Few statements are given regarding nodes in t | vedclass

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Question

(B) For any orbital,the total number of nodes is calculated as $(n - 1)$.Angular nodes are given by the azimuthal quantum number $l$.Radial nodes are

Vedclass · Accepted Answer

$ns-$orbital has $(n + 1)$ nodes.

Vedclass · Answer

(B) For any orbital,the total number of nodes is calculated as $(n - 1)$.Angular nodes are given by the azimuthal quantum number $l$.Radial nodes are given by $(n - l - 1)$.Total nodes $=$ Radial nodes $+$ Angular nodes $= (n - l - 1) + l = n - 1$.For $p_z-$orbital,the electron density is zero in the $xy-$plane,so it is a nodal plane.Therefore,the statement that $ns-$orbital has $(n + 1)$ nodes is incorrect,as it should be $(n - 1)$ nodes.

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

Few statements are given regarding nodes in the orbitals. Mark the statement which is not correct.

Similar Questions

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 :

Which orbital is dumb-bell shaped?

The five valence electrons of $_{15}P$ are represented as shown in the figure. If the spin quantum number of $B$ and $Z$ is $+1/2$,which of the following groups has three identical quantum numbers?

Which of the following is correct for $He$?

For principal quantum number $n = 4$,the total number of orbitals having $l = 3$ is:

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