Explore topic-wise InterviewSolutions in Current Affairs.

This section includes 7 InterviewSolutions, each offering curated multiple-choice questions to sharpen your Current Affairs knowledge and support exam preparation. Choose a topic below to get started.

1.

Give reason : The Density of water is higher than density of ice.

Answer»

Solution :The DENSITY of WATER is higher then density of ice because of the HYDROGEN BOND.
Discussion
2.

Give reason : The hard water is not suitable for washing of clothes.

Answer»

Solution :The HARD water is not suitable for WASHING of clothes because the hard water create PRECIPITATE with SOUP.
Discussion
3.

Give reason : Nitration of benzene is not possible with only HNO_(3).

Answer»

Solution :Benzene possess characteristic STABILITY and `pi`-BONDS on benzene is only BROKEN down with only strong ELECTROPHILE. Only `HNO_(3)`, produce `NO_(2)^(+)` which is not strong electrophile so does `NO_(2)^(+)` which is not strong electrophile so does not perfom NITRATION reaction.
Discussion
4.

Give reason : (i)ZnO crystal on heating acquires the formula Zn_(1+x)O. (ii)There is an increase in conductivity when Silicon is doped withPhosphorus.

Answer»

Solution :(i)On heating ZnO loses oxygen as `ZnO to Zn^(2+) + 1/2 O_2 + 2e^-`
Excess `Zn^(2+)` IONS are entrapped into vacant interstitial sites. Hence, there is metal excess.
(ii)Silicon belongs to Group 14 while phosphorus belongs to Group 15. Hence, in the valence SHELL, P has one EXTRA ELECTRON than Si. When Si is replaced by P atom, due to extra electron, there is increase in conductivity.
Discussion
5.

Give Reason: Lithiu iodide is covalent but potassium iodide is ionic.

Answer»

SOLUTION :Lithium iodide is COVALENT because of the SMALLER atomic size of lithium atom, it distorts to the ELECTRON to the electron cloud of iodine TOWARDS it. Or K has less polarizing power than `Li^(+)`.
Discussion
6.

Give reason : (i)Boron is used as control rods in nuclear reactor. (ii) Graphite is soft and slippery.

Answer»

SOLUTION :(i) Because Boron-10 ISOTOPE has ABILITY to absorb Neutrons.
(ii) Each layer in graphite are held by weak Vanderwaal's FORCE of attraction.
Discussion
7.

Give reason: (i) In stochiometric defects, NaCl exhibits schottky defectand not Frenkel defect. (ii)Silicon on doping with phosphorus forms n-typesemiconductor . (iii)Ferrimagnetic substances show better magnetisation than antiferromagnetic substences.

Answer»

Solution :(i) Schottky defect is shown by those compounds which have small difference in the size of catio and anoin whereas Frenkel defect is shown by those compounds which have large difference in the size of cation and anion. As the difference in the size of `Na^(+) and Cl^(-)`ions is small,hence it shows schoottky defect.
(ii)Silicon belongs to Group 14 whereas PHOSPHORUS blongs to Group 15. s phosphorus has ONE extra electron than silicon, this extra electron makes silicon a semiconductor.Hence, we get n-type semicondutor.
(iii)Ferrimagnetic substances have as unequal NUMBER of electrons with spin in the OPPOSITE direction.Hence, they have a net magnetic moment. Antiferromagnetic substances have equal number of electrons with spin in the opposite direction. Hence, their net magnetic momentis zero. For this reason,ferrimagnetic subtances SHOW better magnetism tha antiferromagnetic substances.
Discussion
8.

Give reason: Graphite is soft and slippery.

Answer»

SOLUTION :Due to the WEAK vande are waal.s FORCE of ATTRACTION between the layers of hexagonal rings.
Discussion
9.

Give reason for the higher melting point value of AlF_(3) (solid) than SiF_(4) (gas).

Answer»

SOLUTION :(i) `AlF_(3)` is an ionic COMPOUND and `SiF_(4)` is a covalent compound.
(ii) Usually covalent compounds exhibit lower melting and boiling points.
(iii) `therefore AlF_(3)` bring ionic in nature has strong coulombic FORCE of attraction between the oppositely charged ions, Thereby posses higher melting point.
Discussion
10.

Give reason for the following : The molecule ofMgCl_(2)is linear while that of stannous chloride is angular.

Answer»

Solution :E.C. of`""_(12)Mg ` is `1s^(2) 2S^(2) 2p^(6) 3s^(2)` in the ground state . Hence , in the excited state, it is `1s^(2) 2s^(2) 2p^(6) 3s^(1) 3p_(x)^(1)` .
It undergoes sp-hybridisation. Therefore, the shape of `MgCl_(2)` is LINEAR . E.C. of Sn is `[KR] 5s^(2)5_(x)^(1) 5p_(y)^(1) `. It
undergoes `sp^(2)`-hybridisation. The two half-filled hybrid orbitals form bonds with Cl-atoms while the third
is occupied by a lone pair . Hence , `SnCl_(2) ` is BENT or V-shaped .
Discussion
11.

Give reason for the following. Windows panes of old bulding become thicker at the bottom than at the top.

Answer»

Solution :GLASS is an amophous substance or SUPERCOOLED liquid. It flows.
Discussion
12.

Give reason for the following. n-butyl bromide has higher bolling point than t-butyl bromide.

Answer»

Solution :For ISOMERIC alkyl halides, the boiling POINT decreases with branching, because with branching the surface area of the alkyl halide decreases and HENCE the magnitude of vander waal's FORCES of attraction decreases.
`therefore`The B.P of n -Butyl BROMIDE `gt`t-Butyl bromide.
Discussion
13.

Give reason for the following statement. Halogens act as good oxidising agents.

Answer»

Solution :Due to HIGH electron gain enthalpy they act as good OXIDISING agents as they can gain ELECTRONS EASILY .
Discussion
14.

Give reason for the following : (i) Zinc is not precipitated as Zn(OH)_(2)on adding NH_(4)OH to a zinc salt solution containing NH_(4)Cl. (ii) BaSO_(4) precipitate is washed with water containing a small amount of H_(2)SO_(4) in gravimetric analysis. (iii) CO_(2) is more soluble in aqueous NaOH solution than in water. (iv) A brown precipitate in a bottle containing aqueous FeCl_(3) solution appears pm standing.

Answer»

Solution :(i) `NH_(4)Cl` suppresses the ionization of `NH_(4)OH` due to common ion effect. Hence, concentration of `OH^(-)` decreases such that the ionic product `[Zn^(2+)][OH^(-)]^(2)` doesnot exceed solubility product. Hence, `Zn(OH)_(2)` is not precipitated.
(ii) `BaSO_(4)` dissociates as : `BaSO_(4)(s)overset(aq)hArrBa^(2+)(aq)+SO_(4)^(2-) (aq)`. Addition of `H_(2)SO_(4)` gives common `SO_(4)^(2-)` ions which suppresses the dissociation of `BaSO_(4)` and thus helps in the complete precipitation.
(iii) In WATER, `CO_(2)` dissolves to formcarbonic acid `(H_(2)CO_(3))`:
`CO_(2)(g)+H_(2)CO_(3)(aq)`
As the reaction is reversible, the solubility is low.
In aqueous `NaOH, H_(2)CO_(3)` reacts with sodium hydroxide as FOLLOWS :
`2NaOH(aq)+H_(2)CO_(3) rarr Na_(2)CO_(3)(aq)+H_(2)O(l)`
As a result, the equillibrium involving dissolution of `CO_(2)` in `H_(2)O` shifts FORWARD , i.e., solubility increases.
(iv) `FeCl_(3)` undergoes hydrolysis forming a brown precipitate of ferric hydroxide
`{:(FeCl_(3)+3H_(2)O,rarr,Fe(OH)_(3)+ 3HCl,,),(,,"Brown ppt.",,):}`
Discussion
15.

Give reason for the following: Ethyl iodide undergoes S_(N)2 reaction faster-than ethyl bromide.

Answer»

SOLUTION :Iodide is a better leaving group because of its larger size than BROMIDE, THEREFORE, ETHYL iodide undergoes `S_(N)2` reaction faster than ethyl bromide.
Discussion
16.

Water can boiled more quickly on the top of a mountain.What is the reason?

Answer»

SOLUTION :BOILING POINT of water is low at HILL station.
Discussion
17.

Give reason for polarity of C-X bond in halo alkane.

Answer»

SOLUTION :Carbon HALOGEN bond is a polar bond as HALOGENS are more electro NEGATIVE than carbon. The carbon atom EXHIBITS a partial positive charge `(delta^(+))` and halogen atom a partial negative charge `(delta^(-))`.
Discussion
18.

Give reason for, 'NaHCO_(3) is known in solid state but Ca(HCO_(3))_(2) is not isolated in solid state.

Answer»

Solution :The sodium bicarbonate `(NaHCO_(3))` is highly stable to HEAT. As the electropositive CHARACTER increases, the stability also increases. So `NaHCO_(3)` is known to EXIST as solid whereas `Ca(HCO_(3))_(2)` is soluble in water. All the carbonates of alkali metals are more soluble in the atmosphere of `CO_(2)` and form bicarbonates. THEREFORE, they cannot be isolated in the solid state.
Discussion
19.

Lithium shows diagonal relationship with Megnesium. Give reason.

Answer»

SOLUTION :Both Lithium and magnesium have small size and high charge density. The electronegativities of Li is 1.0 and Mg is 1.2. They are low and almost same. Their ionic radii are SIMILAR. Hence they show SIMILARITIES which is known as diagonal RELATIONSHIP between first elemetn of a group with the second element in the NEXT higher group.
Discussion
20.

Give reason: Coodination number of Be is 4, but that of Mg is six

Answer»

Solution :Beryllium does not have d subsshell in `n=2` level. It has only 4 VALENCE orbital. HENCE it can have a coordination NUMBER 4.
Mg has d subshels in `n=3` level. Hence it has more valence orbtial. THEREFORE it can have a coordination number 6.
Discussion
21.

Give reason: (a)Why is Frenkel defect found in AgCl but not in NaCl? (b)What is the difference between Phosphorus doped and Gallium doped semiconductors ?

Answer»

Solution :(a)Frenkel DEFECT is found in those ionic compounds in which differences in the SIZE of cation and anion is large. This is so in case of AgCl but not in case of NaCl. Due to SMALL size of silver cations `(Ag^+)` , they can fit into the interstitial sites but `Na^+` ions cannot fit into the interstitial sites
(b)Silicon doped with PHOSPHORUS gives n-type whereas silicon doped with gallium FORMS p-type semiconductors.
Discussion
22.

Give reason : (a) Why is Frenkel defect found in AgCl but not in NaCl ? (b)What is the difference between phosphourus doped and Gallium doped semicoductors ?

Answer»

Solution :(a) FRENKEL defect is found in htose ionic compounds in which idifference in the size of cation and anion is LARGE. This is so in case of AgClbut not in case of NACL. Due to small size of silver cations `( Ag^(+))`they can fit into the interesitial sites but `Na^(+)` ions cannot fit not the interstitial sites.
(b)SILICON doped posphorus gives n-type whereas silicon doped with gallium FORMS p-type semiconductors .
Discussion
23.

Give reagents, inorganic or organic compound needed to convert benzyl bromide into (i) benzyl iodide (ii) benzyl ethyl ether (iii) benzyl alcohol (iv) benzyl cyanide (v) benzyl acetate (vi) (nitromethyl) benzene

Answer»

Solution :(i) NaI (ii) `NaOC_(2)H_(5)`
(iii) NAOH (IV) NaCN
(v) `NaOCOCH_(3)` (VI) `AgNO_(3)`
Discussion
24.

Give reactions/principles of oxygen estimation method. Compound underset(Delta)overset("Stream of" N_(2))rarr O_(2)+ other gases

Answer»

Solution :(A) `2C (ƻ) + O_(2) OVERSET(1373K(O))rarr 2CO` (B) `I_(2)O_(5) + 5CO overset((O))rarr I_(2) + 5CO_(2)`
Discussion
25.

Give reaction when Zn red kept in copper nitrate solution ?

Answer»

Solution :
A strip of Zn is placed in copper nitrate solution blue colour of solution gets disappear and the strip becomes coated with reddish METALLIC and the blue colour of the solution disappears.
When `Cu^(+2)` is reduced into Cu from the solution, it will be deposited on Zn rod so blue colour will be dissappear and `Zn^(+2)` present in solution so it will BECOME colourless.

If hydrogen sulphite gas is passed through thecolourless solution containing `Zn^(+2)` ions, appearance of white zinc sulphide ZNS can be seen on MAKING the solution alkaline with AMMONIA.
Discussion
26.

Give reaction when a Cu rod is kept in AgNO_(3) solution.

Answer»

SOLUTION :When CU ROD dipped in to the solution of `AgNO_(3),Ag^(+)` gets reduced and Ag metal gets deposite on the rod of COPPER.

Here, `Cu_((S))` is oxidised to `Cu_((aq))^(+2)andAg_((aq))^(+)` is reduced to `Ag_((S))` on seeing these Cu rod weight is reducing the solution becomes copper nitrate. THEREFORE, solution is of blue colour.
Discussion
27.

Give reaction of I_2with H_2O_2 in presence of basic medium ?

Answer»

SOLUTION :`I_(2(s)) + H_2O_(2(aq)) + 2OH_((aq))^(-) to 2I_((aq))^(-) + 2H_2O_((l)) + O_(2(G))`
Discussion
28.

Give reaction of acylation of benzene and its mechanism.

Answer»

Solution :Addition reaction of -COR (Acyl group) in benzene ring with -H is known as acylation of benzene. E.g. the reaction of formation of acetophenone from the benzene is GIVEN as follow. These reaction are knoen as Fridal Craft acylation.

This reaction is explained in THREE steps :
(i) Reaction to form electrophilic `CH_(3)overset(+)(C)O`.

Such `CH_(3)overset(+)(C)O` is strong electrophile in nature.
Mechanism of reaction : The above reaction COMPLETED in two steps by aromatic electrophilic substitution reaction.
Slow step -I : Due to strong acylinium ion `(CH_(3)overset(+)(C)O)` it breaks `pi`- ELECTRONS clouds from stable benzene and form CARBOCATION (`sigma`-complex) having one `sp^(3)` hybridized carbon.

This `sigma`- complex is not aromatic but due to stable resonance it exist as unstable intermediate. The resonance of `sigma`-complex are given as follows :

Fast step -II : `AlCl_(4)^(-)` immediately accept the `H^(+)` from the `sigma`-complex (Arenium ion) and form acyl benzene product.
Discussion
29.

Give reaction occur of cathode during electrolysis of C_(2)H_(5)COO^(-)Na^(+).

Answer»

SOLUTION :Positive `NA^(+)` ion is moving towards cathode (NEGATIVE electrode) but reduction of `H_(2)O` is EASY with respect to `Na^(+)`, so on cathode surface `H_(2)` is produced on reduction of water.
(a) `2Na_((aq))^(+) + 2H_(2)O_((l)) + 2e^(-) rarr 2H + 2Na^(+)` (b) and `2overset(.)(H) rarr H_(2)`
(b) `2Na_((aq))^(+) + 2H_(2)O_((l)) + 2e^(-) rarr 2overset(.)(H) uarr + 2Na_((aq))^(+) + 2OH_((aq))^(-)`
`2Na_((aq))^(+) + 2H_(2)O_((l)) + 2e^(-) rarr underset("(Reduction reaction on cathode)")(H_(2) + 2(Na^(+)OH^(-))_((aq)))`
Discussion
30.

Give reaction for the formation of butane from ethane.

Answer»

Solution :`CH_(3)CH_(3)underset(-2HBR)OVERSET(+Br_(2), hv)RARR 2CH_(3)CH_(2)Br underset(-2NaBr)overset(+2Na, "dry ETHER")rarr CH_(2)CH_(2)CH_(2)CH_(3)`
Ethane Bromination Bromoethane WURTZ reactionn-Butane.
Discussion
31.

Give reaction between K_a and K_b

Answer»

SOLUTION :`K_a XX K_b = K_w`
Discussion
32.

Give reaction for following reaction.

Answer»

Solution :(1) Propyne from ETHYNE :

(4) MONO BROMOETHANE from ethyne :

Discussion
33.

Give reaction for the preparation of benzene from benzoic acid and phenol.

Answer»

SOLUTION :
Discussion
34.

Give property comparison between beryllium and other alkaline earth metal elements.

Answer»

SOLUTION :Beryllium, the first member of the Group-2 metals, shows anomalous behaviour as compared to magnesium and rest of the members. Further, it shows diagonal relationship to aluminium which is discussed subsequently.
(i) Beryllium has exceptionally small atomic and ionic sizes and thus does not compare well with other members of the group.Because of high IONISATION enthalpy and small size it forms COMPOUNDS which are largely covalent and get easily hydrolysed.
(ii) Beryllium does not exhibit coordination NUMBER more than four as in its valence shell there are only four orbitals. The remaining members of the group can have a coordination number of six by making use of d-orbitals.
(iii) The oxide and hydroxide of beryllium, unlike the hydroxides of other elements in the group, are amphoteric in nature.
Discussion
35.

Give product of the following reactions.

Answer»


ANSWER :`(##RES_CHM_ORM_II_E01_012_A01##)`
Discussion
36.

Give primary details of alkali and alkali earth metals.

Answer»

Solution :The s-block ELEMENTS of the Periodic Table are those in which the last electron enters the outermost s-orbital.
As the s-orbital can accommodate only two electrons, two GROUPS (1 & 2) belong to the sblock of the Periodic Table.
The general electronic configuration of s-block elements is (noblegas) `ns^(2)`for alkali metals and [noble gas] `ns^(2)` for alkaline earth metals.
Group-1 of the Periodic Table consists of the elements : lithium, sodium, potassium, rubidium, caesium and francium. They are collectively known as the alkali metals. These are so called because they form hydroxides on reaction with water which are strongly alkaline in nature.
The elements of Group 2 include beryllium, magnesium, CALCIUM, strontium, barium and radium. These elements with the EXCEPTION of beryllium are commonly known as the alkaline earth metals.
These are so called because their oxides and hydroxides are alkaline in nature and these metal oxides are found in the earth.s crust.
Discussion
37.

Give primary information and examples of aromatic compounds

Answer»

Solution :In this cyclic compounds (4N + 2) electron are localised on the atoms of the ring. Aromatic compounds are special TWO TYPES of compounds.
(a) Benzenoid compounds: In its structure benzene ring PRESENT : e.g

(b) Non-benzenoid compounds: If in aromatic compounds have heteroatom (N, O, S) in the ring. Such compounds are called heterocylic or non-benzenoid compounds. e.g
Discussion
38.

Give preparation, properties and uses of Calcium oxide or quick lime (CaO).

Answer»

Solution :It is prepared on a commercial scale by HEATING limestone `(CaCO_(3))` in a ROTARY kiln at 1070-1270 K.
`CaCO_(3)overset("heat")hArr CaO+CO_(2)`
The carbon dioxide is removed as soon as it is produced to enable the reaction to proceed to COMPLETION.
Properties : Calcium oxide is a white amorphous solid. It has a melting point of 2870 K. On exposure to atmosphere, it absorbs moisture and carbon dioxide.
`CaO+H_(2)O to Ca(OH)_(2)`
`CaO+CO_(2) to CaCO_(3)`
The addition of limited amount of water breaks the lump of lime. This process is called slaking of lime. Quick lime slaked with soda gives solid sodalime. Being a basic oxide, it combines with acidic oxides at high temperature.
`CaO+SiO_(2) to CaSiO_(3)`
`6CaO+P_(4)O_(10) to 2Ca_(3)(PO_(4))_(2)`
Discussion
39.

Give preparation, properties and uses of sodium chloride (NaCl - Rock salt).

Answer»

Solution : The most abundant source of SODIUM chloride is sea water which contains 2.7 to 2.9% by mass of the salt. Common salt is generally obtained by evaporation of sea water.
Crude sodium chloride, generally obtained by crystallization of brine solution, contains sodium sulphate, calcium sulphate, calcium chloride and MAGNESIUM chloride as impurities. Calcium chloride, `CaCl_(2)`, and magnesium chloride, `MgCl_(2)` are impurities because they are deliquescent (absorb MOISTURE easily from the atmosphere).
To obtain pure sodium chloride, the crude salt is dissolved in minimum AMOUNT of water and filtered to remove insoluble impurities. The solution is then saturated with hydrogen chloride gas. Crystals of pure sodium chloride separate out. Calcium and magnesium chloride, being more soluble than sodium chloride, remain in solution.
Properties: Sodium chloride melts at 1081K. It has a solubility of 36.0g in 100g of water at 273K. The solubility does not increase appreciably with increase in temperature.
Uses : (i) It is used as a common salt or table salt for domestic purpose. (ii) It is used for the preparation of `Na_(2)O_(2)` NaOH and `Na_(2)CO_(3)`.
Discussion
40.

Give preparation, properties and uses of Calcium Hydroxide (Slaked lime)[Ca(OH)_(2)].

Answer»

Solution :Calcium hydroxide is prepared by adding water to quick lime, Cao.
`CaO + H_(2)O to Ca(OH)_(2)`
Properties : It is a white amorphous powder. It is sparingly soluble in water. The aqueous solution is known as lime water and a suspension of slaked lime in water is known as milk of lime.
When carbon dioxide is passed through lime water it turns MILKY due to the formation of calcium carbonate.
`Ca(OH)_(2) + CO_(2) to CaCO_(3) + H_(2)O`
On passing EXCESS of carbon dioxide, the precipitate DISSOLVES to form calcium hydrogen carbonate.
`CaCO_(3) + CO_(2) + H_(2)O to Ca(HCO_(3))_(2)`
Milk of lime reacts with chlorine to form hypochlorite, a constituent of bleaching powder.
` 2Ca(OH)_(2) + 2Cl_(2) to CaCl_(2) + Ca(OCl)_(2) + H_(2)O`
Uses : (i) It is used in the preparation of mortar, a building material. (ii) It is used in white wash due to its disinfectant nature. (iii) It is used in glass making, in tanning industry, for the preparation of bleaching powder and for purification of SUGAR.
Discussion
41.

Give preparation of carbon monoxide.

Answer»

Solution :Direct oxidation of C in limited supply of oxygen or air yields carbon monoxide.
`C+1/2O_2 oversetDeltato CO`
(ii) Reduction of HEAVY metal oxides with carbon gives CO.
`ZnO+C to Zn+ CO`
`Fe_2O_3 + 3C to 2FE + 3CO`
(iii) On small scale PURE CO is prepared by DEHYDRATION of formic acid with concentrated `H_2SO_4` at 373 K.
`underset"Formic acid"(HCOOH) underset("Conc." H_2SO_4)overset(373 K)to CO+ H_2O`
(iv) On commercial scale it is prepared by the passage of steam over hot coke. The mixture of CO and `H_2` thus produced is known as water gas or synthesis gas.
`C_((s)) + H_2O_((g)) overset"473 K - 1273 K"to underset"water gas (SYNTHETIC gas)"(CO_((g)) + H_(2(g)))`
When air is used instead of steam, a mixture of CO and `N_2` is produced, which is called producer gas.
`2C_((s)) +O_(2(g)) + 4N_(2(g))overset(1273 K) to ubrace(2CO_((g)) + 4N_(2(g)))_"Producer gas"`
Water gas and producer gas are very important industrial fuels. Carbon monoxide in water gas or producer gas can undergo further combustion forming carbon dioxide with the liberation of heat.
Discussion
42.

Give preparation and properties of carbon dioxide.

Answer»

Solution :Preparation :
It is prepared by complete combustion of carbon and carbon containing fuels in excess of air.
`C_((s)) + O_(2(g)) oversetDeltato CO_(2(g))`
`CH_(4(g)) + 2O_(2(g)) oversetDeltato CO_(2(g)) + 2H_2O_((g))`
In the laboratory it is conveniently prepared by the action of dilute HCl on calcium carbonate.
`CaCO_3 + 2HCl toCaCl_2+ CO_2 + H_2O`
On commercial scale `CO_2` is obtained by heating limestone and ethenol.
`CaCO_(3) underset"1600 K"oversetDeltato underset"Limestone"(CaO+CO_2)`
`C_6H_12O_6 underset"Enzyme"overset"Zymes"to underset"Ethyl alcohol"(2C_2H_5OH)+2CO_2`
Properties :
(i) It is colourless, odourless gas and about 1.5 times heavier than air.
(II) It is not poisonous but it does not support life of animal and human being but they die in its presence due to lack of `O_2` gas.
(iii) At room temperature and 50 to 60 atm. pressure `CO_2` gas can be liquefied, when LIQUID `CO_2` is allowed to evaporate rapidly i.e. expands rapidly, it is converted into SOLID which is known as dry ice.
(iv) `CO_2` gas is neither combustible nor SUPPORTER of combustion. However, in its presence certain active metals such as Na, K, Mg etc. continue to burn.
`2Mg+CO_2 to 2MgO + C`
(v) `CO_2` turns blue litmus paper red so it is acidic in nature.
(vi) It is sparingly soluble in water but when dissolved in water, carbonic acid (soda water `H_2CO_3`) can be obtained which is dibasic acid and it dissociates in two stages.
`H_2CO_3 + H_2O hArr HCO_(3)^(-) + H_3O^(+)`
`HCO_3^(-) + H_2O hArr CO_3^(-2) + H_3O^+`
(vii) On passing `CO_2` through lime water it turns milky due to formation of insoluble `CaCO_3` and on passing it in excess milkiness disappears due to formation of soluble calcium hydrogen carbonate (calcium bicarbonate)
`underset"Lime water"(Ca(OH)_2) +CO_2 to underset"Insoluble"(CaCO_3) + H_2O`
`CaCO_3 + CO_2 + H_2O to underset"Calcium hydrogen carbonate"(Ca(HCO_3)_2)`
(viii)In the presence of sunlight and chlorophyll of green plants, `CO_2` reacts with water to form glucose. This reaction is called photosynthesis.
`6CO_2 + 6H_2O underset"Chlorophyll"overset"hv"to C_6H_12O_6+6O_2 + 6H_2O`
Discussion
43.

Give preparation and uses of sodium hydrogen carbonate (Baking soda - NaHCO_(3)).

Answer»

Solution :Sodium hydrogen carbonate is known as baking soda because it decomposes on heating to GENERATE bubbles of carbon dioxide (leaving holes in cakes or PASTRIES and making them light and fluffy).
Sodium hydrogen carbonate is made by saturating a solution of sodium carbonate with carbon dioxide. The white crystalline powder of sodium hydrogen carbonate, being less soluble, GETS separated out.
`Na_(2)CO_(3)+H_(2)O+CO_(2) to 2NaHCO_(3)`
Uses : (i) Sodium hydrogen carbonate is a mild antiseptic for skin infections. (II) It is USED in fire extinguishers.
Discussion
44.

Give points of similarities between lithium and magnesium.

Answer»

Solution : The similarity between lithium and magnesium is particularly striking and arises because of their similar SIZES.
Atomic radii : Li (152 pm) , Mg (160 pm)
Ionic radii : `Li^(+)`(76 pm) , `Mg^(2+)` (72 pm)
The main points of similarity are :
(i) Both lithium and magnesium are harder and lighter than other elements in the respective groups.
(ii) Lithium and magnesium react slowly with water. Their oxides and hydroxides are much less soluble and their hydroxides decompose on heating. Both form a nitride, `Li_(3)N` and `Mg_(3)N_(2)`, by direct combination with nitrogen.
(iii) The oxides, `Li_(2)O and MgO` do not COMBINE with excess oxygen to give any superoxide.
(iv) The carbonates of lithium and magnesium decompose easily on heating to form the oxides and `CO_(2)`. SOLID hydrogen carbonates are not formed by lithium and magnesium.
(v) Both LiCl and `MgCl_(2)` are soluble in ethanol.
(vi) Both LiCl and `MgCl_(2)` are deliquescent and crystallise from aqueous solution as HYDRATES, `LiCl* 2H_(2)O`and `MgCl_(2)*8H_(2)O`.
Discussion
45.

Give physical and chemical properties of Orthoboric acid

Answer»

Solution :PHYSICAL properties :
Orthoboric acid, `H_3BO_3` is a white crystalline solid, with soapy touch.
It is sparingly soluble in water but highly solublein hot water.
It is also formed by the hydrolysis (reaction with water or dilute acid) of most boron compounds (halides, hydrides, etc.).
It has a layer structure in which planar `BO_3` units are joined by hydrogen bonds as shown in figure.

Chemical properties: It can be prepared by acidifying an aqueous solution of borax.
`Na_2B_4O_7+ 2HCl + 5H_2O to 2NaCl + 2B(OH_3)` Boric acid is a weak monobasic acid. It is not a protonic acid but acts as a LEWIS acid by accepting electrons from a hydroxyl ION :
On heating, orthoboric acid above 370K forms metaboric acid, `HBO_2`which on further heating yields boric oxide, `B_2O_3` .
`H_3BO_3 OVERSETDELTATO HBO_2 oversetDeltatoB_2O_3`
Discussion
46.

Give physical properties of alkane ?

Answer»

Solution :(i) Alkane are COLOURLESS and odourless.
(ii) Alkanes are insoluble in water.
(iii) Melting point and boiling point to alkane less compared to the carbon having same.
(iv) As increase in molecular WEIGHT boiling point and melting point increases.
(V) As there is increases in ISOMERIC branchedstructure, there is DECREASES in melting point.
Discussion
47.

Give physical properties of B and Al.

Answer»

Solution :Physical properties of B : (i) Boron POSSESSES low melting point, less density and very less electroconductivity. (ii) It is very hard and can tolerate high temperature.
Physical properties of Al: (i) It is silver like white metal. (ii) It possesses high electric and heat conductivity. (III) Al having DOUBLE electro conductivity compared to COPPER.
Discussion
48.

Give periodic trends in valence of elements of oxides and hydrides.

Answer»

Solution :PERIODIC trends of OXIDES and hydrides :

OXIDATION number or hydrogen in hydride is `(-1)" OR " (+1)` . For using of value of oxidation number of hydrogen we can calculate electronegativity of other elements .
Discussion
49.

Give percentage of Pin H_(3)PO_(4).

Answer»




SOLUTION :% of P `=("Mass of P in COMPOUND ")/(" Molecular mass ")XX100`
% of P `=(31)/(98) XX 100 = 31.63%`
Discussion
50.

Give packing efficiency and coordination number of the following crystal structures : (a)body centred cubic (b)cubic close packing

Answer»

SOLUTION :(a)PACKING efficiency of bcc=68%, COORDINATION number =8
(b)Packing efficiency of ccp(fcc)=74%, coordination number =12
Discussion