Water or hydride of oxygen Questions in English

(A) . Water is denser than ice because of the hydrogen bonding interactions and the open cage-like structure of ice.

(D) An amphoteric substance can act as both an acid and a base.
$H_2O$ acts as an acid by donating a proton: $\mathop {H_2O}\limits_{\text{Acid}} \to H^{+} + OH^{-}$.
$H_2O$ acts as a base by accepting a proton: $\mathop {H_2O}\limits_{\text{Base}} + H^{+} \to H_3O^{+}$.
Therefore,water is an amphoteric solvent.

(B) Since the hydride ion $(H^{-})$ is a much stronger base than the hydroxide ion $(OH^{-})$,it will readily abstract a proton $(H^{+})$ from water.
The reaction is as follows:
$H^{-}_{(aq)} + H_2O_{(l)} \to H_2(g) + OH^{-}_{(aq)}$
In this reaction,$H^{-}$ acts as a Bronsted-Lowry base and $H_2O$ acts as a Bronsted-Lowry acid.

(D) The reaction of sulphur trioxide $(SO_3)$ with heavy water $(D_2O)$ is an addition reaction.
$SO_3 + D_2O \to D_2SO_4$
This product is known as dideuterosulphuric acid.

(B) According to the $IUPAC$ nomenclature for hydrides,the systematic name for $H_2O$ is $Oxidane$.
While $Water$ is the common name and $Hydrogen \ oxide$ is a descriptive name,$Oxidane$ is the recognized systematic name for the hydride of oxygen.

(B) Since $H^{-}$ is a stronger base than $OH^{-}$,it will readily accept a proton from water.
$H^{-}$ acts as a Bronsted-Lowry base and $H_2O$ acts as a Bronsted-Lowry acid.
The reaction is: $H^{-}_{(aq)} + H_2O_{(l)} \to OH^{-}_{(aq)} + H_{2(g)}$
This is an acid-base reaction where the stronger base $(H^{-})$ reacts with the acid $(H_2O)$ to form a weaker base $(OH^{-})$ and a weaker acid $(H_2)$.

(B) $D_2O$ (heavy water) is primarily used in nuclear reactors as a moderator.
It is effective at slowing down fast-moving neutrons to thermal energies,which facilitates the fission process.
Additionally,it has a very low cross-section for neutron absorption,making it highly efficient for this purpose.

(C) Heavy water is deuterium oxide $(D_2O)$.
Ordinary water contains a small fraction ($1$ part in $6000$ parts) of heavy water.
On a large scale,it is produced by the repeated electrolysis of ordinary water containing an alkali.
It is also obtained by the repeated fractional distillation and condensation of ordinary water.

(B) Temporary hardness in water is primarily due to the presence of calcium and magnesium bicarbonates.
Adding calcium hydroxide $(Ca(OH)_2)$,also known as lime,precipitates these bicarbonates as insoluble carbonates.
The chemical reaction is:
$Ca(HCO_3)_2 + Ca(OH)_2 \to 2CaCO_3 \downarrow + 2H_2O$
Thus,option $(B)$ is correct.

(A) Heavy water is $D_2O$.
It is composed of oxygen and the heavier isotope of hydrogen,which is deuterium $(D)$.

(B) Hardness in water is primarily caused by the presence of dissolved calcium $(Ca^{2+})$ and magnesium $(Mg^{2+})$ ions.
Temporary hardness is caused by the presence of bicarbonates $(HCO_3^-)$ of calcium or magnesium.
Permanent hardness is caused by the presence of chlorides $(Cl^-)$ and sulfates $(SO_4^{2-})$ of calcium and magnesium.
Among the given options,the pair $(Ca^{2+}, HCO_3^-)$ is responsible for temporary hardness in water.

(B) By boiling,temporary hardness of water can be removed.
$Ca(HCO_3)_2 \xrightarrow{\text{Boil}} CaCO_3 \downarrow + H_2O + CO_2$

(B) Water is known as a universal solvent because it has a high dielectric constant $(82)$,a high liquid range,and the ability to dissolve a maximum number of compounds. Therefore,the statement that it has a very low dielectric constant is incorrect.

(A) . Heavy water,i.e.,$D_2O$,is used as a moderator in nuclear reactors to slow down the velocity of fast-moving neutrons.

(A) Temporary hardness in water is caused by the presence of dissolved bicarbonates of calcium and magnesium,i.e.,$Ca(HCO_3)_2$ and $Mg(HCO_3)_2$.
Permanent hardness is caused by the presence of chlorides and sulphates of calcium and magnesium,i.e.,$CaCl_2, MgCl_2, CaSO_4,$ and $MgSO_4$.

(D) The correct answer is $D$.
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 the water,whereas temporary hardness (caused by bicarbonates) can be removed by boiling.

(B) Hard water is defined as water that contains dissolved salts of calcium $(Ca^{2+})$ and magnesium $(Mg^{2+})$,typically in the form of chlorides,sulfates,or bicarbonates.
$A$ and $C$ contain $Ca^{2+}$ and $Mg^{2+}$ ions,respectively,which contribute to water hardness.
$B$ contains dilute $HCl$,which provides $H^{+}$ ions. While acidic water can be corrosive,it is not classified as 'hard water' because hardness is specifically defined by the presence of calcium and magnesium ions.
Therefore,water containing dilute $HCl$ is not hard water.

(C) Heavy water $(D_2O)$ is used in nuclear reactors as a moderator to slow down the speed of fast-moving neutrons and also as a coolant to remove the heat generated during the nuclear fission process.

(B) The freezing point of heavy water $(D_2O)$ is $3.8 \ ^\circ C$,which is higher than that of ordinary water ($H_2O$,$0 \ ^\circ C$).

(D) The correct option is $(D)$.
Anhydrous $CuSO_4$ is a white powder.
When it reacts with water,it forms hydrated copper$(II)$ sulfate,$CuSO_4 \cdot 5H_2O$,which is blue in colour.
Therefore,adding anhydrous $CuSO_4$ is a standard chemical test to detect the presence of water.

(A) The correct option is $(A)$.
Hardness of water is primarily due to the presence of $Ca^{2+}$ and $Mg^{2+}$ ions.
In the cation exchange process,hard water is passed through a cation exchange resin (containing $R-SO_3H$ or $R-COOH$ groups).
The $Ca^{2+}$ and $Mg^{2+}$ ions are exchanged with $H^+$ or $Na^+$ ions present in the resin,thereby removing the hardness.

(C) Washing soda $(Na_2CO_3)$ removes both temporary and permanent hardness by converting soluble calcium and magnesium salts into insoluble carbonates.
$CaCl_2 + Na_2CO_3 \to CaCO_3(s) + 2NaCl$
$CaSO_4 + Na_2CO_3 \to CaCO_3(s) + Na_2SO_4$
$Mg^{2+} + CO_3^{2-} \to MgCO_3(s)$

(C) Permutit is the technical name for synthetic zeolites,which are hydrated sodium aluminium silicates.
Its chemical formula is represented as $Na_2Al_2Si_2O_8 \cdot xH_2O$.

(D) Heavy water is represented as $D_2O$,where $D$ is deuterium $(_{1}H^{2})$.
The atomic mass of oxygen $(O)$ is $16 \ amu$ and the atomic mass of deuterium $(D)$ is $2 \ amu$.
Therefore,the molecular weight of $D_2O = (2 \times 2) + 16 = 20 \ amu$.

(A) Permanent hardness of water is due to the presence of soluble salts of magnesium and calcium in the form of chlorides and sulphates $(MgCl_2, CaCl_2, MgSO_4, CaSO_4)$.
Temporary hardness is due to the presence of bicarbonates of magnesium and calcium $(Mg(HCO_3)_2, Ca(HCO_3)_2)$.

(C) The correct option is $(C)$.
Heavy water $(D_2O)$ is formed by the combination of the heavier isotope of hydrogen,known as deuterium $(_1H^2 \text{ or } D)$,with oxygen.
The chemical reaction is: $2D_2 + O_2 \to 2D_2O$.

(D) Heavy water is $D_2O$ $(1-c)$.
Temporary hardness is caused by the presence of bicarbonates of $Ca$ and $Mg$ in water $(2-a)$.
Soft water is water that does not contain foreign ions like $Ca^{2+}$ and $Mg^{2+}$ $(3-b)$.
Permanent hardness is caused by the presence of sulphates and chlorides of $Ca$ and $Mg$ in water $(4-d)$.
Therefore,the correct matching is $1-c, 2-a, 3-b, 4-d$.

(B) Hardness of water is caused by the presence of dissolved bicarbonates,chlorides,and sulphates of $Ca^{2+}$ and $Mg^{2+}$ ions.
These ions react with the anions of fatty acids present in soap to form insoluble curdy white precipitates.
Consequently,hard water does not produce lather with soap easily.

(D) Zeolite,which is hydrated sodium aluminium silicate $(Na_2Al_2Si_2O_8 \cdot xH_2O)$,acts as an ion exchanger.
When it is treated with hard water,the $Na^+$ ions present in the zeolite structure are exchanged with the $Ca^{2+}$ and $Mg^{2+}$ ions that cause hardness in water.

(C) Hard water is defined as water that contains dissolved salts of calcium $(Ca^{2+})$ or magnesium $(Mg^{2+})$ ions.
$A$. Potash alum is $K_2SO_4 \cdot Al_2(SO_4)_3 \cdot 24H_2O$,which provides $Al^{3+}$ ions,but hard water is specifically characterized by $Ca^{2+}$ and $Mg^{2+}$ ions. However,in many contexts,the presence of polyvalent cations makes water hard.
$B$. $HCl$ provides $H^+$ ions,which does not cause hardness.
$C$. Calcium nitrate $Ca(NO_3)_2$ provides $Ca^{2+}$ ions,which causes permanent hardness.
Since the definition of hard water is based on the presence of $Ca^{2+}$ and $Mg^{2+}$ ions,option $C$ is the most accurate. However,if the question implies polyvalent cations,$A$ and $C$ are relevant. Given the options,$C$ is the standard answer for hard water.

(D) Ice is a poor conductor of heat,which means it is a good thermal insulator.
Additionally,the density of ice is less than that of liquid water,which is why ice floats on water.
Therefore,both statements $(b)$ and $(c)$ are correct.

(D) The correct answer is $D$. $H_2O$ (ordinary water) has a high tendency to absorb neutrons,which reduces the number of neutrons available for the fission process. $D_2O$ (heavy water) has a much lower neutron absorption cross-section,making it a more efficient moderator in nuclear reactors.

(C) Temporary hardness due to $Mg(HCO_3)_2$ is removed by boiling.
Upon heating,$Mg(HCO_3)_2$ decomposes to form magnesium hydroxide $(Mg(OH)_2)$ precipitate rather than magnesium carbonate $(MgCO_3)$ because $Mg(OH)_2$ is less soluble than $MgCO_3$ in water.
The reaction is: $Mg(HCO_3)_{2(aq)} \xrightarrow{\Delta} Mg(OH)_{2(s)} + 2CO_{2(g)}$.

(C) Permanent hardness due to $Mg^{2+}$ ions is best removed by using a mixture of $Ca(OH)_2$ and $Na_2CO_3$ (Clark's process combined with soda ash treatment).
$Ca(OH)_2$ reacts with $Mg^{2+}$ ions to precipitate $Mg(OH)_2$:
$Mg^{2+} + Ca(OH)_2 \to Mg(OH)_2 \downarrow + Ca^{2+}$
The resulting $Ca^{2+}$ ions (along with existing permanent hardness) are then removed by $Na_2CO_3$:
$Ca^{2+} + Na_2CO_3 \to CaCO_3 \downarrow + 2Na^{+}$
Thus,the combination of $Na_2CO_3 + Ca(OH)_2$ is the most effective method.

(B) $(B)$ Plumbosolvency refers to the ability of water to dissolve lead $(Pb)$ from lead pipes.
Soft water, which lacks dissolved salts like carbonates and sulfates, is more aggressive in dissolving lead compared to hard water.
The presence of $CO_3^{2-}$ and $SO_4^{2-}$ ions in hard water leads to the formation of a protective layer of insoluble lead salts on the inner surface of the pipes, which prevents further dissolution of $Pb$.

(B) The hardness of water is caused by the presence of dissolved salts of calcium and magnesium,such as $Ca^{2+}$ and $Mg^{2+}$ ions (e.g.,chlorides,sulfates,and bicarbonates).
$NaCl$ (sodium chloride) does not contribute to the hardness of water.
Therefore,water containing only $NaCl$ is considered soft water.

(B) The correct answer is $(B)$.
$CaSO_4$ (calcium sulfate) exhibits retrograde solubility,meaning its solubility in water decreases as the temperature increases.

(B) Organic ion exchange resins are designed to remove ionic impurities such as $Na^+$ and $Cl^-$ ions from water.
Table sugar (sucrose) is a non-ionic,covalent compound and does not dissociate into ions in water.
Therefore,the ion exchange resin will remove the common salt $(NaCl)$ but will not remove the dissolved sugar.
As a result,the water will retain the taste of the sugar and will be sweet.

(D) The process of demineralization or deionization of water uses organic ion-exchange resins.
In this process,cation exchange resins remove all cations like $Ca^{2+}$,$Mg^{2+}$,$Na^+$,etc.,and anion exchange resins remove all anions like $Cl^-$,$SO_4^{2-}$,$HCO_3^-$,etc.
Therefore,the water obtained is free from all soluble mineral salts,which is often referred to as deionized or demineralized water.

(A) Soap is a sodium or potassium salt of long-chain fatty acids (e.g.,sodium stearate,$C_{17}H_{35}COONa$).
When soap is added to hard water,it reacts with the calcium $(Ca^{2+})$ and magnesium $(Mg^{2+})$ ions present in the water to form insoluble precipitates (scum),such as calcium stearate.
This process effectively removes the hardness-causing ions from the water,although it is not an economical method for large-scale water softening.

(D) Aqueous $CaCl_2$ is already in a hydrated state and cannot absorb further water or act as a dehydrating agent. Silica gel,$P_2O_5$,and concentrated $H_2SO_4$ are well-known dehydrating or drying agents.

(C) Lead pipes are not suitable for drinking water because lead reacts with water containing dissolved air (oxygen) to form lead$(II)$ hydroxide,$Pb(OH)_2$,which is soluble and toxic to human health.
The reaction is: $2Pb(s) + 2H_2O(l) + O_2(g) \rightarrow 2Pb(OH)_2(aq)$.

(C) Plumbosolvency is the ability of a solvent,notably water,to dissolve lead.
In the public supply of water,this is an undesirable property because lead is toxic.
Therefore,plumbosolvency implies the dissolution of lead in ordinary water.

(A) Oxygen was discovered by Joseph Priestley in Wiltshire,England in $1772$.
The name $Oxygen$ is derived from the Greek word '$Oxygenes$',which means acid-forming.

(A) Oxygen is only slightly soluble in water. Due to its low solubility,it can be collected by the downward displacement of water.

(D) Anhydrous $CuSO_4$ is used to test the presence of water in any liquid.
It is white in its anhydrous form and changes its colour to blue upon hydration due to the formation of $CuSO_4 \cdot 5H_2O$.

(A) $H_2O$ has the highest boiling point among the given hydrides of group $16$ elements.
This is due to the presence of strong intermolecular hydrogen bonding in $H_2O$ molecules,which is absent in $H_2S$,$H_2Se$,and $H_2Te$.

(C) When chlorine water is exposed to sunlight,the following reactions occur:
$H_2O + Cl_2 \xrightarrow{\text{sunlight}} HCl + HClO$
$2HClO \to 2HCl + O_2$
In this process,$Cl_2$ reacts with $H_2O$ to form $HCl$ and $HClO$. The $HClO$ then decomposes to release $O_2$. This indicates that hydrogen has a greater affinity for chlorine than for oxygen,as it prefers to form $HCl$ rather than remaining bonded to oxygen in water.

(A) Chlorine is used as a disinfectant for killing germs in water purification.

(A) The correct answer is $(A)$.
Anhydrous $CuSO_4$ (white) reacts with water to form hydrated copper sulphate $(CuSO_4 \cdot 5H_2O)$,which is blue in colour.
This colour change from white to blue serves as a definitive scientific test for the presence of water in a liquid sample.