Explore topic-wise InterviewSolutions in .

This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.

29651.

The ionization isomer of [Cr(H_2O)_4 Cl(NO_2)]Cl is given as

Answer»

`[CR(H_2O)_4(O_2N)]Cl_2`
`[Cr(H_2O)_4Cl_2]NO_2`
`[Cr(H_2O)_4Cl(ONO)]Cl`
`[Cr(H_2O)_4Cl_2 (NO_2)] .H_2O`

Answer :B
Discussion
29652.

The ionization isomer of [Cr(H_(2)O)_(4)Cl(NO_(2))]Clis

Answer»

`[Cr(H_(2)O)_(4)(O_(2)N)]Cl_(2)`
`[Cr(H_(2)O)_(4)Cl_(2)](NO_(2))`
`[Cr(H_(2)O)_(4)Cl(ONO)]Cl`
`[Cr(H_(2)O)_(4)Cl_(2)(NO_(2))]H_(2)O`

Answer :B
Discussion
29653.

The increasing order of reactivity of the following halides for S_(N)1 reaction is- underset((I))(CH_(3)underset(Cl)underset(|)(C)HCH_(2)CH_(3)), underset((II))(CH_(3)CH_(2)CH_(2)Cl) underset((III))(p-H_(3)CO-C_(6)H_(4)-CH_(2)Cl)

Answer»

(III) lt (II) lt(I)
(II) lt (I) lt (III)
(I) lt (III) lt (II)
(II) lt (III) lt (I)

Solution :
Benzylic carbocation is stabilised by resonance with the `pi`-electorns of benzene RING and the EXTENT of resonance stabilisation is further INCREASED by the presence of electron donating -`OCH_(3)` group in the para-position of benzene ring. We know `2^(@)` carbocation is more stable than `1^(@)` carbocation. so the order of stability of the carbocations is: benzylic carbocation > `2^(@)` carbocation > `1^(@)` carbocation. The feasibility of `S_(N)1` reaction is DIRECTLY proportional to the stability of carbocation. Hence, the increasing order of REACTIVITY of the given halides for `S_(N)1` reaction is (II) < (I) < (III).
Discussion
29654.

The ionization enthalpies of lithium and sodium are 520 kJ mol^(-1) and 495 kJ mol^(-) respectivley . The energy required to convert all the atoms present in 7 mg of lithium vapours and 23 mg of sodium vapours of their respective gaseous cations respectivley are

Answer»

52 J , 49. 5 J
520 J , 495 J
49.5 J, 52 J
495 J , 520 J

Solution :No. of moles of `LI = 7/(1000 xx 7) = 10^(-3)`
No. of moles of `Na = 23/(1000 xx 23) = 10^(-3)`
The amounts of ENERGIES required for `10^(-3)` mole each of Li and Na are `520 kJ xx 10^(-3) and 495 kJ xx 10^(-3)` or 520 J and 495 J respectively.
Discussion
29655.

The ionization enthalpy of selenium is less than that of tellurium. The given statement is true or false?

Answer»

SOLUTION :The IONIZATION ENTHALPY of SELENIUM is more than that of TELLURIUM
Discussion
29656.

The ionization enthalpy of selenium is less than that of tellurium.

Answer»

SOLUTION :The IONIZATION ENTHALPY of SELENIUM is more than that of TELLURIUM
Discussion
29657.

The increasing order of reducing power of the halogen acids is

Answer»

HF LT HCl lt HBR lt HI
HI lt HBr lt HCl lt HF
HBr lt HCl lt HF lt HI
HCl lt HBr lt HF lt HI

Solution :HF lt HCl lt HBr lt HI depends on size of halogen
Discussion
29658.

The ionization energy of the coinage metals fall in the orderCugtAgltAu.

Answer»

Solution :In all the 3 cases an s-electron in the unpaired state is to be removed. In the case of Cu a 4s electron is to be removed which is closer to the nucleus than the 5s electron of AG. So I.P. decreases from Cu to Ag. HOWEVER from Ag to Au the 14 f electrons are added which provide very POOR shielding effect. The nuclear charge is thus enhanced and therefore the OUTER electron of Au is more tightly held and so the IP is HIGH.
Discussion
29659.

The increasing order of reactivity of halogens is:

Answer»

`I_2 LT Br_2 lt Cl_2 ltF_2`
`Cl_2ltF_2ltBr_2ltI_2`
`Cl_2 ltBr_2 ltI_2 lt F_2`
`I_2 lt Cl_2 lt Br_2 lt F_2`

ANSWER :A
Discussion
29660.

The ionization energy of lithium is 520 kJ "mol"^(-1) . Find the amount of energy (in kJ) required to convert 70 mg of lithium atoms in gaseous state into Li+ ions.

Answer»


Solution :70 MG `=70 XX 10^(-7) G = (70 xx 10^(-3))/7 mol = 1 xx 10^(-2)` mol
`therefore` Amount of energy required
`= 1 xx 10^(-2) xx 520 kJ = 5.2 kJ`
Discussion
29661.

The ionization energy of hydrogen atom in the ground state is

Answer»

13.6 MeV
13.6 eV
13.6 Joule
Zero

Solution :theionization ENERGYOF hydrogenatomin THEGROUNDSTATEIS 13.6eV
Discussion
29662.

The increasing order of pK_(a) of following amino acids in aqueous solution is ………………...

Answer»

`ASP LT Gly lt ARG lt LYS`
`Arg lt Lys lt Gly lt Asp`
`Gly lt Asp lt Arg lt Lys`
`Asp lt Gly lt Lys lt Arg`

Solution :`Asp lt Gly lt Lys lt Arg`
Discussion
29663.

The ionization energy of hydrogen atom in the ground state is 1312 kJ "mol"^(-1) . Calculate the wavelength of radiation emitted when the electron in hydrogen atom makes a transition from n = 2 state to n = 1 state (Planck’s constant, h = 6.626 xx 10^(-34) Js, velocity of light, c = 3 xx 10^8 m s^(-1), Avogadro’s constant, N_A = 6.0237 xx 10^23 "mol"^(-1) ).

Answer»

Solution :I.E. of hydrogen atom in the ground state `= 1312 kJ "mol"^(-1)`
Energy of hydrogen atom in the first orbit `(E_1) = -I.E = -1312 kJ "mol"^(-1)`
Energy of hydrogen atom in the nth orbit `(E_n) = (-1312 )/(n^2) kJ "mol"^(-1)`
Energy of hydrogen atom in the SECOND orbit `(E_2) = - (1312)/(2^2) = -328 kJ "mol"^(-1)`
`Delta E = E_2 - E_1 = [-328 - (-1312)] kJ = 984 kJ "mol"^(-1)`
Energy released per atom` = (Delta E)/(N) = (984 XX 10^3 J//"atom")/(6.023 xx 10^23) `
` (Delta E)/(N) = hv = H c/lamda ,therefore lamda = (Nh_1 c)/(Delta E)`
` therefore lamda = (6.626 xx 10^(-34) Js xx 3 xx 10^8 ms^(-1) xx 6.0237 xx 10^23)/(984 xx 10^3 J) = 1.2 xx 10^(-7) m`
Discussion
29664.

The increasing order of the rate of HCN addition to compounds I to IV is(I) HCHO (II)CH_3COCH_3(III) ph COCH_3 (IV)phCOph

Answer»

`A LT B lt C lt D`
`D lt B lt C lt A`
`D lt C lt B lt A`
`C lt D lt B lt A`

ANSWER :C
Discussion
29665.

The ionization constant of phenol is higher than that of ethanol because :

Answer»

phenoxide ION is bulkjer than ethoxide
phenoxide ion is stronger base than ethoxide
phenoxide ion is STABILIZED through delocalization
phenoxide ion is less stable than ethoxide

Solution :`C_(6)H_(5)+H_(2)O harrC_(6)H_(5)O^(-)+H_(3)O^(+)`
The phenoxide ion is stable due to reasonance.
The negative charge is delocalized in the benzene ring which is a stabilizing factor in the phenoxide ion and because of this REASON ionization constant of phenol is higher whereas no resonance is possible in alkoxide IONS (`RO^(-)`) derived from alcohol. The negative charge is localized on oxygen atom in case of alcohols.
Discussion
29666.

The ionization constant of nitrous acid is 4.5 times 10^-4. Calculate the pH of 0.04 M solution nitrite solution and also its degree of hydrolysis.

Answer»

SOLUTION :Sodium nitrite is a salt of WEAK acid, strong base, HENCE,
`K_h=2.22 times 10^-11 K_w//K_a=10^-14//(4.5 times 10^-4)`
`h=sqrt(K_h//C)=sqrt(2.22 times 10^-11//0.04)=sqrt(5.5 times 10^-11)=2.36 times 10^-5`
INITIAL c
After hydrolysis c-ch ch ch
`[OH^-]=ch=0.04 times 2.36 times 10^-5=9.44 times 10^-7`
`pOH=-log(9.44 times 10^-7)=7-0.9750=6.03`
`pH=14-pOH=14-6.03=7.97`
Discussion
29667.

The increasing order of octane number is -

Answer»

n-ALKANE `LT` BRANCHED alkane `lt` CYCLOALKANE `lt` aromatic hydrocarbon
aromatic hydrocarbon `lt` cycloalkane `lt` branched alkane `lt` n-alkane
cycloalkane `lt`branched alkane `lt` aromatic hydrocarbon `lt` n-alkane
aromatic hydrocarbon `lt` cyloalkane `lt` n-alkane `lt` branched alkane

Answer :A
Discussion
29668.

The ionization constant of NH_(4)^(+) in water is 5.6xx10^(-10) at 25^(@)C. The rate constant for the reaction of NH_(4)^(+) and OH^(oplus) to form NH_(3) and H_(2)O at 25^(@)C is 3.4xx10^(10)L mol^(-1)s^(-1). Calculate the rate constant for proton transfer from water of NH_(3).

Answer»

`6.07xx10^(-5)s^(-1)`
`5.6xx10^(10)s^(-1)`
`6.07xx10^(5)s^(-1)`
`6.8xx10^(10)s^(-1)`

ANSWER :C
Discussion
29669.

The ionisation constant of benzoic acid (PhCOOH) is 6.46 xx 10^(-5) and K_(sp) for silver benzoate is 2.5 xx 10^(-3). How many times is silver benzoate more soluble in a buffer of pH 3.19 compared to its solubility is pure water?

Answer»

4
3.32
3.01
2.5

Answer :B
Discussion
29670.

The increasing order of electron affinity of the electronic configurations of element is :- (I) 1s^(2) 2s^(2)2p^(6)3s^(2)3p^(5) (II) 1s^(2) 2s^(2)2p^(3) (III) 1s^(2)2s^(2)2p^(5) (IV) 1s^(2)2s^(2)2p^(6) 3s^(1)

Answer»

`II LT IV lt III lt I`
`I lt II lt III lt IV`
`I lt III lt II lt IV`
`IV lt III lt II lt I`

SOLUTION :Order of E.A.
Discussion
29671.

The ionization constant of ammonium hydroxide is 1.77xx 10^(-5) at 298 K. Hydrolysis constant of ammonium chloride is

Answer»

`5.65 XX 10^(-10)`
`6.50 xx 10^(-12)`
`5.65 xx 10^(-13)`
`5.65 xx 10^(-12)`

Solution :`K_(h) = (K_(w))/(K_(b)) = (1 xx 10^(-14))/(1.77 xx 10^(-5)) = 5.65 xx 10^(-10)`
Discussion
29672.

The ionization constant of acetic acid is 1.74 times 10^-9. Calculate the degree of dissociation of acetic acid in its 0.05 M solution. Calculate the concentration of acetate ions in the solutions and its pH.

Answer»

SOLUTION :`CH_3COOH leftrightarrowCH_3COO^(-) +H^+`
`K_a=([CH_3COO^(-)][H^+])/([CH_3COOH])=[H^+]^2/([CH_3COOH])`
(or) `[H^+]=SQRT(K_a(CH_3COOH))=sqrt((1.74 times10^-5)(5times10^-2))=9.33 times10^-4M`
`[CH_3COO^(-)]=[H^+]=9.33 times10^-4M`
`pH=-log(9.33 times10^-4)=4-0.9699=4-0.97=3.03`
Discussion
29673.

The increasing order of electron affinity of the electronic configuration of element is : (I) 1s^(2)2s^(2)2p^(6)3s^(2)3p^(5) (II) 1s^(2)2s^(2)2p^(3) (III) 1s^(2)2s^(2)2p^(5) (I) 1s^(2)2s^(2)2p^(6)3s^(1)

Answer»

`II LT IV lt III lt I`
`I lt II lt III lt IV`
`I lt III lt II lt IV`
`IV lt III lt II lt I`

ANSWER :A
Discussion
29674.

The ionization constant of a weak acid is 1.6xx10^(-5) and the molar conducitvity at infinite dilutionis 380xx10^(-4)" S " m^(2) mol^(-1). If the cell constant is 0.01 m^(-1), then the conductance of 0.01 M solutionis :

Answer»

`1.52xx10^(-5)" S "`
`1.52" S "`
`1.52xx10^(-3)" S "`
`1.52xx10^(-4)" S "`

Solution :(d) For a WEAK acid `K_(a)=C alpha^(2)`
`K_(a)=C((lambda_(m)^(c ))/(lambda_(m)^(alpha)))^(2)`
`1.6xx10^(-5)=0.01((lambda_(m)^(c ))/(380xx10^(-4)))^(2)`
`16xx10^(-6)=10^(-2)((lambda_(m)^(c ))/(380xx10^(-4)))^(2)`
`(lambda_(m)^(c ))/(380xx10^(-4))=(16xx10^(-4))^(1//2)=4XX10^(-2)`
`lambda_(m)^(c )=(380xx10^(-4))xx(4xx10^(-2))`
`=152xx10^(-5)=1.52xx10^(-3)`
Discussion
29675.

The ionization constant of a certain weak acid is 10^(-4). What should be the [salt] to [acid] ratio if we have to prepare a buffer with pH = 5 using this acid and one of the salts

Answer»

`1:10`
`10:1`
`5:4`
`4:5`

Solution :`PH = -LOG K_(B) + log.(["salt"])/(["acid"])`
`5 = - log 10^(-4) + log.(["salt"])/(["acid"])`
`log.(["salt"])/(["base"]) = 1`
`(["salt"])/(["base"]) = "antilog 1" = 10 : 1`.
Discussion
29676.

The increasing order of diazotization of the following compound is :

Answer»

`(a) LT (B) lt (C) lt (d)`
`(a) lt (d) lt (b) lt (c)`
`(a) lt (d) lt (c) lt (b)`
`(d) lt (c) lt (b) lt (a)`

ANSWER :C
Discussion
29677.

The ionization constant of a weak acid is 1.6 xx 10^(-5) and the molar conductivity at infinite dilution is 380 xx 10^(-4)sm^(2) mol^(-1). If the cell constant is 0.01m^(-1), then conductance of 0.01M acid solution is

Answer»

`1.52 XX 10^(-5) s `
`1.52 s `
`1.52 xx 10^(-3)s`
`1.52 xx 10^(-4)` s

Solution :`K_a = C alpha^(2) , alpha = sqrt((K_a)/(c)) = sqrt((1.6 xx 10^(-5))/(10^(-2))) = sqrt(1.6 xx 10^(-3)) = sqrt(16 xx 10^(-4))`
` alpha = 4 xx 10^(-2) = 0.04 , alpha = (^^_M)/(^^_(M)^(0)) implies ^^_M = alpha = ^^_(M)^(0) = 0.04 xx 380 xx 10^(-4) implies ^^_M(K xx 1000)/(M)`
` K=(^^_M xx M)/(1000) = (0.04 xx 380 xx 10^(-4) xx 0.01)/(1000) , K=0.152 xx 10^(-7) = 1.52 xx 10^(-8)`
`C = (K)/(G) = (1.52 xx 10^(-8))/(0.01) = 1.52 xx 10^(-6) , C = 1.52 xx 10^(-6) S - m^(3) =1.52S CM^(3)`
Discussion
29678.

The ionisation potentials of Li and K are 5.4 and 4.3 eV respectively. The ionization potential of Na will be:

Answer»

9.7 eV
1.1 eV
4.9 eV
cannot be calculated

Answer :C
Discussion
29679.

The increasing order of bond order of O_2 , O_2^+ , O_2^- and O_2^(2-) is

Answer»

`O_2^+ , O_2 , O_2^- , O_2^(2-)`
`O_2^(2-) , O_2^- , O_2, O_2`
`O_2, O_2 , O_2^- , O_2^(2-)`
`O_2^(2-) , O_2^- , O_2 , O_2^+`

Solution :In `O_2^(2-)` and `O_2^-`the added extra electron enters the antibonding orbital which decreases the bond ORDER. In `O_2^+` the electron is removed from antibonding orbital which INCREASES the bond order.
Discussion
29680.

The ionisation potential order for which set is correct ?

Answer»

`CS lt Li lt K`
`Cs gt Li gt B`
`Li gt K gt Cs`
`B gt Li gt K`

SOLUTION :While moving down in a GROUP, EFFECTIVE nuclear attraction decreases due to addition of new orbits. As a RESULT ionisation potential decreases. Hence, the correct order is `LigtKgtCs.`
Discussion
29681.

The ionisation potential of X^- ion is equal to

Answer»

The ELECTRON effinity of X atom
The ELECTRONEGATIVITY of X atom
The IONIZATION POTENTIAL of X atom
None

Answer :A
Discussion
29682.

(I) 1,2-Dihydroxy benzene (II) 1,3-Dihydroxy benzene (III) 1,4-Dihydroxy benzene (IV) Hydroxy benzene The increasing order of boiling points of the above-mentioned alcohols is:

Answer»

`I lt II lt IV ltIII`
`I lt II lt III lt IV`
`IV lt II lt I lt III`
`IV lt I lt II lt III`

SOLUTION :I : Intra MOLECULAR H-Bonding , III : Inter molecular H-Bonding.
Discussion
29683.

The ionisation potential of transition metals is _____ than p-block elements:

Answer»

LESS
More
Equal
None

Answer :A
Discussion
29684.

The ionisation potential of hydrogen from ground state to the first excited state is

Answer»

`-13.6 EV`
13.6 eV
`-3.4 eV`
3.4 eV

Solution :The `1^(ST)` I.P. for HYDROGEN is 13.6 eV.
Discussion
29685.

The ionisation potential of hydrogen atom is 13.6eV. The energy required to remove an electron in the n=2 state of hydrogen atom is

Answer»

27.2eV
13.6eV
6.8eV
3.4eV

Answer :D
Discussion
29686.

The increasing order of boiling points of below mentioned alcohols is(a) 1,2 - dihydroxy benzene (b) 1,3- dihydroxy benzene © 1,4- dihydroxy benzene (d)hydroxy benzene

Answer»

a LT B lt C lt d
a lt b lt d lt c
d lt a lt b lt c
d lt b lt a lt c

ANSWER :C
Discussion
29687.

The ionisation isomer of [Cr(H_(2)O)_(4)Cl(NO_(2))]Cl is

Answer»

`CR(H_(2)O)_(4)(O_(2)N)]Cl_(2)`
`[Cr(H_(2)O)_(4)Cl_(2)](NO_(2))`
`[Cr(H_(2)O)_(4)Cl(ONO)]Cl`
`[Cr(H_(2)O)_(4)Cl_(2)(NO_(2))]H_(2)O`

Solution :Ionisation isomer of `[Cr(H_(2)O)_(4)Cl(NO_(2))]Cl` is `[Cr(H_(2)O)_(4)Cl_(2)]NO_(2)`.
Discussion
29688.

The ionisation isomers of [Cr(H_2O)_4(NO_2) ]Cl is …………….

Answer»

`[CR(H_2O)_4Cl_2(NO)_2]`
` [Cr(H_2O)_4Cl_2]NO_2`
`[Cr(H_2O)_4Cl(ONO) ]`
both (a ) and (B)

ANSWER :B
Discussion
29689.

The increasing order of basicity of the following compounds is:

Answer»

`B LT a lt C lt d`
`b lt a lt d lt c`
`d lt b lt a lt c`
`a lt b lt c lt d`

ANSWER :B
Discussion
29690.

The ionisation enthalpy values of alkali metals indicate that lithium should be poorest reducing agent because of its high ionization energy. However, it is the strongest reducing agent among alkali metals. This may be explained because of:

Answer»

LOW sublimation energy of LITHIUM
reducing power of an element is REVERSE to its IONISATION potential.
high heat of HYDRATION of lithium
high lattice energy of lithium compounds

Answer :C
Discussion
29691.

The increasing order of Ag^(+) ion concentration in I. Saturated solution of AgCl II. Saturated solution of Agl III. 1MAg(NH_(3))_(2)^(+)" in "0.1 M NH_(3) IV. 1MAg(CN)_(2)^(-)" in "0.1 M KCN Given : K_(sp)" of "AgCl=1.0xx10^(-10) K_(sp)" of "Agl=1.0xx10^(-16) K_(d)" of "Ag(NH_(3))_(2)^(+)=1.0xx10^(-8) K_(d)" of "Ag(CN)_(2)^(-)=1.0xx10^(-21)

Answer»

`I lt II lt III lt IV`
`IV lt III lt II lt I`
`IV lt II lt III lt I`
`IV lt II lt I lt III`

Answer :C
Discussion
29692.

The ionisation enthalpy of sodium is 495 kJ "mol"^(-1). How much energy is needed to convert atoms present in 2.3 mg of sodium to sodium ions?

Answer»

24.75 J
49.5 J
99 J
23 J

Solution :I.E. `=495kJ"MOL"^(-1)`
ENERGY needed to ionize(1 mol) 23 g of SODIUM vapiur `=495kJ`
Energy needed to ionize `2.3xx10^(-3)g` of sodium vapour `=495/23xx 2.3xx10^(-3)=49.5J`
Discussion
29693.

The ionisation enthalpy of second period elementsvary with atomic number as The elements present at points B and E are respectively

Answer»

Be,C
B,N
Be,O
Be,N

Solution :The electronic CONFIGURATION of Be = `1s^2 , 2s^2` (fully filled) . The electronic configuration of N=`1s^2, 2s^2, 2p^3` (half filled) . Due to stable electronic configuration of Be and N the IE of these elements is more than elements of the NEXT group .
Discussion
29694.

The increasing order of acidity of the following is _________.

Answer»

`H_2S LT H_2Se lt H_2Te`
`H_2Se lt H_2S lt H_2Te`
`H_2Te lt H_2S lt H_2Se`
`H_2Se lt H_2Te lt H_2S`

ANSWER :A
Discussion
29695.

The ionisation enthalpy of nitrogen is larger than that of oxygen because of:

Answer»

<P>Larger size of N -atom
Greater ATTRACTION of electrons by the nucleus
Stability of half FILLED p- SUBSHELL configuration of N-atom
Greater penetration effect.

Answer :C
Discussion
29696.

The increasing order of acidity among phenol, p-methylphenol, m-nitrophenol and p-nitrophenol is

Answer»

m-nitrophenol,p-nitrophenol, PHENOL, p-methylpenol
p-methylphenol,m-nitrophenol, phenol,p-nitrophenol
p-methylphenol, phenol, m-nitrophenol,p-nitrophenol
Phenol, p-methylphenol, p-nitrophenol, m-nitrophenol

Solution :
Discussion
29697.

The ionisation energy of solid NaCl is 180 kcal, per mol. The dissolution of the solid in water in the from of ions is endothermic to the extent of 1 kcal per mol, If the solvation energies of Na^+ and Cl^- ions are in the rstio 6:5, what is the enthalpy of hydration of sodium ion:

Answer»

`-85.6 kcal/mol`
`-97.5 kcal/mol`
`82.6 kcal/mol`
`+100 kcal/mol`

ANSWER :B
Discussion
29698.

The ionisation energy of hydrogen is:

Answer»

LOWER than ALKALI metals
Lower than halogens
Closer to alkali metals
Closer to halogens

Answer :D
Discussion
29699.

The increasing order of acidic nature of Li_2O, BeO,B_2O_3,CuO is :

Answer»

`Li_2Olt BeOltCuOltB_2O_3`
`BeOltCuOltB_2O_3ltLi_2O`
`Li_2Olt CuOltBeOltB_2O_3`
`B_2O_3lt CuOltBeOltLi_2O`

SOLUTION :On MOVING left to right in periodic table, ACIDIC nature of oxide increases.So, METALLIC oxides are BASIC, non-metallic oxides are acidic and BeO is amphoteric.
Discussion
29700.

The ionisation energy of hydrogen atom is 13.6eV. What will be the ionisation energy of He^(+) and Le^(2+) ions?

Answer»

SOLUTION :54.4 EV, 122.4eV
Discussion