Nickel atom can lose two electrons to form ${\mathrm{Ni}}^{2+}$ ion. The atomic number of nickel is $28$. From which orbital will nickel lose two electrons.

Which of the following orbitals are degenerate?

$3{\mathrm{d}}_{\mathrm{xy}}, 4{\mathrm{d}}_{\mathrm{xy}}, 3{\mathrm{d}}_{{\mathrm{z}}^2}, 3{\mathrm{d}}_{\mathrm{yz}}, 4{\mathrm{d}}_{\mathrm{yz}}, 4{\mathrm{d}}_{{\mathrm{z}}^2}$

The arrangement of orbitals on the basis of energy is based upon their $(\mathrm{n}+\mathrm{l})$ value. Lower the value of $(\mathrm{n}+\mathrm{l})$, lower is the energy. For orbitals having the same values of $(\mathrm{n}+\mathrm{l})$, the orbital with the lower value of $\mathrm{n}$ will have lower energy. 

Based upon the above information, arrange the following orbitals in the increasing order of energy.

$1\mathrm{s}, 2\mathrm{s}, 3\mathrm{s}, 2\mathrm{p}$

The arrangement of orbitals on the basis of energy is based upon their $(\mathrm{n}+\mathrm{l})$ value. Lower the value of $(\mathrm{n}+\mathrm{l})$, lower is the energy. For orbitals having the same values of $(\mathrm{n}+\mathrm{l})$, the orbital with the lower value of $\mathrm{n}$ will have lower energy. 

Based upon the above information, arrange the following orbitals in the increasing order of energy.

$4\mathrm{s}, 3\mathrm{s}, 3\mathrm{p}, 4\mathrm{d}$

The arrangement of orbitals on the basis of energy is based upon their $(\mathrm{n}+\mathrm{l})$ value. Lower the value of $(\mathrm{n}+\mathrm{l})$, lower is the energy. For orbitals having the same values of $(\mathrm{n}+\mathrm{l})$, the orbital with the lower value of $\mathrm{n}$ will have lower energy. 

Based upon the above information, arrange the following orbitals in the increasing order of energy. 

$5\mathrm{p}, 4\mathrm{d}, 5\mathrm{d}, 4\mathrm{f}, 6\mathrm{s}$

The arrangement of orbitals on the basis of energy is based upon their $(\mathrm{n}+\mathrm{l})$ value. Lower the value of $(\mathrm{n}+\mathrm{l})$, lower is the energy. For orbitals having the same values of $(\mathrm{n}+\mathrm{l})$, the orbital with the lower value of $\mathrm{n}$ will have lower energy.

Based upon the above information, arrange the following orbitals in the increasing order of energy. 

$5\mathrm{f}, 6\mathrm{d}, 7\mathrm{s}, 7\mathrm{p}$ 

The arrangement of orbitals on the basis of energy is based upon their $(\mathrm{n}+\mathrm{l})$ value. Lower the value of $(\mathrm{n}+\mathrm{l})$, lower is the energy. For orbitals having the same values of $(\mathrm{n}+\mathrm{l})$, the orbital with the lower value of $\mathrm{n}$ will have lower energy.

Based upon the above information, solve the question given below:

Which of the following orbitals has the lowest energy?  

$4\mathrm{d}, 4\mathrm{f}, 5\mathrm{s}, 5\mathrm{p}$

The arrangement of orbitals on the basis of energy is based upon their $(\mathrm{n}+\mathrm{l})$ value. Lower the value of $(\mathrm{n}+\mathrm{l})$, lower is the energy. For orbitals having the same values of  $(\mathrm{n}+\mathrm{l})$, the orbital with the lower value of $\mathrm{n}$ will have lower energy.

Based upon the above information, solve the question given below:

Which of the following orbitals has the highest energy?  

$5\mathrm{p}, 5\mathrm{d}, 5\mathrm{f}, 6\mathrm{s}, 6\mathrm{p}$