NEET Chemistry Structure of Atom Questions Solved


Which of the following sets of quantum numbers represent an impossible arrangement - 

       n   l    m     s

(A)  3   2   -2   (+)1/2

(B)  4   0   0    (-)1/2

(C)  3   2   -3   (+)1/2

(D)  5   3   0   (-)1/2

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Quantum Numbers

(C) If l=2, m  -3

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How many unpaired electrons are present in Ni2+ cation (atomic number = 28)

(A) 0

(B) 2

(C) 4

(D) 6

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(B) Two unpaired electrons are present in Ni++ (z=28) cation

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Correct set of four quantum numbers for the outermost electron of rubidium (Z=37) is:

(A) 5, 0, 0, 1/2

(B) 5, 1, 0, 1/2

(C) 5, 1, 1, 1/2

(D) 6, 0, 0, 1/2

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Quantum Numbers

(A) Its configuration is 5s1

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Number of unparired electrons in 1S2 2S2 2P3 is -

1. 2

2. 0

3. 3

4. 1

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(c) . In this type of electronic configuration the number of unpaired electrons are 3.

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The quantum numbers of four electrons are given below.

                                  n            l           m          s 

(1) Electron 1             3            0             0           -1/2

(2) Electron 2             4            0             1           1/2

(3) Electron 3            3             2             0           1/2

(4) Electron 4            3             1             0           -1/2

The correct order of decreasing energy of these electrons is - 

(A) Electron 3 > Electron 1 > Electron 4> Electron 2

(B) Electron 4> Electron 2 > Electron 3 > Electron 1

(C) Electron 3 > Electron 2 > Electron 4 > Electron 1

(D) Electron 2 > Electron 4 > Electron 3 > Electron 1

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Quantum Numbers

(C) The correct order of decreasing energy will be electron 3 which has higher energy than 2 and in the order given.

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Which type of radiation is not emitted by the electronic structure of atoms:

(A) Ultraviolet light

(B) X-rays

(C) Visible light

(D) y-rays

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EMT Radiation

(D) y-rays emission occurs due to radioactive change, a nuclear phenomenon.

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Of the following transitions in hydrogen atom, the one which gives an absorption line of lowest frequency is

(A) n=1 to n=2

(B) n=3 to n=8

(C) n=2 to n=1

(D) n=8 to n=3

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Hydrogen Spectra

(B)  Absorption line in the spectra arise when energy is absorbed i.e., electron shifts from lower to higher orbit, out of A and B, B will have the lowest frequency as this falls in the Paschen series.

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If uncertainty in the position of an electron is zero the uncertainty in its momentum will be

(A) < h/4π

(B) > h/4π

(C) zero

(D) infinite

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Heisenberg Uncertainity Principle

(D) x x  h/4π

if  x = 0, then  p will be infinite

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The maximum number of atomic orbitals associated with a principal quantum number 5 is

(A)9

(B)12

(C) 16

(D) 25

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Quantum Numbers

(D) The number of orbitals in a principle shell is n2= 52=25

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The de Broglie wavelength of electron of He+ ion is 3.329Å. If the photon emitted upon de-excitation of this He+ ion is made to hit H atom in its ground state so as to liberate electron from it, what will be the de Broglie's wavelength of photoelectron ?

(A) 0.1518 Å

(B) 6.1518 Å

(C) 2.3518 Å

(D) 4.1218 Å

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de Broglie Equation

(C) 

this is the K.E. of electron Therefore, total energy = -13.58eV

En=-13.6 x (Z2/n2)eV

For He+ion, Z=2 and En = -13.58eV

So, n2= -13.6x4 / -13.58 = n = 2.

Thus, He+ion is in the 1 excited state.

Energy of photon emitted

Energy of photon =I.P. of H + K.E.of photoelectron Thus, K.E. of photoelectron= 40.8-13.6 = 27.2eV

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