Which statement is not correct for n = 5,m = | vedclass

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(B) For a given principal quantum number $n$,the azimuthal quantum number $l$ can take values from $0$ to $n-1$. For $n = 5$,$l$ can be $0, 1, 2, 3, 4

Vedclass · Accepted Answer

$l = 0, 1, 3; s = + \frac{1}{2}$

Vedclass · Answer

(B) For a given principal quantum number $n$,the azimuthal quantum number $l$ can take values from $0$ to $n-1$. For $n = 5$,$l$ can be $0, 1, 2, 3, 4$. The magnetic quantum number $m$ ranges from $-l$ to $+l$. If $m = +3$,then $l$ must be at least $3$ (i.e.,$l = 3$ or $l = 4$). Option $A$ $(l = 4)$ is possible because $m = +3$ is allowed for $l = 4$. Option $C$ $(l = 3)$ is possible because $m = +3$ is allowed for $l = 3$. Option $B$ suggests $l = 0, 1, 3$. While $l = 3$ is valid for $m = +3$,$l = 0$ and $l = 1$ are not valid for $m = +3$ because for $l=0, m=0$ and for $l=1, m=-1, 0, +1$. Therefore,the statement in option $B$ is incorrect.

List-$I$ (Quantum numbers)	List-$II$ (Orbital)
$A$. $n = 2, l = 1$	$I$. $3d$
$B$. $n = 4, l = 0$	$II$. $2p$
$C$. $n = 5, l = 3$	$III$. $4s$
$D$. $n = 3, l = 2$	$IV$. $5f$

Which statement is not correct for $n = 5$,$m = +3$?

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Match List-$I$ with List-$II$ :
List-$I$ (Quantum numbers) List-$II$ (Orbital)
$A$. $n = 2, l = 1$ $I$. $3d$
$B$. $n = 4, l = 0$ $II$. $2p$
$C$. $n = 5, l = 3$ $III$. $4s$
$D$. $n = 3, l = 2$ $IV$. $5f$

Choose the correct answer from the options given below :

The number of radial and angular nodes in $4d$ orbital are,respectively $........$ .

Which of the following electronic configurations has the maximum energy?

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