The magnetic moment of 2.84 BM can be shown by:
(At. no. Ni = 28, Ti = 22, Cr = 24, Co = 27)
1. Ni2+
2. Ti3+
3. Cr3+
4. Co2+
The number of d-electrons in Fe2+ (Z=26) is not equal to the number of electrons in
which one of the following?
1. s-electrons in Mg (Z=12)
2. p-electrons in Cl (Z=17)
3. d-electrons in Fe (Z=26)
4. p-electrons in Ne (Z=10)
Which is the correct order of increasing energy of the listed orbitals in the atom of
titanium?
1. 3s 4s 3p 3d
2. 4s 3s 3p 3d
3. 3s 3p 3d 4s
4. 3s 3p 4s 3d
Calculate the energy in Joule corresponding to light of wavelength 45 nm : (Planck's
constant h = 6.63 x 10-34 Js; speed of light c = 3 x 108 ms-1)
1. 6.67 x 1015
2. 6.67 x 1011
3. 4.42 x10-15
4. 4.42 x 10-18
What is the maximum numbers of electrons that can be associated with the following set
of quantum numbers?
n=3, l = 1 and m=-1
1. 10
2. 6
3. 4
4. 2
From the given equation, certain conclusions are drawn.
\(\mathrm{E}=-2.178 \times 10^{-18} \mathrm{(J)} \frac{\mathrm{z}^2}{\mathrm{n}^2} \)
Choose the incorrect statement:
1. | The negative sign in the equation implies that the energy of an electron bound to the nucleus is lower than it would be if the electron were at an infinite distance from the nucleus. |
2. | The larger the value of n, the greater the orbit radius. |
3. | The equation can be utilized to calculate the change in energy when the electron transitions between orbits. |
4. | For n=1, the electron has a more negative energy than it does for n=6. This means that the electron is more loosely bound in the smallest allowed orbit. |
The maximum number of electrons in a subshell with l = 3 and n = 4 is
1. 14
2. 16
3. 10
4. 12
The correct set of four quantum numbers for the valence electron of rubidium atom (Z =37) is
1. 5, 1, 1, +1/2
2. 6, 0, 0, +1/2
3. 5, 0, 0, +1/2
4. 5, 1, 0, +1/2
The total number of atomic orbitals in fourth energy level of an atom is
1. 16
2. 32
3. 4
4. 8
If the principal quantum number n=6, the correct sequence of filling of electrons will be:
(1) ns→np→(n-1)d→(n-2)f
(2) ns→(n-2)f→(n-1)d→np
(3) ns→(n-1)d→(n-2)f→np
(4) ns→(n-2)f→np→(n-1)d