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.

Assuming ideal behaviour, calculate Boyle's law constant for each of the following gase at 25^(@)C a. 10g of O_(2) in 2 L container b. 8g of CH_(4) in 5 L container

Answer»

Solution :Boyle's law constant is GIVEN by PRODUCT of PV at a given temperature.
Now. PV=nRT
a. 10g of `O_(2)=10//32` mol
THUS, PV`=(10xx0.0831xx298.15)/(32)=7.74`L bar
(B). 18g of `CH_(4)=8//16=0.5` mol
thus `PV=0.5xx0.0831xx298.15=12.38` L bar
Discussion
2.

Assuming fully decomposed, the volume of CO_(2) released at N.T.P on heating 9.85 g of BaCO_(3) (Atomic mass of Ba = 137) will be :

Answer»

0.84 L
2.24 L
4.06 L
1.12 L

Solution :`{:(BaCO_(3),OVERSET("Heat")(rarr),BaO+CO_(2)),(137+12+48,,22.4L),(=197g,,(22.4)/(197)xx9.85),(=9.85g,,=1.12L):}`
Discussion
3.

Assuming Delta H^(0) and Delta S^(0) do not change with temperature the boiling point of liquid ''A'' (the thermodynamics data given below) is

Answer»

300K
13K
150K
50K

Solution :`Delta S = 100JK^(-1) MOL^(-1), Delta H_("ext") = Delta H_(G) - Delta H_(L)`
`100- (-130) = 30kJ`
`T = (30 xx 1000)/(100) = 300 ( :. T = (Delta H)/(Delta S))`
Discussion
4.

Assuming complete ionization same moles of the following compundswill require theleast amount ofacidicfied MNnO_(4) for complete oxidatoin ?

Answer»

`FeSO_(3)`
`FeC_(2)O_(4)`
`FE(NO_(2))_(2)`
`FeSO_(4)`

Solution :Reduction half reaction
`MnO_(4)^(-)+8H^(+)+5e^(-)rarrMn^(2+)+4H_(2)O`
no of electrons involved in the reqction =5
(a) oxidation of `FeSO_(3)`to `Fe_(2)(SO_(4))_(3)`
`Fe^(2+)rarrFe^(3+)+e^(-)`
`SO_(3)^(2-)+H_(2)OrarrSO_(4)^(2-)+2H^(+)+2E^(-)`
(B) oxidation of `FeC_(2)O_(4)`
`Fe^(2+)rarrFe^(3+)+e^(-)`
`C_(2)O_(4)^(2-)rarr2CO_(2)+2e^(-)`
(c ) oxidation of `Fe(NO_(2))_(2)`
`Fe^(2+)rarrFe^(3+)`
`2NO_(2)^(-)+2H_(2)Orarr2NO_(3)+4H^(+)+4e^(-)`
(d) oxidation of `FeSO_(4)`
`Fe^(2+)rarrFe^(3+)+e^(-)`
no of ELECTRONIC involved =1since `FeSO_(4)` gives least number of electons therefore it will required least amount of L `MnO_(4)` for its oxidation
Discussion
5.

Assuming complete dissociation, calculate the pH of the following solutions : (a)0.003 M HCl , (b)0.005 M NaOH , (c)0.002 M HBr , (d)0.002 M KOH

Answer»

SOLUTION :(a)pH of 0.003 M HCl : Complete ionization of HCl, then solution ,
[HCl]=`[H^+]`= 0.003 M = `3.0xx10^(-3)` M
pH =-log `[H^+]`
`=-log (3.0xx10^(-3))={log 3.0+log 10^(-3)}`
=-(0.4771-3.0)=-(-2.5229)
=+2.5229 `approx` 2.52
(B)pH of 0.005 M NaOH : NaOH is a strong BASE , its complete ionization occurs
`[OH^-]`=[NaOH ] =0.005 M = `5.0xx10^(-3)` M
So, pOH = -log `[OH^-]`
`=-log (5.0xx10^(-3))=-(log 5+ log 10^(-3))`
=-0.6990-3.0)=-(-2.3010)
=+2.3010
Now, pH = 14.0-pOH
=14.0- 2.3010
=11.6990 `approx` 11.70
(c)pH of 0.002 M HBR : HBr is strong acid and 100% ionization occur,
`[HBr]=[H^+]=0.002 M = 2.0xx10^(-3)` M
`therefore` pH=-log `[H^+]`
`=-log (2.0xx10^(-3))=-{log 2.0 + log 10^(-3)}`
=-(0.3010-3.00)=-(-2.6990)
=+2.699 `approx` 2.70
(d) pH of 0.002 M KOH : KOH is strong base and complete ionization in solution,
`[OH^-]=[KOH]=0.002 M =2.0xx10^(-3)` M
pOH = -log `[OH^-]`
`=-log (2.0xx10^(-3))`
`=[log 2.0 + log 10^(-3)]`
=-{0.3010 -3.0} = -(-2.6990)
=+2.6990
pH = 14.0-pOH = 14.0-2.699
=11.3010 `approx` 11.30
Discussion
6.

Assuming complete dissociation, calculate the pH of the following solutions : (a) 0.003 M HCl(b) 0.005 M NaOH(c) 0.002 M HBr(d) 0.002 M KOH

Answer»

Solution :(a) `HCl + aq rarr H^(+) + Cl^(-) , :. [H^(+)]=[HCl]=3xx10^(-3)M, pH = - log (3xx10^(-3))=2.52`
(b) `NaOH + aq rarr NA^(+) + OH^(-)= 5xx10^(-3) M, [H^(+)]=10^(-14)//(5xx10^(-3))=2xx10^(-12)M`
`pH = - log (2xx10^(-12))=11.70`
(c) `HBr + aq rarr H^(+) + Br^(-), :. [H^(+)]=2xx10^(-3)M, pH = - log (2xx 10^(-3))=2.70`
(d) `KOH + aq rarr K^(+) + OH^(-) , :. [OH^(-)]=2xx10^(-3)M,[H^(+)]=10^(-14)//(2xx10^(-3))=5xx10^(-12)`
`pH = - log (5xx10^(-12))=11.30`
Discussion
7.

Assuming 2s - 2p mixing is not operative , the paramagnetic species among the following is

Answer»

`Be_(2)`
`B_(2)`
`C_(2)`
`N_(2)`

Solution :If `2S - 2p ` MIXING is not operative , then molecular
orbitals may be ARRANGED in ORDER of energy as follows :
`sigma (1s), sigma^(**) (1s), sigma(2s), sigma^(**) (2s), sigma(2p_(z)), PI(2p_(x))`
`= (2p_(y)), pi^(**) (2p_(x)) = pi^(**) (2p_(y)), sigma^(**) (2p_(z))`
Applying this configuration , `Be_(2), B_(2) and N_(2)`will
be diamgnetic but `C_(2)` will be paramagnetic .
Discussion
8.

Assume that the decomposition of HNO_(3) can be represented by the following equation 4HNO_(3(g)) hArr 4NO_(2(g)) + 2H_(2) O_((g)) + O_(2(g))and the reaction approaches equilibrium at 400K temperature and the copper turnning 0 atm pressure. At cquilibrium partial pressure of HNO_(3) is 2 atm. Calculate Kc in ("mole" //L)^(3) at 400 K.

Answer»

4
8
16
32

Solution :`4HNO_(3(g)) harr 4NO_(2(g))+2H_(2)O_((g))+O_(2(g))`
`P_("Total")=P_(HNO_(3))+P_(NO_(2))+P_(H_(2)O)+P_(O_(2))`
`:. P_(NO_(2))=4_(PO_(2))` and `P_(H_(2)O)=2P_(O_(2))`
`:. P_("Total")=P_(HNO_(3))+7P_(O_(2))`
`IMPLIES 30-2=P_(O_(2)) xx 7, P_(O_(2))=4`
`Kp=(P_(NO_(2))^(4) xx P_(H_(2)O) xx P_(O_(2)))/(P_(HNO_(3))^(4))`
`=((4 xx 4) xx (2 xx 4)^(2) xx 4)/(2^(4))=2^(20)`
`Kp=Kc(RT)^(Delta n_((g)))=Kc(0.0821 xx 400)^(3)`
`implies Kc(0.08 xx 400)^(3) implies Kc=32`
Discussion
9.

Assume that 4g of I_(2) are allowed to react with 4g of Mg metal according to the following equation Mg+I_(2) rarr MgI_(2) Which is the limiting reagent in the reaction ?

Answer»


Solution :The chemical EQUATION is :
`underset(24g)(Mg)+underset(underset(=254g)(2xx127))(I_(2))rarrMgl_(2)`
24 G of Mg metal require `I_(2) = 254 g`
4 g of Mg metal require `I_(2)=((254g))/((24g))XX(4g)=42.3g`
But `I_(2)` actually available = 4g
`:. I_(2)` is the limiting reactant.
Discussion
10.

Assume that 2 xx 10^(-17) J of light energy is needed by the interior of the human eye to see an object. The number of photons of yellow light with lambda = 595.2nm are needed to generate this minimum energy is 12x. Then the value of x is _________ .

Answer»


Solution :`E = n(hc)/(LAMBDA)`
`implies(2xx 10^(-17))/(1.6 XX 10^(-19)) eV = n xx (1240)/(595.2) xx (eV. Nm)/(nm)`
`impliesn = 60 implies 12 x = 60 implies x = 5`
Discussion
11.

Assume ideal gas behavior for all the gasesconsidered and vibratonal degree of freedom to be active. Separated equimolarsample of Ne,O_(2), So_(2) and CH_(4)were subjedted to a two step process as mentioned . Initiallytall areat samestate of temperture and pressure.Step -I:All undergo reversibleabiabatic expansion to attainsame finalvolume therebycausingthedecreasae in theirtemperature.Step -II:Afterstep I, allare given appropriate amount of heat isochoricallyto restorethe originaltemperature . Mark the correct optoin(s).

Answer»

Due to step Ionly , the decrease in temperature will be maximumfor Ne.
During step II , heat given will be minimumfor `CH_(4)`
There will be no NET change in INTERNAL ENERGY for any of the gas after boththe steps of process are completed.
The `(@) P-V` GRAPH of `CH_(4)` and `SO_(2)` wll be same for overall process.

Answer :a,c
Discussion
12.

Assume ideal gas behaviour for all the gases considered and neglect vibrational degrees of freedom. Separate equimolar sample. Separate equimolar samples of Ne, O_(2) , CO_(2) and SO_(2)were subjected to a two process as mentioned. Initially all are at same state of temperature and pressue.Step I rarr All undergo reversible adiabatic expansion to attain same final volume, which is double the original volume thereby causing the decreases in their temperature. Step rarr After step I all are given appropriate amount of heat isochorically to restore the original temperature. Mark the correct option(s) :

Answer»

Due to step I only, the decrease in temperature will be maximum for Ne
During step II, heat given will be minimum for `SO_(2)`
There will be no change in INTERNAL energy for any of the GAS after both the steps of process are competed
The P-V graph of `O_(2)` and `CO_(2)` will be same

Solution :First step is adiabatic (q = 0) so `DeltaU_(1)=w_(1)`
Second step is isochoric (w = 0)

So, `""DeltaU_(2)=q_(2)`
`because` initial and final temp. are same
`therefore""DeltaU_("toal")=DeltaU_(1)+DeltaU_(2)=0`
` "or"w_(1)+q_(2)=0`
Max. work done by the gas, `SO_(2)` is (area) under the curve
so, `SO_(2)` absorbed
`because ""gamma_(SO_(2))lt gamma_(CO_(2))=gamma_(O_(2))=ltgamma_(Ne)`
so, max. decrease in temp. of Ne due to step 1.
Discussion
13.

Assume each reaction is carried out in an open container. For which reaction will Delta H = Delta E?

Answer»

`2CO _((G)) + O _(2 (g)) to 2 CO _(2 (g))` <BR>`H _(2 (g)) + Br _(2 (g)) to 2 HBr _((g))`
`C _((s))+ H _(2 ) O _((g)) to 2 H _(2 (g)) + CO _(2 (g))`
`PCl _(5 (g)) to PCl _(3 (g)) + CL _(2 (g))`

Answer :B
Discussion
14.

Associated colloid among the following is

Answer»

enzymes
protein
cellulose
SODIUM stearate

Solution :Adsorption of krypton an activated charcoal at low temperature means it is adsorbed by physisorptiob i.e. gas molecules are held to the surface by Van der Waal's forces. `Delta S lt 0` because of FREEDOM of movement of gas molecules get restricted when gas is adsorbed. `DeltaH lt 0` because reaction is exothermic due to ATTRACTIVE forces between the gas molecules and SOLID surface.
Discussion
15.

Assining R,S configuration to the given compound :

Answer»

2R, 3S
2S, 3S
2S, 3R
2R, 3R

Solution :N//A
Discussion
16.

Assing the position of the elements with outer electronic configuration ns^2np_4 (n = 3)

Answer»

SOLUTION :`3^rd` PERIOD, GROUP 16
Discussion
17.

Assing the position of the elements with outer electronic configuration (n-2)f7 (n-1)d^1 ns^2 (n=6)

Answer»

SOLUTION :`6^th` PERIOD, GROUP 3 (all f block ELEMENTS are in group 3).
Discussion
18.

Assing the position of the elements with outer electronic configuration (n-1) d^2 ns^2(n = 4)

Answer»

SOLUTION :`4^th period,4^th` GROUP (group NUMBER = number of d ELECTRONS + number of s electrons)
Discussion
19.

Assign the priority order number of the following atoms or groups. (a) -CHO,-CH_(2)OH,-CH_(3),-OH (b) -Ph,-CH(Me)_(2),-H,-NH_(2) (c) -COOH,-Ph,-CHO,-C-=CH,-Ph (d) -CH(Me)_(2),-CH=CH_(2),-CH_(2),-C-=CH,-Ph (e) -CH_(3),-CH_(2)Br,-CH_(2)OH,-CH_(3)Cl (f) -H,-N,(Me)_(2),-Me,-OMe (g) -CH=CH_(2),-Me,-Ph,-Et (h) -CH_(2)-CH_(2)-Br,-Cl,-CH_(2)-CH_(2)-CH_(2)-Br,(Me)_(2)CH- (i) -Cl,-Br,-I,-NH_(2) (j) NH_(2),NO_(2),CH_(2)NH_(2),C-=N

Answer»


ANSWER :[(a) 4,1,2,3 (b) 4,1,2,3 (c) 1,3,2,4 (d) 4,3,2,1 (e) 2,3,4,1 (f) 4,2,3,1 (G) 3,1,4,2 (H) 2,4,1,3 (i) 3,2,1,4 , (j) 2,1,4,3
Discussion
20.

Assign the position of the element having outer electronic configuration (i) ns^(2)np^(4) " for " n = 3 (ii) (n -1) d^(2)ns^(2) " for " n = 4and (iii) (n -2) f^(7) (n -1) d^(1) ns^(2) " for " n = 6,in the periodic table.

Answer»

<P>

Solution :For n = 3, the element belongs to third period. The electronic configuration is `3d^(2) 3d^(4)`.
Since the last electron enters p ORBITAL, therefore, the element belongs to p-blcok,
Group no, of the element,
` = 10 + ` No. of electron in valence shell
` = 10 + 6 = 16`
` :. ` Period = 3 , group = 16
(ii) For n = 4, the element belongs to fourthperiod. The electronic configuration is`3d^(2) 4s^(2)`.
Since the d-sub-shell is incomplete, the element belongs to d-block.
Group no, of the element,
= No. of electron in `(n -2)` subshell ` + ` No. of electron in ns subshell
` = 2 + 2 = 4`
` :. ` Period = 4, group = 4
(III)For n = 6, the element belongs to SIXTH period. The electronic configuration is`4f^(7) 5d^(1) 6s^(2)`.
Since the electron goes to f -orbital, therefore, its belongs to f-block.
All f-block elements belongs to group-3.
`:. `Period = 6, group = 3
Discussion
21.

Assign the position of theelement havingouterelectronicconfiguration , (i) n s^(2) np^(4) for n = 3(ii)(n-1) d^(2)n s^(2) forn=4and (ii) (n-2) f^(7)(n-1)d^(1)n s^(2) for n= 6 in theperiodictable.

Answer»

Solution :(i) n=3 suggeststhat theelementbelongs to thirdperiod . Sincethe lastelectronentersthe p- orbital thereforethe givenelement is p- blockelement .Furthersince thevalenceshellcontains6 (2+electrons thereforegroupnumberof theelement= 10 +no. of electrons in the valenceshell = 10 + 6 =
Thecompleteelectronicconfigurationof the elementis `1 s^(2)2 s^(2)2 p^(6)3 s^(2)3p ^(4)` and theelement is( sulphur )
(II) n=4 suggetsthat theelementlies in the4thperiod.
sincethe d- orbitals are incompletethereforeit isd-blockelement . the groupnumberof the element no. OFD- electrons + no. of s- electrons = 2 + 2 =4 . Thusthe elementslie ingroup 4 and4thperiod . Theconfigurationof the elementis `[Xe] 4 f^(7)5 d^(1) 6 s^(2). ` Theatomicnumber of theelement= 54 + 7 + 1 + 2= 64and thethe element is Gd(gadolinium ).
Discussion
22.

Assign the oxidation number of Mn in MnO_(4)^(-)

Answer»


ANSWER :7
Discussion
23.

Assign reasons for each of the following : (i) + 1 gallium undergoes disproportionation reactions. (ii) Unlike In^(+), Ti^(+) is more stable with respect to disproportionation. (ii) InCl undergoes disproportionationbut TlCl does not. (iv) In (III)is more stablethan In (I) aqueous solution.

Answer»

Solution :(i) Due to inert pair effect, galliumshowsboth +1 and +3 oxidationstates butits +3 oxidationstate is more stable than `+1` oxidationstate.In other words +1 oxidation state. Inother words `+1` gallium is less stable than +3 gallium and hence undergoes disproportionation (self oxidation -reduction) to form gallium metal and the more stable+3 gallium IONS in aqueoussolutionas shownbelow :

(ii) Althroughboth In and TI can show oxidationstates of +1 and +3 , but inert pair effect is more prominent in Tl than in In. Therefore,+1 oxidationstate of Tl is more stablethan its +3 oxidation state while +3 oxiationstate ofln is more stablethan its + 1 oxidationstate.Consequently in aqueous solution, less stable`ln^(+)`undergoesdisproportionation to form more state `In^(3+)` but `+1`thalliumbeing more stabledoes not undergo disproportionationto form +3 THALLIUM.
`3In^(+) (aq) overset("Disproportionation")rarr 2In(s) + In^(3+) (aq)`

(iii) As statedabove, `+3` oxidationstate of In is more stable than its +1 oxidationstate, therefore, `InCl` undergoesdisproportionation in aqueoussolution.
`3In Cl(aq) rarr 2In (s) + In^(3+) (aq) + 3Cl^(-) (aq)`
Since +1 oxidationstate of Tl is morestablethan its +3 oxidationstate,therefore,`TlCl` does notundergo disproportional in aqeuoussolution.

(iv) Same as explained in ANSWER (ii) above.
Discussion
24.

Assign oxidation number to the underlined elements in each of the following species : Naul(B)H_(4)

Answer»

Solution :In `NaBH_(4)`, H is present as `H^(-)` ion because it is a hydride. THEREFORE, oxidation NUMBER of H in it is -1.
`OVERSET(+1)(NA)overset(x)(B)overset(-1)(H_(4))`
`(+1)+(x)+[(-1)xx4]=0`
or `x=+4-1=+3`
`:.` The oxidation number of B in `NaBH_(4)` is +3.
Discussion
25.

Assign oxidation number to the underlined elements in each of the following species : NaHul(S)O_(4)

Answer»

Solution :`overset(+1)(NA)overset(+1)(H)overset(X)(S)overset(-2)(O_(4))`
`(+1)+(+1)+(x)+[(-2)xx4]=`
or `x=+8-2=+6`
Therefore, the OXIDATION number of S in `NaHSO_(4)` is + 6.
Discussion
26.

Assign oxidation number to the underlined elements in each of the following species : NaH_(2)ul(P)O_(4)

Answer»

Solution :LET the oxidation number of P be x. Writing the oxidation NUMBERS of all elements present, we have
`overset(+1+1)(NaH_(2))overset(x-2)(PO_(4))`
The sum of oxidation numbers of all atoms in a NEUTRAL molecule is zero. Hence,
`(+1)+[(+1)xx2]+(x)+[(-2)xx4]=0`
or `x=+8-3=+5`
Thus, the oxidation number of P in `NaH_(2)PO_(4)` is +5.
Discussion
27.

Assign oxidation number to the underlined elements in each of the following species : KAl(ul(S)O_(4))_(2).12H_(2)O

Answer»

Solution :`overset(+1)(A)overset(+3)(Al)overset(X-2)((SO_(4))_(2)).12overset(+1-2)(H_(2)O)`
`(+1)+(+3)+[(x)+(-2)xx4]xx2+[(+1)xx2+(-2)]xx12=0`
or `+4+2x-16+24-24=0`
or `2x=+16-4+12`
or x=+6
`:.` The oxidation NUMBER of S in `KAI(SO_(4))_(2).12H_(2)O` is +6.
Discussion
28.

Assign oxidation number to the underlined elements in each of the following species : H_(4)ul(P)_(2)O_(7)

Answer»

Solution :`overset(+1)(H_(4))overset(X-2)(P_(2)O_(7))`
`(+1)xx4+(x)xx2+[(-2)xx7]=0`
or `2x=+14-4=+10`
or x=+5
`:.` The oxidation state of P in `H_(4)P_(2)O_(7)` is +5.
Discussion
29.

Assign oxidation number to the underlined elements in each of the following species : K_(2)ul(Mn)O_(4)

Answer»

SOLUTION :`overset(+1)(K_(2))overset(x)(Mn)overset(-2)(O_(4))`
`(+1)xx2+(x)+[(-2)xx4]=0`
ror `x=+8-2=+6`
`:.` The OXIDATION state of P in `K_(2)MnO_(4)` is +6.
Discussion
30.

Assign oxidation number to the underlined elements in each of the following species : H_(2)ul(S_(2))O_(7)

Answer»

Solution :`overset(+1)(H_(2))overset(x)(S_(2))overset(-2)(O_(7))`
`[(+1)xx2]+[(x)xx2]+[(-2)xx7]=0`
or `2x=+14-2=+12`
or x=+6
`:.` The oxidation NUMBER of S in `H_(2)S_(2)O_(7)` is +6.
Discussion
31.

Assign oxidation number to the underlined elements in each of the following species : Caul(O)_(2)

Answer»

Solution :`OVERSET(+2)(Ca)overset(x)(O)`
`(+2)+(x)=0`
or x=-2
`:.` The oxidation STATE of O in CAO is -2.
Discussion
32.

Assign oxidation number to the underlined elements in each of the following species : (a) NaH_(2)underlinePO_(4) (b) NaHunderlineSO_(4) ( c) H_(4)underlineP_(2)O_(7) (d) K_(2)underline(Mn)O_(4) (e) CaunderlineO_(2) (f) NaunderlineBH_(4) (g) H_(2)underlineS_(2)O_(7) (h) KAl(underlineSO_(4))_(2)*12H_(2)O

Answer»

Solution :(a) `NaH_(2)underlinePO_(4)to1(Na)+2(H)+(P)+4(O)=0`
`therefore1(+1)+2(+1)+(P)+4(-2)=0`
`therefore1+2+P-8=0`
`thereforeP=+5`
(b) `NaHunderlineSO_(4)to(+1)+(+1)+(S)+4(-2)=0`
`therefore+2+S-8=0`
`thereforeS=+6`
( C) `H_(4)underlineP_(2)O_(7)to4(+1)+2(P)+7(-2)=0`
`therefore4+2P-14=0`
`therefore2P-10=0`
`thereforeP=+5`
(d) `K_(2)UNDERLINE(Mn)O_(4)to2(+1)+Mn+4(-2)=0`
`therefore+2+Mn-8=0`
`thereforeMn=+6`
(e) `CaunderlineO_(2)toCa+2(-1)=0`
`thereforeCa-2=0`
`thereforeCa=+2`
(F) `NaunderlineBH_(4)to1(+1)+B+4(-1)=0`
`thereforeB=+3`
(g) `H_(2)underlineS_(2)O_(7)to2(+1)+2(S)+7(-2)=0`
`thereforex=+6`
(h) `KAl(underlineSO_(4))_(2)*12H_(2)Oto1+3+2(S)+8(-2)+12(2xx1-2)=0`
`therefore2(S)+4-16`
`thereforeS=+6`
OR
As `H_(2)O` is NEUTRAL therefore total charge remain zero.
`therefore1+3+2(S)+8(-2)=0`
`thereforeS=+6`
Discussion
33.

Assign oxidation number to the underlined element in each of the following species NaH_(2)PO_(4) (b) NaHSO_(4) (c )H_(4)P_(2)O_(7) (d)K_(2)MnO_(4) (e )CO_(2) (f) NaBH_(4) (g)H_(2)S_(2)O_(7) (h)KAI(SO_(4))_(2)12H_(2)O

Answer»

Solution :(a) Let the oxidation number of p be x writing the oxidation number of each atom above its symbol we
sum of oxidation numbr of various atoms in `NaH_(2)PO_(4)=1(+1)+2 (+1)+1(x)+4(-2)=x-5` but the sum of oxdation numberof various atoms in `NaH_(2)PO_(4)` (neutral) is ZERO
Thus the oxidation nuber of p in `NaH_(2)PO_(4)=5`
(b) `overset(+1)NA overset (x)H overset(x)S overset(-2) O_(4) therefore +1(+1)+x+4(+1)-2=0 or x=_6`
Thus the oxidation number of S in `NaHSO_(4)=+5`
(C ) `overset(+1)H_(4)overset(+1)P_(2)overset(x)S overset(-2)O_(4)therefore4(+1)+2(x)+7(-2)=0 or x=+5`
(d)`K_(2)Mn O_(4)^(-2) therefore 2(+1)+1(x)+7(-2)=0 or x =+6`
thus the oxidation numbr of Mn in `K_(2)MnIO_(4)=+7`
(e ) let hte oxidation number of O be x since ca is an alkaline earth METAL therefore its oxidation number is +2 thus
`CaO_(2)therefore +2+2(x)=0 or x=-1`
(f) In `NaBH_(4)` H is present as hydride ion therefore its oxidation number ois -1 thus
`Na BH_(4) therefore2(+1)+x+(-1)=0 or x=+3`
thus the oxidation number of B in `NaBH_(4)=+3`
(g) `H_(2)S_(2)O_(7)^(2-) therefore 2(+1)+2(x)+7(-2)=0 or x_6`
thus the oxidation number of s in `H_(2)S_(2)O_(7)=+6`
(h)`K AI(SO_(4))12(H_(2)O or +1+3+2x+8(-2)+12(2xx1-2)or x=+6`
alternatively since `H_(2)O` is a neutral molecule therefore sum of oxidation numberof S
`therefore+1+3+2+x-16=0or x=+6`
thus the oxidation number of S in KaI `(SO_(4))12H_(2)O=+6`
Discussion
34.

Assign oxidation number to the underlined elements in each of the following species

Answer»

SOLUTION :`+3`(H is PRESENT as -1)
Discussion
35.

Assign E and Z configuration ?

Answer»


Answer :[Z-I,II,III,VI,VII; E-IV,VIII,IX,X,XI,XII]
Discussion
36.

Assetion: An acidifiedd queous solution of KCIO_(3) when boiled with iodine produces KIO_(3) Reason : KCIO_(3) is an oxidising agent while KIO_(3) is not

Answer»

If both assetion and RESON are corect and REASON is CORRECT explanation for ASSERTION
If both assetion and reason are correct but reason is not correct explanation for assertion
If asertion is correct but reason is incorrect
If both assetion and reason are incorrect

SOLUTION :Reason is the correct explantion for asssertion
Discussion
37.

Assetion : An electrochemical cell can be set up only if redox reaction is sponttaneous. Reason: A reaction is spontaneous if free energy change (triangleG) is negative

Answer»

If both ASSETION and RESON are CORECT and reason is correct EXPLANATION for ASSERTION
If both assetion and reason are correct but reason is not correct explanation for assertion
If asertion is correct but reason is incorrect
If both assetion and reason are incorrect

Solution :Reason is correct explanation for assertion.
Discussion
38.

Assertion:The packing efficiency is maximum for the fcc structure. Reason: The coordination number is 12 in fcc structures.

Answer»

ASSERTION and reason both are correct statements and reason is correct explanation for assertion.
Assertion and reason both are correct statements but reason is not correct explanation for assertion
Assertion is correct statement but reason is WRONG statement
Assertion is wrong statement but reason is correct statement.

SOLUTION :Correct explanation. Fcc is a close PACKED structure.
Discussion
39.

Assertion.The bond enthalpy of C-H bondin CH_(4) is nearly 416 k J mol^(-1) Reason . First , second, third and fourth C-H bonds in CH_(4) have same bond enthalpy.

Answer»

If both A and R are TRUE,andR is the trueexplanation of A.
If both A and R are true,but R is not the true explanation of A.
If A is true, but R is false
If both A and R are false.

Answer :C
Discussion
40.

Assertion:Temporary hardness can be removed by the addition of lime Ca(HCO_(3))_(2) in hard water is converted to insoluble CaCO_(3) on moderate heating.

Answer»

Solution :(B) Correct explanation. Lime REACTS with `CA(HCO_(3))_(2)` to from insouble `CaCO_(3)`.
`Ca(OH_(3)_(2)+CaO to 2CaCO_(3)+H_(2)O`
Discussion
41.

Assertion:Oxidation state of oxygen in OF_(2) and Na_(2)O is +2 and -2 respectively.

Answer»

If both ASSERTION and REASON is the CORRECT EXPLANATION of assertion.
If both assertion and reason is the correct explanation of assertion.
If both assertion and reason is the correct explanation of assertion.
If both assertion and reason is the correct explanation of assertion.

Answer :B
Discussion
42.

Assertion:Number of raidal and angular nodes for 3p orbital are1,1 respecticvely. Reason:Number of radial and angular nodes depends only on principle quantum number.

Answer»

both assertion and REASON are true and reason is the CORRECT explanation of assertion.
both assertion and reason are true but reason is not the coreect explanation of assertion.
assertion is true but reason is false
both assertion and reason are false

Solution :asssertion is true but reason is false
No.of radial node=n-l-1
No.of ANGULAR node=lfor3p orbital
No.of angular node=l=1
No. of radial node=n-l-1=3-1-1=1
Discussion
43.

Assertionj: The following structures (I) and (II) canot e the major contributors to the real structure of CH_(3)COOCH_(3) CH_(3)-underset(I)underset(+)overset( : overset(..)(C)^(-): )overset(|)(C)-underset(..)overset(..)(O)harrunderset(II)(CH_(3)-overset( : overset(..)(O)^(-): )overset(|)(C)=underset(..)overset(+)(O)-CH_(3)) Reason: Both the structures involve charge separation and structure (I) contains a carbon atom with an incomplete octet.

Answer»

If both ASSERTION and REASON are TRUE and reason is the CORRECT EXPLANATION of assertion
If both assertion and reason are true but reason is not the correct explanation of assertion
If assertion is true but reason is false.
If both assertion and reason are false.

Answer :A
Discussion
44.

Assertion.If the length of the unit cell of LiCl having NaCl structure is 5.14 Å, the ionic radius of Cl^- ion 1.82 Å Reason. Anion-anion contact is retained in LiCl structure because anions constitute the lattice.

Answer»

If both assertion and reason are true, and reason is the true explanation of the assertion.
If both assertion and reason are true, but reason is not the true explanation of the assertion.
If assertion is true , but reason is false
If both assertion and reason are false.

SOLUTION :If anion-anion contact is retained, then is shown in the adjoininig

Interionic distance of LiCl =`a/2`
`="5.14 Å"/2`=2.57 Å
`BC=sqrt(AB^2+AC^2)=sqrt((2.57)^2+(2.57)^2)`
=3.63 Å
Radius of `CL^-` ion =`1/2BC =1/2xx3.63`=1.81 Å
Hence , R is the CORRECT explanation of A.
Discussion
45.

Assertion::In the electrolysis of water containing 15 to 20 percent H_(2)SO_(4) SO_(4)^(2-) ions are not discharged at anode. Reason: The discharge potentail of SO_(4)^(2-) ions is higher than that OH^(-) ions.

Answer»

SOLUTION :(a) REASON is the CORRECT EXPLANATION for ASSERTION.
Discussion
46.

Assertion.Hexagonal close packing is more closely packed than cubic close packing. Reason.Hexagonal close packing has a coordination number of 12 whereas cubic close packing hasa coordination number of 8

Answer»

If both assertion and REASON are true, and reason is the true explanation of the assertion.
If both assertion and reason are true, but reason is not the true explanation of the assertion.
If assertion is true , but reason is FALSE
If both assertion and reason are false.

SOLUTION :Correct A . Hexagonal close packing and cubic close packing are EQUALLY close packed (space OCCUPIED =74%)
Correct R. Both have a coordination number of 12.
Discussion
47.

Assertion.CsCl has body-centred cubic arrangement Reason.CsCl has one Cs^+ ion and 8 Cl^- ions in its unit cell.

Answer»

If both assertion and reason are true, and reason is the true explanation of the assertion.
If both assertion and reason are true, but reason is not the true explanation of the assertion.
If assertion is true , but reason is FALSE
If both assertion and reason are false.

Solution :Correct R. CSCL has one `Cs^+` ION and one `CL^-` ion in its UNIT cell
Discussion
48.

Assertion:Addition of bromine to trans-2-butene yields meso-2, 3 -dibromobutane. Reason:Addition of bromine to an alkene is an electrophilic addition.

Answer»

Both Assertion and REASON are true and Reason is the CORRECT explanation of Assertion.
Both Assertion and Reason are true and Reason is notcorrect explanation of Assertion.
Assertion is true but Reason is false
Assertion is false but Reason is true.

Answer :B
Discussion
49.

Assertion(A): When is subjected to dehydrohalogenation using NaOEt/EtOH, the product contains no deuterium . Reason(R): E_(2) follows anti elimination.

Answer»

If both A and R are true and R is the correct explanation of A.
If both A and R are true and R is not the correct EXPLANATIO of A.
If A is true but R is FALSE
If A is false and R is true.

Answer :A
Discussion
50.

Assertion(A) : Neutral molecule SnCI_(4) can act as an electrophile. Reason (R) : It has vaccant 'd' orbitals which can accommadate the electrons from others.

Answer»

Both(A) and (R) are CORRECT and (R) is the correct EXPLANATION of (A).
Both (A) and (R) are correct but (R) is not the correct explanation of (A).
(A) is correct but (R) is WRONG
(A) is wrong but (R) is correct

Solution :Both(A) and (R) are correct and (R) is the correct explanation of (A).
Discussion