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(D) The energy of atomic orbitals in multi-electron atoms is determined by the $(n+l)$ rule.$1$. According to the $(n+l)$ rule, the orbital with the l

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$E < B < A < D < C$

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(D) The energy of atomic orbitals in multi-electron atoms is determined by the $(n+l)$ rule.$1$. According to the $(n+l)$ rule, the orbital with the lower $(n+l)$ value has lower energy.$2$. If the $(n+l)$ values are equal, the orbital with the lower value of $n$ has lower energy.Calculating $(n+l)$ for each orbital:$A: n=3, l=2 \implies n+l = 3+2 = 5$$B: n=4, l=0 \implies n+l = 4+0 = 4$$C: n=6, l=1 \implies n+l = 6+1 = 7$$D: n=5, l=1 \implies n+l = 5+1 = 6$$E: n=2, l=1 \implies n+l = 2+1 = 3$Comparing the $(n+l)$ values: $3 (E) Therefore, the order of increasing energy is $E < B < A < D < C$.

List-$I$ (Quantum number)	List-$II$ (Orbital)
$a. n=2, \ell=1$	$i. 2s$
$b. n=3, \ell=2$	$ii. 3s$
$c. n=3, \ell=0$	$iii. 2p$
$d. n=2, \ell=0$	$iv. 3d$

Arrange the following atomic orbitals of multi-electron atoms in order of increasing energy.
$A$. $n=3, l=2, m=+1$
$B$. $n=4, l=0, m=0$
$C$. $n=6, l=1, m=0$
$D$. $n=5, l=1, m=+1$
$E$. $n=2, l=1, m=+1$
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Match List-$I$ with List-$II$ :
List-$I$ (Quantum number) List-$II$ (Orbital)
$a. n=2, \ell=1$ $i. 2s$
$b. n=3, \ell=2$ $ii. 3s$
$c. n=3, \ell=0$ $iii. 2p$
$d. n=2, \ell=0$ $iv. 3d$

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For $ns$ orbital,the magnetic quantum number has value

The five valence electrons of $P^{15}$ are labeled as $A, B, X, Y, Z$ as shown in the figure. If the spin quantum number of $B$ and $Z$ is $+\frac{1}{2}$,identify the group of electrons that share three identical quantum numbers.

The $p_x$ orbital can accommodate:

The number of orbitals in the fourth principal quantum number $(n = 4)$ is:

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Arrange the following atomic orbitals of multi-electron atoms in order of increasing energy.$A$. $n=3, l=2, m=+1$$B$. $n=4, l=0, m=0$$C$. $n=6, l=1, m=0$$D$. $n=5, l=1, m=+1$$E$. $n=2, l=1, m=+1$Choose the correct answer from the options given below: