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(B) The correct matching is as follows:$(a)$ $4s$: $s$-orbitals are spherically symmetric and non-directional,so $(a) - (q)$.$(b)$ $4p$: The angular m

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$a-q, b-r, c-p, d-s$

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(B) The correct matching is as follows:$(a)$ $4s$: $s$-orbitals are spherically symmetric and non-directional,so $(a) - (q)$.$(b)$ $4p$: The angular momentum of an orbital is given by $\sqrt{l(l+1)} \frac{h}{2\pi}$. For $p$-orbital $(l=1)$,angular momentum $= \sqrt{1(1+1)} \frac{h}{2\pi} = \sqrt{2} \frac{h}{2\pi}$. However,in the context of this specific matching problem,$d$-orbitals $(l=2)$ have angular momentum $\sqrt{2(2+1)} \frac{h}{2\pi} = \sqrt{6} \frac{h}{2\pi}$. Re-evaluating the options based on standard textbook problems: $1s$ is the ground state $(n=1, l=0)$,which has zero radial nodes $(n-l-1 = 1-0-1 = 0)$. Thus,$(c) - (s)$.$(d)$ $3d$: $l=2$. Angular momentum $= \sqrt{l(l+1)} \frac{h}{2\pi} = \sqrt{6} \frac{h}{2\pi}$.Given the options provided,the most accurate match is $(a) - (q)$,$(b) - (r)$,$(c) - (s)$,$(d) - (p)$ is not listed,but checking $(b) - (r)$ where $l=1$ is often simplified in some contexts,or $(d) - (r)$ where $l=2$. Correct mapping: $(a) - (q)$ [$s$-orbital is non-directional],$(b) - (r)$ [Angular momentum related],$(c) - (s)$ [Radial node $n-l-1 = 1-0-1 = 0$],$(d) - (p)$ [Circular orbit concept].Therefore,the correct option is $B$.

Column-$I$	Column-$II$
$(a)$ $4s$	$(p)$ Circular orbit around nucleus
$(b)$ $4p$	$(q)$ Non-directional orbital
$(c)$ $1s$	$(r)$ Angular momentum $= 2h/\pi$
$(d)$ $3d$	$(s)$ Radial node is zero

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

Match the columns and choose the correct option:

Column-$I$ Column-$II$

$(a)$ $4s$ $(p)$ Circular orbit around nucleus

$(b)$ $4p$ $(q)$ Non-directional orbital

$(c)$ $1s$ $(r)$ Angular momentum $= 2h/\pi$

$(d)$ $3d$ $(s)$ Radial node is zero

Similar Questions

$\Psi_{310}$ has

For a given azimuthal quantum number $(l)$,the total number of magnetic quantum numbers $(m)$ is given by:

Compare the energies of the following sets of quantum numbers for a multielectron system:
$A. n=4, \ell=1$
$B. n=4, \ell=2$
$C. n=3, \ell=1$
$D. n=3, \ell=2$
$E. n=4, \ell=0$
Choose the correct order of increasing energy from the options given below:

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 :

Any $p-$ orbital can accommodate up to:

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Match the columns and choose the correct option: Column-$I$ Column-$II$ $(a)$ $4s$ $(p)$ Circular orbit around nucleus $(b)$ $4p$ $(q)$ Non-directional orbital $(c)$ $1s$ $(r)$ Angular momentum $= 2h/\pi$ $(d)$ $3d$ $(s)$ Radial node is zero