NEET Chemistry Questions: Structure of Atom

Pratcice NEET questions from all capters from huge question bank for free. All MCQs are based on NCERT syllabus. To practice from a specific subject and chapter, select a subject below. Login to practice in a structured way with explanations, bookmarks, lists, notes etc. Click here to Login or Sign up for free.

Physics Chemistry Botany Zoology

Question bank:

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 ?

$ 3 \rightarrow 2 $
$ 5 \rightarrow 2 $
$ 4 \rightarrow 1 $
$ 2 \rightarrow 5 $

The orbital angular momentum for an electron revolving in an orbit is given by $ \sqrt { l ( l+1 )} . { h \over 2 \pi } $ This momentum for an s-electron will be given by

$ + { 1 \over2 } { h \over 2 \pi } $
zero
$ { h \over 2 \pi } $
$ \sqrt 2 { h \over 2 \pi } $

The emission spectrum of hydrogen is found to satisfy the expression for the energy change “E (in Joules) such that $ “E = 2·18 \times 10^{–18} J where n_1 = 1,2, 3 ..........and n_2 = 2,3 ,4 $ The spectral lines correspond to Paschen series it

$ n_1 = 1 and n_2 = 2, 3, 4 $
$ n_1 = 3 and n_2 = 4, 5, 6 $
$ n_1 = 1 and n_2 = 3, 4, 5 $
$ n_1 = 2 and n_2 = 3, 4, 5 $

Among the following series of transition metal ions, the one where all metal ions have same 3d electronic configuration is

$ Ti^{2+} , V^{3+}, Cr ^ {4+} , Mn ^ {5+} $
$ Ti^{2+} , V^{2+}, Cr ^ {3+} , Mn ^ {4+} $
$ Ti^{+} , V^{4+}, Cr ^ {6+} , Mn ^ {7+} $
$ Ti^{4+} , V^{3+}, Cr ^ {2+} , Mn ^ {3+} $

For d-electron, the orbital angular momentum is

$ \sqrt { 6 } h / 2 \pi $
$ \sqrt { 2h / 2 \pi } $
$ h / 2 \pi $
$ 2h / \pi $

Time taken for an electron to complete one revolution in the Bohr orbit of hydrogen atom is

$ { 4 \pi ^2 m r ^2 \over nh } $
$ { nh \over 4 \pi ^2 mr } $
$ {2 \pi mr \over n ^2 h ^2 } $
$ { h \over 2 \pi mr }$

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

n = 4, l= 3, m = +4, s = + 1 /2
n = 3, l= 2, m = -2, s = + 1 /2
n = 4,l = 3, m = +1, s = + 1 /2
n = 4, l= 4, m = -4, s = - 1 /2

The wavelength of radiation emitted when in a hydrogen atom electron falls from infinity to stationary state 1, would be (Rydberg constant  1·09 × 107m–1)

91nm
$ 9·1 \times 10^{–8} nm $
406 nm
192 nm

The relationship between energy E, of the radiation with a wavelength 8000 Å and the energy of the radiation with a wavelength 16000 Å is

$ E_1 = 6 E_2 $
$ E_1 = 2 E_2 $
$ E_1 = 4 E_2 $
$ E_1 = 1/2 E_2 $

The energy of second Bohr orbit of the hydrogen atom is $ –328 kJ mol^{–1} $ , hence the energy of fourth Bohr orbit would be

$ –41 kJ mol^{–1} $
$ –1312 kJ mol^{–1} $
$ –164 kJ mol^{–1} $
$ –82 kJ mol^{–1} $

Which of the following statements in relation to the hydrogen atom is correct?

3s-orbital is lower in energy than 3p-orbital
3p-orbital is lower in energy than 3d-orbital
3s and 3p orbitals arc of lower energy than 3d orbital
3s, 3p and 3d orbitals all have the same energy.

The most probable radius (in pm) for finding the electron in He+ is

0·0
52·9
26·5
105·8

If the speed of electron in Bohr first orbit of hydrogen atom be x, then speed of the electron in 3rd orbit is

x/9
x/3
3x
9x

If wavelength of photon is $ 2·2 \times 10 ^{– 11} m , h= 6·6 \times 10^{–34} Js$ , then momentum of photon is

$ 3\times 10^{–23} kg/s $
$ 1·452 \times 10^{–44} kg/s $
$ 3·33 \times 10^{22} kg/s $
$ 6·89 \times 10^{43} kg/s. $

The wave number of first line of Balmer series of hydrogen is $ 15200 cm^{–1}. The wave number of the first Balmer line of Li^{1+} $ ion is

$ 15200 cm^{–1}$
$ 60800 cm^{–1} $
$ 76000cm^{–1} $
$ 136,800cm^{–1.} $