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201.

The emission of electrons from a metal surface exposed rto light radaition of appropriate wavelength is called photoelectroic effect .The emmited electron are called photo=-weklectron work function of threshold energy may be defined as the minimum amount of energy required to ejercted electron from a most surface .According to Einstein Maximum kinetic energy of ejected electron = Aborbed energy - Work function `(1)/(2) mv_(max)^(2) = h(v) - h(v_(n)) = hv [(1)/(lambda) -0 (1)/(lambda_(n))]` Where `v_(n)` and `lambda_(0)` are thereshold frequency and threshold wavelength respectively Sopping potential : it is the miximum potential at which the photoelectric current becomes zero if `V_(0)` is the stopping potential `eV_(0) = h(v- v_(0))` Which graph is correct ?A. B. C. D.

Answer» Correct Answer - C
202.

A photon of `3000 Å` is obserbed by a gas and then re-emmited as two photon .One photon in red (`7600 Å)` what would be the wavelength of the other photon ?

Answer» `hv = hv_(1) + hv_(2)`
`or (hv)/(lambda) = (hv)/(lambda_(1)) + (hv)/(lambda_(2))`
`or (1)/(3000) = (1)/(7600) + (1)/(lambda_(2))`
`or (1)/(3000) = (1)/(7600) + (1)/(lambda_(2))`
`or lambda_(2) = (228000)/(46) = 4956.5 Å`
203.

The emission of electrons from a metal surface exposed rto light radaition of appropriate wavelength is called photoelectroic effect .The emmited electron are called photo=-weklectron work function of threshold energy may be defined as the minimum amount of energy required to ejercted electron from a most surface .According to Einstein Maximum kinetic energy of ejected electron = Aborbed energy - Work function `(1)/(2) mv_(max)^(2) = h(v) - h(v_(n)) = hv [(1)/(lambda) -0 (1)/(lambda_(n))]` Where `v_(n)` and `lambda_(0)` are thereshold frequency and threshold wavelength respectively Sopping potential : it is the miximum potential at which the photoelectric current becomes zero if `V_(0)` is the stopping potential `eV_(0) = h(v- v_(0))`The folloeing figure indicates the energy livels of a certain atom .When the system moves from `2E` level to E lvel a photon of wavelength `lambda` is emitted .The wavelength of the photon produced during the transition from level `4E//3` to level E is A. `(lambda)/(3)`B. `(3lambda)/(4)`C. `(4 lambda)/(3)`D. `3 lambda`

Answer» Correct Answer - D
204.

The emission of electrons from a metal surface exposed rto light radaition of appropriate wavelength is called photoelectroic effect .The emmited electron are called photo=-weklectron work function of threshold energy may be defined as the minimum amount of energy required to ejercted electron from a most surface .According to Einstein Maximum kinetic energy of ejected electron = Aborbed energy - Work function `(1)/(2) mv_(max)^(2) = h(v) - h(v_(n)) = hv [(1)/(lambda) -0 (1)/(lambda_(n))]` Where `v_(n)` and `lambda_(0)` are thereshold frequency and threshold wavelength respectively Sopping potential : it is the miximum potential at which the photoelectric current becomes zero if `V_(0)` is the stopping potential `eV_(0) = h(v- v_(0))`The stopping potential as a function on electron frtequency is plotted for two photoelectric surface A abd B The graph show that the work function of A is A. Greater than that of BB. Smaller than that of BC. Same as that of BD. Such that no comparison can be done from given graph

Answer» Correct Answer - B
205.

The emission of electrons from a metal surface exposed rto light radaition of appropriate wavelength is called photoelectroic effect .The emmited electron are called photo=-weklectron work function of threshold energy may be defined as the minimum amount of energy required to ejercted electron from a most surface .According to Einstein Maximum kinetic energy of ejected electron = Aborbed energy - Work function `(1)/(2) mv_(max)^(2) = h(v) - h(v_(n)) = hv [(1)/(lambda) -0 (1)/(lambda_(n))]` Where `v_(n)` and `lambda_(0)` are thereshold frequency and threshold wavelength respectively Sopping potential : it is the miximum potential at which the photoelectric current becomes zero if `V_(0)` is the stopping potential `eV_(0) = h(v- v_(0))` Whaich of the following is the graph between the frequency (V) of the incident radiation and the stopping potential (v) ?A. B. C. D.

Answer» Correct Answer - C
206.

Assertion (A) : Nuclide `AI_(13)^(30)` is less stable than `Ca_(20)^(40)` Reason (R ) : Nuclide having odd number of proton and neuctrons are generally unstableA. If both (A) and (R ) correct and (R ) is the correct explanation for (A)B. If both (A) and (R ) correct and (R ) is the correct explanation for (A)C. If (A) is correct but (R ) is incorrectD. If (A) is incorrect but (R ) is correct

Answer» Correct Answer - A
`_(13)AI^(30)` is nuclide and it is less stable than `Ca_(20)^(40)` The given reason is correct explanation
207.

The five d-orbitals are designated as `d_"xy", d_"yz", d_"xz",d_(x^2-y^2)` and `d_z^2` . Choose the correct statementA. The shapes of the first three orbitals are similar but that of the fourth and fifth orbitals are differentB. The shapes of all five d-orbitals are similarC. The shapes of the first four orbitals are similar but that of the fifth orbital is differentD. The shapes of all five d-orbitals are different

Answer» Correct Answer - C
First four orbitals contain four lobes, while fifth orbital consists of only two lobes. The lobes of `d_"xy"` orbital lie between x and y axis. Similar is the case for `d_"yz"` and `d_"zx"` .Four lobes of `d_(x^2-y^2)` orbital are lying along x and y axis while two lobes of `d_z^2` orbital are lying along z-axis
208.

The ratio between potential energy and total energy of an electron in H-atom according to Bohr atomA. `1:-1`B. `1:1`C. `1:2`D. `2:1`

Answer» Correct Answer - D
`(P.E)/(T.E) = ((-e^(2))/(r))/((-e^(2))/(2r)) =2`
209.

STATEMENT-1: Half-filled and fully-filled degenerate orbitals are more stable. STATEMEHNT-2: Extra stabillity is due to the symmetrical distribution of electrons and exchange energy.A. Both (A) and (R) are true and (R) is the correct explanation of (A)B. Both (A) and (R) are true and (R) is not the correct explanation of (A)C. (A) is true but (R) is falseD. (A) is false but (R) is true

Answer» Correct Answer - A
2P orbitals are degenerate.
210.

STATEMENT-1: Half-filled and fully-filled degenerate orbitals are more stable. STATEMEHNT-2: Extra stabillity is due to the symmetrical distribution of electrons and exchange energy.A. If both the statement are TRUE and STATEMENT-2 is the correct explanation of STATEMENT-3B. If both the statements are TRUE but STATEMENT-2 is NOT the correct explanation of STATEMENT-2C. If STATEMENT-1 is TRUE and STATEMENT-2 is FALSED. If STATEMENT-1 is FALSE and STATEMENT-2 is TRUE

Answer» Correct Answer - A
Electronic configuration concept
211.

A compound of vanadium possesses a magnetic moment of `1.73BM`. The oxidation state of vanadium in this compounds is:

Answer» Correct Answer - 4
`mu_(s) = sqrt(n(n+2))` where `n = 1`
212.

Total number of orbitals associated with thrid shell will be.....A. 2B. 4C. 9D. 3

Answer» Correct Answer - C
Total no of orbital in a shell `=n^(2)`
213.

Which of the following statements (regarding an atom of H ) are correct?A. Kinetic energy of the electron is maximum in the first orbitB. potential energy of the electron is maximum in the first orbitC. radius of the second orbit is four times the radius of the first orbitD. Various energy levels are equally spaces.

Answer» Correct Answer - A::C
214.

Assertion (A) :F atom has less electron than `CI^(Theta)` atom Reason (R ) : Additional electrons are repelled more effectively by `3p` electron in `CI` atom than by` 2p` electron in F atomA. If both (A) and (R) are correct and (R) is the correct reason for (A).B. If both (A) and (R) are correct but (R) is not the correct explanation for (A).C. If (A) is correct but (R) is incorrect.D. If (A) is incorrect but (R) is correct.

Answer» Correct Answer - C
215.

Calculate the uncertainty in the velocity of a cricket ball of mass `150g`, if the uncertainty in its position in of the orer of `1Å`. `(h=6.6xx10^(-34)kg " "m^(2) s^(-1))`

Answer» `Deltax*mDeltav=(h)/(4pi)`
`Deltav=(h)/(4piDeltax*m)`
`=(6.6xx10^(-34))/(4xx3.143xx10^(-10)xx0.150)`
`=3.499xx10^(-24)ms^(-1)`.
216.

Which has a higher energy, a photon of violet light with wavelength 4000Ã… or a photon of red light with wavelength 7000Ã…? [`h=6.62xx10^(-34)Js`]

Answer» We know that `E=hv=h.(c)/(lamda)`
Given `h=6.62xx10^(-34)Js,c=3xx10^(8)ms^(-1)`
For a photon of vilet light,
`lamda=4000"Ã…"=4000xx10^(-10)m`
`E=6.62xx10^(-34)xx(3xx10^(8))/(4xx10^(-7))=4.96xx10^(-19)J`
For a photon of red light,
`lamda=7000` Ã… `=7000xx10^(-10)m`
`E=6.62xx10^(-34)xx(3xx10^(8))/(7000xx10^(-10))=2.83xx10^(-19)J`.
217.

The balmer series occurs between the wavelength of `[R = 1.0968 xx 10^7 m^-1]`.

Answer» For Balmer series,
`(1)/(lamda)=R[(1)/(2^(2))-(1)/(n^(2))]`
where, `n=3,4,5, . . Infty`
To obtain the limits for Balmer series n=3 and `n=infty` respectively.
`lamda_(max)(n=3)=(1)/(R[(1)/(2^(2))-(1)/(3^(2))])=(36)/(5R)`
`=(36)/(5xx1.0968xx10^(7))m=`6564Ã…
`lamda_("min")(n=infty)=(1)/(R[(1)/(2^(2))-(1)/(infty^(2))])=(4)/(R)`
`=(4)/(1.0968xx10^(7))m`
=3647Ã….
218.

Calculate the wavelength, wave number and frequency of photon having an energy to three electron volt. `(h=6.62xx10^(-27) erg-sec)`.

Answer» We know that,
`E=h*v`
`v=(E)/(h)" "(1eV=1.602xx10^(-12)erg)`
`=(3xx(1.602xx10^(-12)))/(6.62xx10^(-27))`
`=7.26xx10^(14)s^(-1)`
`=7.26xx10^(14)Hz`
`lamda=(c)/(v)=(3xx10^(10))/(7.26xx10^(14))=4.132xx10^(-5)cm`
`overline(v)=(1)/(lamda)=(1)/(4.132xx10^(-5))=2.42xx10^(4)cm^(-1)`.
219.

What is the mass of a photon of sodium light with a wavelength of `5800`Ã…? `(h=6.63xx10^(-27)erg-sec,c=3xx10^(10)cm//sec)`

Answer» `lamda=(h)/(mc)`
or `m=(h)/(hc)`
So, `m=(6.63xx10^(-27))/(5890xx10^(-8)xx3xx10^(10))`
`=3.752xx10^(-33)g`.
220.

Which one of the following pairs of ions have the same electronic configuration?A. `Cr^(+3),Fe^(+3)`B. `Fe^(+3),Mn^(+2)`C. `Fe^(+3),CO^(+3)`D. `Sc^(+3),Cr^(+3)`

Answer» Correct Answer - B
221.

Light of wavelength 12818Ã… is emitted when the electron of a hydrogen atom drops from 5th to 3rd orbit. Find the wavelength of the photon emitted when the electron falls from 3rd to 2nd orbit.

Answer» We know that,
`(1)/(lamda)=R[(1)/(n_(1)^(2))-(1)/(n_(2)^(2))]`
When `n_(1)=3 and n_(2)=5`
`(1)/(12818)=R[(1)/(9)-(1)/(25)]=(16R)/(9xx25)`
or `12818=(9xx25)/(16xxR)` . . .(i)
When `n_(1)=2 and n_(2)=3`
`(1)/(lamda)=R[(1)/(4)-(1)/(9)]=(5R)/(36)` . . . (i)
`lamda=(36)/(5R)`.
dividing (ii) by eqn. (i),
`(lamda)/(12818)=(36)/(5R)xx(16R)/(9xx25)=(64)/(125)`
`lamda=(64)/(125)xx12818=6562.8` Ã…
222.

The ratio of the velocity of light and the velocity of electron in the first orbit of a hydrogen atom. `[Given h = 6.624 xx 10^-27 erg-sec , m = 9.108 xx 10^-28 g, r = 0.529 xx 10^-8 cm]`.

Answer» `v=(h)/(2pimr)`
`=(6.624xx10^(-27))/(2xx3.14xx9.108xx10^(-28)xx0.529xx10^(-8))` ltbr `=2.189xx10^(8)cm//sec`
`(c)/(v)=(3xx10^(10))/(2.189xx10^(8))=137`.
223.

Which of the following is the electronic configuration of an atom in its first excited state if that atom is boelectronic with `O_(2)` ?A. `[Ne]3s^(2)3p^(4)`B. `[Ne]3s^(2)3p^(3)3d^(1)`C. `[Ne]3s^(1)3p^(3)`D. None of these

Answer» `16` electron `= 1s^(2) 2s^(2) 2p^(6)3s^(2)3p^(6)` (Two unpaired electrons)
Excited state `= [Ne]3s^(2)3p^(3) 3d^(6)`
Four unpaired electron)
224.

Which of the following is the electronic configuration of `H_(2)PO_(4)` ?A. `[Ne]`B. `[Ne]3s^(2)3p^(3)3d^(1)`C. `[Ne]3s^(1)3p^(3)`D. None of these

Answer» Correct Answer - `[Ne]`
`[Ne]`
225.

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

Answer» Ionisation energy of H atom `= 13.6 eV`
Ionisation energy of `He^(o+) = IE "for" H xx Z^(2) = 13.6 xx 4 = 54.4 eV`
Ionisation energy for `Li^(2+) = IE for H xx Z^(2) = 13.6 xx 9 = 122.4 eV`
226.

The total number of electrons presnt in ` 1` mL Mg Given density of ` ._(12) Mg^(24) = 1. 2 g//mL `.A. `0. 6 N`B. `6 N`C. ` 2N`D. 3N`

Answer» Correct Answer - A
` 1 mL -= 1.2 mg`, Also ` 24 g Mg` has ` 12 N` electrons .
227.

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

Answer» `E_(1)` for `He^(o+) = E_(1) H xx Z^(2)`
`13.6 xx 4 =54.4 eV`
`E_(1)` for `Li^(2+) = E_(1)` for `H xx Z^(2)`
`13.6 xx 9 = 122.4 eV`
228.

What is the minimum product of the uncertainty in position and the uncertainty in momentum of a moving electron?

Answer» Correct Answer - A
The minimum product of two uncertainties is equal to `h//4pi`
229.

If the electron is to be located within ` 5 xx 10^(-5)` what will be the uncertainty in the velocity ?

Answer» The electron is to exist within `5 xx 10^(-5) Å`
Let `Delta x = 5 xx 10^(-5) xx 10^(-10) m = 5 xx 10^(-15) m`
According to Heisembery uncrtainty principle
`Delta x - Delta p ge (h)/(4pi)`
`Delta x - m Delta v ge (h)/(4pi)`
`Delta x ge (h)/(4pi Delta x m)`
` ge (6.62 xx 10^(-34) kg m^(2)s^(-1))/(4 xx 3.14 xx 5 xx 10^(-15) m xx 9.1 xx 10^(-31) kg)`
` ge 16 xx 10^(19) m s^(-1)`
The uncertainty in the velocity of the electrn is `1.62 xx 10^(19) m s^(-1)`
230.

What is the minimum product of the uncrtainty in position and the uncertainty in momentum of a miving electron ?

Answer» The minimum product of two uncertainties is equal to `h//4pi`
231.

The ionisation energy of `He^(o+) is `19.6 xx 10^(-19) J "atom "^(-1)` .Calculate the energy of the first stationary state of `li^(2+)`

Answer» `E_(1)` for `Li^(2+) = E_(1)` for `H xx Z^(2) = E_(1)` for `H xx 9`
`E_(1)` for `He^(o+) = E_(1)` for `H xx Z^(2) = E_(1)` for `H xx 4`
` :. E_(1)` for` Li^(2+) = E_(1)` for `He^(o+) xx (9)/(4)`
`19.6 xx 10^(-19) xx (9)/(4)`
`= 44.1 xx 10^(-19) J "atom"^(-1)`
232.

Threshold frequency of metal is `f_(0)`. When light of frequency `v = 2f_(0)` is incident on the metal plate, velocity of electron emitted in `V_(1)`. When a plate frequency of incident radiation is `5f_(0), V_(2)` is velocity of emitted electron, then `V_(1):V_(2)` isA. `1:4`B. `1:2`C. `2:1`D. `4:1`

Answer» Correct Answer - B
`K.E. = h (v-v_(0)) Ϸ (1)/(2)mv^(2) = h(v-v_(0))`
`v^(2) = (2h(v-v_(0)))/(m), v_(1)^(2) = (2h(2f_(0)-f_(0)))/(m)` ...(1)
`v_(2)^(2) = (2h(5f_(0)-f_(0)))/(m)`...(2)
233.

The discovery of neutron becomes very late because.A. Neutrons are present in nucleusB. Neutrons are highly unstable particlesC. Neutrons are chargelessD. Neutrons do not move

Answer» Correct Answer - C
This is because chargeless particles do not undergo any deflection in electric or magnetic field unlike proton and electron.
234.

Calculate the wavelength and energy of radiation emitted for the electron transition from infinity to stationary state of the hydrogen atom

Answer» Correct Answer - A::B::C::D
`(1)/(lambda) = R [(1)/(n_(1)^(2)) - (1)/(n_(2)^(2))]`
` = 109678 cm^(-1)[(1)/(1^(2)) - (1)/(oo^(2))]`
`= 109678 cm^(-1)`
`lambda = (1)/(109678 cm^(-1)) = 9.118 xx 10^(-6) cm`
`E = (hc)/(lambda) = (6.62 xx 10^(-34) xx 3 xx 10^(8))/(9.118 xx 10^(-8)) = 2.178 xx 10^(18) J`
235.

The ionisation energy of `He^(o+)` is `19.6 xx 10^(-19) J "atom"^(-1)`. Calculate the energy of the first stationary state of `Li^(2+)`

Answer» Correct Answer - A::B::D
IE of an ion = IE of H atom `xx Z^(2)`
IE of H atom = `(IE of an ion)/(Z^(2))= (IE of an He^(o+))/(Z^(2))`
`= (19.6 xx 10^(-18))/(4) J = 4.9 xx 10^(-18) J`
`IE of Li^(2+) = 4.9 xx 10^(-18) J xx 3^(2) = 4.41 xx 10^(-17) J`
236.

Which of the following statements about the electron is incorrect?A. It is negatively charged particleB. The mass of electron is equal to the mass of neutronC. It is basic constituent of all atomD. It is a constituent of cathode rays

Answer» Correct Answer - B
mass of electron is less than that of neutron.
237.

The proton and neutron are collectively called asA. DeutronB. PositronC. MesonD. Nucleon

Answer» Correct Answer - D
Nucleus consists of proton and neutron both are called as nucleon.
238.

In an oil drop experiment , the following charge (in orbitrary units) were found on a series of oil droplets `2.30 xx 10^(-15),6.90 xx 10^(-15) xx 1.38 xx 10^(-14)``,5.75 xx 10^(-15), 3.45 xx 10^(-15),,1.96 xx 10^(-14),`The magnitude of charge on the electron (in the same unit) isA. `1.15 xx 10^(-15)`B. `2.30 xx 10^(-15)`C. `0.575 xx 10^(-15)`D. `1.69 xx 10^(-14)`

Answer» Correct Answer - A
The magnitude of the charge be smallest and other charged should be integral multiple of that smallest charges so, in this problem, smallest charge is `2.30 xx 10^(-15)` , but other charged are not integral multiple of this charge
So smaller charge is `1.15 xx 10^(-15)` because other charges are integral multiple of this charge
239.

In an oil drop experiment , the following charge (in orbitary units) were found on a series of all droplets `2.30 xx 10^(-15),6.90 xx 10^(-15) xx 1.38 xx 10^(-14),5.75 xx 10^(-15), 3.45 xx 10^(-15),,1.96 xx 10^(-15),`.Calculate the magnitude of the charge on the electron

Answer» `1.15 xx 10^(-15)` is the largest number that divides all the listed charge every .The smaller of the cahrge listed is `2.30 xx 10^(-15)` but this cahrge does not divide into all the other an even number of times `2.30 xx 10^(-15)` must represent tthe charge of two electron
240.

Light of wavelength `lambda` shines on a metal surface with initail X and the metal emit Y electron per second of average Z what will happen to Y and Z if X is doubled ?A. y will be doubled and z will become halfB. y will remains same and z will be doubledC. Both y and z will be doubledD. y will be doubled but z will remain same

Answer» Correct Answer - D
no of photo electrons ejected `alpha` intensity of radiation but KE is independent of intensity of radiation.
241.

Calculate the total number of proton neutron and electgron is `_(35)Be^(40)`

Answer» Correct Answer - D
Number of electron `= 35`, Number of proton `= 35` Number of neutron `= (80 - 35) = 45`
242.

The number of electrons,protons, neutron in a species are `18,16 `and `16` respectiveluy Assigs proper symbols

Answer» Correct Answer - A::B::C
Atomic number`1 = 16`
The species contain `2` electron more than the protons and hence carries `2` units negative charge .The element with atomic number `16` is sulphur (S)
Threfore , The symbol of the species in `_(16)^(32) S^(2-)`
243.

Light of wavelength `lambda` shines on a metal surface with intensity X and the metal emit Y electron per second of average energy Z what will happen to Y and Z if X is doubled ?A. Y will be doubled and Z will become halfB. Y will remain same and Z will be doubledC. Both Y and Z will be doubledD. Y will be doubled but Z will be remain same

Answer» When intensity is doubled the number of electrons emitted per second is also doubled but the average energy of photoelectrons emitted remain the same
244.

If the atomic weight oc C and SI are `12` and `28` respectively , then what is the ratio of the number of neatrons in them?A. ` 3: 7`B. ` 7 : 3`C. ` 3: 4`D. ` 6 : 28`

Answer» Correct Answer - A
Carbon is `_6 C^(12)` and silicon is ` _(14)Si^(28)`.
245.

The set of quantum numbers n=3, l=0, m=0, s=-1/2 belongs to the elementA. MgB. NaC. NeD. F

Answer» Correct Answer - A
`Mg_12` have `1s^2 2s^2 2p^6 3s^2` electronic configuration
n=3,l=0,m=0,`s=-1/2`
246.

The hydrogen -like species ` Li^(2+)` is in a spherically symmetric state ` S_1` whth one radisal node . Upon absorbing light the ion undergoes transitoj ot a state ` S_2` has one radial node and its energy is equal to the groun sate energy of hhe hydrogen atom. The state ` S_1` is :A. ` 1s`B. ` 2s`C. ` 2p`D. ` 3s`

Answer» Correct Answer - B
For ` s_1` (spherically suymmetrical ) state , number of bnodes =1 ltbRgt ` 1 = n-1`
` :. N=2`
i.e It is ` 2s`
For ` s_2` radial node =1
` E_(s_2) = (- 13.6 xx z^2)/n^2 =E_H` inground state =- 13. 6 ltbr. ` E= (- 13. 6 xx 9)/n^2 rArr n=3` ltbRgt So, statem `S_1` is ` 2s` and ` s_2` is ` 3p`.
247.

Calculate the number of electrons , protons and neatrons in the following species a. `._6C^(13)` b. `._6C^(12)`

Answer» a. `_(6)C^(12)`
Number of protons = Alomic number = Number of electronns = 6
Number of neutron = Mass number - Atomic number `= 13 - 6= 7`
b. `_(6)C^(13)`
Number of protons = Number of electronns = Alomic number = 6
Number of neutron = Mass number - Atomic number `= 12 - 6= 6`
248.

Which of the following set of quantum numbers represents the highest energy of an atom ?A. `n = 3, l = 1 , m = 1, s = + 1//2`B. `n = 3, l = 2, m = 1, s = + 1//2`C. `n = 4, l = 0, m = 0, s = + 1//2`D. `n = 3, l = 0, m = 0, s = + 1//2`

Answer» Correct Answer - B
(b) Because it has the maximum value of `n + 1`.
249.

Which of the following set of quantum number belongs to highest energy.A. `n = 4, l = 0, m = 0, s = + (1)/(2)`B. `n = 3, l = 0, m = 0, s = + (1)/(2)`C. `n = 3, l = 1, m = 1, s = + (1)/(2)`D. `n = 3, l = 2, m = 1, s = + (1)/(2)`

Answer» Correct Answer - D
(d) Orbitals are `4s, 3 s, 3 p` and `3 d`. Out of these `3d` has highest energy.
250.

Which ion has the maximum magnetic moment ?A. ` Mn^(3+)`B. ` Cu^(2+)`C. `Fe^(3+)`D. `V^(3+)`

Answer» Correct Answer - C
Magnetic moment ` = sqrt ( [n ( n + 2 ) ] ) ` where ( n) is number of unpaired electons . ` Fe^(3+) ` has five unpaired electrons .