Test: 35 Year JEE Previous Year Questions: Structure of Atom


29 Questions MCQ Test Chemistry 35 Years JEE Main & Advanced Past year Papers | Test: 35 Year JEE Previous Year Questions: Structure of Atom


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Attempt Test: 35 Year JEE Previous Year Questions: Structure of Atom | 29 questions in 60 minutes | Mock test for JEE preparation | Free important questions MCQ to study Chemistry 35 Years JEE Main & Advanced Past year Papers for JEE Exam | Download free PDF with solutions
QUESTION: 1

In a hydrogen atom, if energy of an electron in ground state is 13.6. ev, then that in the 2nd excited state is [2002]

Solution:

2nd excited state will be the 3rd energy level.

QUESTION: 2

Uncertainty in position of a minute particle of mass 25 g in space is 10–5 m. What is the uncertainty in its velocity (in ms–1) ? (h = 6.6 x 10-34 Js) [2002]

Solution:

TIPS/Formulae :

= 2.1 x 10-28 ms-1

QUESTION: 3

The number of d-electrons retained in Fe2+ [2003] (At. no. of Fe = 26) ion is

Solution:

Fe++ (26 – 2 = 24) = 1s2 2s2 2p6 3s2 3p6 4s0 3d6 hence no. of d electrons retained is 6. [Two 4s electron are removed]

QUESTION: 4

The orbital angular momentum for an electron revolving in an orbit is given by  This momentum for an s-electron will be given by [2003]

Solution:

TIPS/Formulae : For s-electron, ℓ = 0

∴Orbital angular momentum =

QUESTION: 5

Which one of the following groupings represents a collection of isoelectronic species ?(At. nos. : Cs : 55, Br : 35)[2003]

Solution:

N3–, F and Na+ contain 10 electrons each.

QUESTION: 6

In Bohr series of lines of hydrogen spectrum, the third line from the red end corresponds to which one of the following inter-orbit jumps of the electron for Bohr orbits in an atom of hydrogen [2003]

Planck’s constant, h = 6.63 × 10–34 Js

Solution:

The lines falling in the visible region comprise Balmer series. Hence the third line from red would be n1 =2,
n2 = 5 i.e. 5 → 2.

QUESTION: 7

The de Broglie wavelength of a tennis ball of mass 60 g moving with a velocity of 10 metres per second is approximately [2003]

Solution:

QUESTION: 8

Which of the following sets of quantum numbers is correct for an electron in 4f orbital ? [2004]

Solution:

The possible quan tum numbers for 4f electron are n = 4, ℓ = 3, m = – 3, –2 –1, 0, 1 , 2 , 3 and s =  

Of various possiblities only option (a) is possible.

QUESTION: 9

Consider the ground state of Cr atom (X = 24). The number of electrons with the azimuthal quantum numbers, ℓ = 1 and 2 are, respectively [2004]

Solution:

Electronic configuration of Cr atom (z = 24) = 1s2, 2s2 2p6, 3s2 3p6 3d5, 4s1
when ℓ = 1, p - subshell,
Numbers of electrons = 12
when ℓ = 2, d - subshell,
Numbers of electrons = 5

QUESTION: 10

The wavelength of the radiation emitted, when in a hydrogen atom electron falls from infinity to stationary state 1, would be (Rydberg constant = 1.097×107 m–1) [2004]

Solution:

TIPS/Formulae :

λ = 91.15 x 10 -9 m ≈ 91nm

QUESTION: 11

Which one of the followin g sets of ion s represents the collection of isoelectronic species? [2004]

(Atomic nos. : F = 9, Cl = 17, Na = 11, Mg = 12, Al = 13, K = 19, Ca = 20, Sc = 21)

Solution:

 

each contains 18 electrons.

QUESTION: 12

In a multi-electron atom, which of the following orbitals described by the three quantum members will have the same energy in the absence of magnetic and electric fields? [2005]
(A) n = 1, l = 0, m = 0
(B) n = 2, l = 0, m = 0
(C) n = 2, l = 1, m = 1
(D) n = 3, l = 2, m = 1
(E) n = 3, l = 2, m = 0

Solution:

The energy of an orbital is given by (n + l) in (d) and (c). (n + l) value is (3 + 2) = 5 hence they will have same energy, since there n values are also same.

QUESTION: 13

Of the following sets which one does NOT contain isoelectronic species? [2005] 

Solution:

Calculating number of electrons

Hence the species in option (b) are not isoelectronic.

QUESTION: 14

According to Bohr's theory, the angular momentum of an electron in 5th orbit is [2006]

Solution:

Angular momentum of an electron in nth orbital is given by ,

For n = 5, we have

Angular momentum of electron  

QUESTION: 15

Uncertainty in the position of an electron (mass = 9.1 × 10–31 kg) moving with a velocity 300 ms–1, accurate upto 0.001% will be [2006]

(h = 6.63 × 10–34 Js)

Solution:

Given m = 9.1 × 10–31kg , h = 6.6 × 10–34Js

= 0.003ms–1

From Heisenberg's uncertainity principle

 = 1.92 x 10-2 m

QUESTION: 16

Which one of the following sets of ions represents a collection of isoelectronic species? [2006]

Solution:

(a)

(b)  Li+ = 3+1= 4e,  Na+ = 11–1 = 10e,      
Mg++  = 12–2=10e      
Ca++ = 20 – 2 = 18e–       (not isoelectronic)

(c) K+ = 19 – 1= 18e, Cl =17 + 1  = 18e,      
Ca++ = 20 – 2 =18e, Sc3+ = 21–3 = 18e
                                          (isoelectronic)

(d) Ba++ 56 – 2 = 54e,  Sr++ 38–2 = 36e–      
K+= 9–1 = 18e,  Ca++= 20–2 = 18e
                                       (not  isoelectronic)

QUESTION: 17

Which of the following sets of quantum numbers represents the highest energy of an atom? [2007]

Solution:

(a) n = 3, ℓ = 0 means 3s-orbital and n + ℓ = 3
(b) n = 3, ℓ = 1 means 3p-orbital n + ℓ = 4
(c) n = 3, ℓ = 2 means 3d-orbital n + ℓ = 5
(d) n = 4, ℓ = 0 means 4s-orbital n + ℓ = 4
Increasing order of energy among these orbitals is 3s < 3p < 4s < 3d
∴ 3d has highest energy..

QUESTION: 18

Which one of the following constitutes a group of the isoelectronic species? [2008]

Solution:

Species having same n umber of el ectr on s ar e isoelectronic calculating the number of electrons in each species given here, we get.
CN (6 + 7 + 1 = 14); N2 (7 + 7 = 14);
O22–(8 + 8 +2 = 18) ; C22– (6 + 6 + 2 = 14);
O2 (8 + 8 + 1 = 17) ; NO+ (7 + 8 – 1 = 14)
CO (6 + 8 = 14) ; NO (7 + 8 = 15)
From the above calculation we find that all the species listed in choice (b) have 14 electrons each so it is the correct answer.

QUESTION: 19

The ionization enthalpy of hydrogen atom is 1.312 × 106 J mol–1. The energy required to excite the electron in the atom from n = 1 to n = 2 is [2008]

Solution:

(ΔE), The energy required to excite an electron in an atom of hydrogen from n = 1 to n = 2 is ΔE  (difference in energy E2 and E1) Values of E2 and E1 are, E1 = – 1.312 × 106 J mol–1

= –3.28 × 105 J mol–1

ΔE is given by the relation,

∴ ΔE = E2 – E1  = [–3.28 × 105]– [–1.312 × 106 ] J mol–1

= (–3.28 × 105 + 1.312 × 106) J mol–1

= 9.84 × 105 J mol–1

Thus the correct answer is (d)

QUESTION: 20

Calculate the wavelength (in nanometer) associated with a proton moving at 1.0 × 103 ms –1. (Mass of proton = 1.67 × 10–27 kg and  h = 6.63 × 10–34 Js)

Solution:

 = 3.97 × 10–10 meter  = 0.397 nanometer

QUESTION: 21

In an atom, an electron is moving with a speed of 600 m/s with an accuracy of 0.005%. Certainity with which the position of the electron can be located is ( h = 6.6 × 10–34 kg m2s–1, mass of electron, em = 9.1 × 10–31 kg) : [2009]

Solution:

According to Heisenberg uncertainty principle.

Here  = 0.03

So,

  = 1.92 × 10–3 meter

QUESTION: 22

The energy required to break one mole of Cl – Cl bonds in Cl2 is 242 kJ mol–1. The longest wavelength of light capable of breaking a single Cl – Cl bond is (c = 3 × 108 ms–1 and NA = 6.02 × 1023 mol–1). [2010]

Solution:

Energy required to break one mole of Cl – Cl bonds in Cl2

= 0.4947 × 10–6 m  = 494.7 nm

QUESTION: 23

Ionisation energy of He+ is 19.6 × 10–18 J atom–1. The energy of the first stationary state (n = 1) of Li2+ is [2010]

Solution:

Given I1 = – 19.6 × 10–18 , Z1 = 2, n1 = 1 , Z2 = 3 and n2 = 1

Substituting these values in equation (ii).

= – 4.41 × 10–17 J/atom

QUESTION: 24

A gas a bsor bs a ph oton of 355 n m an d emits at two wavelengths. If one of the emissions is at 680 nm, the other is at:[2011]

Solution:

Energy of absorbed photon = Sum of the energies of emitted photon

= 1.346 x106

or λ2= 1/1.346 × 106 = 743 × 10–9m = 743 nm

QUESTION: 25

The electrons identified by quantum numbers n and l :
(A) n = 4, l = 1
(B) n = 4, l = 0
(C) n = 3, l = 2
(D) n = 3, l = 1
can be placed in order of increasing energy as : [2012]

Solution:

(a) 4 p   (b)  4 s (c) 3 d (d)   3 p

Accroding to Bohr Bury's (n + l) rule, increasing order of energy (D) < (B) < (C) < (A).
Note : If the two orbitals have same value of (n + l) then the orbital with lower value of n will be filled first.

QUESTION: 26

Energy of an electron is given by E = – 2.178 ×             

Wavelength of light required to excite an electron in an hydrogen atom from level n = 1 to n = 2 will be : (h = 6.62 × 10 –34 Js and c = 3.0 × 108 ms–1) [JEE M 2013]

Solution:

 = 1.214 × 10–7m

QUESTION: 27

The correct set of four quantum numbers for the valence electrons of rubidium atom (Z = 37) is: [JEE M 2014]

Solution:

The electronic configuration of Rubidium (Rb = 37) is

1s22s2 2 p6 3s23 p6 3d10 4s2 4 p6 5s1
Since last electron enters in 5s orbital Hence n = 5, l = 0, m = 0, s = 

QUESTION: 28

 Which of the following is the energy of a possible excited state of hydrogen ? [JEE M 2015]

Solution:

Total energy = 

where n = 2, 3, 4 ....
Putting n = 2

 = -3.4eV

QUESTION: 29

 A stream of electrons from a heated filaments was passed two charged plates kept at a potential difference V esu. If e and m are charge and mass of an electron, respectively, then the value of h/λ (where λ is wavelength associated with electron wave) is given by: [JEE M 2016]

Solution:

As electron of charge 'e' is passed through 'V' volt, kinetic energy of electron will be eV

Wavelength of electron wave 

 

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