite is used to soften hard water $(4-r)$.
Therefore,the correct matching is $(1-s, 2-p, 3-q, 4-r)$.

(A) $(1-c, 2-d, 3-a, 4-b)$

(N/A) The correlations are as follows:
$A$ (Synthesis gas): $9$ (Synthesis of methanol),$10$ (Mixture of $CO$ and $H_2$)
$B$ (Dihydrogen): $4$ (Reducing agent),$7$ $(Zn + NaOH)$,$8$ $(Zn + dil. H_2SO_4)$
$C$ (Heavy water): $6$ (Prolonged electrolysis of water)
$D$ (Calgon): $1$ $(Na_2[Na_4(PO_3)_6])$,$3$ (Softening of water)
$E$ (Hydrogen peroxide): $2$ (Oxidising agent),$4$ (Reducing agent)
$F$ (Salt like hydrides): $5$ (Stoichiometric compounds of $s-$block elements)

(A-4, B-3, C-2, D-1) The correct matches are:
$A-4$: Atomic hydrogen $(H)$ is used in atomic hydrogen welding and cutting.
$B-3$: Dihydrogen $(H_2)$ is used in the hydroformylation of olefins to produce aldehydes.
$C-2$: Water $(H_2O)$ can be reduced to dihydrogen by $NaH$ $(NaH + H_2O \rightarrow NaOH + H_2)$.
$D-1$: Hydrogen peroxide $(H_2O_2)$ is known as perhydrol.
Therefore,the correct sequence is $A-4, B-3, C-2, D-1$.

(A-5, B-4, C-2, D-1, E-3) $A-5, B-4, C-2, D-1, E-3$

(A-2,4; B-3; C-1,3) $A-2, 4; B-3; C-1, 3$.
$A$. Hydrogen peroxide $(H_2O_2)$ is known as perhydrol (at high concentrations) and is also used as a rocket propellant.
$B$. Calgon method uses sodium hexametaphosphate $(Na_6P_6O_{18})$ to remove hardness.
$C$. Permanent hardness of water is removed by using zeolites (ion exchange method) or sodium hexametaphosphate (Calgon method).

(B) The correct matches are:
$a. CO_{(g)} + H_{2(g)} \rightarrow iii. \text{Synthesis gas}$
$b. \text{Temporary hardness of water} \rightarrow i. Mg(HCO_3)_2 + Ca(HCO_3)_2$
$c. B_2H_6 \rightarrow ii. \text{An electron deficient hydride}$
$d. H_2O_2 \rightarrow iv. \text{Non-planar structure}$
Thus, the correct sequence is $a-iii, b-i, c-ii, d-iv$.

(D) $P$. $Mg(HCO_3)_2 + 2Ca(OH)_2 \rightarrow Mg(OH)_2 + 2CaCO_3 + 2H_2O$ (Produces $Mg(OH)_2$ and $CaCO_3$)
$Q$. $BaO_2 + H_2SO_4 \rightarrow H_2O_2 + BaSO_4$ (Produces $H_2O_2$)
$R$. $Ca(OH)_2 + MgCl_2 \rightarrow Mg(OH)_2 + CaCl_2$ (Produces $Mg(OH)_2$)
$S$. $BaO_2 + 2HCl \rightarrow BaCl_2 + H_2O_2$ (Produces $BaCl_2$ and $H_2O_2$)
$T$. $Ca(HCO_3)_2 + Ca(OH)_2 \rightarrow 2CaCO_3 + 2H_2O$ (Produces $CaCO_3$)
Matching:
$I (H_2O_2)$ is produced in $Q$ and $S$.
$II (Mg(OH)_2)$ is produced in $P$ and $R$.
$III (BaCl_2)$ is produced in $S$.
$IV (CaCO_3)$ is produced in $P$ and $T$.
Given the options,the best match is $I$ $\rightarrow Q; II$ $\rightarrow R; III$ $\rightarrow S; IV$ $\rightarrow P$.

(C) $i$. $2Al_{(s)} + 6HCl_{(aq)} \rightarrow 2AlCl_{3(aq)} + 3H_{2(g)}$ (Liberates $H_2$)
$ii$. $2Al_{(s)} + 2NaOH_{(aq)} + 6H_2O_{(l)} \rightarrow 2Na[Al(OH)_4]_{(aq)} + 3H_{2(g)}$ (Liberates $H_2$)
$iii$. $B_2H_{6(g)} + 6H_2O_{(l)} \rightarrow 2H_3BO_{3(aq)} + 6H_{2(g)}$ (Liberates $H_2$)
$iv$. $2F_{2(g)} + 2H_2O_{(l)} \rightarrow 4HF_{(aq)} + O_{2(g)}$ (Liberates $O_2$,not $H_2$)
Therefore,reactions $i, ii,$ and $iii$ liberate $H_2$.

(D) $(i)$ $2Al_{(s)} + 6HCl_{(aq)} \rightarrow 2AlCl_{3(aq)} + 3H_{2(g)}$
$(ii)$ $2Al_{(s)} + 2NaOH_{(aq)} + 6H_2O_{(l)} \rightarrow 2Na[Al(OH)_4]_{(aq)} + 3H_{2(g)}$
$(iii)$ $2NaBH_4 + I_2 \rightarrow B_2H_6 + 2NaI + H_{2(g)}$
In all three reactions,hydrogen gas $(H_2)$ is liberated.

(C) In deep-sea diving,$O_2$ is diluted with $He$ instead of $N_2$.
This is because $He$ has very low solubility in blood even under high pressure,which prevents the formation of bubbles in the blood (a condition known as bends) when the diver returns to the surface.