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.

Determine which of the following reactions at constant pressure represent systems that do work on the surrounding environment I. Ag^(+) (aq) + Cl^(-) (aq) rarr AgCl(s) II. NH_(4)Cl(s) rarr NH_(3) (g) + HCl (g) III. 2NH_(3)(g) rarr N_(2)(g) + 3H_(2)(g)

Answer»

I
III
II and III
I and II

Solution :W by system `gt 0 rArr` expansion `rArr DELTA NG gt 0 rArr II, III`
Discussion
2.

Determines the Miller indices of the shaded plane. Coordinates of the corners of the plane

Answer»

SOLUTION :A PLANE parallel to the plane between (0,0,0) , (1,1,0) and (0,1,1) will intersect the AXES at x=a, y=-b and z=c
Thus, WEISS indices are 1,-1, 1
Reciprocals of Weiss indice are 1,-1, 1
Hence, Miller indices are `(1bar1 1)` or `(bar1 1 bar1)`
Discussion
3.

Determine which of the following compounds is a trans isomer.

Answer»




Solution :`{:("POSITION",,"CIS",,"TRANS"),("1, 2",,"ae/ea",,"aa/ee"),("1, 3",,"aa/ee",,"ae/ea"),("1, 4",,"ae/ea",,"aa/ee"):}`
Discussion
4.

Determine which of the following compounds is a trans isomer ?

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SOLUTION :
Discussion
5.

Determine whether each of the followings compounds is a cis isomer or a trans isomer.

Answer»


Answer :(a) CIS (b) cis (c) cis (d) TRANS (E) trans (f) trans
Discussion
6.

Determine whether each of the followings compounds is a cis isomer or a trans isomer.

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ANSWER :(a) cis (b) cis (C) cis (d) TRANS (e) trans (F) trans
Discussion
7.

Determine whether each of the following alkenes can exist ac c is trans isotmers (a) 1 chloropropene (b) 2-chloropropene

Answer»

Solution :
in 2- chloropropeneit DEVIATES fromthe rulethat is`At ` LEASTONE samegroupis presenton the twobondedcarbonatom "Therefore-2-chloropropenecannotexistas CIS- trainsisomers.
Discussion
8.

Determine the volume of M//8 KMnO_(4) solution required to react completely with 25.0 cm^(3) of M//4 FesO_(4) solution in acidic medium

Answer»

Solution :The balanced ionic equation for the reaction is `MnO_(4)^(-)+5Fe^(2+)+8 H^(+) rarr Mn^(2+)+5Fe^(3+)+4 H_(2)O`
From the balced equation it is evident that 1 mole of `KmnO_(4)=5` moles of `FeSO_(4)`
Applying molarity eqaution to the balced redox equation we have
`(M_(1)V_(1))/(n_(1))(KMnO_(4))=(M_(2)V_(2))/(n_(2))(FeSO_(4)) or (1xxV)/(8xx1)=1/4xxx25/5 or V_(1)=(1xx25xx8)/(4xx5)=10.0 cm^(3)`
THUS th volume of `M/8 KnnO_(74)` solution required =10.0 ML
Discussion
9.

Determine the volume (in mL) of dilute nitric acid (d=1.08 g mL^(-1), 19% HNO_(3) by wt.) that can be prepared by diluting 48 mL of conc. HNO_(3) (d=1.44 g mL^(-1), 76% by wt.).

Answer»


ANSWER :256
Discussion
10.

Determine the volume of M/10 KMnO_4 solution required to react completely with 25.0 mL of M/5 oxalic acid solution.

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SOLUTION :The balanced CHEMICAL equation is
`2KMnO_4+5(COOH)_2+3H_2SO_4rarr K_2SO_4+2MnSO_4+10CO_2+8H_2O`
From the balanced equation , it is CLEAR that
2 mol of `KMnO_4 = 5 "mol of "(COOH)_2`
`[(M_1V_1)/n_1]_(KMnO_4)=[(M_2V_2)/n_2]_((COOH)_2)`
`1/10xxV_1/2=1/5xx(25.0)/5`
or `V_1=(25.0xx10xx2)/(5xx5)=20.0mL`
Discussion
11.

Determine the values of all the four quantum numbers of the 8^(th) electron in o-atom and 15^(th) electron in Cl atom.

Answer»

Solution :(i) `o(Z=8)1s^(2)2s^(2)2p_(x)^(2)2p_(y)^(1)2p_(z)^(1)`
Four quantum numbers for `2o_(z)^(1)` electron in or=xygen ATOM.
n=principal quantum number=2
l=azimuthal quantum number =1
m=magnetic quantum number=+1
s=spin quantum number`=-(1/2)`
(ii)` Cl(Z+17)1s^(2)2s^(2)2p_(x)^(2)2p_(y)^(1)2p_(z)^(1)`
Four quantum numbers for `15^(TH)`electron in CHLORINE atom:
`n=3,l=1,m=+1,s=+(1/2)`
Discussion
12.

Determinethe value of DeltaH and DeltaU for the reversible isothermal evaporationof 90.0 gof water at 100^(@)C . Assume that water vapour behave as an ideal gas and heatof evaporationof water is 540 cal g^(-1)( R = 2.0 cal mol^(-1)K^(-1))

Answer»


Solution :`DeltaH = 540 xx 90 cal = 48600 cal, DeltaU = DeltaH - Deltan_(G)RT`,
`90 g H_(2) O = 5 ` MOLES , `Deltan_(g) = 5 ` for MOLESOF `H_(2)O(l) rarr 5` moles of `H_(2)O(g)`
Discussion
13.

Determine the values of all the four quantum numbers of the 8^(th) electron in O-atom and 15^(th) electron in CI atom .

Answer»

SOLUTION :Electronic configuration of oxygen=

`:. 8^(th)` electron PRESENT in `2p_(x)` ORBITAL and the quantum number are
`n = 2, l = 1 , m_1 = "either" + 1 "or" - 1 and s = -1//2`
Electronic configuration of chlorine =

`15^(th)` electron present in `3P_z` orbital and the quantum numbers are `n = 3, l = 1, m_1= ` either + 1or - 1 and ms = + 1//2.
Discussion
14.

Determine the solubilities of silver chromate, barium chromate, ferric, hydroxide, lead, chloride and mercurous iodide at 298 K. Given K_(sp) values : Ag_(2)CrO_(4)=1.1xx10^(-12), BaCrO_(4)=1.2xx10^(-10), Fe(OH)_(3)=1.0xx10^(-38), PbCl_(2)=1.6xx10^(-5), Hg_(2)I_(2)=4.5xx10^(-29) Determine also the molarities of individual ions.

Answer»


Solution :`{:(Ag_(2)CrO_(4)HARR 2"Ag"^(+)+CrO_(4)^(2-),K_(sp)=(2s)^(2)(s)=4s^(3)"," :. s^(3)=K_(sp)//4,,),(s""2s"" s""=(1.1xx10^(-12))//4=0.275xx106(-12)=2.75xx10^(-13),,):}`
`:. 3 log s = log (2.75xx 10^(-13))=-13+0.4393=-12.5607`
or` log s = - 4.1869 = bar(5).8131`
`:. s= 6.5xx10^(-5) ` mol `L^(-1)`
`[Ag^(+)]=2xx6.5xx10(-5)=13.0xx10^(-5)M = 1.30 xx 10^(-4)M`
`[CrO_(4)^(2-)] = 6.5xx10^(-5)M`
For other salts, PROCEED exactly in similar manner .
Discussion
15.

Determine the solubilities of silver chromate , barium chromate , ferric hydroxide , lead chloride and mercurous constants given in Table 7.9 . Determine also the molarities of individual ions. (i)K_(sp)(Ag_2CrO_4)=1.1xx10^(-12) (ii)K_(sp)(BaCrO_4)=1.2xx10^(-10) (iii)K_(sp)(Fe(OH)_3) = 1.0xx10^(-38) (iv)K_(sp) (PbCl_2) = 1.6xx10^(-5) (v) K_(sp) (Hg_2Cl_2)=1.3xx10^(-18) (vi)K_(sp) (Hg_2I_2)=4.5xx10^(-29)

Answer»

Solution :(i)Solubilities of `Ag_2CrO_4` :
`{:(,Ag_2CrO_(4(s)) to , 2Ag_((aq))^(+) + , CrO_4^(2-)),("Molarity of equili. :",S,2S,S):}`
`K_(sp)=[Ag^+]^2[CrO_4^(2-)]`
`therefore 1.1xx10^(-12) = (2S)^2 (S) =4S^3`
`therefore S=((1.1xx10^(-12))/4)^(1/3)`
`=0.6503xx10^4 "MOL L"^(-1)`
`=6.503xx10^(-5) "mol L"^(-1)`
So, `[Ag^+]=2S=2 (6.503xx10^(-5))=1.3xx10^(-4)` M
`[CrO_4^(2-)]=S=6.503xx10^(-5) "mol L"^(-1)`
(ii)Solubilities of `BaCrO_4` :
`{:(BaCrO_(4(s)) to , Ba_((aq))^(2+)+, CrO_(4(aq))^(2-)),("Molarity at equilibrium",S,S):}`
`K_(sp)=[Ba^(2+)]=[CrO_4^(2-)]=(S)(S)=S^2`
`therefore S=sqrtK_(sp)=sqrt(1.2xx10^(-10))=1.095xx10^(-5)` M
`therefore [Ba^(2+)]=[CrO_4^(2-)] =S=1.095xx10^(-5)` M
(iii) Solubilities of `Fe(OH)_3` :
`{:("Equilibrium", Fe(OH)_(3(s)) hArr, Fe_((aq))^(3+) + , 3OH_((aq))^(-)),("Solubility/ At equilibrium M", "S M","S M","3 S M"):}`
`K_(sp)=[Fe^(3+)][OH^-]^3 = (S)(3S)^3`
`therefore 1.0xx10^(-38) = 27S^4 = 100xx10^(-40)`
`therefore S=((100xx10^(-40))/27)^(1/4)`
`=(3.704xx10^(-40))^(1/4)`
`=1.3872xx10^(-10) approx 1.39xx10^(-10) "mol L"^(-1)`
So, `[Fe^(3+)]=S=1.39xx10^(-10) "mol L"^(-1)`
`[OH^-]=3S=3(1.39xx10^(-10))`
`=4.17xx10^(-10) "mol L"^(-1)`
(iv) Solubilities of `PbCl_2` :
`{:(PbCl_2 hArr, Pb_((aq))^(2+) + , 2Cl_((aq))^(-)),("S M", "S M" , "2S M"):}`
`K_(sp)=[Pb^(2+)][Cl^-]^2 = (S) (2S^2)=4S^3`
`therefore S=(K_(sp)/4)^(1/3) = ((1.6xx10^(-5))/4)^(1/3)= (16/4 xx 10^(-6))^(1/3)`
`therefore S=1.5874xx10^(-2)` M
So, `[Pb^(2+)]=S=1.5874xx10^(-2) "mol L"^(-1)`
`[Cl^-] = 2S=2 (1.5874xx10^(-2))`
`= 3.1748xx10^(-2) "mol L"^(-1)`
(v)Solubilities of `Hg_2Cl_2` :
`{:(Hg_2Cl_(2(s)) hArr, Hg_(2(aq))^(2+) + , 2Cl_((aq))^(-)),(S, "S M", "2S M"):}`
`therefore S=(K_(sp)/4)^(1/3) = (1.3/4 xx 10^(-18))^(1/3)`
`=0.6876xx10^(-6) M = 6.876xx10^(-5)`
So, `[Hg_2^(2+)]=S=6.876xx10^(-5)` M
`[Cl^-]=2S=2(6.876xx10^(-5))`
`=1.375xx10^(-4)` M
(vi) Solubilities of `Hg_2I_2` :
`{:(Hg_2I_(2(s)) hArr , Hg_(2(aq))^(2+) + , 2I_((aq))^(-)),(S, "S M", "2S M"):}`
`K_(sp)=[Hg_2^(2+)][I^-]^2 = (S)(2S)^2 =4S^3`
`therefore S=root3(K_(sp)/4)=((4.5xx10^(-29))/4)^(1/3)=(11.25xx10^(-30))^(1/3)`
`=2.241xx10^(-10) "mol L"^(-1)`
`therefore [Hg_2^(2+)]=(S)=2.241xx10^(-10) "mol L"^(-1)`
`[I^-]^2 =2(S)=2(2.41xx10^(-10))`
`= 4.82xx10^(-10) "mol L"^(-1)`
Discussion
16.

Determine the shapes of the given molecules or ions: (1) POCl_(3) (2) CH_(3)^(-)

Answer»

SOLUTION :(1) P-atom in `POCl_(3)` molecule is attached to the O-atom by a double bond and to three Cl-atoms by three SINGLE bonds. The number of electrons in the valence shell of P-atom =5+2+1+1+1=10. out of these 10 electrons or 5 electron pairs, 1 electron pair is involved in formation of a `pi`-bond which have no role in determining the SHAPE of the molecule. according to VSEPR theory, four electron pairs are oriented tetrahedraly, i.e., shape of the molecule is tetrahedral.
(2) C-atom in `CH_(3)^(-)` ion is bonded to three H-atoms by three single bonds. the number of electrons in the valence shell of C-atom =4+3+1 (for the negative CHARGE)=8. out of these 8 electrons or 4 electron pairs, there are three bond-pairs and one lone-pair. According to VSEPR theory, the ion is pyramidal.
Discussion
17.

Determine the probable formula of an acid salt which is an oxidising agent from the following data.Its equivalent weight as an acid is 390 and as an oxidising agent is 32.5. It contains 10% of potassium and 65% of iodine.

Answer»

Solution :The acid salt has an equivalent weight `=390`
`10%` of this is `39g-=1` mole of potassium. Hence the simplest formula contains one K and one H. Also 390 g
Contains `65%` of iodine i.e., `(390xx65)/(100)g=253.5g`
Since the atomic weight of iodine is 127, the simplest formula is `KHI_(2)O_(x).nH_(2)O`.
Further. `390 g` correspond to `(390)/(32.5)=12`, equivalents of oxidising AGENT since `IO_(3)^(ɵ)+6H^(o+)+6e^(-)I^(ɵ)+3H_(2)O`,
one `IO_(3)^(ɵ)-=6` equivalents
The formula is `KHI_(2)O_(6).nH_(2)O`
molar mass `=[390+18n]=390impliesn=0`
Therefore, the probeble formula`=KH(IO_(3))_(2)`
Discussion
18.

Determine the precentage of water of crystallisation in pure sample of green vitriol (FeSO_(4).7H_(2)O)

Answer»


Solution :MOLECULAR mass of `FeSO_(4).7H_(2)O=56+32+64+126=278.0u`
`:.` Percentage of WATER of CRYSTALLISATION `= ((126u))/((278u))xx100=45.32%`.
Discussion
19.

Determine the pH of the solution that results from the addition of 20.00 mL of 0.01 M Ca (OH)_(2)to 30.00 mL of 0.01 M HCl

Answer»

`11.30`
`10.53`
`2.70`
`8.35`

Solution :`20.0 ML "of" 0.01 M CA (OH)_(2) "has" Ca(OH)_(2)`
`=20xx0.01` MILLIMOLES = 0.2 millimole
`:. OH^(-)` ions present `=2xx0.2` millimole
`=0.4` millimole
30.0 mL of 0.01 M HCl has HCl
`=30xx0.01` millimoles = 0.3 millimole
`:. H^(+)` ions present = 0.3 millimole
0.3 millimole `H^(+)` will neutralize 0.3 millimoe of `OH^(-)`. Hence, `OH^(-)` ions present after mixing = 0.1 millimole
Volume of solution = 20 +30 = 50 mL
Hence, molarity of `OH^(-)` ions `=(0.1)/(50)=2xx10^(-3)M`
`pOH = - LOG (2xx10^(-3))=3-0.30=2.70`
`:. pH = 14 - 2.70 = 11.30`
Discussion
20.

Determine the oxidtion number of O in the following : OF_(2),Na_(2)O_(2),Na_(2)O,KO_(2),KO_(2),KO_(3) and O_(2)F_(2)

Answer»

SOLUTION :O.N ofO=+2 in `OF_(2)-1 in Na_(2)O_(2)-2 in Na_(2)O,-1//2 in KO_(2),-1//3 in KO_(3) and +1 in O_(2)F_(2)`
Discussion
21.

Determinethe oxidation number of followingelements given in bodl letters. (i)CuH, (ii) Na_(2)S_(3)O_(6) (iii) N_(2)O, (iv) Ba_(2)XeO_(6) (v) C_93)O_(2), (vi) V(BrO_(2))_(2) (vii) Ca(clO_(2))_(2), (viii) Cs_(4)Na(HV_(10)O_(28)) (ix) LiAlH_(4) (x) K[Co(C_(2)O_(4))_(2)(NH_(3))_(2)] (xi) [Ni(CN)_(4)]^(2), (xii) Na_(2)S_(2) (xiii) Fe(CO)_(5), (xiv) [OCN]^(-) (xv) S_(2)O_(4)^(2-), (xvi) S_(2)O_(8)^(2-)

Answer»


Answer :(i) `-1`, (II) `+10//3`, (iii) `+1`, (iv) `+8`, (v) `+4//3`, (vi) `+2`, (vii) `+3`, (VIII) `+5`, (ix) `+3`, (X) `+3`, (xi) `+2`, (xii) `-1`, (xiii) `0`, (xiv) `-2,+4,-3`, (xv) `+3`, (XVI) `+6`
Discussion
22.

Determine the oxidation number of following underline elements: a. HCulN b. Hul(N)C c. Hul(N)O_(3) d. KulO_(2) e. ul(Fe_(3))O_(4) f. Kul(I_(3)) g. ul(N_(3))Hh. ul(Fe)(CO)_(5) i. ul(Fe_(0.94))O j. ul(N)H_(2)NH_(2) k. ul(Fe)SO_(4)(NH_(4))_(2)SO_(4)6H_(2)O l. ul(N)OCI m. NOul(Cl)O_(4) n. Na_(2)[ul(Fe)(CN)_(5)NO] o. [ul(Fe)(NO)(H_(2)O)_(5)]SO_(4) p. Na_(ul(2S_(4))O_(6) q. (CH_(3))_(2)ul(SO) r. Naul(S_(2))O_(3) s. CaOul(Cl_(2)).

Answer»

Solution :a. `HCN`: The evaluation cannot be made directly in some cases, e.g., `HCN`, by using rules proposed EARLIER as we have no rule for the evaluation of oxidation number of both `N` and `C`. In all such cases, evaluation of oxidation number should be made using indirect concept or using fundamentals by which following rules have been formed:
i. Each covalently bond contribures one unit of oxidation number.
ii. Covalently bonded atoms with less electronegativity acquire positive oxidation number whereas other atoms with more electronegativity acquire negative oxidation number.
iii.In case of coordinate bond, give `+2` value for oxidation number to the atom from which coordinate bond is directed to a more electronegative atom.
If coordinate bond is directed from a more electronegative to a less electronegative atom, then neglect the contribution of coordinate bond of both atoms in which coordinate bond exists.
Thus, `H-C-=N`
Three bond `N` atom implies more electronegative
`1+a+3(-1)=0`
Oxidation number of `N=3(-1)= -3`
Oxidation number of `C, a= +2`
b. `Hul(N)C: H-C-=N`
Oxidation number of `H= +1`
Oxidation number of `N`
`{:("[=-2","+(-1)" ,"+0]=-3"),({:("[For covalent"),("bond with C]"):},{:("[For covalent"),("bond with H"):},{:("[No contribution"),("for coordinate"),("bond]"):}):}`
According to fundamental concept `= -3`
`:. 1 + (-3) +a = 0 rArr a = +2`
c. `underline(N)O_(3)` : By rules , `1 + a + 3(-2) = 0 rArr a=+5`
By fundamental approach
`+N3`
Oxidation number of `H = +1`
Oxidation number of `N`
`{:("=+1","+(+2)" ,"+(+2)=+5"),({:("[ covalent"),("bond with O]"):},{:("[Two covalent"),("bond with O,"),("N being less"),("electronegative"),("than O]"):},{:("[coordinate"),("bond]"):}):}`
d. `Kunderline(O_(2))`: A superoxide of `K`
Oxidation number of `K = +1`
Oxidation number of `O = a`
`1+2(a) = 0` and `a = -1//2`
e. `underline(Fe_(3))O_(4)`: `3(a) + 4(-2) = 0 rArr a = +8//3`
`Fe_(3)O_(4)` is a mixed oxide of `FeO.Fe_(2)O_(3)`.
Therefore, `Fe` has two oxidation numbers `+2` and`+3`, respectively. However, factually speaking, oxidation number in `Fe_(3)O_(4)` is an average of two values, i.e., `+2` and `+3`.
Therefore, average oxidation number `= (+2+(xx3))/(3) = +(8)/(3)`
`underline(I_(3))`: `1+3(a) = a rArr a = -1//3`
Since `KI_(3)` is `KI + I_(2)`, therefore, `I` has two oxidation numbers `-1` and `0`, respectively.
However, oxidation number of `I` and `KI_(3)` is an average of two values `-1` and `0`.
Therefore, average oxidation number
`= (-1 + 2(0))/(3) = -(1)/(3)`
g.`underline(N_(3))H`:`3(a) + 1 = 0 rArr a = -1//3`
h. `underline(Fe)(CO)_(5)`: Sum of oxidation number of `CO = 0`
`:. a+5(0) = 0 rArr a = 0`
i. `underline(Fe_(0.94))O`: `0.94(a) + (-2) = 0 rArr a = 200//94`
j. `underline(N)H_(2)underline(N)H_(2)`: Both `N` have same nature, therefore each `n` has oxidation number `-2`.
`underline(Fe)SO_(4)(NH_(4))_(2)SO_(4).6H_(2)O`:
Oxidation number of `Fe = a`
Sum of oxidation number for `(NH_(4))_(2)SO_(4) = 0`
Sum of oxidation number of `H_(2)O = 0`
Sum of oxidation number of `SO_(4)^(2-) = -2`
`:. a + (-2) + 0 + 6(0) = 0 rArr a = +2`
l. `underline(N)OCl`: `Cl-N=O` or use `NO^(o+)Cl^(Ө)`
Oxidation number of `N =- +1` (for covalent bonds with `Cl`)
Oxidation number of `N = +2` (for two covalent bonds with `O`)
Therefore, oxidation number of `N` in `NOCl = +3`
m. `NOunderline(Cl)O_(4)`: The COMPOUND may be written as `NO^(+)ClO_(4)^(Ө)` for `ClO_(4)^(Ө)`.
For `ClO_(4)^(Ө)`, let the oxidation number of `Cl =a`.
`:. a + 4(-2) = -1rArr a = +7`
n. `Na_(2)[underline(Fe)(CN)_(5)NO]` : `NO`in ion complex has `H^(o+)` nature
`:. 2 xx 1 + [a + 5(-1) + (+1)] = 0`
`:. a = +2`
o. `[underline(Fe)(NO)(H_(2)O)_(5)]SO_(4)`:
`a + 1 + 5 xx 0 + (-2) = 0`
`rArr a = +1`
p. `Na_2S_(4)O_(6)`: `2(+1) + 4a + 6(-2) = 0`
`:. a = +5//2`
Here also, this value is the average oxidation number of `S`. The structure of `Na_(2)S_(4)O_(6)` is
`Na-O-underset(O)underset(darr)overset(O)overset(uarr)(S)-S-S-underset(O)underset(darr)overset(O)overset(uarr)(S)-O-Na`
Thus, oxidation nuber of each `S` atom formingcoordiante bond is `+5`, whereas oxidation number of each `S` atom involved in pure covalent bonding is zero. Therefore, average oxidation number
`= (+5+5+0+0)/(4) = +(5)/(2)`
q. Dimethyl sulphoxide or `(CH_(3))_(2)underline(S)O`:
Oxidation number of `CH_(3) = +1`
Oxidation number of `O = -2`
`:. 2(+1) + a+(-2) = 0 rArr a = 0`
r. `Na_(2)underline(S_(2))O_(3)`: `2 xx 1 + 2 xx a + 3(-2) = 0 rArr a = +2`
Here too, it is the average oxidation number.
The sturcture of `Na_(2)S_(2)O_(3)` is
`Na-S-underset(O)underset(darr)overset(O)overset(uarr)(S)-O-Na`
The oxidation number of `S` involved in coordinate bond, i.e., DONOR `S` atom, is `+5`. The oxidation number of other `S` atom is `-1`.
s. `Caunderline(Cl_(2))`: In bleachingpowder, two `Cl` atoms are as `Ca(Cl)Cl`, i.e., one as `Cl^(Ө)` having oxidation number `-1` and other as `OCl^(Ө)` having oxidation number `+1`.
Discussion
23.

Determine the oxidation number of the underlined elements in the following species. ul(Si)H_(4),ul(B)H_(3),ul(B)F_(3),ul(Br)O_(4)^(-) and Hul(P)O_(4)^(2-)

Answer»

SOLUTION :`-4,-3,+3,+,7,+5`
Discussion
24.

Determine the oxidation numbr of C in the following : C_(2)H_(6),C_(4)H_(10),CO,cO_(2)and HCO_(3)^(-)

Answer»

Solution :O.N of C =-3 in `C_(2)H_(6)-2.5 in C_(4)H_(10)+2 in CO+4 in 4 in CO_(2) and +4 in HCO_(3)^(-)`
Discussion
25.

Determine the oxidation number of the element in the bold in the following species (i)SiH_(4),(ii)BH_(3),(iii)Bf_(3),(iv)S_(2)O_(3)(2-),(v)BrO_(4)^(-),(vi)HPO_(4)^(2-)

Answer»

SOLUTION :`(i) SiH_(4) :overset(x)Sioverset(+1)H_(4) X+4(+1)=0or x=-4`
`(II)BH_(3) :overset(x)Boverset(+1)H_(3) X+3(+1)=0or x=-3`
`(iii)Bf_(3) : overset(x)Boverset(-1)F_(3)X+3(-1)=0 or X=+3 `
`(iv)S_(2)O_(3)^(2-) : overset(x)S_(2)overset(-2)O_(3)2(x)+(-2)=-2OR X=+7`
`(v) BrO_(4)^(-): overset(x)Broverset(-2)O_(4)X=4(-2)=-1or X=+7`
`(vi)HPO_(4)^(2-) : overset(+1)Hoverset(x)Poverset(-2)O_(4)(+1)+X+4(-2)=-2 or X=+5`
Discussion
26.

Determine the oxidation number of the atom in bold in the following species : BH_3,BF_3,BrO_4^(-) , HPO_4^(2-),S_2O_3^(2-),SiH_4

Answer»


Answer :`-3,+3,+7,+5,+2,-4`
Discussion
27.

Determine the oxidation number of O in the following: OF_(2),Na_(2)O_(2) and CH_(3)COOH.

Answer»

SOLUTION :`+2,-1,0`
Discussion
28.

Determine the oxidation number of O in the following: Na_2O_2 and CH_3COOH

Answer»

SOLUTION :OXIDATION NUMBER of O=+1 in `Na_2O_2` and -2 in `CH_2 COOH`.
Discussion
29.

Determine the oxidation number of Cl in HCl , HClO, ClO_4^(-) and Ca (OCl) Cland ClO_2 .

Answer»


ANSWER :`-1,+1,+7,0,(+2-2-2x=0:. X=0)+4`
Discussion
30.

Determine the oxidation number of all the atoms in the following well known oxidants KMnO_(4),K_(2)Cr_(2)O_(7) and LiAIH_(4)

Answer»

Solution :O.N OFS CHANGES FORM -2 in `H_(2)S` and +4 in `SO_(2)` to ZERO in elemental suphur
Discussion
31.

Determine the oxidation number of C in the following: CO,CO_(2),HCO_(3)^(-),C_(2)H_(6),C_(4)H_(10).

Answer»

SOLUTION :`+2,+4,+4,-3,-2.5`
Discussion
32.

Determine the oxidation number of C in the following : C_2H_6,C_4 H_10,CO,CO_2 and HCO_3

Answer»

Solution :OXIDATION number of C=-3 in `C_2H_6,-2.5 i N C_4H_10+2 i n CO,+4 i n CO_2and+4i n HCO_3^-`
Discussion
33.

Determine the oxidation number of all the atoms in KClO_4.

Answer»

SOLUTION :K=+1, Cl=+7, O= -2.
Discussion
34.

Determine the number of moles of Agl which may be dissolved in 1.0 litre of 1.0M CN^(-)solution ,K_(sp)for Agl and K_ffor [ Ag (CN)_2]^(-)" are "1.2 xx 10 ^(-17)M^(2) and 7. 1 xx 10 ^(-19)respectively. If answer is 4.9 xx 10 ^(x)then x =__________?

Answer»


Solution :`Agl hArr Ag^(+) I^(-),KSP`
` (Ag^(+) +2CN^(-)hArr [Ag(CN)_2]^(-),K_f)/( underset( 1-2s) (Agl)+2CN^(-)hArr [Ag(CN)_2]^(-)+I^(-),K =KspK_f) `
` ((s)/(1-2s) )^(2)=Ksp, K_f =1.2 xx 10^(-17)xx 7.1 xx 10 ^(19) `
` rArr (s)/( 1-2s)=SQRT(8.52 xx 10 ^(2)) =29`
` 59S =29 rArr S =0.49 rArr S = 4.9 xx 10 ^(-1) `
Discussion
35.

Determine the number of moles of AgI which may be dissolved in 1.0 M CN^(-) solution. K_(sp) for AgI and K_(c)for Ag(CN)_(2)^(-) are 1.2xx10^(-7) M^(2) and 7.1xx10^(19) M^(-2) respectively.

Answer»

Solution :Suppose number of moles of AGI which may be dissolved in 1.0 litre of 1.0 M `CN^(-)` solution = x. Then
`{:(,AgI (s) ,+ ,2CN^(-) (AQ) ,HARR,[Ag(CN)_(2)]^(-),+,I^(-)),("Initial moles " ,x,,1,," "0,,0),("Moles after reaction",0,,1-2x,," "x,,x),(,,,,,,,),(,,,,,,,):}`
`K_(eq)=([Ag(CN)_(2)]^(-)[I^(-)])/([CN^(-1)]^(2))=(x^(2))/((1-2x)^(2))`...(i)
Further,`AgI (s) hArr Ag^(+) (aq) + (Aq) + I^(-) (aq)`
`K_(sp) = [Ag^(+)][I^(-)] = 1.2xx10^(-17) ` (Given)...(ii)
`Ag^(+) (aq) + 2 (CN^(-) (aq)) hArr [Ag(CN)_(2)]^(-)`
`K_(c)=([Ag(CN)_(2)]^(-))/([Ag^(+) ] [ CN^(-) ] ^(2) ) = 7.1xx10^(19) ` (Given)
From eqn. (i) , (ii) and (iii)
`K_(eq)=K_(sp)xxK_(c)=(1.2xx10^(-17))xx(7.1xx10^(19))= 8.52xx10^(2)`
`:. (x^(2))/((1-2x)^(2)) = 8. 52xx10^(2) or (x)/(1-2x) = 29.2 or x = 29.2- 58.4x or x = 0.49 ` mole
Discussion
36.

The simultaneous solubility of Ag CN (K_(f)=2.5 xx 10^(-16)) and AgCl (Ksp =1.6 xx 10^(-10 ) )in 1. 0M NH_3 (aq)are respectively :[Given ,K_f [Ag(NH_3)_2^(+) ]= 10^(7) ]

Answer»


Solution :`AGL hArr Ag^(+) I^(-),Ksp`
` (Ag^(+) +2CN^(-)hArr [Ag(CN)_2]^(-),K_f)/( underset( 1-2s) (Agl)+2CN^(-)hArr [Ag(CN)_2]^(-)+I^(-),K =KspK_f) `
` ((s)/(1-2s) )^(2)=Ksp, K_f =1.2 XX 10^(-17)xx 7.1 xx 10 ^(19) `
` rArr (s)/( 1-2s)=sqrt(8.52 xx 10 ^(2)) =29`
` 59S =29 rArr S =0.49 rArr S = 4.9 xx 10 ^(-1) `
Discussion
37.

Determine the molecular formula of the compound which contains H = 6.67 %, C = 40 % and the rest is oxygen. 0.6 g of the compound occupy 224 cc at N.T.P

Answer»


Solution :STEP I. Percentage of oxygen `= 100 - (6.67 + 40) = 100-46.67 = 53.33`
Step II. Empirical formula of the compound
`{:("Element","Percentage","Atomic mass","Gram atoms (Moles)","Atomic ratio (Molar ratio)","SIMPLEST WHOLE no. ratio"),("C",40.00,12,(40.0)/(12)=3.33,(3.33)/(3.33)=1.0,1),("H",6.67,1,(6.67)/(1)=6.67,(6.67)/(3.33)=2.0,2),("O",53.33,16,(53.33)/(16)=3.33,(3.33)/(3.33)=1.0,1):}`
Empirical formula of the compound `= CH_(2)O`
Step III. Molecular formula of the compound
224 cc of the vapours of the compound at N.T.P occupy mass = 0.6 g
22400 cc of the vapours of the compound at N.T.P occupy mass `= 0.6//224 xx 22400 = 60.0 g`
`:.` Molecular mass of the compound = 60 g = 60 u
Empirical formula mass `= 12 + 2 xx 1 + 16 = 30 u , n = ("Molecular mass")/("Empirical formula mass")=((60u))/((30u))=2`
`:.` Molecular formula `= 2 xx CH_(2)O=C_(2)H_(4)O_(2)`.
Discussion
38.

Determine the molecular formula of an oxide of iron in which the mass per cent of iron and oxygen are 69.9 and 30.1 respectively.

Answer»



Solution :
Empirical formula `=Fe_(2)O_(3)`
`=2xx56+3xx16`
`=160`
`n=("MOLECULAR mass")/("Empirical formula mass")`
`=(160)/(160)=1`
Molecular formula `=n xx ` emplirical formula
Molecular formula `= n xx` emplirical formula
`=1xx(Fe_(2)O_(3))`
`=Fe_(2)O_(3)`
Discussion
39.

Determine the molecular formula of an oxide of iron in which the mass per cent of iron and oxygen are 69.9 and 30.1 respectively. Given that the molar mass of the oxide is 159.8 g mol^(-1).

Answer»

Solution :For calculation of empirical FORMULA, Empirical formula (`Fe_2O_3`) mass = (`2 XX 55.85) + (3 xx 16.00) = 159.7 g mol^(-1)`
`N=("Molar mass")/("Empirical formula mass") = 159.8/159.7=1`
Here, MOLECULAR formula of the given OXIDE = `Fe_(2)O_(3)`.
Discussion
40.

Determine the miller indices of the shaded plane. Coordinates of the corner of the plane

Answer»

SOLUTION :A PLANE parallel to the plane between ( 0,0,0) , ( 1,1,0) and ( 0,1,1)will interest the axes at X =a, y = -b , and c =cthus, wesis indics are 1, -1 and 1 .
Resciprocals of Weiss indices are 1,-1 and 1,
Hence, MILLER indices are `( 1 bar1 1 ) or (bar1 1 bar1) `
Discussion
41.

Determine the following for the fourth shell of an atom. (d) The maximum number of electrons that can be contaied in each subshell.

Answer»

Solution :Each orbital can contain a maximum of two electrons.
4s - 1 orbital - 2 electrons
4p - 3 orbital - 6 electrons
4d - 5 orbital - 10 electrons
4s - 7 orbital- 14 electrons.
Discussion
42.

Determine the following for the fourth shell of an atom. (c ) The number of orbitals in each subshell.

Answer»

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Solution :The number of ORBITALS in s,p, d and F subshell is 1,3,5 and 7 respectively.
Discussion
43.

Determine the following for the fourth shell of an atom. (a) The number of subshells.

Answer»

Solution :The number of SUBSHELLS is the same as the number USED to designate the SHELL ie., n = 4 so contains 4 SUB shells.
Discussion
44.

Determine the following for the fourth shell of an atom. (b) The designation for each subshell.

Answer»

SOLUTION :The subshells in the FOURTH SHELL are designated as 4s. 4P . 4d and 4f.
Discussion
45.

Determine the empirical formula of an oxide of iron which has 69.9% iron and 30.1% oxygen by mass.

Answer»

Solution :
`:.` The EMPIRICAL FORMULA is `Fe_(2)O_(3)` .
Discussion
46.

Determine the empirical formula of an oxide of iron, which has 69.9% iron and 30.1% dioxygen by mass.

Answer»

SOLUTION :CALCULATION of EMPIRICAL formula:

`THEREFORE` Empirical formula `=Fe_(2)O_(3)`.
Discussion
47.

Determine the empirical formula of an oxide of Iron which has 69.9% Iron and 30.1% dioxygen by mass. [Atomic mass of Fe=56, O=16]

Answer»

SOLUTION :
Discussion
48.

Determine the empirical formula of an oxide of iron which has 69.9% iron and 30.1% dioxygen by mass.

Answer»



Solution :
EMPIRICAL formula `=Fe_(2)O_(3)`
Discussion
49.

Determine the empirical formula of an oxide of iron which contains 69.9%iron and 30.1% oxygen by mass.

Answer»

SOLUTION : EMPIRICAL FORMULA =`Fe_2O_2`
Discussion
50.

Determine the empirical formula of a compound containing K=24.75%, Mn =34.77% and rest is oxygen.

Answer»

SOLUTION :
EMPIRICAL formula of the COMPOUND = `KMnO_(4)`
Discussion