94420 + Interview Questions in Chemistry in Class 12 Page 606 InterviewSolution

30251.	The halogen that absorbs a light of maximum wavelength is ............
Answer» `F_(2)`<BR>`Cl_(2)` `Br_(2)` `I_(2)` Solution :`I_2` APPEARS violet due to absorption of orange-yellow light.

Discussion

30252.	The halogen having Greenish - yellow gas reacts with hot and concentrated NaOH solution, and give products. The oxidation state of that halogen changes from
Answer» 0 to -1 0 to `+5` `-1` to `+1` 0 to "-1" and " `+5`" states Answer :D

Discussion

30253.	The halogen having smallest covalent radius is:
Answer» I Cl Br F Answer :D

Discussion

30254.	The halogen having five vacant orbitals in the outermost shell belongs to
Answer» 3RD PERIOD 4th period 2ND period 5th period Solution :FIVE vacant orbitals (3d-level) are present in CL `(1s^2 2s^2 2p^6 3s^2 3p^5)` which belongs to 3rd period.

Discussion

30255.	The halogen-halogen bond length is longest for
Answer» `F_2` `Cl_2` `Br_2` `I_2` SOLUTION :BOND LENGTH INCREASES down the GROUP from `F_2` to `I_2`

Discussion

30256.	The halogen forming largest number of inter-halogens is
Answer» F Cl Br I Answer :A

Discussion

30257.	The halogen derivative used as an antiseptic is:
Answer» `CHCl_(3)` `CHBr_(3)` `"CCl"_(4)` `CHI_(3)` Answer :D

Discussion

30258.	The halogen atom is on the sphybridesed carbon which itself is attached to an aromatic ring, is called as
Answer» Allylic halide Benzylhalide Perhalo alkane ARYL halide ANSWER :B

Discussion

30259.	The halogen atom is least reactive in:
Answer» Chlorobenzene Chloroethane 2-chloropropane 3-chloropropane Answer :A

Discussion

30260.	The halogen atom in haloalkanes can be easily replaced by neucleophile(s) such as
Answer» `CN^(-)` `NO_(2)^(-)` `OR^(-)` NONE of these Answer :A::B::C

Discussion

30261.	The halogen atom in aryl halides is
Answer» o-and p-directing m-directing o, m and p-directing only m-directing Answer :A::C::D

Discussion

30262.	The halogen atom in aryl halide is
Answer» <P>o and p DIRECTING META directing o, p, and m-directing only o-directing ANSWER :A

Discussion

30263.	The halogen atom from haloalkane can be replaced by nucleophile such as
Answer» `NH_(2)^(-)` `CN^(-)` `RO^(-)` all of these Answer :D

Discussion

30264.	The halogen acid (X) is
Answer» HF HCl HI `Cl_2` (AQ.) ANSWER :B::C

Discussion

30265.	The haloform theat gives of acetone with sodium hyporbomite yields.
Answer» ACETIC ACID propionic acid ACETALDEHYDE ISOPROPYL alcohol Answer :A

Discussion

30266.	The halides of transition elements become more covalent with increasing oxidation state of the metal . Why ?
Answer» Solution :As the oxidation state increases, size of the TRANSITION metalions DECREASES. Applying Fajan's rule, as the size of the metal ion decreases, covalent character of the BOND formed increases ( discussed in `+1` in the CHAPTER ).

Discussion

30267.	The halides of transition elements become more covalent with increasing oxidation state of the metal. Why?
Answer» SOLUTION :As the OXIDATION state of the element increases, its CHARGE increases. According to Fajan.s Rules, as the charge of the metal ion increases covalent CHARACTER increases because the positively charged cation attracts the electron CLOUD on the anion towards itself.

Discussion

30268.	The halides of group III elements behave as Lewis acids. The acceptor ability is maximum for the halides of :
Answer» B Al Ga In. Answer :A

Discussion

30269.	The halide which will not react with benzene in presence of anhydrous AICI_3 is
Answer» `CH_3CHClCH_3` `C_6 H_5 CH_2 CL` `C_6 H_5 Cl` `CH_3 CH_2 CH_2 Cl` ANSWER :C

Discussion

30270.	The halide which is inert to water is
Answer» `F_(2)` `Cl_(2)` `Br_(2)` `I_(2)` Solution :Sometimes, when MAXIMUM covalency is obtained, the halides becomes INERT to water.Thus, (or similarly `CCI_4)` is stable. This is because `SF_6` is coordinately SATURATED and sterically hindered. Thus, `SF_6` is inert to water, because of kinetic RATHER than thermodynamic factors: (a) `PCl_(5)+4H_(2)O to H_(3)PO_(4)+5H^(+)Cl^(-)` (b) `SiCl_(4)+4H_2O to SI(OH)_(4)+4H^(+)Cl^(-)` `(c) BCl_(3)+3H_(2)O to B(OH)_(3)+3H^(+)Cl^(-)`

Discussion

30271.	The halide which is inert to water is :
Answer» `PCl_(5)` `SiCl_(4)` `BCl_(3)` `SF_(6)` Solution :Sometimes, when MAXIMUM covalency is obtained, the halides become inert to WATER, thus `SF_(6)`(or similarly `C Cl_(4)`)is stable. This is because `SF_(6)`is coordinately SATURATED and sterically hindered. Thus, `SF_(6)`is inert to water, because of KINETIC rather than thermodynamic factor. A. `PCl_(5) +4H_(2)O to H_(3)PO_(4) +5HCl` B. `SiCl_(4) +4H_(2)O to Si(OH)_(4) +4HCl` C. `BCl_(3)+3H_(2)O to B(OH)_(3) +3HCl`

Discussion

30272.	The halide that cannot actas lewis acid is
Answer» `BF_3` `SnCl_4` `C-Cl_4` `SF_4` ANSWER :C

Discussion

30273.	The halide ion easiest to oxidise is :
Answer» `F^(-)`<BR>`CL^(-)` `Br^(-)` `I^(-)` ANSWER :D

Discussion

30274.	The half time of first order decomposition of nitramide is 2.1 hour at 15^(@)C. NH_(2)NO_(2)(aq)to N_(2)O(g)+H_(2)O(l) If 6.2 g of NH_(2)NO_(2) is allowed to decompose, calculate (i) time taken for NH_(2)NO_(2) to decompose 99% and (ii) volume of dry N_(2)O produced at this point, measured at STP.
Answer» SOLUTION :(i) `k=(0.693)/(t_(1//2))=(0.693)/(2.1" hr")=0.33" hr"^(-1)` `x=99%" of "a=0.99" a"` `t=(2.303)/(k)log""(a)/(a-x)=(2.303)/(0.33" hr"^(-1))log""(a)/(a-0.99a)=(2.303)/(0.33)log10^(2)=13.96" hours"` (ii) Amount decomposed `=99%" of "6.2" G "=(99)/(100)xx6.2" g "=6.138" g"` 1 mol `NH_(2)NO_(2)(62" g")" PRODUCE "N_(2)O" at STP"=22.4" L"` 6.138 g will produce `N_(2)O` at STP `=(22.4)/(62)xx6.138" L"=2.2176" L"`

Discussion

30275.	The halflife of I^(131) is 8 days . Given a sample of I^(131) at t = 0 , we can assert that :
Answer» No NUCLEUS will decay at t = 4 days No nucleus will decay before t=8 days All nucleus will decay before t = 16 days A given nucleus may decay at any time after ( t= 0) Solution :RADIOACTIVE decay FOLLOWS 1ST order kinetics .

Discussion

30276.	The half-time of the first-order decomposition of nitramide is 2.1 hour at 15^@C ,NH_2NO_2(aq) to N_2O(g) + H_2O(l)If 6.2 g of NH_2NO_2is allowed to decompose, calculate (i) the time taken for 99% decomposition, and(ii) the volume of dry N_2Oproduced at this point measured at STP.
Answer» SOLUTION :(i) 21 HOURS (II) 2.2176 LITRES

Discussion

30277.	The half reaction time required to decrease initial cocentration from 40% to 20 % is 20 minute.What time will be taken to decrease initial concentration from 10% to 5%?
Answer» 20 minute 5 minute 10 minute 60 minute Answer :B

Discussion

30278.	The halfperiod of a reaction is 60 seconds. Calculate the rate constant.
Answer» SOLUTION :`t_(1/2)`=60seconds RATE CONSTANT`K=0.693/(t_1/2)`=`0.693/(60"SEC")`=0.01155s`sec^-1`

Discussion

30279.	The half-life period of a reaction is 60 s. Calculate its rate constant.
Answer» SOLUTION :`t_(1/2)`=60seconds RATE CONSTANT`K=0.693/(t_1/2)`=`0.693/(60"SEC")`=0.01155s`sec^-1`

Discussion

30280.	The half period of a radioactive element is 20 years. If a sample of this nucleide has an initial activity of 20000 dis/min, the activity after 80 years would be:
Answer» 2500 dis/min 5000 dis/min 20,000 dis/min 1250 dis/min. Answer :D

Discussion

30281.	The half lives of two radioactive nuclides A and B are 1 and 2 min. respectively. Equal weights of A and B are taken respectively and allowed to disintegrate for 4 min. What will be the ratio of weights of A and B disintegrated
Answer» `1 : 1` `5 : 4` `1 : 2` `1 : 3` Solution :`{:("For A","For B"),(T = 4 min,T = 4 min),(t_(1//2) = 1MIN,t_(1//2) = 2 min),(N_(0) = x("say"),N_(0)= x("say")),(n = 4//1 = 4,n = 4//2 = 2),(N = (N_(0))/(2^(n)) = (x)/(2^(4)),N = (N_(0))/(2^(n)) = (x)/(2^(2))),(N = (x)/(16),N = (x)/(4)),("A disintegrated","B disintegrated"),(x - (x)/(16) = (15)/(16)x,x - (x)/(4) = (3x)/(4)),("Ratio of A B disintegrated",),((15)/(16)x : (3x)/(4) rArr 5 :4,):}`

Discussion

30282.	The half-life period t_(1//2) of a radioactive element is N years. The period of its complete decays is
Answer» `N^(2)` YEARS 2N years `(1)/(2)N^(2)` years INFINITY Solution :The `t_(1//2)` of a radioactive element = N years `:.` The period of its complete decay is infinity

Discussion

30283.	The half life periods of four isotopes are given : I = 6.7 years II = 8000 years III = 5760 years IV=2.35xx105 years. Which of these is most stable ?
Answer» I II III IV Solution :LONGER the half LIFE PERIOD, more will be the STABILITY of the element.

Discussion

30284.	The half-life period, t_(1//2) is related to the order n and initial concentration a according to the equation
Answer» `f_(1//2)PROPA/n` `f_(1//2)propan` `f_(1//2)propa^(n+1)` `f_(1//2)PROP(1)/(a^(n-1))` Solution :For `n^(TH)` order reaction, `t_(1//2) prop (1)/(a^(n-1))`

Discussion

30285.	The half-life period of Uranium is 4.5 billion years. After 90 billion years, the number of moles of Helium liberated from the following nuclear reaction will be ._(92)U^(238) rarr ._(90)Th^(234) + ._(2)He^(4)
Answer» 0.75 mole 1.0 mole 11.2 mole 22.4 mole Solution :`N = (N_(0))/(2^(n)) = N = (N_(0))/(2^(2)) = (1)/(4)` `:.` CONVERTED moles `= 1 - (1)/(4) = (3)/(4) = 0.75`

Discussion

30286.	The half-life period of zero order reaction A to product is given by :
Answer» `([A]_(0))/(k)` `(0.693)/(k)` `([A]_(0))/(2k)` `(2[A]_(0))/(k)` ANSWER :C

Discussion

30287.	The half life period of the reaction : N_(2) O_(5) rarr 2NO_(2) +(1)/(2) O_(2) is 24 hrs at 30^(@) C. Starting with 10 g of N_(2) O_(5) how many grams of N_(2) O_(5) will remain after a period of 96 hours ?
Answer» `1.25 g ` `0.63 g` `1.75 g ` `0.5 g` Solution :No. of half life PERIODS =4 Amount LEFT `=([A]_(0))/(2^(n))=(10)/(2^(4))=0.63 g `

Discussion

30288.	The half life period of the reaction H_(2)+I_(2)to2HI is proportional to ……
Answer» INITIAL CONCENTRATION the SQUARE of initial concentration inverse of initial concentration independent of initial concentration ANSWER :C

Discussion

30289.	The half life period of Radon is 3.8 years. After how many days will one twentieth of Radon sample be left over ?
Answer» Solution :`lambda=(0.693)/(t_(1//2))=(0.693)/(3.8)=0.182day^(-1)` LET the initial AMOUNT of radon be `N_(0)` and amount left after `t` DAYS be `N` which is equal to `(N_(0))/(20)` `t=(2.303)/(lambda)log_(10).(N_(0))/(N)` `=(2.303)/(0.182)log_(10).(N_(0))/(N_(0)//20)=(2.303)/(0.182)log_(10)20` `=16.54` days

Discussion

30290.	The half life period of radium is 1580 years. It remains 1.16 after the years
Answer» 1580 years 3160 years 4740 years 6320 years Solution :We KNOW that `N = N_(0) XX ((1)/(2))^(2)` Here, `N = (1)/(16), N_(0) = 1`, by putting the VALUES `(1)/(16) = 1 xx ((1)/(2))^(n) "" :. n = 4` we know that `T = t_(1//2) xx n` `:. T 1580 xx 4 = 6320` years

Discussion

30291.	The half-life period of radioactive element is 140 days. After 560 days, 1 g of element will reduce to
Answer» `1/2`G `1/4`g `1/8`g `1/16`g Solution : As we KNOW, `T= nxxt_(1//2)` where, `t_(1//2)` = half-life period `N=(T)/(t_(1//2))=(560)/(140)=4` Now, `N_1=N_0(1/2)^n=1xx(1/2)^4=1/16g`

Discussion

30292.	The half-life period of Pb^(210) is 22 years. If 2 g of Pb^(210) is taken, then after 11 years how much of Pb^(210) will be left
Answer» 1.414 g 2.428 g 3.442 g 4/456 g Solution :`N_(t) = N_(0) ((1)/(2))^(2) [ :' t_(1//2) = 22 " YEARS, " T = 11 " years" , N_(0) = 2, N_(t) = ?]` `T = t_(1//2) xx N, 11 = 22 xx n or n = (11)/(22) = (1)/(2)` `:. N_(t) = 2g xx ((1)/(2))^(1//2) = 1.414g`

Discussion

30293.	The half-lifeperiod of ………… is 60days.The radioactivity after 180 days will be
Answer» 0.25 0.125 `33.3%` `3.0%` Solution :RADIOACTIVITY after time,t `(N_t) = (N_0)/((2)^(n)) and n = t/(t^(1//2))` So, `n = 180/60 = 3, N_t = (N_0)/(8) = 0.125` Hence, radioactivity after 180 days = 12.5%.

Discussion

30294.	Half life period of ""_(53)I""^(125) is 60 days. Percentage of radioactivity preent after 180 days isa).5b).75c).36d).125
Answer» `50%` `20.5%` `12.5%` `25%` Solution :`t_(1//2)=60 "days", "and" t=180 "days"` Therefore, `t//t_(1//2) = (180)/(60)=3.0 " Or"N=(N_(0))/(2^(3.0))=(1)/(8)N_(0)=(1)/(8)XX100` Precent of `N_(0)=12.5%`

Discussion

30295.	The half life period of first order reaction is 10 seconds. What is the time required for 99.9% completion of that reaction ?
Answer» 20 seconds 1000 seconds 100 seconds 99 seconds Solution :`10xxt_(1//2)=t_(99.9%)` `:. t_(99.9%) = 10xx10sec=100sec`

Discussion

30296.	The half life periods of four reactions labelled by A,B, C & D are 30 sec, 4.8 in, 180 sec and 16 minrespectively. Thefastest reaction is
Answer» A B C D ANSWER :A

Discussion

30297.	The half life period of C^(14) is 5760 years. For a 200 mg of sample of C^(14), the time taken to change to 25 mg is
Answer» 11520 YEARS 23040 years 5760 years 17280 years. Solution :d) `R_(0)=200mg`, R=25mg, `t_(1//2)=5760` years, Suppose required years =x Suppose no. of HALF lifes=n `(R/R_(0)) = (1/2)^(n)` or `(25/250) = (1/2)^(n)` `(1/8) = (1/2)^(n)` or `(1/3)^(3) = (1/2)^(n)` n=3 Now, `n=x/(t_(1//2)) = 3 XX 5760 Y` `=17280` years.

Discussion

30298.	The half life period of a zero order reaction is independent of initial concentration
Answer»

Discussion

30299.	The half-life period of a zero order reaction is given by
Answer» `([A]_(0))/(k)` `(2.303k)/([A]_(0))` `([A]_(0))/(2k)` `(2[A]_(0))/(k)` Answer :A::B::C

Discussion

30300.	The half life period of a subtance in a certain enzyme-catalysed reaction is 138s. The time required for the concentration of the substance to fall from 1.28 mgL^(-1) is:
Answer» 414s 552 s 690 s 276 s Solution :C) Amount left after n HALF life peridos `[A]_(0)` `=[A]_(0)/(2^(n))` `2^(n) = [A]_(0)/[A]= (1.28 mg L^(-1))/(0.04 mg L^(-1))` `[2]^(n) = 32=[2]^(5)` No. of `t_(1//2)` periods =5 Time REQUIRED = `138 xx 5=690s`

Discussion

Explore topic-wise InterviewSolutions in .

The halogen that absorbs a light of maximum wavelength is ............

The halogen having Greenish - yellow gas reacts with hot and concentrated NaOH solution, and give products. The oxidation state of that halogen changes from

The halogen having smallest covalent radius is:

The halogen having five vacant orbitals in the outermost shell belongs to

The halogen-halogen bond length is longest for

The halogen forming largest number of inter-halogens is

The halogen derivative used as an antiseptic is:

The halogen atom is on the sphybridesed carbon which itself is attached to an aromatic ring, is called as

The halogen atom is least reactive in:

The halogen atom in haloalkanes can be easily replaced by neucleophile(s) such as

The halogen atom in aryl halides is

The halogen atom in aryl halide is

The halogen atom from haloalkane can be replaced by nucleophile such as

The halogen acid (X) is

The haloform theat gives of acetone with sodium hyporbomite yields.

The halides of transition elements become more covalent with increasing oxidation state of the metal . Why ?

The halides of transition elements become more covalent with increasing oxidation state of the metal. Why?

The halides of group III elements behave as Lewis acids. The acceptor ability is maximum for the halides of :

The halide which will not react with benzene in presence of anhydrous AICI_3 is

The halide which is inert to water is

The halide which is inert to water is :

The halide that cannot actas lewis acid is

The halide ion easiest to oxidise is :

The half time of first order decomposition of nitramide is 2.1 hour at 15^(@)C. NH_(2)NO_(2)(aq)to N_(2)O(g)+H_(2)O(l) If 6.2 g of NH_(2)NO_(2) is allowed to decompose, calculate (i) time taken for NH_(2)NO_(2) to decompose 99% and (ii) volume of dry N_(2)O produced at this point, measured at STP.

The halflife of I^(131) is 8 days . Given a sample of I^(131) at t = 0 , we can assert that :

The half-time of the first-order decomposition of nitramide is 2.1 hour at 15^@C ,NH_2NO_2(aq) to N_2O(g) + H_2O(l)If 6.2 g of NH_2NO_2is allowed to decompose, calculate (i) the time taken for 99% decomposition, and(ii) the volume of dry N_2Oproduced at this point measured at STP.

The half reaction time required to decrease initial cocentration from 40% to 20 % is 20 minute.What time will be taken to decrease initial concentration from 10% to 5%?

The halfperiod of a reaction is 60 seconds. Calculate the rate constant.

The half-life period of a reaction is 60 s. Calculate its rate constant.

The half period of a radioactive element is 20 years. If a sample of this nucleide has an initial activity of 20000 dis/min, the activity after 80 years would be:

The half lives of two radioactive nuclides A and B are 1 and 2 min. respectively. Equal weights of A and B are taken respectively and allowed to disintegrate for 4 min. What will be the ratio of weights of A and B disintegrated

The half-life period t_(1//2) of a radioactive element is N years. The period of its complete decays is

The half life periods of four isotopes are given : I = 6.7 years II = 8000 years III = 5760 years IV=2.35xx105 years. Which of these is most stable ?

The half-life period, t_(1//2) is related to the order n and initial concentration a according to the equation

The half-life period of Uranium is 4.5 billion years. After 90 billion years, the number of moles of Helium liberated from the following nuclear reaction will be ._(92)U^(238) rarr ._(90)Th^(234) + ._(2)He^(4)

The half-life period of zero order reaction A to product is given by :

The half life period of the reaction : N_(2) O_(5) rarr 2NO_(2) +(1)/(2) O_(2) is 24 hrs at 30^(@) C. Starting with 10 g of N_(2) O_(5) how many grams of N_(2) O_(5) will remain after a period of 96 hours ?

The half life period of the reaction H_(2)+I_(2)to2HI is proportional to ……

The half life period of Radon is 3.8 years. After how many days will one twentieth of Radon sample be left over ?

The half life period of radium is 1580 years. It remains 1.16 after the years

The half-life period of radioactive element is 140 days. After 560 days, 1 g of element will reduce to

The half-life period of Pb^(210) is 22 years. If 2 g of Pb^(210) is taken, then after 11 years how much of Pb^(210) will be left

The half-lifeperiod of ………… is 60days.The radioactivity after 180 days will be

Half life period of ""_(53)I""^(125) is 60 days. Percentage of radioactivity preent after 180 days isa).5b).75c).36d).125

The half life period of first order reaction is 10 seconds. What is the time required for 99.9% completion of that reaction ?

The half life periods of four reactions labelled by A,B, C & D are 30 sec, 4.8 in, 180 sec and 16 minrespectively. Thefastest reaction is

The half life period of C^(14) is 5760 years. For a 200 mg of sample of C^(14), the time taken to change to 25 mg is

The half life period of a zero order reaction is independent of initial concentration

The half-life period of a zero order reaction is given by

The half life period of a subtance in a certain enzyme-catalysed reaction is 138s. The time required for the concentration of the substance to fall from 1.28 mgL^(-1) is: