For s-orbitals,since psi (orbital) is indepen | vedclass

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Question

(C) For $s-$orbitals,the wave function $\psi$ depends only on the distance $r$ from the nucleus and is independent of the angular coordinates $(	heta

Vedclass · Accepted Answer

both $(a)$ and $(b)$ are correct

Vedclass · Answer

(C) For $s-$orbitals,the wave function $\psi$ depends only on the distance $r$ from the nucleus and is independent of the angular coordinates $(	heta, \phi)$.Since the probability density is given by $\psi^2$,if $\psi$ is independent of angles,then $\psi^2$ is also independent of angles.An orbital that is independent of angles is spherically symmetric,meaning the electron density is the same in all directions at a given distance $r$ from the nucleus.Therefore,both statements $(a)$ and $(b)$ are correct.

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 $s-$orbitals,since $\psi$ (orbital) is independent of angles,the probability $(\psi^2)$ is

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

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

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

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

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

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

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