The energy of a 2p orbital in a multi-electro | vedclass

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(B) For multi-electron atoms,the energy of an orbital depends on both the principal quantum number $(n)$ and the azimuthal quantum number $(l)$.Accord

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More than that of $2s$ orbital

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(B) For multi-electron atoms,the energy of an orbital depends on both the principal quantum number $(n)$ and the azimuthal quantum number $(l)$.According to the $(n+l)$ rule,the energy of an orbital increases as the value of $(n+l)$ increases.For $2s$ orbital: $n=2, l=0$,so $(n+l) = 2+0 = 2$.For $2p$ orbital: $n=2, l=1$,so $(n+l) = 2+1 = 3$.Since the $(n+l)$ value for $2p$ is greater than that for $2s$,the $2p$ orbital has higher energy than the $2s$ orbital.

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

The energy of a $2p$ orbital in a multi-electron atom is

Similar Questions

What is the atomic number of the element with the maximum number of unpaired $4p$ electrons?

The magnetic quantum number $m_l$ is determined by the azimuthal quantum number $l$. If $l = 2$,what are the type and number of orbitals indicated?

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

Choose the correct answer from the options given below :

What is the atomic number of an element whose $M^{2+}$ ion has the electronic configuration $[Ar] \, 3d^8$?

What is the number of protons in the nucleus of an element that has the maximum number of unpaired electrons in the $2p$ subshell?

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