Question

Which of the following statements is/are correct?

• A. The electronic configuration of $F{e}^{2+}$ is (atomic number = 26) is $\left[Ar\right]3d^6$
• B. According to Pauli's exclusion principle, the maximum number of electrons in an orbital cannot exceed 2
• C. In Vanadium atom, 12 electrons can have a spin of $+\frac{1}{2}$ and the other 11 electrons can have a spin of $-\frac{1}{2}$ (Z = 23)
• D. All of the above

Answer 1

The correct option is D All of the above

Electronic configuration of Fe is $\left[Ar\right]3d^{6}4s^2$ and that of $F{e}^{2+}$ is $\left[Ar\right]3d^{6}4s^0$.

According Pauli's exclusion principle, no two electrons can have all the four quantum number same. Hence, statement 3 is correct.
Only two electrons can exist in a orbital with opposite spin which limit maximum number of electrons to 2.

Vanadium has an atomic number of 23 hence it will have 23 electrons. So the spin quantum number can be either clockwise or anticlockwise. Therefore it will have 12 electrons of same spin (clockwise) and 11 electrons of other spin (anti-clockwise) or vice -versa.