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Question

(A) $1$. Angular nodes are equal to the azimuthal quantum number $(l)$. For $d$-orbitals,$l=2$,so they have $2$ angular nodes. Thus,$A$ matches $P$ $(

Vedclass · Accepted Answer

$A-P, B-Q, C-S, D-R$

Vedclass · Answer

(A) $1$. Angular nodes are equal to the azimuthal quantum number $(l)$. For $d$-orbitals,$l=2$,so they have $2$ angular nodes. Thus,$A$ matches $P$ $(4d_{x^2-y^2})$.$2$. The $d_{z^2}$ orbital has zero nodal planes (it has two conical nodal surfaces). Thus,$B$ matches $Q$ $(3d_{z^2})$.$3$. Radial nodes are calculated by the formula $(n - l - 1)$. For $3s$,$n=3, l=0$,so radial nodes $= 3-0-1 = 2$. Thus,$C$ matches $S$ $(3s)$.$4$. Angular nodes for $f$-orbitals are $l=3$. Thus,$D$ matches $R$ $(4f)$.Therefore,the correct match is $A-P, B-Q, C-S, D-R$.

Column-$I$	Column-$II$
$A$. The orbital which has two angular nodes	$P$. $4d_{x^2-y^2}$
$B$. The $d$ orbital with zero nodal plane	$Q$. $3d_{z^2}$
$C$. The orbital with two radial nodes	$R$. $4f$
$D$. The orbital with three angular nodes	$S$. $3s$

Match the column:

Column-$I$ Column-$II$

$A$. The orbital which has two angular nodes $P$. $4d_{x^2-y^2}$

$B$. The $d$ orbital with zero nodal plane $Q$. $3d_{z^2}$

$C$. The orbital with two radial nodes $R$. $4f$

$D$. The orbital with three angular nodes $S$. $3s$

Correct match will be $:-$

Similar Questions

Find the number of electrons in the ground state of $Pd$ $(Z = 46)$ with $l = 2$.

The order of increasing energies of the orbitals follows:

Which of the following statements is not correct for an electron that has the quantum numbers $n = 4$ and $m_l = 2$?

Identify the incorrect statement from the following.

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Match the column: Column-$I$ Column-$II$ $A$. The orbital which has two angular nodes $P$. $4d_{x^2-y^2}$ $B$. The $d$ orbital with zero nodal plane $Q$. $3d_{z^2}$ $C$. The orbital with two radial nodes $R$. $4f$ $D$. The orbital with three angular nodes $S$. $3s$ Correct match will be $:-$