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.

201.

If an electron and proton when in motion have the same wavelength associated with each of them, which would be moving faster and why ?

Answer» For electron and proton, h and `lambda` are constants. Therefore, `m_(e)v_(e)=m_(p)v_(p) or (v_(e))/(v_(p))=m_(p)//m_(e)`. Since mass of proton is more than that of electron, velocity of the electron will be more. Therefore, electron 0will be moving faster than proton.
202.

Calculate the number of protons, neutrons and electrons in `._(35)^(80)Br`.

Answer» Here, Z = 35, A = 80
`:.` No. of protons = Atomic No. = 35
No. of neutrons `= A - Z = 80 - 35 = 45`
As the atom is neutral, No. of electrons = No. of protons = 35
203.

The configuration `1s^(2)2s^(2)2p^(5)3s^(1)` shownA. excited state of `O_(2)`B. excited state of neonC. excited state of fluorineD. ground state of fluorine atom

Answer» Correct Answer - B
Acccording to the Aufbau approcah, the ground state eletron configuration of `10` electrons should be
`1s^(2)2s^(2)2p^(6)`
Thic corresponds to the configuration of `Ne`. Thus, the given configuration is the excited state configuration of neon. On excitation, one of teh `2p` electrons moves to higher enegry subshell `3s`.
204.

The electronic configuration of a dipositive ion `M^(2+)` is 2, 8, 14 and its atomic mass is 56. what is the number of neutrons in its nucleus ?

Answer» Total number of electrons in `M^(2)+"ion"=2 +8+14=24`
Total number of electrons in M=24+2=26
Atomic number of M=26
Number of neutrons in M=56-26=30
205.

Calculate the number of proton, neutrons and electrons in `._(19)^(39) K` .

Answer» `p= 19, e^(-) = 19 n =20`
206.

Many elements have non-intergral atomic masses because (1)the constituents neutrons, protons, and electrons, commbine to give fractional masses (2) they have isotopes (3) their isotopes have nonintergal masses (4) their isotopes have different massesA. `(i),(ii),(iii),(iv)`B. `(ii),(iii),(iv)`C. `(ii),(iv)`D. `(iii),(iv)`

Answer» Correct Answer - B
Many element occur in nature as mixtures of isotopes. The atomic mass (weight) of such an element is the weighted average of the masses of its isotopes. Thus, atmoic masses are fractional numbers, not integers. For example, `{:(Isot ope,%Abundance,Mass(am u),,),(._(12)^24Mg,78.99,23.98504,,),(._(12)^25Mg,10.00,24.98584,,),(._(12)^(26)Mg,11.01,25.98259,,):}`
Atomic mass (weight)
`=(0.7899)(23.98504u)+(0.1000)(24.98584) +(0.1101) (25.98259u)`
`=18.946u+2.4986u+ 2.860u`
`= 24.30u` (to four significant figures)
207.

The elements P (Atomic weight 39) and Q (Atomic weight 80) contain 20 and 45 neutrons respectively their nucleus. Give their electronic arrangements separately.

Answer» `Z (P) = 19, Z (Q) = 35`
208.

Calculate the frequency of infrared radiations having wavelength, `3 xx 10^(6)nm`.

Answer» Correct Answer - `10^(11) s^(-1)`
`lambda = 3 xx 10^(6) nm = (3 xx 10^(6)) xx 10^(-9) m = 3 xx 10^(-3) m, " " V = (c)/(lamda) = (3 xx 10^(8) ms^(-1))/(3 xx 10^(-3) m) = 10^(11) s^(-1)`
209.

The wavelength of an electron moving with velocity of `10^(7)ms^(-1)` isA. `7.27xx10^(-11)`mB. `3.55xx10^(-11)`mC. `8.25xx10^(-4)`mD. `1.05xx10^(-16)`m

Answer» Correct Answer - A
`lambda=h/(hv)=(6.626xx10^(-34)Js)/(9.11xx10^(-31)kgxx10^(7)ms^(-1))=7.27xx10^(-11)m`
210.

Write down the actual value of charge and mass of an electron ?

Answer» Charge `= 1.602 xx 10^(-19)` coulombs, Mass `= 9.11 xx 10^(-31)kg`
211.

The position of both, an electron and a helium atom, is known within 1.0 nm. Further, the momentum of the electron is known within `5.0 xx 10^(-26) kg ms^(-1)`. The minimum uncertainty in the measurement of momentum of helium atom isA. `50 kg ms^(-1)`B. `5.0 xx 10^(-26) kg ms^(-1)`C. `80 kg ms^(-1)`D. `80 xx 10^(-26) k ms^(-1)`

Answer» Correct Answer - B
`Deltax Delta p =` const. As `Delta x` is same, `Delta p` will be same
212.

In a photoelectric effect experiment irradiation of a metal with light of frequency `5.2 xx 10^(14) s^(-1)` yields electrons with maximum kinetic energy `1.3 xx 10^(-19) J`. Calculate the threshold frequency `(v_(0))` for the metal.

Answer» We known that, `hv=hv^(0)+KE" "` or `" " v^(0)=v-(KE)/(h)`
`KE =1.3 xx 10^(-19)J , v=5.2 xx 10^(14)s^(1) , h=6.626 xx 10^(-34)Js`
`:." "` Threshold frequency `(v^(0))=(5.2 xx 10^(14)s^(-1))-((1.3 xx 10^(-19)J))/((6.626 xx 10 ^(-34)Js))`
`=(5.2 xx 10^(14)s^(-1))-(1.96 xx 10^(14)s^(-1))=3.24 xx 10^(14 s^(-1)`
213.

We do not see a car moving as a wave on the road. Explain.

Answer» According to de Broglie relation `: lambda=(h)/(mv)` i.e. `lambda prop (1)/(m)`. The mass of the car is very large and its wavelength `(lambda)` or wave character is negligible. Therefore, we do not see a car moving like a wave.
214.

Match the following.

Answer» `Ararr4Brarr4Crarr2,3Drarr1,3`
A. Photon has particle nature as well as wave nature it exhibits both momentum and wavelength
B. Electron also has particle nature as well as wave nature. Thus, it also exhibits both momentum and wavelength.
C. `psi^(2)` represents probability density of electron and always has positive values.
D. Principal quantum numb er `n=4` for N-shell. ltbr. K L M N
n=1 2 3 4
It always has positive values.
215.

The number of electrons present in 3d of `Cu^(o+)` isA. 20B. 10C. 16D. 24

Answer» Correct Answer - 2
216.

The number of unpaired electrons in magnesium atom is

Answer» Correct Answer - 1
217.

How are `d^(xy)` and `d_(x^(2)-y^(2))` orbitals related to each other ?

Answer» Both `d_(xy)` and `d_(x^(2)-y^(2))` orbitals are degenerate orbitals (have same energy). However, `d_(xy)` orbital lies in the xy plane while `d_(x^(2)-y^(2))` orbital is along x and y axes.
218.

The maximum number of electrons that can be accomoddated in `d_(x^(2)-y^(2))` orbital isA. 10B. 5C. 2D. 1

Answer» Correct Answer - 3
219.

How many electrons with `l=2` are there in an atom having atomic number 54 ?A. 3B. 10C. 14D. 20

Answer» Correct Answer - D
220.

The correct sequence of energy of orbitals of multielectron species isA. `4p lt 3d lt 4s`B. `4s lt 4p lt 3d`C. `4s lt 3d lt 4p`D. `3d lt 4s lt 4p`

Answer» Correct Answer - 3
221.

What is the maximum number of electrons that can be associated with a following set of quantum numbers ? `(n = 3, l = 1 and m = -1)`.A. 2B. 10C. 6D. 4

Answer» Correct Answer - A
It is the correct answer.
222.

Electrons will first enterinto the orbital with the set of quantum numbersA. `n =5, l = 0`B. `n = 4, l = 1`C. `n = 3, l =2`D. any of these

Answer» Correct Answer - C
According to the `(n +l)` rule, the lowest the value of `(n+l)` for an orbital the lower is its enegry. Here, all the orbitals have equal values of `(n +l)`.
If two or more orbitals have the same value of `(n+l)`, the orbital with lowest value of `n` has lower enegry. Thus, orbital `(n = 3,l = 2)` is having the lower enegry.
223.

The ratio of the frequency corresponding to the third line in the Lymas series of hydrogen atomic spectrum to that of the first line in Balmer series of `Li^(2+)` spectrum isA. `(4)/(5)`B. `(5)/(4)`C. `(4)/(3)`D. `(3)/(4)`

Answer» Correct Answer - D
For Lyman series, `n_(1) = 1`. For 3rd line of Lyman series `n_(2) = 4`
For hydrogen, Z = 1. Hence,
`v_(H) = (c)/(lamda) = c R_(H) Z^(2) ((1)/(n_(1)^(2)) - (1)/(n_(2)^(2)))`
`= c R_(H) (1)^(2) ((1)/(1) - (1)/(4^(2))) = (15)/(16) R_(H.c)`
for lithium, Z = 3
For 1st line of Balmer series, `n_(1) = 2, n_(2) = 3`
`:. v_(Li) = c.R_(H) (3)^(2) ((1)/(2^(2)) - (1)/(3^(2)))`
`= c.R_(H) xx 9 xx (5)/(36) =(5)/(4) c.R_(H)`
`:. (v_(H))/(v_(Li)) = ((15//16) c R_(H))/((5//4) c R_(H)) = (15)/(16) xx (4)/(5) = (3)/(4)`
224.

Which of the following are not electromagnetic waves? (i) sound waves (ii) radiowaves (iii) `X` rays (iv) water wavesA. `(ii),(iii)`B. `(i),(iv)`C. `(i),(ii)`D. `(iii),(iv)`

Answer» Correct Answer - B
They are mechanical waves as they need a medium to travel.
225.

Calculate the energy of photon of light having frequency of `2.7 xx 10^(13) s^(-1)`

Answer» `E= 1.78 xx 10^(-20)J`
226.

What will be the mass of a particle if uncertainty in its position is `10^(-8)` m and velocity is `5.26xx10^(-25)ms^(-1)`?A. 0.01 kgB. 0.1 kgC. 1 kgD. 10 kg

Answer» Correct Answer - A
`Deltax.mDeltav=h/(4pi)
10^(-8)mxxmxx5.26xx10^(-25)ms^(-1)=(6.626xx10^(-34)Js)/(4xx3.14)
m=(6.626xx10^(-34)Js)/(4xx3.14xx10^(-8)mxx5.26xx10^(-25)ms^(-1))=0.01kg`
227.

Write the electronic configuration of `._(9)F^(19), ._(16)S^(32)` and `._(18)Ar^(38)` and then point out the element with : (i) Maximum nuclear charge (ii) minimum number of neutrons (iii) highest mass number (iv) maximum number of unpaired electrons.

Answer» `._(9)F^(19)=1s^(2) 2s^(2) 2p_(x)^(2) 2p_(y)^(2)2p_(z)^(1), ._(16)S^(32)=1s^(2) 2s^(2) 2p^(6) 3s^(2) 3p_(x)^(2)3p_(y)^(1) 3p_(z)^(1), ._(18)Ar^(38)=1s^(2) 2s^(2) 2p^(6) 3s^(2) 3p^(6)`
(i) Max. nuclear charge `=18` in `._(18)Ar^(38)` (ii) Minimum no. of neutrons `=10` in `._(9)F^(19)`
(iii) Maximum no. of unpaired electrons `=2` in `._(16) S^(32)`
228.

If travelling at same speeds, which of the following matter waves have the shortest wavelength?A. ElectronB. Alpha particle `(He^(2+))`C. NeutronD. Proton

Answer» Correct Answer - B
`lamda = (h)/(mv)`. For same value of v, larger the value of mass m, shorter is the wavelength, `lamda`. Hence `alpha`-particles have the largest mass
229.

Identify the pairs which are not of isotopes ?A. `._(16)^(12)X, ._(6)^(13)Y`B. `._(17)^(35)X, ._(17)^(37)Y`C. `._(16)^(14)X, ._(7)^(14)Y`D. `._(4)^(8)X, ._(5)^(8)Y`

Answer» Correct Answer - C::D
(c) and (d) are not isotopes because in each, atom have different atomic number
230.

How many electrons are present in the ion `._(26)^(56)Fe^(2+)` ?

Answer» `._(26)Fe` has 26 electrons. Hence, `Fe^(2+)` has `26 - 2 = 24` electrons.
231.

Consider the electronic configurations : (i) `1^(2) 2s^(1)` (ii) `1s^(2) 3s^(1)` (a) Name the element corresponding to (i) (b) Does (ii) correspond to the same or different element ? (c) How can (ii) be obtained from (i) ? (d) Is it easier to remove one electron from (ii) or (i) ? Explain.

Answer» (a) The element corresponding to (i) is Lithium (Li). (b) This electronic configuration represents the same element in the excited state. (c) By supplying energy to the element when the electron jumps from the lower energy 2s-orbital to the higher energy 3s-orbital.
(d) It is easier to remove an electron from (ii) than from (i) since inthe former case, the electron is present in a 3s-orbital which is away from the nucleus and hence is less strongly attracted by the nucleus than an electron in the 2s-orbital.
232.

Which of the following configurations represents the most electronegative element?A. `1s^(2)2s^(2)2p^(6)3s^(1)`B. `1s^(2)2s^(2)2p^(5)`C. `1s^(2)2s^(2)2p^(6)3s^(2)3p^(5)`D. `1s^(2)2s^(2)2p^(4)`

Answer» Correct Answer - B
Z = 9 is the atomic number of fluorine which is the most electronegative element.
233.

Which one of the following configurations represents a noble gas?A. `1s^(2)2s^(2)2p^(6)3s^(2)3p^(6)3d^(10)4s^(2)4p^(6)4d^(10)5s^(2)`B. `1s^(2)2s^(2)2p^(6)3s^(2)3p^(6)3d^(10)4s^(2)4f^(14)5s^(2)`C. `1s^(2)2s^(2)2p^(6)3s^(2)3p^(6)3d^(10)4s^(2)4p^(6)4d^(10)5s^(2)5p^(6)`D. `1s^(2)2s^(2)2p^(6)3s^(2)3p^(6)3d^(10)4s^(2)4p^(6)4d^(10)5s^(2)5p^(3)`

Answer» Correct Answer - C
The outermost shell configuration of the element shows 8 electrons hence it is a noble gas.
234.

The orbital angular momentum of an electron in a d-orbital is:A. B. C. D.

Answer» Correct Answer - C
Angular momentum `=sqrt(l(l+1))h` .
For d orbitl , l=2
Angular momentum `=sqrt(2(2+1))h=sqrt(6)`.
235.

What will be the wavelength of oxygen molecule in picometers moving with a velocity of `660 m s^(-1) (h = 6.6 xx 10^(-34) kg m^(2) s^(-1))`

Answer» Correct Answer - 18.8 pm
`m = (32 xx 10^(-3) kg)/(6.02 xx 10^(23)), 1` pm = `10^(-12)m`
236.

What is the lowest value of n that allows g orbitals to exist?A. 6B. 7C. 4D. 5

Answer» Correct Answer - D
For g orbital, value of l is 4.
Since l = n - 1, n should be 5.
237.

Which of the following expression respresents the electron probability function `(D)`?A. `4 pi rdr Psi^(2)`B. `4 pi r^(2)dr Psi`C. `4 pi r^(2)dr Psi^(2)`D. `4 pi dr Psi`

Answer» Correct Answer - C
is the correct answer.
238.

The total number of electrons present in all `s` orbitals, all the `p` orbitals, and all the `d` orbitals of cesium ion are, respectively,A. 8, 26, 10B. 10, 24, 20C. 8, 22, 24D. 12, 20, 22

Answer» Correct Answer - B
The electronic configuration of `Cs^(+)`(Z=55) is :
`1s^(2)2s^(2) 2p^(6) 3s^(2) 3p^(6)3d^(10)4s^(2)4p^(6)4d^(10)5s^(2)5p^(6)`.
No. of electrons in s-orbitals =10
No. of electrons in p-orbitals = 24
No. of electrons in d-orbitals =20
239.

The atomic number of an element is `35`. What is the total number of eletrons present in all the `p` orbitals of the ground state of that element?A. `6`B. `11`C. `17`D. `23`

Answer» Correct Answer - C
Accoridng to the Aufbau apporach, the ground state dectron configuraiton is
`1s^(2)2s^(2)2p^(6)3s^(2)3p^(6)3s^(10)4s^(2)3p^(5)`
Thus, the number of electron present in all the `p` orbitals is
`6 +6 +5 = 17`.
240.

The total number of electrons present in all `s` orbitals, all the `p` orbitals, and all the `d` orbitals of cesium ion are, respectively,A. `8, 26, 10`B. `10, 24, 20`C. `8, 22, 24`D. `12, 20, 22`

Answer» Correct Answer - B
The atomic numbe rof cseium is `55`. Therefore, its ground state electron configuration is
`1s^(2)2s^(2)2p^(6)3s^(6)3d^(10)4s^(2)4d^(6)4d^(10)5s^(2)5p^(6)6s^(1)`
or simply, `[Xe] 6s^(1)`
`Cs^(+)` ion will have `Xe` configuration.
Eletron present in the `s` subshell `= 5xx2 = 10`
Electrons present in the `p` subshell `= 4 xx 6 = 24`
Electron present in the `d` subshell `= 10 xx2 = 20`
241.

Ordinary 'lead' pencils actually are made of a form of carbon called graphite. If a pencil line is `0.35 mm` wide and the diameter of a carbon atom is `1.5 xx 10^(-10)m`, how many `C` atom wide is the line?A. `2.3xx10^(6)`atomsB. `7.8xx10^(6)`atomsC. `5.6xx10^(6)`atomsD. `9.7xx10^(6)`atoms

Answer» Correct Answer - A
Number of `C` atoms
`=("Width of the pencil line in meters")/("Diameter of a single "`C`" atom in meters")`
`= 0.35mm xx (1m)/(1000mm)xx (1C "atom")/(1.5xx10^(-10)m)`
`= 2.3 xx 10^(6)C` atoms
242.

The allowed values of magnetic quantum nunber `(m_(l))` are the integers fromA. `.-lto +(l-1)`B. `-(l - 1)`to +l`C. `-l` to +l`D. s

Answer» Correct Answer - C
Within a subshell, the value of `m_(i)` depends on the values of the angular momenutm quantum number `l`. For acertain value of `l`, there are `(2l +1)` intergral values of `m_(l)` as follows: `-l, (-l +1),..,0,….(+l+1), +l`. For `l = 0 (s` subshell), the allowed `m_(l)` quantum number is `0` only, there is only one orbital in the `s` subshell. For `l = 1 (p` subshell), `m_(l) =- 1, 0`, and `+l`, there are three different orbitals in the `p subshell. these orbitals have the same enegry and identical shape but different orientations in space.
243.

(i) Calculate the number of electrons which will together weight one gram. (ii) Calculate the mass and charge of one mole of electrons

Answer» (i) Mass of one electron `= 9.11 xx 10^(-31) kg, " i.e., " 9.11 xx 10^(-31) kg = 1` electron
`:. 1 g " i.e., " 10^(-3) kg = (1)/(9.11 xx 10^(-31)) xx 10^(-13)` electrons = `1.098 xx 10^(27)` electrons
(ii) Mass of one electron `= 9.11 xx 10^(-31) kg`
`:.` Mass of one mole of electrons `= (9.11 xx 10^(-31)) xx (6.022 xx 10^(23)) = 5.486 xx 10^(-7) kg`
Charge on one electron `= 1.602 x 10^(-19)` coulomb
`:.` Charge on one mole of electrons `= (1.602 xx 10^(-19)) xx (6.022 xx 10^(23)) = 9.65 xx 10^(4)` coulombs
244.

Isoelectronic species are :A. `F^(-), O^(2-)`B. `F^(-), O`C. `F^(-), O^(+)`D. `F^(-), O^(2+)`.

Answer» Correct Answer - A
Both have 10 electrons.
245.

Assertion. All isotopes of a given element show the same type of chemical behaviour Reason. The chemical properties of an atom are controlled by the number of electrons in an atom.A. If both assertion and reason are true, and reason is the true explanation of the arrertionB. If both assertion and reason are true, but reason is not the true explanation of the assertionalC. If assertion is true, but reason is falseD. If both assertion and reason are false

Answer» Correct Answer - A
R is the correct explanation of A
246.

Why Pauli exlusion principle is called exlusion principle ?

Answer» This is because according to this principle, if one electron in an atom has some particular values for the four quantum numbers, then all the other electrons in that atom are exclused from having the same set of values
247.

A photon of light with wavelength 700 nm has energy E. A photon of light of what wavelength would correspond to energy 2E ?

Answer» Correct Answer - 350 nm
`E=(hc)/(lambda) or E prop (1)/(lambda) `(`:.` h and c are both constants)
`(E_(1))/(E_(2))=(lambda_(2))/(lambda_(1))` Now `lambda_(1)=700 nm, E_(1)=E, E_(2)=2E , (E)/(2E) = (lambda_(2))/(700) or lambda_(2)=(700)/(2)=350 nm`.
248.

A transition element `X` has a configuration `[Ar] 3d^(4)` in its ` + 3` oxidation state. Its atomic number is

Answer» Configuration of the ion `=[Ar] 3d^(4)`
Oxidation state of the ion `=+3`
`:. ` The electronic configuration of the atom `=[Ar] 3d^(5)4s^(2)`
249.

Calculate the Einstein when the frequency of photon is `10^(10)` kHz.

Answer» Correct Answer - `39.87xx10^(2)J`
`v=10^(10) kH=10^(13)Hz=10^(13)s^(-1) , h=6.62xx10^(-34) J s, N_(0)=6.022xx10^(23)`.
Einstein = Energy of 1 mole of photons `=N_(0)xxhxxv`
`=6.022xx10^(23)xx(6.62xx10^(-34)Js)xx(10^(13)s^(-1))=39.87xx10^(2)J`
250.

How many unpaired electrons are present in pd(Z=46) ?

Answer» The electronic configuration of the element palladium (Z=46) is `[Kr]^(36) 4d^(10)5s^(0)`. This means that it has no unpaired electron.