100525 + Interview Questions in GENERAL QA IN GENERAL AWARENESS Page 47 InterviewSolution

2301.	Find the sum of first 10 multiples of 6.
Answer» And 55x6= 330

Discussion

2302.	A thin conducting rod of length `l=5m` is moved such that its end B moves along the X-axis while end A moves along the Y-axis A uniform magnetic field `B=6hat(k)` T exists
Answer» Correct Answer - `21` `varepsilon=vec(B)(vec(V)_(cm)xxvec(L))` `=(6hat(k))((3/2hat(i)-4/3hat(j))xx(4hat(i)+3hat(j)))=21`

Discussion

2303.	A ring of mass M and radius R is at rest at the top of an incline as shown. The ring rolls down the plane without slipping. When the ring reaches bottom, its angular momentum about its centre of mass is: A. `MR sqrt(gh)`B. `MR sqrt((gh)/(2))`C. `MR sqrt(2gh)`D. none
Answer» Correct Answer - A By work - energy theorem, `Mgh = (1)/(2) Mv^(2) + (1)/(2) I omega^(2) = (1)/(2) M (R omega)^(2) + (1)/(2) (MR^(2)) omega^(2)` `:. omega = sqrt((gh)/(R))` Angular momentum about C.M. `L = (MR^(2)) omega = MR sqrt(gh)`

Discussion

2304.	A ring of mass m can slide on the vertical smooth rod. Ring is connected by block with string as shown in figure. Pulley is of mass-less and smooth the system is released from reast. Initial tension in string isA. `T=Mg`B. `TgtMg`C. `TltMg`D. `T=Mg+Mg`
Answer» Correct Answer - A A

Discussion

2305.	A uniform solid cylinder having mass M and radius R is pulled by a horiozntal force F acting through the center as shown. The cylinder rolls to the right without slipping. What is the magnitude of the force of friction between the cylinder and the ground?A. F/4B. F/3C. F/2D. 2F/3
Answer» Correct Answer - B `F-f=Ma,fR=(MaR)/(2)` `impliesf=F//3`

Discussion

2306.	A stretched wire of length 260 cm is set into vibrations. It is divided into three segments whose frequencies are in the ratio 2 : 3 : 4. Their lengths must beA. 80 cm, 60 cm, 120 cmB. 120 cm, 80 cm, 60 cmC. 60 cm, 80 cm, 120 cmD. 120 cm, 60 cm, 80 cm
Answer» Correct Answer - B The frequency produced by a streched wire is given by `f = (p)/(2l) = sqrt((T)/(m)) " "……(i)` where p = number of loops formed in vibrations, l = length of wire T = tension in wire , and m = mass per unit lenght of wire So, form Eq, (i) we get `f prop (1)/(l)` Hence , the ratio of frequencies of three segment is `f_(1) : f_(2): f_(3) = 2 : 3:4` `rArr " "l_(1) : l_(2): l_(2) =(1)/(2): (1)/(3):(1)/(4)` ` = 6: 4: 3` The lenght of wire L = 260 cm , so `l_(1) =(6)/(13) xx 260 = 120 cm ` ` l_(2) = (4)/(13) xx 260 = 80cm` and `l_(3) =(3)/(13) xx 260= 60 cm`

Discussion

2307.	A bar magnet has length 3 cm, cross-sectional area `2 cm^(3)` and magnetic moment `3 Am^(2)`. The intensity of magnetisation of bar magnet isA. `3 xx 10^(5) A//m`B. `4xx 10^(5) A//m`C. `2 xx 10^(5) A//m`D. `1 xx 10^(5) A//m`
Answer» Correct Answer - D Given `l = 5 cm = 5 xx 10^(-2)m`, `a = 2 cm^(2) = 2 xx 10^(-4) m^(-2)` `a = 2 cm^(2) = 2 xx 10^(-4) m^(2)` and `M = 1 Am^(-2) ` The magnetisation of the bar magnet is given by `I =(M)/(V)` where V = volume of bar magnet `= a xx l` `rArr " " i = (M)/(a xx l)` Substituinting given values , we get `I =(1)/(2xx 10^(-4) xx 5 xx 10^(-2)) =(1)/(10^(-5))` ` = 1xx 10^(5) A//m`

Discussion

2308.	Force F and density D are related as `F=(alpha)/(beta+sqrtd)`, Then find the dimensions of `alpha` and `beta`A. `[L^((1)/(2))M^((3)/(2))T^(-2)]`B. `[L^(-1)M^((1)/(2))T^(-2)]`C. `[L^(-1)M^((3)/(2))T^(-2)]`D. `[L^(-(1)/(2))M^((1)/(2))T^(-2)]`
Answer» Correct Answer - A The dimessions of force `[F] = [ML T^(-2)]` and density `[d] = [ML^(-3) T^(0)]` From the given relation, `F =(y)/(sqrt(d))` `rArr " " y = F sqrt(d)` Substituting the above dimesions, we get ` [y] = [F] [d]^(1//2) = [ML T^(-2)] [ML^(-3) T^(0)]^(1//2)` ` = [M^(3//2) L^(-1//2)T^(-2)]`

Discussion

2309.	Force F and density D are related as `F=(alpha)/(beta+sqrtd)`, Then find the dimensions of `alpha` and `beta`
Answer» `\|alpha\|=[Fsqrt(d)]=[MLT^(-2)]=M^(3//2)L^(-1//2)T^(-2),[beta]=[d]^(1//2)=[M^(1)L^(-3)T^(0)]^(1//2)=[M^(1//2)L^(-3//2)T^(0)]`

Discussion

2310.	The dimensions of `1/2 epsilon_(0)E^(2) (epsilon_(0)=` permittivity of free space, `E=` electric field) is
Answer» `[epsilon_(0) E^(2)]=[("Energy")/("Volume")]=(ML^(2)T^(-2))/(L^(3))=ML^(-1)T^(-2)`

Discussion

2311.	The dimesional formula for impulse is ______
Answer» Impulse`=`force`xx`time`=[MLT^-2][T]=[MLT^-1]`

Discussion

2312.	A conducting rod of length `1m` aligned along the unit vector `((3)/(5)hatj+(4)/(5)hatj`), is moving with a velocity `vecv=(2hati)m//s`. The motion take place in a uniform magnetic field `vecB=(hatj-2hatk)`tesla. Find the magnitude of induced emf in across the rod in volt.
Answer» `\|epsilon\|=\|vecv*(vecBxxvecl)\|` `=(2hati)\|(hati,hatj,hatk),(0,1,-2),(0,3//5,4//5)\|=4`

Discussion

2313.	The quantities `A and B` are related by the relation `A//B = m`, where `m` is the linear mass density and `A` is the force , the dimensions of `B` will beA. pressureB. workC. latent heatD. None of these
Answer» Correct Answer - C `m=`linear density`=`mass per unit length`=[(M)/(L)]` `A=`force`=[MLT^-2]` because `[B]=([A])/([m])=([MLT^-2])/([ML^-1])=[L^2T^-2]` This is same dimesnsion as than of latent heat

Discussion

2314.	The dimesional formla of angylar velocity is ______
Answer» Angular velocity `=(theta)/(t)`,`[omega]=([M^0L^0T^0])/([T])=[T^-1]` `M^(0)L^(0)T(-1)`

Discussion

2315.	The dimensions of power are _____
Answer» `Power=(wo rk don e)/(time)=[(ML^2T^-2)]/([T])=[ML^2T^-3]` `M^(1)L^(2)T^(-3)`

Discussion

2316.	Find the quantum number n corresponding to nth excited state of `He^(++)` ion if on transition to the ground state the ion emits two photons in succession with wavelength 108.5 nm and 30.4 nm. The ionization energy of the hydrogen atom is 13.6 eV.
Answer» The energy of the `He^(+)` ion in the`nth` state `=(13.6Z^(2))/(n^(2))eV` Here `Z=Z` `:. E_(n)=-(54.4)/(n^(2))eV` The energy in the ground state `=-54.4eV` `:. 54.4(1-(1)/(n^(2)))=(hc)/(1.6xx10^(-19)) ((1)/(lambda_(1))+(1)/(lambda_(2)))=52.35eV`. `:. 1-(1)/(n^(2))=(52.35)/(54.4)` `rArr n=5`

Discussion

2317.	If `y=x^2+2x-3`,`y-x` graph isA. B. C. D.
Answer» Correct Answer - C `y=x^2+2x-3`, If we compare with `y=ax^2+bx+c` The coefficient of `x^2` is positive hence parabola opens up At `x=0`,`y=-3` At `y=0`,`x=(-2+-sqrt(4+12))/(2xx1)` `=(-2+-4)/(2)=-1+-2=-3`,1. Hence, option ( c) is correct.

Discussion

2318.	Describe the principle of extraction of each of the following : (a) `Sn` from `SnO_(2)` (b) `Zn` frin `ZnO` ( c) `Cr` from `Cr_(2) O_(3)`.
Answer» (i) `Sn` from `SnO_(2)` : `SnO_(2)` is crushed, concentrared, roasted, and washed. It is smelted with powdered anthracite in blast furnace to obtain tin. `SnO_(2) + 2e^(Ө) rarr Sn + 2CO` (ii) `Zn` from `ZnO` : Zinc oxide is mixed with powdered coke and heated to `1673 K` in a fire day retort when it is reduced to zinc metal. `ZnO +C overset(1673 K)rarrZn + CO` (iii) `Cr` from `Cr_(2) O_(3)` : `Cr` is obtained from `Cr_(2)O_(3)` byaluminothermy process. In this process, a mixture of aluminium powder and `Cr_(2)O_(3)` is ignited in a closed crucible by inserting a burning magnesium ribbon into the ignition mixture consisting of magnesium power and barium peroxide. `2Al + Cr_(2)O_(3) rarr Al_(2)O_(3) + 2Cr`.

Discussion

2319.	Calculate the number of unpaired electrons in the following gaseous ions: `Mn^(3+),Cr^(3+),V^(3+)` and `Ti^(3+)`. Which one of these is the most stable in aqueous solution?
Answer» `Mn^(3+3d^(4)=4` unpaired electrons, `Cr^(3+)=3d^(3)=3` unpaired electrons `V^(3+)=3d^2=2` unpaired electrons, `Ti^(3+)=3d^1=1` unpaired electron. `Cr^(3+)` is most stable out of these in aqueous solution bacause it has half filled `t_(2g)^(3)`.

Discussion

2320.	Which metal in the first series of transition metals exhibits+1 oxidation state most frequently and why?
Answer» Cu has the electronic configuration `3d^(10)4s^(1)`. It can easily lose `4s^1` electron to give the stable `3d^(10)` configuration hence it shows `+1` oxidation state.

Discussion

2321.	How would you account for the following. (i). Of the `d^4` species. `Cr^(2+)` is strongly reducing while manganese (III) is strongly oxidising. (ii). Cobalt (II) is stable in aqueous solution but in the presence of complexing reagents it is easily oxidised. (iii). The `d^1` configuration is very unstable in ions.
Answer» (i). `E^(ɵ)` value for `((Cr^(3+))/(Cr^(2+)))` is negative `(-0.41V)` whereas `E^(ɵ)` value for `((Mn^(3+))/(Mn^(2+)))` is positive `(+1.57V)`. Hence `Cr^(2+)` ions can easily undergo oxidation to give `Cr^(3+)` ions and therefore, act as strong reducing agent whereas `Mn^(2+)` can easily undergo reduction to give `Mn^(2+)` and hence act as oxidising agent. (ii). `Co^(3+)` has greater tendecny to form coordination complexes than `Co^(2+)`. Hence in the presence of ligands `Co^(2+)` changes to `Co^(3+)` i.e., is easily oxidised. (iii) The ions with `d^1` configuration have the tendency to lose the only electron present in d-subshell to acquire stable `d^0` configuration. Hence, they are unstable and undergo oxidation or disproportionation.

Discussion

2322.	(a) Deduce the number of 3d electrons in the following ions : Cu2+, Sc+3(b) Why do transition metals form alloy ?(c) Why Zn+2 salts are white ?
Answer» Cu⁺² : 9 electrons Sc⁺³ : 0 electron (b) Transition metals have similar atomic radii. (c) Absence of unpaired electron.

2322.

(a) Deduce the number of 3d electrons in the following ions : Cu2+, Sc+3(b) Why do transition metals form alloy ?(c) Why Zn+2 salts are white ?

Answer»

Cu⁺² : 9 electrons
Sc⁺³ : 0 electron

(b) Transition metals have similar atomic radii.
(c) Absence of unpaired electron.

Discussion

2323.	Write the correct IUPAC name of 2-amino-1, 3,7-trihydroxy hept-4-ene-1, 7-dione?A. 6-Amino-5-hydroxy hept -3-ene -1, 7-dioiacidB. 2-Amino-3-hydroxy hept-4-ene -1, 7-dioiacidC. 6-Amino-5-hydroxy hept-4-ene-1,7-dicarboxylic acidD. 6-Amino-5-hydroxy hept -3-ene-1,7-dicarboxylic acid
Answer» Correct Answer - 1 `underset("6-amino-5-hydroxy hept-3-ene-1,7-dioicacid")(HO-underset(O)underset(\|\|)(overset(7)(C))-underset(6)(overset(NH_(2))overset(\|)(C)-underset(OH)underset(\|)(overset(4)(C))-overset(4)(C)=overset(3)(C)-overset(2)(C)-underset(O)underset(\|\|)(overset(1)(C))-OH)`

Discussion

2324.	Compare the general characteristics of the first series of the transition metals with those of the seconds and third series metals in the respective vertical columns. Give special emphasis on the following points. (i). Electronic configuration and (ii). Oxidation states. (iii) ionisation enthalpies and (iv) Atomic.
Answer» (i). Electronic configuration: The elements in the same vertical column generally have similar electronic configuration However the first series shows only two exceptions, `i.e.,Cr=3d^(5)ds^(1)` and `Cu=3d^(10)4s^(1)` but second series shows more exceptions e.g., `Mo(42)=4d^(5)5s^(1),Tc(43)=4d^(6)5s^(1),Ru(43)=4d^(7)5s^(1),Rh(44)=4d^(9)5s^(1),Pd(46)=4d^(10)5s^(0),Ag(47)=4d^(10)5s^(1)`. Similarly in the third series, `W(74)=4d^(4)6s^(2),Pt(78)=5d^(9),6s^(1)` and `Au(79)=5d^(10)6s^(1)`. Thus in the same vertical column in a number of cases, the electronic configuration of the three series are not similar. (ii). Oxidation states: The elementws in the same vertical column generally show similar oxidation states. The number of oxidation states shown by the elements in the middle of each series is maximum and minimum at the extreme ends. (iii). Ionisation enthalpies: `IE_1` in each series generally increase gradually along the series though some exception are observed in each series. `IE_1` of some elements in the second `(4d)` series are higher while some of them have lower value than the element of `3d` series in the same vertical column. However, the `IE_1` of third `(5d)` series are higher than those of `3d` and `4d` series. This is because of poor shielding of nucleus by `4f` electrons in the `5d` series. (iv). Atomic sizes: In general ions of the same charge or atoms in a given series hwo progeressively decrease in radius with increasing atomic number though the decrease is quite small. But the size of the of the atoms of the atoms of `4d` series is larger than the corresponding elements of the `3d` series whereas those of corresponding elements of the `5d` series nearly the same as those of `4d` series because of lanthanoid contraction. (v). Enthalpies of atomisation: The metals of the second and third series have greater enthalpies of atomisation than the corresponding elements of the first series. This is due to much more frequent metal-metal bonding in compounds of heavy transition metals.

Discussion

2325.	Write the electronic configuration of the elements with the atomic number `61,91,101` and `109`.
Answer» `Z=61` (promethium, Pm) `=4f^(5)5d^(0)6s^(2)` `Z=91` (Protactium Pa) `=5f^(2)6d^(1)7s^(2)` `Z=101` (Mendelevium Md) `=5f^(13)6d^(0)7s^(2)` `Z=109` (Meitnerium Mt) `=5f^(14)6d^(7)7s^(2)`

Discussion

2326.	Compare the chemistry of the actinoids with that of lanthanoids with reference to:(i) electronic configuration(ii)oxidation states and (iii) chemical reactivity.
Answer» Electronic configuration The general electronic configuration for lanthanoids is [Xe]⁵⁴ 4^f0-14 5d^0-1 6s² and that for actinoids is [Rn]⁸⁶5f1^-14 6d^0-1 7s^2. Unlike 4f orbitals, 5f orbitals are not deeply buried and participate in bonding to a greater extent. Oxidation states The principal oxidation state of lanthanoids is (+3). However, sometimes we also encounter oxidation states of + 2 and + 4. This is because of extra stability of fully-filled and half-filled orbitals. Actinoids exhibit a greater range of oxidation states. This is because the 5f, 6d, and 7s levels are of comparable energies. Again, (+3) is the principal oxidation state for actinoids. Actinoids such as lanthanoids have more compounds in +3 state than in +4 state. Chemical reactivity In the lanthanide series, the earlier members of the series are more reactive. They have reactivity that is comparable to Ca. With an increase in the atomic number, the lanthanides start behaving similar to Al. Actinoids, on the other hand, are highly reactive metals, especially when they are finely divided. When they are added to boiling water, they give a mixture of oxide and hydride. Actinoids combine with most of the non-metals at moderate temperatures. Alkalies have no action on these actinoids. In case of acids, they are slightly affected by nitric acid (because of the formation of a protective oxide layer).

Discussion

2327.	`A_(2)B` molecules`(" molar mass " = 259.8 g//"ml")` crystallises in a hexagonal lattice as shown in figure .The lattic constants were `a=5 Å` and `b=8 Å` . If denstiy of crystal is `5 g//cm^(3)` then how many molecules are contained in given unit cell ? `(" Use" N_(A) = 6xx10^(23))`
Answer» Correct Answer - 2 Volume of unit cell `=a^2 sin 60xxb` `=173.2xx10^(-24) cm^3` Mass of unit cell `=173.2xx10^(-24)xx5xx6xx10^(23)` No of molecules in unit cell `=519.6/259.8=2`

Discussion

2328.	On dissolving moderate amount of sodium metal in liquid ammonia at low temperature, which of the following does not occur ?A. Blue coloured solution is obtainedB. `Na^(+)` ions are formed in the solutionC. Liquid `NH_(3)` becomes good conductor of electricityD. Liquid ammonia remains diamagnetic
Answer» Correct Answer - D

Discussion

2329.	How many of following anions have `1.5` bond order ? `IO_(4^(-)), CH_(3)COO^(-), CO_(3^(2-)), NO_(3^(-)), BrO_(2^(-)), NO_(2^(-)), HCOO^(-), SO_(3^(2-)), ClO_(2^(-)), SO_(4^(2-))` and `PO_(4^(3-))`
Answer» Correct Answer - 5 `CH_(3)-COO^(-)` `NO_(2^(-))` `HCOO^(-)` `ClO_(2)^(-)` `BrO_(2^(-))`

Discussion

2330.	With what neutral molecule is ClO−isoelectronic? Is that molecule a Lewis base?
Answer» ClO− is isoelectronic to ClF. Also, both species contain 26 electrons in all as shown. Total electrons ClO− = 17 + 8 + 1 = 26 In ClF = 17 + 9 = 26 ClF acts like a Lewis base as it accepts electrons from F to form ClF₃.

Discussion

2331.	Mixture of 10 moles of `Fe_(2)S_(3), 20` moles of `H_(2)O` and 30 moles of `O_(2)` react with `5%` yield in the given reaction : `Fe_(2)S_(3) + H_(2)O + O_(2) rarr Fe(OH)_(3) + S` Then moles of `Fe(OH)_(3)` that can be produced is -A. `(10)/(3)`B. `(20)/(3)`C. `20`D. 10
Answer» Correct Answer - B `{:(Fe_(2)S_(3),+,3H_(2)O,+(3)/(2)O_(2),overset(50%)(rarr),2Fe(OH)_(3),+,3S,),(10 " mole ",,20 " mole",,30 " moles",,,,):}`H_(2)O` is the limiting reagent. moles of `Fe(OH)_(3) = (2)/(3) xx 20 xx 0.5 = (20)/(3)` moles

Discussion

2332.	How many number of statements (s) are correct ? (i) `XeF_(6)` and `IF_(5)` both are distorted octahedral (ii) Bond angle of `BeCl_(2)` is greater than `NO_(2)^(+)` (iii) All bond length are equal in `NO_(3)^(-)` (iv) Bond order of `CO_(3)^(2-)` and `NO_(3)^(-)` is equal (v) Dipole moment of `CH_(3)Cl gt CH_(3)F` (vi) EA of `Cl gt EA` of F (vii) Radius of `Zr ~~` radius of Hf (viii) Ionisation energy of `Tl` is greater than Al (ix) (x) Ionorganic benzene and benzene both have `sp^(2)` hybridised atoms.
Answer» Correct Answer - 8 (i), (iii), (iv), (v), (vi), (vii), (viii),(x)

Discussion

2333.	Statement (I) : Automotive Petrol engines require Petrol of Octane number between 85-95. Statement (II) : Automotive Diesel engines require Diesel oil of Cetane number between 85-95.(a) Both Statement (I) and Statement (II) are individually true and Statement (II) is the correct explanation of Statement (I) (b) Both statement (I) and Statement (II) are individually true but Statement (II) is NOT the correct explanation of Statement (I) (c) Statement (I) is true but Statement (II) is false (d) Statement (I) is false but Statement (II) is true
Answer» (b) Both statement (I) and Statement (II) are individually true but Statement (II) is NOT the correct explanation of Statement (I) In petrol engine octane no. used now a days in range of 85-95, so, statement I is correct. While disel having cetone no. 40 is used as regular fuel and in premium fuel and in biodiesel it is 55. So, statement II is wrong.

2333.

Statement (I) : Automotive Petrol engines require Petrol of Octane number between 85-95. Statement (II) : Automotive Diesel engines require Diesel oil of Cetane number between 85-95.(a) Both Statement (I) and Statement (II) are individually true and Statement (II) is the correct explanation of Statement (I) (b) Both statement (I) and Statement (II) are individually true but Statement (II) is NOT the correct explanation of Statement (I) (c) Statement (I) is true but Statement (II) is false (d) Statement (I) is false but Statement (II) is true

Answer»

(b) Both statement (I) and Statement (II) are individually true but Statement (II) is NOT the correct explanation of Statement (I)

In petrol engine octane no. used now a days in range of 85-95, so, statement I is correct. While disel having cetone no. 40 is used as regular fuel and in premium fuel and in biodiesel it is 55. So, statement II is wrong.

Discussion

2334.	Figure15.1 shows the two crop fields [Plots A and B] have been treated by manures and chemical fertilizers respectively, keeping other environmental factors same. Observe the graph and answer the following questions.(i) Why does plot B show sudden increase and then gradual decrease in yield?(ii) Why is the highest peak in plot A graph slightly delayed?(ii) What is the reason for the different pattern of the two graphs?
Answer» (i) With addition of chemical fertilizer there is sudden increase in yield due to release of nutrients N,P,K etc in high quantity. The gradual decline in the graph may be due to continuous use and high quantity of chemicals which kills microbes useful for replenishing the organic matter in the soil. This decreases the soil fertility. (ii) Manures supply small quantities of nutrients to the soil slowly as it contains large amounts of organic matter [Hint: importance of organic matter can be included]. It enriches soil with nutrients thereby increasing soil fertility continuously. (iii) The difference in the two graphs indicate that use of manure is beneficial for long duration in cropping as the yield tends to remain high when the quantity of manure increases. In case of Plot B the chemical fertilizers may cause various problems when used continuously for long time. Loss of microbial activity reduces decomposition of organic matter and as a result soil fertility is lost that affects the yield.

2334.

Figure15.1 shows the two crop fields [Plots A and B] have been treated by manures and chemical fertilizers respectively, keeping other environmental factors same. Observe the graph and answer the following questions.(i) Why does plot B show sudden increase and then gradual decrease in yield?(ii) Why is the highest peak in plot A graph slightly delayed?(ii) What is the reason for the different pattern of the two graphs?

Answer»

(i) With addition of chemical fertilizer there is sudden increase in yield due to release of nutrients N,P,K etc in high quantity. The gradual decline in the graph may be due to continuous use and high quantity of chemicals which kills microbes useful for replenishing the organic matter in the soil. This decreases the soil fertility.

(ii) Manures supply small quantities of nutrients to the soil slowly as it contains large amounts of organic matter [Hint: importance of organic matter can be included]. It enriches soil with nutrients thereby increasing soil fertility continuously.

(iii) The difference in the two graphs indicate that use of manure is beneficial for long duration in cropping as the yield tends to remain high when the quantity of manure increases.

In case of Plot B the chemical fertilizers may cause various problems when used continuously for long time. Loss of microbial activity reduces decomposition of organic matter and as a result soil fertility is lost that affects the yield.

Discussion

2335.	Statement-1 : In ZnS zinc blende structure `Zn^(2+)` from FCC while alternate tetrahedral voids are occupied by `S^(2-)`. Statement-2: Positions of `Zn^2+ and S^(2-)` in zinc blende structure are similar.A. Statement-1 is True, Statement-2 is True, Statement-2 is a correct explanation for Statement-1B. Statement-1 is True, Statement-2 is True, Statement-2 is NOT a correct explanation for Statement-1C. Statement-1 is True, Statement-2 is False.D. Statement-1 is False, Statement-2 is True.
Answer» Correct Answer - A In zinc blende `S^(2-)` from FCC while alternate tetrahedral voids are occupied by `Zn^(2+)`

Discussion

2336.	Statement-1: In general, boiling points rise with increasing molecular weights of the hybrides of `14^(th)` group elements because Statement-2: The additional mass requires higher temperature for rapid movement of the molecules and the larger number of electrons in the heavier molecules provide larger London forces.A. Statement-1 is True, Statement-2 is True, Statement-2 is a correct explanation for Statement-1B. Statement-1 is True, Statement-2 is True, Statement-2 is NOT a correct explanation for Statement-1C. Statement-1 is True, Statement-2 is False.D. Statement-1 is False, Statement-2 is True.
Answer» Correct Answer - A Statement-1 and statement-2 are True, Statement-2 is a correct explanation of statement-1

Discussion

2337.	What is Saheb looking for in the garbage dumps? Where is he and where has he come from?
Answer» Saheb is looking for any precious thing which he cannot afford to buy. Things like a rupee, silver coin or a pair of shoes. He has come to the garbage dump in the writer’s neighbourhood. He lives in Seemapuri in Delhi and has come from Dhaka.

Discussion

2338.	The thermal stability of the hybrides of oxygen family is in order :A. `H_2Po lt H_2Telt H_2Se lt H_2SltH_2O`B. `H_2Polt H_2O lt H_2Te lt H_2Se lt H_2S`C. `H_2S lt H_2O lt H_2Te lt H_2Se lt H_2Po`D. `H_2O lt H_2S lt H_2Te lt H_2Te lt H_2Se lt H_2Po`
Answer» Correct Answer - A Down the group (H-E) bond dissociation enthalpy decreases as (H-E) bond length increase because of increases in size of cation. Thus thermal stability of hybrides also decreases.

Discussion

2339.	Rate constant for the reaction `NO_2+COtoNO+CO_2` are `1.3M^(-1)S^(-1)` at 700 K and `23 M^(-1)S^(-1)` at 800 K. What will be the rate constant at 750 K ? (`ln23/1.3=2.873, e^(1.53)=4.629,e^(1.34)=3.822`)A. `12.15 M^(-1) S^(-1)`B. `24.3 M^(-1) S^(-1)`C. `6.0 M^(-1) S^(-1)`D. `20.0 M^(-1) S^(-1)`
Answer» Correct Answer - C `"ln"23/1.3=(Ea)/R(1/700-1/800)` `implies (Ea)/R="ln"23/1.3xx(700xx800)/100` then , `"ln"k_(750)/1.3="ln"23/1.3xx(700xx800)/100(1/700-1/750)` `"ln"k_(750)/1.3=2.873xx5600xx50/(700xx750)implies "ln"k_(750)/1.3=1.532` `implies k_(750)=1.3xx4.629=6 M^(-1) s^(-1)`

Discussion

2340.	An aqueous solution of `NaCI` freezes at `-0.186^(@)C`. Given that `K_(b(H_(2)O)) = 0.512K kg mol^(-1)` and `K_(f(H_(2)O)) = 1.86K kg mol^(-1)`, the elevation in boiling point of this solution is:A. 0.0585 KB. 0.0512 KC. 1.864 KD. 0.0265 K
Answer» Correct Answer - B `(DeltaT_f)/(DeltaT_b)=K_f/K_b DeltaT_b=(K_bxxDeltaT_f)/K_f` `DeltaT_b=(0.512xx0.186)/1.86=0.0512`

Discussion

2341.	choose the correct option
Answer» option b : of its encashment value it's enchasement

Discussion

2342.	In which of the following complexes more than one type of structural isomers is possible :A. `[Co(NH_3)_5(NO_2)]Cl`B. `[Co(NH_3)_5(H_2O)](NO_2)_3`C. `[Pt(NH_3)_4 ] Pt (SCN)_4`D. `[Cr(dmg)_3]`
Answer» Correct Answer - A,B,C A`to`lingage , ionisation isomerism B`to`hydrate, linkage isomerism C`to` linkage , Coordination isomerism (B)STP refers to a temperature of 273 K & Nermst eqn. mentioned is at 298 K (D)`lambda_(mBa(OH_2))^(oo)=lambda_(mBa^(2+))^(oo)+2lambda_(mOH^(-))^(oo)`

Discussion

2343.	Which of the following statement is incorrect :A. `O_2` is paramagnetic , `O_3` is also paramagneticB. `O_2` is paramagnetic , `O_3` is diamagneticC. `B_2` is paramagnetic , `C_2` is also paramagneticD. Different observation is found in their bond length when NO `to NO^(+)` and `COtoCO^+`
Answer» Correct Answer - A,C,D `O_2`=paramagnetic (2 unpaired electrons in antibonding `pi`-orbital) `O_3`=diamagnetic `B_2`=paramagetic (2 unpaired electrons) `C_2`=diamagnetic molecule. NO `to NO^(+)`(bond length decreases ) CO `to CO^(+)`(bond length decreases )

Discussion

2344.	Which of the following molecules or ions may act as a bidentate ligand ?A. `CO_3^(2-)`B. `NO_3^-`C. `CH_3C-=N`D. `C_2O_4^(2-)`
Answer» Correct Answer - A,B,D `CO_3^(2-)` and `NO_3^(-)` are flexidentate ligands having 1 or 2 donor oxygen atoms `C_2O_4^(2-)-2` donor oxygen atoms. `CH_3C=N-1` donor nitrogen atom.

Discussion

2345.	How many moles of AgCl will be precipitated when an excess of AgNO3 is added to a molar solution of [Cr (H2O)5Cl] Cl2? a). 1 mol of AgCl b). 2 moles of AgCl c). 3 moles of AgCl d). 4 moles of AgCl
Answer» b). 2 moles of AgCl

Discussion

2346.	EDTA is a a). Bidentate ligand b). Tridentate ligand c). Tetradentate ligand d). Hexadentate ligand
Answer» d). Hexadentate ligand

Discussion

2347.	Give an example of hexadentate ligand. Write its one use.
Answer» EDTA^4-(Ethane1,2-diamminetetraacetate ion).It is used in the treatment of lead poisoning.

Discussion

2348.	Extraction of metal fron the ore cassiterlate involves (i) carbon reduction of an oxide are self reduction of a sulphide or (iii) removal of copper imputiry (iv) removel of iron inpurityA. (i),(ii),(iii),(iv)B. (ii),(iii),(iv)C. (i),(iv)D. (i),(iii) and (iv)
Answer» Correct Answer - d T in does not occur free in nuture be chef source is the ore cascherite `SaO_(2)` also known as tin same ,The extraction of tin metal from cauterls `SaO_(2)` is condocted through following stages (i) Concentration ,which is done by hydraulic washing and magnitic sepration (ii) Reasting to oxides sulphur to `SO_(2)` and aresenic to `AI_(2)O_(3)` (iii) Smelting to reduce its oxide to be metal `SaO_(2) + 2C rarr Sn + 2CO` Tin so obtained is crude and contain iros impurities its reflsaing is done in two stage (a) Liquation which consite in heating the crate metal on the sloping hearth of a reverately farmace .The on metal flows down having behind iron copper tangestan, etc (b) polling which involves storing of the mothen metal taken in a big pot by logs of work .Any tin oxide left unreducal is now reduced by tyhe hydrocathons of wood.

Discussion

2349.	In self-reduction, the reducing species isA. SB. `S^(2-)`C. `O^(2-)`D. `SO_(2)`
Answer» Correct Answer - B S is more electronegative

Discussion

2350.	In self -reduction used in extraction of copper, the reducing agent isA. SB. `O_(2-)`C. `S_(2)`D. `SO_(2)`
Answer» Correct Answer - c `2Cu_(2)S + 3O_(2) rarr 2Cu_(2)O + 2SO_(2)` `Cu_(2) + 2Cu_(2)O rarr 6Cu + SO_(2)`

Discussion

Explore topic-wise InterviewSolutions in .

Find the sum of first 10 multiples of 6.

A thin conducting rod of length `l=5m` is moved such that its end B moves along the X-axis while end A moves along the Y-axis A uniform magnetic field `B=6hat(k)` T exists

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If `y=x^2+2x-3`,`y-x` graph isA. B. C. D.

Describe the principle of extraction of each of the following : (a) `Sn` from `SnO_(2)` (b) `Zn` frin `ZnO` ( c) `Cr` from `Cr_(2) O_(3)`.

Calculate the number of unpaired electrons in the following gaseous ions: `Mn^(3+),Cr^(3+),V^(3+)` and `Ti^(3+)`. Which one of these is the most stable in aqueous solution?

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(a) Deduce the number of 3d electrons in the following ions : Cu2+, Sc+3(b) Why do transition metals form alloy ?(c) Why Zn+2 salts are white ?

Write the correct IUPAC name of 2-amino-1, 3,7-trihydroxy hept-4-ene-1, 7-dione?A. 6-Amino-5-hydroxy hept -3-ene -1, 7-dioiacidB. 2-Amino-3-hydroxy hept-4-ene -1, 7-dioiacidC. 6-Amino-5-hydroxy hept-4-ene-1,7-dicarboxylic acidD. 6-Amino-5-hydroxy hept -3-ene-1,7-dicarboxylic acid

Write the electronic configuration of the elements with the atomic number `61,91,101` and `109`.

Compare the chemistry of the actinoids with that of lanthanoids with reference to:(i) electronic configuration(ii)oxidation states and (iii) chemical reactivity.

`A_(2)B` molecules`(" molar mass " = 259.8 g//"ml")` crystallises in a hexagonal lattice as shown in figure .The lattic constants were `a=5 Å` and `b=8 Å` . If denstiy of crystal is `5 g//cm^(3)` then how many molecules are contained in given unit cell ? `(" Use" N_(A) = 6xx10^(23))`

On dissolving moderate amount of sodium metal in liquid ammonia at low temperature, which of the following does not occur ?A. Blue coloured solution is obtainedB. `Na^(+)` ions are formed in the solutionC. Liquid `NH_(3)` becomes good conductor of electricityD. Liquid ammonia remains diamagnetic

How many of following anions have `1.5` bond order ? `IO_(4^(-)), CH_(3)COO^(-), CO_(3^(2-)), NO_(3^(-)), BrO_(2^(-)), NO_(2^(-)), HCOO^(-), SO_(3^(2-)), ClO_(2^(-)), SO_(4^(2-))` and `PO_(4^(3-))`

With what neutral molecule is ClO−isoelectronic? Is that molecule a Lewis base?

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What is Saheb looking for in the garbage dumps? Where is he and where has he come from?

The thermal stability of the hybrides of oxygen family is in order :A. `H_2Po lt H_2Telt H_2Se lt H_2SltH_2O`B. `H_2Polt H_2O lt H_2Te lt H_2Se lt H_2S`C. `H_2S lt H_2O lt H_2Te lt H_2Se lt H_2Po`D. `H_2O lt H_2S lt H_2Te lt H_2Te lt H_2Se lt H_2Po`

Rate constant for the reaction `NO_2+COtoNO+CO_2` are `1.3M^(-1)S^(-1)` at 700 K and `23 M^(-1)S^(-1)` at 800 K. What will be the rate constant at 750 K ? (`ln23/1.3=2.873, e^(1.53)=4.629,e^(1.34)=3.822`)A. `12.15 M^(-1) S^(-1)`B. `24.3 M^(-1) S^(-1)`C. `6.0 M^(-1) S^(-1)`D. `20.0 M^(-1) S^(-1)`

An aqueous solution of `NaCI` freezes at `-0.186^(@)C`. Given that `K_(b(H_(2)O)) = 0.512K kg mol^(-1)` and `K_(f(H_(2)O)) = 1.86K kg mol^(-1)`, the elevation in boiling point of this solution is:A. 0.0585 KB. 0.0512 KC. 1.864 KD. 0.0265 K

choose the correct option

In which of the following complexes more than one type of structural isomers is possible :A. `[Co(NH_3)_5(NO_2)]Cl`B. `[Co(NH_3)_5(H_2O)](NO_2)_3`C. `[Pt(NH_3)_4 ] Pt (SCN)_4`D. `[Cr(dmg)_3]`

Which of the following statement is incorrect :A. `O_2` is paramagnetic , `O_3` is also paramagneticB. `O_2` is paramagnetic , `O_3` is diamagneticC. `B_2` is paramagnetic , `C_2` is also paramagneticD. Different observation is found in their bond length when NO `to NO^(+)` and `COtoCO^+`

Which of the following molecules or ions may act as a bidentate ligand ?A. `CO_3^(2-)`B. `NO_3^-`C. `CH_3C-=N`D. `C_2O_4^(2-)`

How many moles of AgCl will be precipitated when an excess of AgNO3 is added to a molar solution of [Cr (H2O)5Cl] Cl2? a). 1 mol of AgCl b). 2 moles of AgCl c). 3 moles of AgCl d). 4 moles of AgCl

EDTA is a a). Bidentate ligand b). Tridentate ligand c). Tetradentate ligand d). Hexadentate ligand

Give an example of hexadentate ligand. Write its one use.

Extraction of metal fron the ore cassiterlate involves (i) carbon reduction of an oxide are self reduction of a sulphide or (iii) removal of copper imputiry (iv) removel of iron inpurityA. (i),(ii),(iii),(iv)B. (ii),(iii),(iv)C. (i),(iv)D. (i),(iii) and (iv)

In self-reduction, the reducing species isA. SB. `S^(2-)`C. `O^(2-)`D. `SO_(2)`

In self -reduction used in extraction of copper, the reducing agent isA. SB. `O_(2-)`C. `S_(2)`D. `SO_(2)`