For how many electrons in an atom is n = 3, m | vedclass

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(A) For a given principal quantum number $n = 3$,the possible values of the azimuthal quantum number $l$ are $0, 1, 2$.For $l = 0$ ($3s$ orbital),$m_l

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(A) For a given principal quantum number $n = 3$,the possible values of the azimuthal quantum number $l$ are $0, 1, 2$.For $l = 0$ ($3s$ orbital),$m_l = 0$.For $l = 1$ ($3p$ orbital),$m_l = -1, 0, +1$.For $l = 2$ ($3d$ orbital),$m_l = -2, -1, 0, +1, +2$.The condition $m_l = 1$ is satisfied by the $3p$ orbital $(m_l = 1)$ and the $3d$ orbital $(m_l = 1)$.Each orbital can hold a maximum of $2$ electrons with opposite spins.Therefore,the $3p$ orbital with $m_l = 1$ can hold $2$ electrons,and the $3d$ orbital with $m_l = 1$ can hold $2$ electrons.Total number of electrons = $2 + 2 = 4$.

List-$I$ (Element)	List-$II$ (Electronic Configuration)
$A. N$	$I. [Ar] 3d^{10} 4s^2 4p^5$
$B. S$	$II. [Ne] 3s^2 3p^4$
$C. Br$	$III. [He] 2s^2 2p^3$
$D. Kr$	$IV. [Ar] 3d^{10} 4s^2 4p^6$

For how many electrons in an atom is $n = 3, m_l = 1$ possible?

Similar Questions

Given below are two statements :
Statement $I$ : $A$ spectral line will be observed for a $2p_x \rightarrow 2p_y$ transition.
Statement $II$ : $2p_x$ and $2p_y$ are degenerate orbitals.
In the light of the above statements,choose the correct answer from the options given below :

If an electron has a spin quantum number of $+ \frac{1}{2}$ and a magnetic quantum number of $- 1$,it cannot be present in an

Match List-$I$ with List-$II$.

List-$I$ (Element) List-$II$ (Electronic Configuration)

$A. N$ $I. [Ar] 3d^{10} 4s^2 4p^5$

$B. S$ $II. [Ne] 3s^2 3p^4$

$C. Br$ $III. [He] 2s^2 2p^3$

$D. Kr$ $IV. [Ar] 3d^{10} 4s^2 4p^6$

Choose the correct answer from the options given below:

Which orbital is closer to the nucleus: $2p$ or $3p$?

Which set has the same number of unpaired electrons in their ground state?

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