Water or hydride of oxygen Questions in English

(C) The physical properties of $H_2O$ and $D_2O$ differ due to the isotope effect.
$D_2O$ is heavier than $H_2O$,which leads to stronger intermolecular forces in $D_2O$.
Consequently,$D_2O$ has a higher boiling point $(101.4 \ ^\circ C)$ compared to $H_2O$ $(100 \ ^\circ C)$.
Similarly,the temperature at maximum density is $11.6 \ ^\circ C$ for $D_2O$ and $4 \ ^\circ C$ for $H_2O$.
The bond dissociation energy for $D-O$ is higher than $H-O$ due to the stronger bond formed by the heavier isotope.
The dielectric constant of $H_2O$ ($78.39$ at $25 \ ^\circ C$) is higher than that of $D_2O$ ($78.06$ at $25 \ ^\circ C$).

(A) Permanent hardness of water is caused by the presence of soluble salts of magnesium and calcium in the form of chlorides $(Cl^-)$ and sulphates $(SO_4^{2-})$.
$HCO_3^-$ ions are responsible for temporary hardness.
Therefore,$SO_4^{2-}$ is a correct option for permanent hardness.

(D) Hardness in water is primarily caused by the presence of dissolved salts of calcium and magnesium,specifically their chlorides,sulfates,and bicarbonates.
$MgCO_3$ (magnesium carbonate) is practically insoluble in water.
Since it does not dissolve to release $Mg^{2+}$ ions in significant amounts,it does not contribute to the hardness of water.

(D) Water softening involves the removal of calcium $(Ca^{2+})$ and magnesium $(Mg^{2+})$ ions from hard water.
$A$. Washing soda $(Na_2CO_3)$ is used to remove permanent hardness by precipitating calcium and magnesium ions as carbonates.
$B$. Calgon $(Na_2[Na_4(PO_3)_6])$ is used to sequester calcium and magnesium ions,preventing them from forming precipitates.
$C$. Sodium zeolite $(Na_2Al_2Si_2O_8 \cdot xH_2O)$ is an ion-exchange resin used to replace $Ca^{2+}$ and $Mg^{2+}$ ions with $Na^+$ ions.
$D$. $CaCO_3$ (calcium carbonate) is a source of hardness in water,not a softening agent. Therefore,it is not used for softening water.

(C) The process of preparing pure and demineralised water (also known as deionised water) involves the removal of all dissolved mineral salts,including cations and anions.
This is achieved using ion-exchange resins.
Specifically,a cation-exchange resin (in $H^+$ form) removes cations like $Ca^{2+}$ and $Mg^{2+}$,while an anion-exchange resin (in $OH^-$ form) removes anions like $Cl^-$ and $SO_4^{2-}$.
Therefore,an acid-base resin system is used for this purpose.

(D) Saline hydrides like $CaH_2$ react with water to produce $H_2$ gas.
The reaction is: $CaH_2(s) + 2H_2O(l) \rightarrow Ca(OH)_2(aq) + 2H_2(g)$.
$MgH_2$ and $NaH$ are also saline hydrides,but the question specifically asks for the compound that produces $H_2$ in an aqueous solution among the given options,and $CaH_2$ is a standard example of this reaction.

(D) Permanent hardness is the hardness that cannot be removed by boiling.
It is usually caused by the presence of soluble calcium sulphate,magnesium sulphate,or chlorides and nitrates of $Ca$ and $Mg$ in the water,which do not precipitate out as the temperature increases.
Therefore,the statement that permanent hardness can be removed by boiling is incorrect.

(C) The reaction between calcium carbide $(CaC_2)$ and heavy water $(D_2O)$ is a hydrolysis reaction.
$CaC_2 + 2D_2O \to Ca(OD)_2 + C_2D_2$
Thus,the products are calcium deuteroxide $(Ca(OD)_2)$ and deuteroacetylene $(C_2D_2)$.
Among the given options,$Ca(OD)_2$ is the correct product.

(B) Heavy water $(D_2O)$ is used as a moderator in nuclear reactors to slow down fast-moving neutrons,not as a coolant.
Chemical reactions of $D_2O$ are analogous to $H_2O$:
$1$. $SO_3 + D_2O \rightarrow D_2SO_4$
$2$. $CaC_2 + 2D_2O \rightarrow C_2D_2 + Ca(OD)_2$
$3$. $Al_4C_3 + 12D_2O \rightarrow 3CD_4 + 4Al(OD)_3$
Therefore,the statement that it is used as a coolant is incorrect.

(C) Permanent hardness is caused by the presence of dissolved chlorides and sulfates of $Ca^{2+}$ and $Mg^{2+}$ ions.
Boiling is only effective for removing temporary hardness,which is caused by the presence of $HCO_3^-$ (bicarbonate) ions.
Methods like treatment with washing soda $(Na_2CO_3)$,Calgon's method,and the ion exchange method are used to remove permanent hardness.

(B) Heavy water is $D_2O$.
In $D_2O$,there are two deuterium atoms $(D)$ and one oxygen atom $(O)$.
Each $D$ atom has $1$ proton,$1$ electron,and $1$ neutron.
Each $O$ atom has $8$ protons,$8$ electrons,and $8$ neutrons.
Total number of protons $= (2 \times 1) + 8 = 10$.
Total number of electrons $= (2 \times 1) + 8 = 10$.
Total number of neutrons $= (2 \times 1) + 8 = 10$.
Thus,the numbers of protons,electrons,and neutrons are $10, 10, 10$ respectively.

(C) The temporary hardness of water is caused by the presence of dissolved magnesium hydrogen carbonate $Mg(HCO_3)_2$ or calcium hydrogen carbonate $Ca(HCO_3)_2$.
These bicarbonates decompose upon heating,leading to the precipitation of insoluble carbonates.
The reaction for calcium hydrogen carbonate is:
$Ca(HCO_3)_2 \rightarrow CaCO_3 \downarrow + H_2O + CO_2$

(A) The hardness of water is expressed in terms of $CaCO_3$ equivalents.
$n_{eq}(CaCO_3) = n_{eq}(Ca(HCO_3)_2) + n_{eq}(Mg(HCO_3)_2)$
Given that the n-factor for bicarbonate salts is $2$:
$\frac{w}{100} \times 2 = \frac{0.81}{162} \times 2 + \frac{0.73}{146} \times 2$
$\frac{w}{50} = 0.005 \times 2 + 0.005 \times 2$
$\frac{w}{50} = 0.01 + 0.01 = 0.02$
$w = 0.02 \times 50 = 1.0 \, g$
Hardness in $ppm = \frac{\text{Mass of } CaCO_3 \text{ in } g}{\text{Mass of water in } g} \times 10^6$
Since $100 \, mL$ of water has a mass of $100 \, g$:
$\text{Hardness} = \frac{1.0}{100} \times 10^6 = 10,000 \, ppm$

(C) Temporary hardness in water is caused by the presence of magnesium or calcium bicarbonates.
For magnesium bicarbonate,the reaction upon boiling is:
$Mg(HCO_3)_{2(aq)} \xrightarrow{\Delta} Mg(OH)_{2(s)} + 2CO_{2(g)}$.
Here,$X$ is $Mg(HCO_3)_2$ and $Y$ is $Mg(OH)_2$.
Thus,the correct option is $C$.

(D) To soften hard water,the zeolite method is employed.
Sodium zeolite,represented as $Na_{2}Al_{2}Si_{2}O_{8} \cdot xH_{2}O$,is used in this process.
It possesses the unique property of exchanging its sodium ions $(Na^{+})$ with the hardness-causing cations present in hard water,specifically $Ca^{2+}$ and $Mg^{2+}$ ions.

(C) The molecular mass of $D_2O$ $(20 \ g/mol)$ is greater than $H_2O$ $(18 \ g/mol)$.
Due to stronger hydrogen bonding in $D_2O$,its melting point $(3.8 \ ^\circ C)$ and boiling point $(101.4 \ ^\circ C)$ are higher than those of $H_2O$ ($0 \ ^\circ C$ and $100 \ ^\circ C$ respectively).
$H_2O$ is more reactive than $D_2O$ due to the stronger $O-D$ bond compared to the $O-H$ bond.
Therefore,the statement $B.P. = H_2O > D_2O$ is incorrect.

(D) Hardness in water is caused by the presence of soluble salts of calcium and magnesium,such as bicarbonates,chlorides,and sulfates.
$Mg(HCO_3)_2$,$CaCl_2$,and $MgSO_4$ are all soluble in water and contribute to its hardness.
$MgCO_3$ (magnesium carbonate) is practically insoluble in water.
Since it does not dissolve in water,it does not contribute to the hardness of water.

(B) Chlorine is a powerful oxidizing agent and has strong anti-microbial properties.
When added to water,it reacts to form hypochlorous acid $(HOCl)$,which effectively kills bacteria,viruses,and other pathogenic microorganisms,making the water safe for consumption.

(A) The dipole moment of water is $1.85 \ D$,which is not zero.
Because the molecule is bent,the bond dipoles do not cancel each other out.
Therefore,the molecule is polar and has a net dipole moment.
$H_2O$ can act as a Lewis base due to the presence of lone pairs on the oxygen atom.
It exhibits an abnormally high boiling point due to intermolecular hydrogen bonding.
Thus,the statement that the molecule has $\mu = 0$ is incorrect.

(D) In the structure of ice, each oxygen atom is tetrahedrally surrounded by four other oxygen atoms at a distance of $276 \ pm$.
These oxygen atoms are linked by hydrogen bonds.
Each oxygen atom is covalently bonded to two hydrogen atoms and hydrogen-bonded to two other hydrogen atoms from neighboring water molecules.
Therefore, each oxygen atom is surrounded by two hydrogen atoms via covalent bonds and two hydrogen atoms via hydrogen bonds, totaling four hydrogen atoms.
Option $D$ is incorrect because each oxygen atom is surrounded by two covalent bonds and two hydrogen bonds, not four hydrogen bonds.

(C) Calcium hydride reacts with water to form calcium hydroxide and hydrogen gas.
The chemical equation for the reaction is:
$CaH_2 + 2H_2O \rightarrow Ca(OH)_2 + 2H_2$

(B) In the hydrides of group $16$ elements,the boiling point generally increases down the group due to an increase in molecular mass and van der Waals forces.
However,$H_2O$ has an exceptionally high boiling point due to strong intermolecular hydrogen bonding.
Comparing the remaining hydrides $(H_2S, H_2Se, H_2Te)$,$H_2S$ has the lowest molecular mass and the weakest van der Waals forces.
Therefore,$H_2S$ has the lowest boiling point.

(A) Anhydrous $HF$ or $H_2F_2$ is a bad conductor of electricity.
This is due to the very high bond dissociation energy of the $H-F$ bond,which prevents the molecule from ionizing effectively.

(A) Polyphosphates,such as sodium hexametaphosphate $(Na_6P_6O_{18})$,are used to remove $Ca^{2+}$ and $Mg^{2+}$ ions from hard water.
They form stable,soluble complexes with these metal cations,effectively sequestering them and preventing them from forming insoluble precipitates (like carbonates or sulfates) that cause hardness.
The reaction is represented as:
$2Ca^{2+} + Na_2[Na_4(PO_3)_6] \rightarrow Na_2[Ca_2(PO_3)_6] + 4Na^+$
Since the resulting complex is soluble in water,the hardness is removed without forming a precipitate.

(D) Water softening by Clark's process uses calcium hydroxide $(Ca(OH)_2)$,which is also known as lime.
This process is used to remove temporary hardness from water caused by calcium bicarbonate.
The chemical reaction is:
$Ca(HCO_3)_2 + Ca(OH)_2 \rightarrow 2CaCO_3(s) + 2H_2O(l)$.

(D) Fluorine $(F_2)$ is a very strong oxidizing agent. It reacts with water to oxidize it to oxygen $(O_2)$.
The chemical reaction is as follows:
$2F_2(g) + 2H_2O(l) \rightarrow 4HF(aq) + O_2(g)$

(D) The reaction is $NaH + H_2O \longrightarrow NaOH + H_2 \uparrow$.
In this reaction,the underlined atom is Hydrogen $(H)$ in $NaH$.
Upon hydrolysis,$NaH$ reacts with water to form $NaOH$ and $H_2$ gas.
$H_2$ is a diatomic molecule and is not an oxyacid.
Therefore,the product is neither an oxyacid with an $-ic$ suffix nor an $-ous$ suffix.
Thus,the correct option is $D$.

(B) The hydrolysis of $Al_2S_3$ is represented by the reaction: $Al_2S_3 + 6H_2O \rightarrow 2Al(OH)_3 + 3H_2S$.
$H_2S$ (hydrogen sulfide) is a well-known acidic gas.
In contrast,$Li_2NH$ releases $NH_3$ (basic),$CaC_2$ releases $C_2H_2$ (neutral/weakly acidic),and $CaNCN$ releases $NH_3$ (basic).

(C) $Ca_3P_2 + 6H_2O \to 3Ca(OH)_2 + 2PH_3 \uparrow$ ($PH_3$ is a basic gas).
$(b)$ $AlN + 3H_2O \to Al(OH)_3 + NH_3 \uparrow$ ($NH_3$ is a basic gas).
$(c)$ $Al_2S_3 + 6H_2O \to 2Al(OH)_3 + 3H_2S \uparrow$ ($H_2S$ is an acidic gas).
$(d)$ $CaH_2 + 2H_2O \to Ca(OH)_2 + 2H_2 \uparrow$ ($H_2$ is a neutral gas).

(C) Hydrated sodium aluminium silicate $(Na_2 Al_2 Si_2 O_8 \cdot x H_2 O)$ is known as permutit or zeolite.
During the water softening process,the $Na^+$ ions in the permutit are replaced by $Ca^{2+}$ or $Mg^{2+}$ ions present in the hard water.
When all the $Na^+$ ions are replaced by $Ca^{2+}$ or $Mg^{2+}$ ions,the permutit is said to be exhausted.
Therefore,exhausted permutit is $Ca Al_2 Si_2 O_8 \cdot x H_2 O$ (or the corresponding magnesium salt).

(C) Polymetaphosphates,such as sodium hexametaphosphate $(Na_6P_6O_{18})$,are commonly used as water softeners.
They function by sequestering the metal cations responsible for water hardness,such as $Ca^{2+}$ and $Mg^{2+}$.
These polyphosphates form stable,soluble complexes with these cations,thereby preventing them from reacting with soap to form scum.
Therefore,the correct mechanism is the formation of soluble complexes with the cations of hard water.

(D) $1$. $D_2O$ (deuterium oxide) has a higher molecular mass $(20 \ g/mol)$ compared to $H_2O$ $(18 \ g/mol)$. Consequently,at the same temperature,$D_2O$ exhibits a higher density than $H_2O$.
$2$. The temperature of maximum density for $H_2O$ is $4 \ ^\circ C$ $(277 \ K)$,whereas for $D_2O$,it is $11.6 \ ^\circ C$ $(284.6 \ K)$. Thus,the temperature of maximum density is higher for $D_2O$.
$3$. Regarding viscosity,$D_2O$ is actually more viscous than $H_2O$ at the same temperature due to stronger hydrogen bonding in $D_2O$.
$4$. Therefore,both statements $A$ and $B$ are correct.

(D) Temporary hardness is caused by the presence of bicarbonates of $Ca^{2+}$ and $Mg^{2+}$ and can be removed by boiling water.
Permanent hardness is caused by the presence of soluble salts of magnesium and calcium in the form of chlorides and sulphates in water.
Permanent hardness cannot be removed by boiling water; it requires chemical treatment (e.g.,washing soda,ion exchange resins).
Therefore,the statement that permanent hardness can be removed by boiling is incorrect.

(B) In ice,each water molecule is linked to four other water molecules by hydrogen bonds in a three-dimensional tetrahedral arrangement. This creates an open cage-like structure with large empty spaces. When ice melts,these hydrogen bonds break and the molecules occupy these empty spaces,resulting in a higher density for liquid water compared to ice.

(A) Temporary hardness of water is caused by the presence of bicarbonates of calcium and magnesium.
Slaked lime $(Ca(OH)_{2})$ is used in Clark's process to remove this hardness by converting soluble bicarbonates into insoluble carbonates.
The chemical reaction is:
$Ca(OH)_{2} + Ca(HCO_{3})_{2} \rightarrow 2 CaCO_{3} \downarrow + 2 H_{2}O$

(A) Hardness is expressed in terms of $CaCO_3$ equivalents.
Formula: $\text{Hardness} = \text{Mass of salt} \times \frac{\text{Molar mass of } CaCO_3}{\text{Molar mass of salt}}$.
$1$. For $MgSO_4$ $(M = 120 \ g/mol)$: $\text{Hardness} = 30 \times \frac{100}{120} = 25 \ ppm$.
$2$. For $MgCl_2$ $(M = 95 \ g/mol)$: $\text{Hardness} = 19 \times \frac{100}{95} = 20 \ ppm$.
Total hardness = $25 + 20 = 45 \ ppm$.

(A) Calgon is the commercial name for sodium hexametaphosphate,which has the chemical formula $Na_2[Na_4(PO_3)_6]$.
It is used in water softening to sequester calcium and magnesium ions by forming soluble complexes.

(A) The hardness of water is caused by the presence of calcium $(Ca^{2+})$ and magnesium $(Mg^{2+})$ ions. Calgon (sodium hexametaphosphate,$Na_6P_6O_{18}$) is a well-known method used for water softening. It works by sequestering these ions into soluble complexes,thereby removing the hardness of water.

(D) The correct matches are:
$1.$ Heavy water is $D_2O$ $(1-c)$.
$2.$ Temporary hardness is caused by bicarbonates of $Ca$ and $Mg$ $(2-a)$.
$3.$ Soft water contains no foreign ions $(3-b)$.
$4.$ Permanent hardness is caused by sulphates and chlorides of $Ca$ and $Mg$ $(4-d)$.
Thus,the correct sequence is $1-c, 2-a, 3-b, 4-d$.

(C) Heavy water $(D_2O)$ is prepared by the exhaustive electrolysis of water.
Since $H_2O$ is electrolyzed faster than $D_2O$,the remaining water becomes enriched in $D_2O$ after multiple stages of electrolysis.
In practice,$0.5 \ M \ NaOH$ solution is used as the electrolyte with nickel electrodes to facilitate this process.

(C) The hardness of water is caused by the presence of dissolved calcium and magnesium salts.
To express the total hardness of water,it is standard practice to convert the concentration of all dissolved calcium and magnesium salts into an equivalent amount of $CaCO_3$.
This is done because the molecular weight of $CaCO_3$ is $100 \ g/mol$,which simplifies calculations.
Therefore,the degree of hardness is expressed as $ppm$ (parts per million) by weight of $CaCO_3$ equivalent.

(C) Clark's method is used for the removal of temporary hardness of water.
In this process,a calculated amount of lime,i.e.,calcium hydroxide $(Ca(OH)_2)$,is added to the hard water.
This reacts with calcium bicarbonate to precipitate calcium carbonate: $Ca(HCO_3)_2 + Ca(OH)_2 \rightarrow 2CaCO_3 \downarrow + 2H_2O$.

(B) Temporary hardness is caused by the presence of magnesium and calcium bicarbonates. It can be removed by adding calcium hydroxide ($CaO$ or lime),which precipitates the bicarbonates as carbonates.
$Ca(HCO_3)_2 + Ca(OH)_2 \rightarrow 2CaCO_3 \downarrow + 2H_2O$.
Permanent hardness is caused by the presence of soluble chlorides and sulfates of magnesium and calcium. It can be removed by adding sodium carbonate $(Na_2CO_3)$,which converts these into insoluble carbonates.
$CaCl_2 + Na_2CO_3 \rightarrow CaCO_3 \downarrow + 2NaCl$.

(D) Zeolite is represented as $NaAlSiO_4$.
When hard water containing $Ca^{2+}$ and $Mg^{2+}$ ions is passed through a bed of zeolite,the sodium ions $(Na^+)$ in the zeolite are replaced by $Ca^{2+}$ and $Mg^{2+}$ ions.
The reaction is: $Na_2Al_2Si_2O_8(s) + M^{2+}(aq) \rightarrow MAl_2Si_2O_8(s) + 2Na^+(aq)$ (where $M = Ca, Mg$).
This process effectively removes the hardness-causing ions from the water.

(C) Polyphosphates,such as sodium hexametaphosphate $(Na_6P_6O_{18})$,are commonly used as water softeners.
They work by sequestering the metal ions responsible for water hardness,such as $Ca^{2+}$ and $Mg^{2+}$.
These ions are $cationic$ in nature.
The polyphosphate anions form stable,water-soluble complexes with these $Ca^{2+}$ and $Mg^{2+}$ ions,thereby preventing them from reacting with soap or forming scale.
Therefore,they form soluble complexes with cationic components.

(A) Permanent hardness of water is caused by the presence of dissolved chlorides and sulfates of calcium and magnesium.
It can be removed by adding washing soda $(Na_2CO_3)$.
The reaction is: $Ca^{2+} + Na_2CO_3 \rightarrow CaCO_3 \downarrow + 2Na^+$.
$CaCO_3$ precipitates out,thereby removing the hardness.

(D) The presence of water in a liquid can be detected using anhydrous copper sulfate $(CuSO_4)$.
Anhydrous copper sulfate is white in color.
When it comes in contact with water,it absorbs water to form hydrated copper sulfate $(CuSO_4 \cdot 5H_2O)$,which is blue in color.
This color change from white to blue confirms the presence of water.

(A) In the structure of ice,each oxygen atom is surrounded by $4$ hydrogen atoms.
Two of these hydrogen atoms are covalently bonded to the oxygen atom,while the other two are linked through hydrogen bonding.
This arrangement results in a tetrahedral geometry around the oxygen atom.

(B) Alane is the chemical compound with the formula $AlH_3$.
It is a polymeric hydride of aluminum,often represented as $(AlH_3)_n$ in its solid state due to its structure involving bridging hydrogen atoms.
Therefore,the most accurate representation of its chemical nature as a polymeric species is $(AlH_3)_n$.

(C) The term 'plumbo solvency' refers to the ability of lead to dissolve in ordinary water,especially when the water contains dissolved oxygen and carbon dioxide. This process leads to the contamination of water with lead ions,which is toxic.