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29701.

The increasing order of acid strength is ...

Answer»

`CH_3COOH GT Cl_2CHCOOH gt "CCl"_3 CCOOH gt ClCH_2COOH`
`Cl_3"CC"OOH gt Cl_2CH COOH gt ClCH_2COOH gt CH_3COOH`
`CH_3COOH LT Cl_2CHCOOH lt "CCl"_3"CC"OOH lt ClCH_2COOH`
`Cl_2CH - COOH lt "CCl"_3 COOH lt ClCH_2COOH lt CH_3COOH`

SOLUTION :`Cl_3"CC"OOH gt Cl_2CH COOH gt ClCH_2COOH gt CH_3COOH`
29702.

The ionisation energy of hydrogen atom (in the groundstate) is x kJ. The energy required for an electron to jump from 2^(nd) orbit to the 3^(rd)orbit will be

Answer»

x/6
5x
7.2x
5x/36

Solution :`(I.E)_H=E_oo- E_1=-E_1=x`
PUT `E_1=-K/n^2=-K/1^2=-K, THEREFORE K=x`
`DeltaE=E_3-E_2=-K/3^2 - (-K/2^2)=K5/36=5/36x`
29703.

The increasing order of acid strength of hydrogen halide in water is

Answer»

HFgt HCL GT HBR gt HI
HCl gtHFgt HBr gt HI
HlgtHBrgtHClgtHF
HClgtHBrgtHFgtHI

Answer :C
29704.

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

Answer»

SOLUTION :`54.4eV, 122.4eV`
29705.

The increasing order of acid strength HClO_4 ,HClO_3 , HClO_2 HClO is

Answer»

HCLO LT `HClO_2`lt `HClO_3` lt `HClO_4`
`HClO_4`lt`HClO_3`lt`HClO_2`ltHClO
`HClO_4`lt`HClO_2`lt`HClO_3`ltHClO
None of these

Answer :A
29706.

The ionisation energy of He^(+) " is " 19.6 xx 10^(-18)J per atom. Calculate the energy of the first stationary state of Li^(2+)

Answer»

Solution :The ionisation energy of `He^(+) " is " 19.6 xx 10^(-18)J` per ATOM
`therefore` energy of the first ORBIT of `He^(+) (Z=2)= 19.6 xx 10^(-18)J`
`therefore` enegy of the first orbit of `H^(+) (Z=1) = (19.6 xx 10^(-18))/(4)J`
`therefore` energy of the first obit of `Li^(2+) (Z=3) = (19.6 xx 10^(-18))/(4) xx 9`
`=4.41 xx 10^(-17)J`
29707.

The ionisation energy of H atom is x kJ. The energy required for the electron to jump from n=2 to n=3 will be

Answer»

5X
36x/f
5x/36
9x/4

Solution :I.E. `=E_(OO)-E_(1)`
`=0-(-1311.8/1)=1311.8=x`
Energy required `=-1311.8/(3^(2))-(-1311.8/(2^(2)))`
`-x/9+x/4` or `(5x)/36`.
29708.

The increasing order basicity among the following is

Answer»

Y lt X lt Z
Y lt Z lt X
X lt Z lt Y
X lt Y lt Z

Answer :A
29709.

The ionisation energy of dioxygen (O_2) is very close to that of Xenon. Also F and O have the highest electronegativity and consequently can oxidise Xe among rare gases. So Xe forms a large number of compounds with F and O. Xe and F_2 are mixed and reacted at different temperatures to give XeF_2, XeF_4 and XeF_6. Xe also forms an unstable gaseous XeO_4 and soild XeO_3 which Is a very powerful explosive at higher temperatures .Some of the rare gases form clathrates or cage compounds by being entrapped in the cages of cystals laiitice of water, phenol of quinols. Helium can form intersitital compound with transition metals. Bigger members of rare gases do not form such compounds because of their large size. XeF_6 cannot be prepard by the method :

Answer»

`Xe+3F_2 underset"50 ATM"overset"475-532 K"to XeF_6`
`XeF_2+2F_2 overset"500 K"to XeF_6`
`XeF_4 + F_2 overset"475 K"to XeF_6`
`XeO_3+6HF underset(DELTA)overset("475 K")to XeF_6+3H_2O`

Answer :D
29710.

The ionisation energy of gaseous Na atoms is 495.8 KJ mol^(-1). The lowest possible frequency of light that can ionise a Na atom is

Answer»

`1.24xx10^(15) s^(-1)`
`4.76 XX 10^(14) s^(-1)`
`1.24xx10^(12) s^(-1)`
`3.15 xx 10^(14) s^(-1)`

Solution :E=hv or v=`E/h`
`=(495.8xx10^(3) "J/atom")/(6.02xx10^(23))xx(1)/(6.62xx10^(-34)JS)`
`=1.24xx10^(14)s^(-1)`
29711.

The increasing electron releasing tendencies of Cu , Ag , Fe and Zn are in the order :

Answer»

Ag , CU , FE , ZN
Cu , Ag , Fe , Zn
An , Cu , Fe , Ag
Fe , Zn , Cu , Ag

Answer :A
29712.

The ionisation energy of dioxygen (O_2) is very close to that of Xenon. Also F and O have the highest electronegativity and consequently can oxidise Xe among rare gases. So Xe forms a large number of compounds with F and O. Xe and F_2 are mixed and reacted at different temperatures to give XeF_2, XeF_4 and XeF_6. Xe also forms an unstable gaseous XeO_4 and soild XeO_3 which Is a very powerful explosive at higher temperatures .Some of the rare gases form clathrates or cage compounds by being entrapped in the cages of cystals laiitice of water, phenol of quinols. Helium can form intersitital compound with transition metals. Bigger members of rare gases do not form such compounds because of their large size. He and Ne do not from any clathrates because :

Answer»

(A) He and NE are very large in size .
(B) Being NEUTRAL they cannot from any POLAR BOUNDS with the host molecules.
(C) Being too small , they cannot be entrapped in the cagesof water,phenol or quinol.
(D) clathrates with He and Ne are HIGHLY expolsive.

Answer :C
29713.

The ionisationenergy of Ga is higher than that of Al because of _______

Answer»

more EFFECTIVE NUCLEAR charge of Ga
smaller atomic size of Ga
larger size of Ga
both (a) and (B)

ANSWER :A::B::D
29714.

The increasing basicity order of the following compounds is : (a) CH_3CH_2CH_2""(B)CH_3CH_2overset(CH_2CH_3)overset(|)(NH) H_(3)C-overset(CH_3)overset(|)N-CH_3""Ph-overset(CH_3)overset(|)(N)-H

Answer»

(A) lt (B) lt (D) lt (C )
(A) lt (B) lt(C ) lt (D)
(D) lt (C ) lt (A) lt (B)
(D) lt (C ) lt (B) lt (A)

Solution :For `Ph=overset(CH_(2))overset("|")"N "-H`, lone pair of N is involved in resonance, so it is LEAST basic
`CH_(3)CH_(2)NH_(2)""pK_(b)=3.29`
`{:(CH_(3)CH_(2)-NH""pK_(b)=3.00),("|"),(""CH_(2)CH_(3)):}`
`H_(3)C-UNDERSET(CH_(3))underset("|")"N "-CH_(3)""pK_(b)=4.22`
29715.

The increase in the value of molar conductivity of acetic acid with dilution is due to

Answer»

DECREASE in interionic FORCES
increase in DEGREE of ionisation
increase in self ionisation of water
none of these

Solution :increase in degree of ionisation
29716.

The ionisation energy of dioxygen (O_2) is very close to that of Xenon. Also F and O have the highest electronegativity and consequently can oxidise Xe among rare gases. So Xe forms a large number of compounds with F and O. Xe and F_2 are mixed and reacted at different temperatures to give XeF_2, XeF_4 and XeF_6. Xe also forms an unstable gaseous XeO_4 and soild XeO_3 which Is a very powerful explosive at higher temperatures .Some of the rare gases form clathrates or cage compounds by being entrapped in the cages of cystals laiitice of water, phenol of quinols. Helium can form intersitital compound with transition metals. Bigger members of rare gases do not form such compounds because of their large size. Xenon fromthe larges number of compounds only with oxygen and fuorine because: (i) oxygen and fluorine have very high electronegativity. (ii) lonisation energy of Xe is the largest among rare gases.(iii) lonisationenergy of Xe is low compared to those of other rare gases. (iv) low disscoiation energy of fluorine molecule compared to those of CI_2 and Br_2

Answer»

(A) (I) (ll) (ILL)
(B) (I), (Ill), (LV)
(C) (Ill), (lV)
(D) (I), (lV)

ANSWER :B
29717.

The ionisation energy of an element is

Answer»

The same as the electron AFFINITY of the element
Equal in magnitude but of opposite sign to the electron affinity of the element
The ENERGY released when an electron is added to an atom of the element
The energy required to REMOVE the outermost electron of an atom of the element

Answer :D
29718.

The ionisation constant of overset(oplus)NH_(4)ion in water5.6xx10^(-10)" at "25^(@)CThe rate constant for the reaction of overset(oplus)NH_(4) and oversetThetaOHion to formNH_(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 toNH_(3)

Answer»

Solution :Given `NH_(4)^(+)+H_(2)OhArrNH_(3)+H_(3)O^(+)`
`K_(a) = ([NH_(3)][H_(3)O^(+)])/([NH_(4)^(+)][H_(2)O]) (K_(a)= 5.6xx10^(-10))""...[1]`
`NH_(4)^(+)+BAR(O)Hunderset(k_(b))OVERSET(k_(f))hArrNH_(3)+H_(2)O`
`""(k_(f)=3.4xx10^(10)"L.mol"^(-1).s^(-1))""...[2]`
`K_(eq)=(k_(f))/(k_(b))=([NH_(3)][H_(2)O])/([NH_(4)^(+)][OH^(-)])`
`=([NH_(3)][H_(3)O^(+)])/([NH_(4)^(+)][H_(2)O])xx([H_(2)O]^(2))/([H_(3)O^(+)][OH^(-)])=(K_(a))/(K_(W))`
`thereforek_(b)=(K_(w))/(K_(a))xxk_(f)=(1.0xx10^(-14))/(5.6xx10^(-10))xx3.4xx10^(10)=6.07xx10^(5)`
29719.

The increase in solubility of iodine in an aqueous solution of potassium iodide is due to the formation of ................... .

Answer»

SOLUTION :`KI_(3)`
29720.

The ionisation constant of phenol is higher than that of ethanol because

Answer»

phenoxide ion is BULKIER than ethoxide.
phenoxide ion is stronger base than ethoxide
phenoxide ion is stablized through DELOCALISATION.
phenoxide ion is less STABLE than ethoxide ion.

ANSWER :C
29721.

The ionisation constant of NH_4^+ in water is 5.6xx10^(-10) at 25^@C the rate constant for the reaction of NH_4^+ and OH^- to form NH_3 and H_2O at 25^@C IS 3.4xx10^10L mol/sec Find the rate constant for proton transfer from water to NH_3?

Answer»

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

ANSWER :A
29722.

The ionisation constant of HF, HCOOH ,HCN at 298 K are 6.8 times 10^-4.1.8 times 10^-4 and 4.8 times 10^-9. Respectively. Calculate the ionisation constant of the corresponding conjugate base.

Answer»

Solution :(i) HF, CONJUGATE base F
`K_b=K_w//K_a=(1 times 10^-4)/(6.8 times 10^-4)=1.47 times 10^-11= 1.5times 10^-11`
(ii) for `HCOO^-`
`K_b=(1 times 10^-14)/(1.8 times 10^-4)=5.6 times 10^-11`
(iii) for `CN^-`
`K_b=(1 times 10^-14)/(4.8 times 10^-4)=2.8 times 10^-6`
29723.

The ionisation constant of ammonium hydroxide is 1.77xx10^(-5) at 298 K. Hydrolysis constant of ammonium chloride is :

Answer»

`6.5xx10^(-12)`
`5.65xx10^(-13)`
`5.65xx10^(-12)`
`5.65xx10^(-10)`

SOLUTION :`K_(h)=(K_(w))/(K_(B))=(10^(-14))/(1.77xx10^(-5))=5.6xx10^(-10)`
29724.

The ionisation constant of 0.2 M formic acid is 1.8xx10^(-4). Calculate its percentage ionisation.

Answer»

SOLUTION :FORMIC acid is a weak acid. Let is be represented as HA.
`underset(C(I-alpha))(HA) hArr underset(C alpha)( H^(+))+underset(C alpha)(A^-)`
ACCORDING to Ostwald's dilution law,
`K_a=C alpha^2. `
`therefore alpha=sqrt(K_a)/(C )`
`K_a=1.8xx10^(-4), C=0.2 M=2xx10^(-1) M`
` therefore alpha =sqrt((1.8xx10^(-4))/(2xx10^(-1)))=sqrt(9xx10^(-4)3xx10^(-2))`
Percentage of IONISATION `= 100 alpha `
`=10^2 xx3xx10^(-2)=3`
29725.

The increase in rate constant of a reaction is more when the temperature increases from

Answer»

290K - 300K
300K - 310K
310K - 320K
320K - 330K

Answer :A
29726.

The ionisation constant for NH_(4)^(+) in water is 5.0xx10^(-10)" at " 25^(@)C. The rate constant for the reaction of NH_(4)^(+) and OH^(-) to form NH_(3) and H_(2)O" at " 25^(@)C is 3.0xx10^(10)" Lmol"^(-1)s^(-1). The rate constant for proton transfer from water to NH_(3) is x xx 10^(5). The value of 'x' is

Answer»


ANSWER :6
29727.

The ionisation constant for acetic acid is 1.8xx10^(-5). At what concentration will it be dissociated to 2% ?

Answer»

1 M
0.018 M
0.18 M
0.045 M

Solution :`c=(K_(a))/(ALPHA^(2))=(1.8xx10^(-5))/((2xx10^(-2))^(2))`
`=4.5xx10^(-2)=0.045 M`
29728.

The ionic radius of._(57)La^(3+) is 1.06Å. Which one of the following given values will be closest to the ionic radius of ._21Lu^(3+)?

Answer»

1.06Å
1.56Å
0.85Å
1.35Å

Solution :IONIC RADIUS of `LU^(3+)` is smaller than that of `LA^(3+)`because of lanthanide contraction across the period.
29729.

Theionic radius ofCl^(-)ion is1.81 Å . The inter-ionic distances of NaCl andNaF are2.79 Årespectively. The ionic radiusof F^(-)ionwill be

Answer»

`0.98 Å`
`0.80 Å`
`1.33 Å`
`2.29 Å`

Solution :`d_(NACL)=r_(NA).^(+)+r_(Cl-)`
`2.79=r_(Na)+1.81`
or`r_(Na)+=2.79-1.81 Å =0.98 Å`
`d_(NAF)=r_(Na).^(+)+r_(F).^(-)`, i.e.,`2.31=0.98+r_(F).-`or
`r_(F).- =2.31-0.98=1.33 Å`
29730.

The increase in rate constant of a chemical reaction with increasing temperature is(are) due to the fact(s) that

Answer»

the NUMBER of collisons AMONG the reactant MOLECULES increases with increasing TEMPERATURE.
the activation energy of the reactions decreases with increasing temperature.
the corresponding of the reactant molecules increases with increasing temperature.
the number of rectant molecules acquiring the activation energy increases with increasing temperature.

SOLUTION :(a,d) Both the statements are correct.
29731.

The ionic radii of Rb^+ andI^- are 1.46overset@A and 2.16overset@A. The most probable type of structure exhibited by it is :

Answer»

CSCL TYPE
NaCl type
ZnS type
`CaF_2`type

Answer :B
29732.

The increase in boiling point of a solution containing 0.6 g urea in 200 g water is 0.50^(@)C, find the molal elevation constant.

Answer»

`1.0"K kg mol"^(-1)`
10 K kg `mol^(-1)`
10 K kg `mol^(-1)`
10 K kg mol

ANSWER :C
29733.

The increase in boiling points of noble gases from He to Xe is due to the

Answer»

decreases in ionization energy
increases in polarizability
increase in ELECTRON affinity
increase in ATOMIC volume.

Solution :As we go down the GROUP from He to Xe, the size of atom increases, due to which ease of polarization of noble GAS atom (or molecule) increases. Xe is 20 times more polarizable than He. The increasing polarizability has its effect on strengthening intermolecular interactions (dispersion forces), which in-turn increases the boiling and MELTING points. Increasing intermolecular attractions also make their liquefaction easier.
29734.

The incorrect statemetns amgong the following is

Answer»

The entropy of the universe REMAINS CONSTANT
HEAT cannot be completely converted into work
The absoltute entropy of a perferctly cystaline solid at absolute zero temperature is zero
The total energy of an isolates system remains contant

Answer :A
29735.

The incorrect structure of glycine at given pH are :

Answer»

`H_3overset(o+)NCH_2-UNDERSET(O)underset(||)C-OH" at " PH = 2.0`
`H_3overset(o+)NCH_2-underset(O)underset(||)C-O^(-) " at " pH = 6.0`
`H_2NCH_2-underset(O)underset(||)C-O^(-) " at " pH = 9`
`H_2NCH_2-underset(O)underset(||)C-OH" at " pH =12`

ANSWER :D
29736.

The Incorrect statements(s) among the following is/are :-

Answer»

The FIRST ionisation energy of calcium is more than first ionisation energy of Gallium
The second ionisation energy of copper is greater than that of potassium
The third ionisation energy of MG is greater than the third ionisation energy of Al
The `IE_(1)` of `Mg^(+)` is less than the `IE_(1)` of `NA^(+)`

Answer :B
29737.

The incorrect statement(s) regarding oxides of group-16 elements is/are

Answer»

Reducing property of dioxide DECREASES from `SO_(2) " to " TeO_(2)`
All these elements from oxides of the `EO_(2) and EO_(4)` types
Selenium and TELLURIUM do not FORM `SeO_(3) and TeO_(3)`
`SO_(2)` is an oxidising agent while `TeO_(2)` is a reducing agent

Answer :B::C::D
29738.

The incorrect statement(s) about Cr^(2+) and Mn^(3+)is (are) [Atomic numbers of Cr = 24 and Mn = 25]

Answer»

`CR^(2+)` is a reducing agent
`Mn^(3+)` is an oxidizing agent
Both `Cr^(2+) and Mn^(3+)` EXHIBIT `d^(4)` ELECTRONIC configuration
When `Cr^(2+)`is used as a reducing agent, the chromium ion ATTAINS `d^(5)` electronic configuration

Answer :D
29739.

The incorrect statements among the following is/are (I) NCl_5 does not exist while PCl_5 does (II) Lead prefers to form tetravalent compounds in the carbonate ion (III) The three C-O bonds are not equal in the carbonate ion (IV) Both O_2^(+) and NO one paramagnetic

Answer»

I,III, and IV
I and IV
II and III
I and III

Answer :C
29740.

The incorrect statement(s) among the following is/are

Answer»

the first ionization potential of AL is less than the first ionization potential of Mg
the SECOND ionization potential of Mg is greater than the second ionization potential of Na
the first ionization potential of Na is less than the first ionization potential of Mg
The third ionization potential of Mg is greater than the third ionization potential of Al.

SOLUTION :`IE_2` of Mg is lower than that of Na because in case of `Mg^(2+)` 3s-ELECTRON has to be REMOVED while in case of `Na^(+)`, an electron from the stable inert gas configruation `(Ne : 2s^2 2p^6)` has to be removed.
29741.

The incorrect statement(s) among the following is/are?

Answer»

In aqueous solution `NA` has stronger reducing character than `LI`.
`LiH` is thermally more stable than `NaOH`
`Li_(2)CO_(3)` is thermally more stable than `Na_(2)CO_(3)`.
`KO_(2)` is paramagnetic in nature.

Solution :(i) `Li` has stronger reducing character than `Na` in aqueous solution.
(ii) `Li_(2)CO_(3)` is thermally less stable than `Na_(2)CO_(3)`.
29742.

The incorrect statement (s) regarding 2M MhCl_(2) aqueous solution is/are (d_("solution") = 1.09 gm//ml)

Answer»

Molality of `CL^(-)` is 4.44 m
Mole fraction of `MgCl_(2)` is exactly 0.035
The conc. Of `MgCl_(2)` is 19% w/v
The conc. of `MgCl_(2) " is " 19 xx 10^(4)` ppm

Answer :B::D
29743.

The incorrect statement with respect to saccharin is

Answer»

Artificial sweetening agent
About 550 times as sweet as CANE sugar
Excreted from the BODY in urine
Chemical NAME is para-sulphobenzimide

Answer :D
29744.

The incorrect statement with respect to valence bond theory

Answer»

The electrons in the metal orbitals may undergo regrouping even against Hund's rule.
The electrons in an incompletely filled orbitals give rise to the RESULTANT MAGNETIC moment.
LIGAND orbitals OVERLAP the vacant metal orbitals to form a strong coordinate covalent bond to the extent possible.
The NUMBER of electrons lost is equal to the coordination number of the metal

Answer :D
29745.

The incorrect statement (s) regarding overline(*)(C)X_(3)

Answer»

If electronegativity of SURROUNDING element X is LESS than 2.5, then central carbon atom used almost 33% s-character in their HYBRID bonding orbital
If electronegativity of surrounding element X is less than 2.5, then central carbon atom used almost 25% s-character in their hybrid bonding orbital
If X is F, then species should be polar and pyramidal
If X is H, then species should be polar and planar

Solution :` (##BLJ_VKJ_ORG_CHE_C03_E02_006_S01.png" width="80%">
29746.

The incorrect statement regarding transition elements is

Answer»

They have low tensile STRENGTH
Their melting and BOILING points are very high
They are good conductors of heat
Their DENSITY is very high.

Solution :They have low tensile strength
29747.

The incorrect statement regarding to Noble gases is

Answer»

Their electronegetive values are zero
They are HELD together by VANDERWAALS forces
They OCCUPY the peaks in the graphs of ionisation potential and atomic NUMBER
Their boiling points decrease from He to Xe

Answer :D
29748.

The incorrect statement regarding structure of ozone is ________

Answer»

Bond angle is less than `120^(@)`
It is linear
The TWO oxygen-oxygen bond LENGTH in ozone are identical
Both (b) and (c).

ANSWER :A::B
29749.

The incorrect statement regarding O(SiH_(3))_(2) and OCl_(2) molecule is/are

Answer»

The strength of back BONDING is more in`O(SiH_(3))_(2)` molecule than `Ocl_(2)` molecule
`Si-O-Si` bond ANGLE in `O(SiH_(3))_(2)` is greater than `Cl-O-Cl` bond angle in `OCl_(2)`
The nature of back in both molecules is `2p_(x)-3d_(pi)`
Hybridisation of centralO-atom in both molecules is same

Solution : hybridisation of O-atom : `sp^(2)` due to more extent of `(2p_(pi)-3d_(pi))` back bonding.
Hybridisation of O-atom remains `sp^(3)`, becouse of LESS effective `(2p_(pi)-3d_(pi))` back bonding due to d-orbital resonance
29750.

The Incorrect statement regarding Nelson cell process

Answer»

`H_(2)andNaOH` are the by products in this process
At anode `2Cl^(-)to2e^(-)+Cl_(2)` LIBERATED
`2H_(2)O+2e^(-)toH_(2)+2OH^(-)` at cathode
Hydrogen chloride is oxidised to give chlorine

Answer :D