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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(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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(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 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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(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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(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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(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/4
(B) > h/4
(C) zero
(D) infinite
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(D) x x p
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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(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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(C)
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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