The number of d-electrons in Fe²⁺ (Z = 26) is not equal to the number of :
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)

The angular momentum of electrons in d orbital is equal to 
1. $\sqrt{6}$ $\sout{h}$
2. $\sqrt{2}$$\sout{h}$
3. $2\sqrt{3}$ $\sout{h}$
4. 0 $\sout{h}$

What is the maximum number of orbitals that can be identified with the following quantum numbers?
n = 3, l = 1, m = 0
1. 1
2. 2
3. 3
4. 4

Calculate the energy in corresponding to light of wavelength 45 nm :
(Planck's constant h = 6.63 × 10^–34 Js: speed of light c = 3 × 10⁸ ms^–1)
1. 6.67 x 10¹⁵ J
2. 6.67 x 10¹¹ J
3. 4.42 x 10^-15 J
4. 4.42 x 10^-18 J

Based on equation E = –2.178 × 10^–18J, $\frac{{\mathrm{Z}}^2}{{\mathrm{n}}^2}$ certain conclusions are written. Which of them is not correct ?
1. Larger the value of n, larger is the orbit radius.
2. Equation can be used to calculate the change in energy when the electron changes orbit.
3. For n = 1, the electron has a more negative energy than it does for n = 6 which means that the electron is more loosely bound in the smallest allowed orbit.
4. The negative sign in equation simply means that the energy of electron bound to the nucleus is lower than what it would be if the electrons were at an infinite distance from the nucleus.

Maximum number of electrons in a sub shell 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 a rubidium atom (Z =37) is 

1. $5,1,1,+\frac{1}{2}$

2. $6,0,0,+\frac{1}{2}$

3. $5,0,0,+\frac{1}{2}$

4. $5,1,0,+\frac{1}{2}$