The atomic number of the element (in ground s | vedclass

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Question

(A) The electronic configuration of the given elements in their ground state is as follows:$1.$ For atomic number $8$ $(O)$: $1s^2 2s^2 2p^4$. Number

Vedclass · Accepted Answer

$15$

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(A) The electronic configuration of the given elements in their ground state is as follows:$1.$ For atomic number $8$ $(O)$: $1s^2 2s^2 2p^4$. Number of unpaired $3p$ electrons = $0$.$2.$ For atomic number $10$ $(Ne)$: $1s^2 2s^2 2p^6$. Number of unpaired $3p$ electrons = $0$.$3.$ For atomic number $12$ $(Mg)$: $1s^2 2s^2 2p^6 3s^2$. Number of unpaired $3p$ electrons = $0$.$4.$ For atomic number $15$ $(P)$: $1s^2 2s^2 2p^6 3s^2 3p^3$. The $3p$ subshell has $3$ orbitals,each containing $1$ unpaired electron according to Hund's rule.Thus,the element with atomic number $15$ has the maximum number of unpaired $3p$ electrons ($3$ unpaired electrons).

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

The atomic number of the element (in ground state) having the maximum number of unpaired $3p$ electrons is:

Similar Questions

The electronic configuration,which obeys Hund's rule for the ground state of carbon atom is

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 :

Given below are the quantum numbers for $4$ electrons.
$A. \; n = 3, l = 2, m_l = 1, m_s = +1/2$
$B. \; n = 4, l = 1, m_l = 0, m_s = +1/2$
$C. \; n = 4, l = 2, m_l = -2, m_s = -1/2$
$D. \; n = 3, l = 1, m_l = -1, m_s = +1/2$
The correct order of increasing energy is

Which of the following sets of species has the $3^{rd}$ principal energy level $(n=3)$ completely filled?

For hydrogen-like atoms,the average distance of an electron from the nucleus for the orbitals belonging to the $4^{th}$ orbit is minimum for:

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