Question

Arrange the orbitals in the increasing order of their energy $3p_x,2s,4d_{xy},3s,4p_z,3p_y,4s$

• A. $2s<3s=3p_x=3p_y<4s=4p_z=4d_{xy}$
• B. $2s<3s<3p_x=3p_y<4s=4p_z=4d_{xy}$
• C. $2s<3s<3p_x=3p_y<4s<4p_z=4d_{xy}$
• D. $2s<3s<3p_x=3p_y<4s<4p_z<4d_{xy}$

Answer 1

The correct option is D $2s<3s<3p_x=3p_y<4s<4p_z<4d_{xy}$

Orbitals are arranged with increasing energy on the basis of $(n+l)$ rule. The higher the value of $(n+l)$ for an orbital, the higher is its energy. Similarly, the lower the value of (n+l) for an orbital the lower is its energy.
If two orbitals possess same $(n+l)$ value, the orbital with lower $(n+l)$ value will have the lower energy.
Let us find the energies of different orbitals.
For 2s orbital:
n=2, l=0
$\therefore (n+l)=(2+0)=2$
For 3s orbital:
n=3, l=0
$\therefore (n+l)=(3+0)=3$
For $3p_x$ orbital:
n=3, l=1
$\therefore (n+l)=(3+1)=4$
For $3p_y$ orbital:
n=3, l=1
$\therefore (n+l)=(3+1)=4$
For 4s orbital:
n=4, l=0
$\therefore (n+l)=(4+0)=4$
Both $3p_y$, $3p_x$ and 4s orbitals have same $(n+l)$ value, but the n value of 3p orbital is lower than 4s orbital hence, it will have lower energy.
For $4d_{xy}$ orbital:
n=4, l=2
$\therefore (n+l)=(4+2)=6$.
For $4p_z$ orbital:
n=4, l=1
$\therefore (n+l)=(4+1)=5$
The order of increasing energies of orbitals is
$2s<3s<3p_x=3p_y<4s<4p_z<4d_{xy}$