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.

Considering the chemical properties, atomic weight of the element 'Be' was corrected based on

Answer»

Valency
Configuration 
Density
Atomic VOLUME 

ANSWER :A
Discussion
2.

Considering the basic strength of amines in aqueous solution, which one has the smallest pK_(b) value ?

Answer»

`(CH_(3))_(2)NH`
`CH_(3)NH_(2)`
`(CH_(3))_(3)N`
`C_(6)H_(5)NH_(2)`

Answer :A
Discussion
3.

Considering the atomic number and position in the periodic table, arrange the following elements in the increasing order of metallic character : Si , Be, Mg, Na, P.

Answer»

Solution :Metallic character increases down a group and decreases along a period as we MOVE from LEFT to right. HENCE the order of INCREASING metallic character is : `P lt Si lt Be lt Mg lt Na`.
Discussion
4.

Consideringthe atomicnumberand positionin theperiodicarrange the followingelements in theincreasingorder tometalliccharacter:Si, Be, Mg, Na, P

Answer»

<P>

Solution :Arrangingthe elements intodifferentgroupsand periods in orderof theirincreasingnumberwe have.

We knowthat themetalliccharacterincreasesdown agroupandalonga periodas wemove fromleftto right. THEREFORE, Na is themostmetallicelementfollowedby Mg and Si, WhileP is theleast metallicelement. AmongBe andMg , Mg is moremetallicthan Be. thereforethe overall increasing ORDER to metalliccharacter is `P LT Si lt Be lt Mg lt Na`
Discussion
5.

Considering greater polarization in LiCl compared to that in NaCl, which of the statements you would expect to be wrong?

Answer»

LiCl has lower MELTING point that NaCl
LiCl dissolves more in organic solvents
LiCl will IONIZE in WATER more than NaCl
Fused LiCl would be LESS conducting than fused NaCl

Answer :C
Discussion
6.

Considering entropy (S) as a thermodynamic parameter, the criterion for the spontaneity of any process is

Answer»

`DeltaS_("system") -DeltaS_("surrounding") gt 0`
`DeltaS_("system") gt 0`
`DeltaS_("SURROUNDINGS") gt 0`
`DeltaS_("system") + DeltaS_("surrounding") gt 0`

Answer :D
Discussion
7.

Considering entropy (s) as a thermo dynamic parameter, the criterion for the spontanity of any process is

Answer»

`Delta S_("sys") + Delta S_("SURR") GT 0`
`Delta S_("sys') - Delta S_("surr") gt 0`
`Delta S_("sys') gt 0` only
`Delta S_("surr") gt 0` only

Solution :`Delta S_("universe") gt 0 RARR Delta S_("sys") + Delta S_("surr") gt 0`
Discussion
8.

Consider three one-litre flasks labeled A, B and C filled with the gases NO, NO_(2) and N_(2)Orespectively, each at 1 atm and 273 K. In which flask do the molecules have the highest average kinetic energy?

Answer»

Flask C 
All are the same 
Flask A 
None 

Solution :`K in PROP T`. HENCE all have same KE.
Discussion
9.

Consider this reaction. 2NO(g)+Cl_(2)(g) hArr 2NOCl(g)""DeltaH=-78.38"kJ" What conditions of temperature and pressure will produce the highest yield of NOCl at equilibrium?

Answer»

<P>`{:("T","P"),("HIGH","high"):}`
`{:("T","P"),("high","LOW"):}`
`{:("T","P"),("low","high"):}`
`{:("T","P"),("low","low"):}`

ANSWER :C
Discussion
10.

Consider thiol anion (RS^(Θ)) and alkoxy anion (RO^(Θ)). Which of the following statement is correct ?

Answer»

`RS^(Θ)` is less basic and less nucleophilic than `RO^(Θ)`
`RS^(Θ)` is less basic but more nucleophilic than `RO^(Θ)`
`RS^(Θ)` is more basic and more nucleophilic than `RO^(Θ)`
`RS^(Θ)` is more basic but less nucleophilic than `RO^(Θ)`

Solution :Since O-H BOND is stronger than S-H bond, therefore, `RO^(-)` has a GREATER TENDENCY to ACCEPT a proton than `RS^(-)` and hence `RO^(-)` is a stronger base than `RS^(-)`. Conversely, since S is less electro-negative than O, therefore, `RS^(-)` is more willing to donate a pair of ELECTRONS than `RO^(-)` and hence `RS^(-)` is more nucleophilic than `RO^(-)`. Thus, option (b) is correct.
Discussion
11.

Consider the van der Waals' constants, a and b, for the following gases:{:(gas, Ar, Ne, Kr, Xe),(a//(atm " dm"^6 mol^(-2)),1.3,0.2,5.1,4.1),(b//(10^(-2)dm^3 mol^(-1)),3.2,1.7,1.0,5.0):} Which gas is expected to have the highest critical temperature?

Answer»

Xe
Ne
KR
Ar.

Solution :Kr
Discussion
12.

Consider the titratonof potassium dichromate solution with acidfied mohr salt solution using dmiethylamine as incdicator the number of moles of mohr salt required per mole of dichromate

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3
4
5
6

Solution :Mohr 's salt `[FeSO_(4).(NH_(4))_(2)SO_(4).6H_(2)O]` is a reducing agent involving 1`E^(-)` change while `K_(2)Cr_(2)O_(7)` is an oxidising agent invoving 6 `e^(-)` change i.e
`Fe^(2+)RARRFE^(3+)+e^(-)]xx6`
`Cr_(2)O_(7)^(-)+14H^(+)+6e^(-)rarr2Cr^(3+)+7H_(2)O`
`6Fe^(2+)+Cr_(2)O_(7)^(2-)+14H^(+)rarr6Fe^(3+)+2Cr^(3+)+7H_(2)O`
thus 1 mole of `Cr_(2)O_(7)^(2-)` ION will oxidise 6 MOLES of Mohr 's salt
Discussion
13.

Consider the titration of potassium dichromate solution with acidified Mohr's salt solutionusing diphenylamine as indicator. The number of moles of Mohr's salt required per mole of dichromate is :

Answer»

3
4
5
6

Solution :`Cr_(2)O_(7)^(2-)+6Fe^(2+) +14H^(+) to 2Cr^(3+) +6Fe^(3+)+7H_(2)O`
1 MOLE of dichromate oxidises SIX MOLES of ferrous ion present in Moh's salt
Discussion
14.

Consider the titration of 0.1000 M NH_(3)(K_(b) = 1.76 xx 10^(-5)) with 0.1000 M HNO_(3). The pH at the equivalence point is

Answer»

between `2.0` and `4.0`
between `4.5` and `6.5`
APPROXIMATELY `7.0`
between `7.5` of `9.5`

SOLUTION :same s Q.No. `14`
Discussion
15.

Consider the three solutions of 1 M concetration (1) Sodium acetate (CH_3COONa) (2)Acetic acid +Sodium acetate (CH_3COOH+ CH_3COONa) (3)Acetic acid (CH_3COOH) pH of these solutions will lie in the following sequence:

Answer»

<P>` 3 lt 2 lt 1 `
` 2 lt 1 lt 3 `
` 1 lt 2 lt 3 `
` 3 lt 1 lt 2 `

Solution :For `CH_3 COONa , PH GT 7`
(2) For buffer `, pH =P^(KA) `
(3)For acid, pH =` (P^(Ka))/( 2) `
Discussion
16.

Jagadeesh: How can we predict the nature of a salt, sodium acetate (CH_3COONa)? Leela: Explained the doubt of Jagadeesh by asking some questions. Here their conversation is given in incomplete sentence. Frame the questions and fill in it. Leela: ...................................................................? Jagadesh : CH_3COONa (Sodium acetate) is basic in nature.

Answer»

<P>` 3 LT 2 lt 1 `
` 2 lt 1 lt 3 `
` 1 lt 2 lt 3 `
` 3 lt 1 lt 2 `

SOLUTION :For `CH_3 COONa , pH gt 7`
(2) For buffer `, pH =P^(KA) `
(3)For acid, pH =` (P^(Ka))/( 2) `
Discussion
17.

Consider the system at equilibrium: NH_(4)HS(s) hArr NH_(3)(g)+H_(2)S(g)""DeltaHgt0 Factors which favour the formation of more H_(2)S(g) include which of the following? (P) adding a small amount of NH_(4)HS(s) at constant volume (Q) increasing the pressure at constant temperature (R) increasing the temperature at constant pressure

Answer»

<P>P only
R only
P and Q only
P and R only

Answer :B
Discussion
18.

Consider the system at equilibrium: 2SO_(2)(g) + O_(2)(g) hArr 2SO_(3)(g) for which DeltaHlt0. Which change(S) will increase the yield of SO_(3)(g)? (P) Increasing the temperature (Q) Increasing the volume of the container

Answer»

<P>P only
Q only
Both P and Q
Neither P nor Q

Answer :D
Discussion
19.

Consider the structure of Al_(2)Me_(6) compound and find the value of x/y Where, x = total number of atoms that are sp^(3) hydridised and y = total number of 3c2e bonds.

Answer»


ANSWER :4
Discussion
20.

Consider the statements : I . Bond length in N_(2)^(+) is 0.002 Å greater than in N_(2) II. Bond length in NO^(+) is 0.09 Å less than in NO III. O_(2)^(2-) hasshorter bond length than O_(2) which of the following statements are ture ?

Answer»

I and II
II and III
I, II and III
I and III

Solution :Bond order of `N_(2)` is 3 and that of `N_(2)^(+)` is 2.5 .
Greater the bond order , shorter is the bond length
HENCE order of NO is 2.5 and that of `NO^(+)` is 3 THUS , bond length of `NO^(+)` is shorter . Hence ,II is TRUE
Bond order of `O_(2)`is 2 while that ` O_(2)^(2-)` is 1. Thus,
bond length of `O_(2)` is shorter . Hence , III is not ture .
Discussion
21.

Consider the same expansion, to a final volume of 10 liters conducted reversibly.

Answer»

SOLUTION :We have `Q= -W = 2.303 xx 10 "LOG" (10)/(2)`
`= 16.1` litre-atm.
Discussion
22.

Consider the same expansion, but this time against a constant external pressure of 1 atm.

Answer»

<P>

Solution :We have `q=-2 = p_(ex) (8) = 8` litre - ATM
Discussion
23.

Consider the same exapansion, but this time against a constant external pressure of 1 atm.

Answer»

Solution :We have `q=-w=p_(EX)(10-2)=1(8)=8 `litre ATM.
Discussion
24.

Consider the redox reaction 2S_(2)O_(3)^(2-)+I_(2)rarrS_(4)O_(6)^(2-)+2I^(-)

Answer»

`S_(2)O_(3)^(2-)+I_(2)rarrS_(45)O_(6)^(2)+2I^(-)`
`S_(2)O_(3)^(2-)` gets reduces to `S_(4)O_(6)^(2-)`
`S_(2)O_(3)^(2-)` gets oxidised to `S_(4)O_(6)^(2-)`
`I_(2)` gets REDUCED to `I^(-)`

Answer :b,c
Discussion
25.

Consider the reactions:(P)S ("rhombic") +(3)/(2)O_(2)(g) to SO_(3)(g), ""DeltaH_(1)(Q) S("Monoclinic")0 +(3)/(2)O_(2)(g) to SO_(2)""DeltaH_(2)(R) S("rhiombic")+O_(3)(g) to SO_(3)(g)""DeltaH_(3) (S)S("monoclince")+O_(3)(g) to SO_(3)(g)""DeltaH_(4)

Answer»

`DeltaH_(1)ltDeltaH_(2)LT DeltaH_(4)`(magnitude only)
`DeltaH_(1)ltDeltaH_(3)lt DeltaH_(4)`(magnitude only)
`DeltaH_(1)ltDeltaH_(2)=DeltaH_(3)lt DeltaH_(4)`(magnitude only)
`DeltaH_(1) + DeltaH_(4)=DeltaH_(2)lt DeltaH_(3)`

Answer :a,B,d
Discussion
26.

Consider the reactions (i) H_2O_2 + 2HI to I_2 + 2H_2O (ii) HOCl + H_2O_2 to H_3O^(+) + Cl^(-) + O_2 Which of the following statements is correct about H_2O_2 with reference to these reactions ? Hydrogen peroxide is ……….

Answer»

an oxidising AGENT in both (i) and (ii)
an oxidising agent in (i) and reducing agent in (ii)
a reducing agent in (i) and oxidising agent in (ii)
a reducing agent in both (i) and (ii)

Solution :
In above reduction `H_2O_2` ACTS as a Reducing agent and CONVERT HOCL into `Cl^(-1)`
Discussion
27.

Consider the reactions given below. On the basis of these reactions find out which of the algebric relations given in options( a) to (d) is correct ? (i) C(g) + 4H(g) rarrCH_(4)(g), Delta_(r) H= x kJ mol^(-1) (ii) C( graphite, s) + 2H_(2)(g) rarrCH_(4)(g), Delta_(r)H= y kJ mol^(-1)

Answer»

`x=y`
`x=2y`
`x gt y`
`x lt y`

Solution :Same BONDS are FORMED in REACTION(i) and(ii) but not bounds are broken in reaction (i) whereas bondsin the reactant molecules are broken in reaction (ii) . As energyis abosrbed when bondsare broken, energyreleased in reaction (i) is greater than that in reaction (ii) ,i.e., `x gty `.
Discussion
28.

Consider the reactions given below. On the basis of these reactions find out which of the algebric relations given in options (a) to (d) is correct ? (i) C_((g))+4H_((g))rarrCH_(4(g)),Delta_(r)H=x" kJ mol"^(-1) (ii) C_("(graphite, s)")+2H_(2(g))rarrCH_(4(g)),Delta_(r)H="y kJ mol"^(-1)

Answer»

x = y
x = 2y
x gt y
x lt y

Solution :In eqn (i), no BOND is being broken while in eqn (ii), 2 H - H bonds are broken. So, in eqn (ii) some of the ENERGY is used up to BREAK the bonds.Thus, `x gt y`.
Discussion
29.

Consider the reactions given below. On the basis of these reactions find out which of the algebraic relations given in options (A) to (D) is correct? (1) C_((g)) + 4H_((g)) to CH_(4(g)) , Delta_(r) H= x "kJ mol"^(-1) (2) C_(("graphite") ) + 2H_(2(g)) to CH_(4(g)) , Delta_(r) H = " y kJ mol"^(-1)

Answer»

`x=y`
`x=2y`
`x gt y`
`x LT y`

SOLUTION :Same bonds are FORMED in reaction (1) and (2) but no bonds are broken in reaction (1) whereas bonds in the REACTANT molecules are broken in reaction (2).
As energy is absorbed when bonds are broken, energy released in reaction (1) is greater than that in reaction (2) hence, `x gt y`
Discussion
30.

Consider the reactions C_(6)H_(5)CHO(l)+2Cu^(2)+(aq)+5OH^(-)(aq)to no change observed What interference do you draw about the behaviour of Ag^+ & Cu^(2+) from these reactions?

Answer»

SOLUTION :This reaction indicates that explain `CU^(2+)` ion is not capable of OXIDISING `C_(6)H_(5)CHO` This explain `Cu^(2+)`ion is a weaker oxidising agent than `Ag^(+)`iom.
Discussion
31.

Consider the reactions : (a) H_(3)PO_(2) (aq) + 4 AgNO_(3) (aq) +2 H_(2)O (1) to H_(3) PO_(4) ( aq) +4 Ag(s) + 4HNO_(3)(aq) (b) H_(3)PO_(2) (aq) + 2Cu SO_(4)(aq) + 2H_(2)O(1) to H_(3) PO_(4)(aq) + 2 Cu(s) + H_(2)SO_(4)(aq) (c) C_(6)H_(5) CHO(1) + 2[Ag (NH_(3))_(2)]^(+) (aq) + 3 OH^(-) (aq) to C_(6)H_(5)COO^(-)(aq) + 2Ag(s) + 4NH_(3)(aq) + 2 H_(2)O(1) (d) C_(6)H_(5)CHO(1) + 2Cu^(2+) (aq) + 5 OH^(-) (aq) toNo change observed . What inference do you draw about the behaviour of Ag^(+) and Cu^(2+) from these reactions ?

Answer»

Solution :These reactions SUGGEST that `Ag^(+)` ION is a stronger oxidising agent than `Cu^(+)` ion. It is evident from the following facts.
(i) Reactions (a) and (b) suggest that both `Ag^(+)` and `Cu^(2+)` ions can oxidise `H_(3)PO_(3)` to `H_(3)PO_(4)`. Thus, both are oxidising agents.
(ii) REACTION (c) suggests that `[Ag(NH_(3))_(2)]^(+)` can oxidise `C_(6)H_(5)CHO` to `C_(6)H_(5)COOH`, but reaction (d) indicates that `Cu^(2+)` ions are unable to do so.
Hence, it may be concluded that although both `Ag^(+)` and `Cu^(2+)` act as oxidising agents, yet `Ag^(+)` is a stronger oxidising agent than `Cu^(2+)`.
Discussion
32.

Consider the reactions : (a) H_(3)PO_(2(aq))+4AgNO_(3(aq))+2H_(2)O_((l))toH_(3)PO_(4(aq))+4Ag_((s))+4HNO_(3(aq)) (b) H_(3)PO_(2(aq))+2CuSO_(4(aq))+2H_(2)O_((l))toH_(3)PO_(4(aq))+2Cu_((s))+H_(2)SO_(4(aq)) ( c) C_(6)H_(5)CHO_((l))+2[Ag(NH_(3))_(2)]_((aq))^(+)+3OH_((aq))^(-)toC_(6)H_(5)COO_((aq))^(-)+2Ag_((s))+4NH_(3(aq))+2H_(2)O_((l)) (d) C_(6)H_(5)CHO_((l))+2Cu_((aq))^(2+)+5OH_((aq))^(-)to No change observed. What inference do you draw about the behaviour of Ag^(+)andCu^(2+) from these reactions ?

Answer»

Solution :In REACTION (a) and (b) `Ag^(+)` oxidises `C_(6)H_(5)CHO" to "C_(6)H_(5)COO^(-)`, but in reaction (d) `Cu^(+2)` COULD not oxidises `C_(6)H_(5)CHO`.
Therefore, we can say that `Ag^(+)` is strong oxidizing agent then to `Cu^(+)`.
Discussion
33.

Consider the reactions : (a) 6CO_(2(g))+6H_(2)O_((l))toC_(6)H_(12)O_(6(aq))+6O_(2(g)) (b) O_(3(g))+H_(2)O_(2(l))toH_(2)O_((l))+2O_(2(g)) Why it is more appropriate to write these reactions as : (a) 6CO_(2(g))+12H_(2)O_((l))toC_(6)H_(12)O_(6(aq))+6H_(2)O_((l))+6O_(2(g)) (b) O_(3(g))+H_(2)O_(2(l))toH_(2)O_((l))+O_(2(g)) Also suggest a technique to investigate the path of the above (a) and (b) redox reactions.

Answer»

SOLUTION :(a) Process of photosynthesis is very complex so it occurs in two steps.
Step - 1 : `H_(2)O` decompose in presence of chlorophyll and `H_(2)andO_(2)` is obtained.
Step - 2 : `H_(2)` reduce `CO_(2)` in `C_(6)H_(12)O_(6)` and some molecules of `H_(2)O` are obtained.

We must have to write this equation for total reaction of photosynthesis reaction. So, `12H_(2)O` USED for production of 1 MOLE carbohydrate and `6H_(2)O` produce during this reaction.
(b) `O_(2)` write two times because `O_(2)` is obtained from two DIFFERENT reactants.

For deciding reaction of path (a) `H_(2)O^(18)orH_(2)O` is used.
For deciding reaction of path (b) `H_(2)O_(2)^(18)orO_(3)^(18)` is used.
Discussion
34.

Consider the reactions (A) H_(2)O_(2)+ 2HI to I_(2)+2H_(2)O (B) HOCl + H_(2)O to H_(3)O^(+)+Cl^(-) + O_(2) Which of the following statements is correct about H_(2)O_(2) with reference to these reactions ? Hydrogen perioxide is …......... .

Answer»

an oxidising AGENT in both (A) and (B)
an oxidising agent in (A) and reducing agent in (B)
a reducing agent in (A) and oxidising agent in (B)
a reducing agent in both (A) and (B)

Solution :The rule of thumb for oxidising-reducing hebaviour of `H_(2)O_(2)` is :` H_(2)O_(2) ` ACTS as a reducing agent if `O_(2)` is EVOLVED and `H_(2)O_(2)` acts as an oxidising agent if `O_(2)` is not evolved.
Thus, in reaction (A), `H_(2)O_(2)` acts as an oxidising agent since `O_(2)` is not evolved but in reaction (B), it acts a reducing agent because `O_(2)` is evolved , i.e., option (b) is correct.
Discussion
35.

Consider the reactions : (a) 6CO_2(g) +6H_2O(l) rarr C_6H_(12)O_6(aq) +6O_2(g) (b) O_s(g) +H_2O_2(l) rarr H_2O(l) +2O_2(g) Why it is more appropriate to write these reactions as : (a) 6CO_2(g) +12H_2O(l) rarr C_6H_(12)O_(6)(aq)+6H_2O(l)+6O_2(g) (b) O_3 (g) +H_2O_2(l) rarr H_2O(l)+O_2(g)+O_2(g) Also suggest a technique to investigate the path of the above (a) and (b) redox reactions .

Answer»

Solution :It is believed that the PHOTOSYNTHESIS reaction occurs. In two steps . In the first step , `H_2O` decomposes to give `H_2 and O_2` in the pressure of CHLOROPHYLL and the `H_2` produced reduces `CO_2 " to " C_6H_(12)O_6` in the second step , During the second step , some `H_2O` molecules are also produced and therefore , the reaction occurs as :
`{:("(i)"12H_(2)O(l)rarr12H_(2)(g)+6O_(2)(g)),("(II)"6CO_(2)(g)+12H_(2)(g)rarrC_(6)H_(12)O_(6)(s)+6H_(2)O(l)),(bar("(iii) "6CO_(2)(g)+12H_(2)O(l)rarrC_(6)H_(12)O_(6)(s)+6H_(2)O(l)+6O_(2)(g)"")):}`
Therefore , it is more appropriate to write the reaction for photosynthesis as (iii) because it MEANS that 12 moleules of `H_2O` are used per molecule of carhydrate and `6H_2O` molecules are produced per molecule of carbohydrate during the PROCESS.
(ii) `O_2` is written two times in the product which suggests that `O_2` is being obtained from the two reactants as
`{:(""O_3(g) rarrO_2(g)+O(g)),(H_2O_2(l)+O(g)rarrH_2O(l)+O_2(g)),(bar((O_3)(g)+H_2O_2(l)rarrH_2O(l)+O_2(g)+O_2(g))):}`
The path of the reaction can be studies by using `H_2O^(16)` in reaction (a) or by using `H_2O_2^(18) " or" O_3^(18)` in reaction (b).
Discussion
36.

Consider the reactions : (a) 6 CO_(2) (g) + 6 H_(2)O (1) to C_(6) H_(12) O_(6) (aq) + 6 O_(2)(g) (b) O_(3)(g) + H_(2)O_(2)(1) + 2 O_(2)(g) Why it is more appropriate to write these reactions as : (a) 6 CO_(2) (g) + 12 H_(2)O(1) to C_(6)H_(12)O_(6) (aq) + 6 H_(2)(1) + 6 O_(2)(g) (b) O_(3)(g)+ H_(2)O_(2)(1)to H_(2)(1) + O_(2)(g) + O_(2)(g) Also suggest a technique to investigate the path of the above (a) and (b) redox reactions .

Answer»

Solution :This reaction EXPRESSES the process of photosynthesis which is a very complicated process and occurs in several steps. In this reaction, 12H_(2)O` molecules first decompose to give `H_(2)` and `O_(2)` in the presence of chlorophyll and the `H_(2)` thus produced reduces `CO_(2)` to `C_(6)H_(12)O_(6)`. Thus, in a simple way, the reaction can be expressed as follows.
`{:(""12H_(2)O(l)to12H_(2)(g)+6O_(2)(g)""....(i)),(6CO_(2)(g)+12H_(2)(g)toC_(6)(g)+12H_(2)(g)toC_(6)H_(12)O_(6)(s)+6H_(2)O(l)......(ii)),(bar(6CO_(2)(g)+12H_(2)O(l)toC_(6)H_(12)O_(6)(s)+6H_(2)O(l)+6O_(2)(g))........(iii)):}
Therefore, it is more appropriate to WRITE the reaction as (iii). This representation involves the participation of `12H_(2)O` molecules and liberation of `6H_(2)O` molecules in the reaction. (b) The given reaction actually takes PLACE as follows:
`{:(""O_(3)(g)toO_(2)(g)+O(g)""......(i)),(H_(2)O_(2)(l)+O(g)toH_(2)O(l)+O_(2)(g)""......(ii)),(bar(O_(3)(g)+H_(2)O_(2)(l)toH_(2)O(l)+O_(2)(g)+O_(2)(g))""......(iii)):}`
The representation of the reaction as (iii) SUGGESTS that one `O_(2)` molecule is obtained from `O_(3)` while the other from `H_(2)O_(2)`. Therefore, it is a more appropriate representation of the reaction.
The path of the REACTIONS (a) and (b) can he investigated by tracer technique, i.e., by using `H_(2)O^(18)` in reaction (a) and by using `H_(2)O_(2)^(18)` (or `O_(3)^(18)` ) in reaction (b).
Discussion
37.

consider the reactions: 6CO_(2)(g)+6H_(2)O(l)toC_(6)H_(12)O_(6)(aq)+6O_(2)(g) O_(3)(g)+H_(2)O_(2)(l)toH_(2)O(l)+2O_(2)(g) Why it is more approporiate to write these reactions as: 6CO_(2)(g)+12H_(2)O(l)toC_(6)H_(12)O_(6)(aq)+6H_(2)(l)+6O_(2)(g) O_(3)(g)+H_(2)O_(2)(l)toH_(2)O(l)+O_(2)(g)+O_(2)(g) Also suggest a technique to investigate the path of the above 1 and 2 redox reactions.

Answer»

Solution :`6CO_(2)(g)+12H_(2)O(l)toC_(6)H_(12)O_(6)(s)+6O_(2)(g)+6H_(2)O(l)`
Thus ,equation (3) will be more APPROPRIATE for representing photosynthesis reaction because it gives the ACTUAL stoichiometryof the REACTANTS and the products involved in the given reaction.
Discussion
38.

Consider the reactions : 2S_(2)O_(3(aq))^(2-)+I_(2(s))toS_(4)O_(6(aq))^(2-)+2I_((aq))^(-) 2S_(2)O_(3(aq))^(2-)+2Br_(2(l))+5H_(2)O_((l))to2SO_(4(aq))^(2-)+4Br_((aq))^(-)+10H_((aq))^(+)

Answer»

Solution :In `S_(2)O_(3)^(-2)`, OXIDATION number of S is +2. Where in `S_(4)O_(6)^(-2)` is +2.5. In `SO_(4)^(-2)`, oxidation number of S is +6.
`Br_(2)` is STRONG oxidising agent than `I_(2)`. So `Br_(2)` is convert `S_(2)O_(3)^(-2)` into `SO_(4)^(-2)`. In which oxidation number of oxygen is `S_(2)O_(3)^(-2)` is (+2) and `SO_(4)^(-2)` is (+6). So, it is oxidise (+2) into (+6).
While `I_(2)` is weak oxidizing agent. Therefore, it oxidised `SO_(3)^(-2)` (S = +2) to `S_(4)O_(6)^(-2)` (S = +2.5). Therefore, it gives different reactions.
Discussion
39.

Consider the reactions : 2S_(2)O_(3)^(2-)(aq)+l_(2)(s)toS_(4)O_(6)^(2-)(aq)+2l^(-)(aq) S_(2)O_(3)^(2-)(aq)+2Br_(2)(l)+5H_(2)O(l)to2SO_(4)^(2-)(aq)+4Br^(-)(aq)+10H^(+)(aq) Why does the same reductant, thiosulphate react differently with iodine and bromine?

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Solution :The oxidation numbers of S in the INVOLVED species are as follows:
`S_(2)O_(3)^(2-):+2,S_(4)O_(6)^(2-):+2.5,SO_(4)^(2-):+6`
Bromine is a stronger oxidising agent than `l_(2)`. Therefore, it oxidises `S_(2)O_(3)^(2-)` (O.N. of S=+2) to `SO_(4)^(2-)` (O.N. of S=+6) in which is in a higher oxidation state. `l_(2)` being a weaker oxidising agent is able to oxidise `S_(2)O_(3)^(2-)` to `S_(4)O_(6)^(2-)` (O.N. of S = 2.5) in which S is in a lower oxidation state. This is why `S_(2)O_(3)^(2-)` REACTS differently with bromine and IODINE.
Discussion
40.

Consider the reaction where K_(p)=0.5 at a particular temperature PCl_(5)(g)hArrPCl_(3)(g)+Cl_(2)(g) If the three gases are mixed in a container so that the partial pressure of each gas is initially 1 atm, then which one of the following is true

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more `PCl_(3)` will be PRODUCED
more `Cl_(2)` will be produced
more `PCl_(5)` will be produced
NONE of these

Answer :C
Discussion
41.

Consider the reaction where K_P = 0.5 at a particular temperature PCl_5(g) hArr PCl_3(g) + Cl_2(g) if the three gases are mixed in a container so that the partial pressure of each gas is initially1 atm,then which one of the following is true?

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more `PCl_3` will be produced
more `Cl_2` will be produced
more `PCl_5` will be produced
NONE of these

Solution :`K_P = 0.5`
`Q = (P_(PCl_5).P_(Cl_2))/(P_(PCl_5))`
`Q = (1xx1)/1`
`Q GT K_P`
`:.` Reverse reaction is favoured, i.e., more `PCl_5` will be produced.
Discussion
42.

Consider the reaction where , K_P = 0.5at a particular temperaturePCl_(5)(g) hArr PCl_3 (g) + Cl_2 (g) if the three gases are mixed in a container so that the partial pressure of each gas is initially 1 atm, then which one of the following is true.

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<P>more `PCl_3` will be PRODUCED
more `Cl_2` will be produced
more `PCl_3` will be produced
none of these

Solution :`K_P = 0.5`
`Q=(P_(PCl_3) * P_(Cl_2))/(P_(PCl_5))`
`Q= (1xx1)/1`
`Q gt K_p`
`THEREFORE ` REVERSE reaction is favoured , i.e. more `PCl_5` will be produced.
Discussion
43.

Consider the reaction where K_(p) = 0*497 "at 500 K PCl_(5) (g) hArr PCl_(3) (g) + Cl_(2) (g)If the three gases are mixed in a rigid container so that the partial pressure of each gas is initially1 atm, which is true ?

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More `PCl_(5)` will be PRODUCED
More `PCl_(3)` will be produced
Equilibrium will be established when 50% of the REACTION is complete
None of the above

Solution :`Q_(p) = (p_(PCl_(3))xxp_(Cl_(2)))/(p_(PCl_(5)))=(1 atm xx1 atm )/(1atm )=1`
As `Q_(p) gt K_(c)` , equilibrium will go in the BACKWARD direction , i.e., more `PCl_(5)` will be produced .
Discussion
44.

Consider the reaction The end product of the reaction is

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SOLUTION :Aldehudes and KETONES REACT with thiols to form thioacetals which react which react with Raney `Ni` to yiedl hydrocarbons:

The REDUCTION of thioacetal is takes place in neutral solution and can be used for compounds that are senstitive to both acids and bases.
Discussion
45.

Consider the reaction sequence below :

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ANSWER :A
Discussion
46.

Consider the reaction of water with F_(2) and suggest , in terms of oxidation and reduction, which species are oxidised /reduced ?

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Solution :`underset("Oxidant")(2Fe_(2)(g))+ underset("Reductant")(2H_(2)O(l))to O_(2)(g) + 4H^(+)(aq) + 4F^(-)(aq)`
or `underset("Oxidant")(3Fe_(2)(g))+ underset("Reductant")(3H_(2)O(l)) to O_(3)(g) + 6H^(+)(aq) + 6F^(-)(aq)`
In these reactions, water acts as a reducing AGENT and HENCE itself gets oxidised to EITHER `O_(2)` or `O_(3)` while `F_(2)` acts as an oxidising agent and hence is itself reduced to `F^(-)` ion.
Discussion
47.

Consider the reaction of water with F_(2) and suggest in terms of oxidation and reduction whichh species are oxidised/reduced

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SOLUTION :
(I) Water `(H_(2)O)` is oxidised to `O_(2)`.
(ii) Fluorine `(F_(2))` is REDUCED to `F^(-)` ions or HF
Discussion
48.

Consider the reaction of water with F_2 and suggest in terms of oxidation and reduction, which species are oxidised reduced.

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Solution :`2F_(2(g)) + 2H_(2)O_((l)) to O_(2(g)) + 4H_((AQ))^(+) + 4F_((aq))^(-)`Reduction/oxidation
In above reaction water act as a REDUCING agent so, it accept oxygen while fluorine act a oxidising agent. It release Fluorine ion `(F^-)` .
Discussion
49.

Consider the reaction NH_(4)COONH_(2)(s) hArr 2NH_(3)(g) + CO_(2)(g)at a certain temperature, the equilibrium pressure of the system is 0.318atm. Find K_(p) of the decomposition of ammonium carbonate.

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SOLUTION :`P_("total") = 3P "" :. P = 0.318//3 = 0.106 `
` kp = 4P^(3) = 4.76 xx 10^(-3)`
Discussion
50.

Consider the reaction, N_(2)(g)+3H_(2)(g) hArr 2NH_(3)(g) . Explain the effect of pressure on this equilibrium reaction.

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Solution :`N_(2)(g) + 3H_(2)(g) HARR 2NH_(3)(g)`
In the above equlibirum, if the pressure is increased, the volume will decrease, The system responds to this effect by reducing the number of gas molecules , i.e, It favours the FORMATION of ammonia. If the pressure is reduced, the volume will increase. It favours the decomposition of ammonia.
Discussion