For the given question, we need to consider the electron configuration of the Argon atom (\( \text{Ar} \)) and find out how many electrons have \( m_l = 1 \). The electron configuration for Argon (\( \text{Ar} \)) is as follows: \[ 1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^6 \] Now, let's focus on the \( 2p \) and \( 3p \) orbitals because the question specifically asks about the electrons with \( m_l = 1 \).
The \( 2p \) orbital has three possible values for \( m_l \): \( m_l = -1, 0, 1 \). The number of electrons in the \( 2p \) orbital is 6, and these electrons are distributed over the three values of \( m_l \), with two electrons having \( m_l = 1 \).
The \( 3p \) orbital also has three possible values for \( m_l \): \( m_l = -1, 0, 1 \). The number of electrons in the \( 3p \) orbital is 6, and these electrons are distributed over the three values of \( m_l \), with two electrons having \( m_l = 1 \).
Thus, there are a total of 4 electrons with \( m_l = 1 \), two electrons from the \( 2p \) orbital and two electrons from the \( 3p \) orbital.
Thus, the correct answer is (B) 4.
Which of the following is/are correct with respect to the energy of atomic orbitals of a hydrogen atom?
(A) \( 1s<2s<2p<3d<4s \)
(B) \( 1s<2s = 2p<3s = 3p \)
(C) \( 1s<2s<2p<3s<3p \)
(D) \( 1s<2s<4s<3d \)
Choose the correct answer from the options given below:
The energy of an electron in first Bohr orbit of H-atom is $-13.6$ eV. The magnitude of energy value of electron in the first excited state of Be$^{3+}$ is _____ eV (nearest integer value)
Correct statements for an element with atomic number 9 are
A. There can be 5 electrons for which $ m_s = +\frac{1}{2} $ and 4 electrons for which $ m_s = -\frac{1}{2} $
B. There is only one electron in $ p_z $ orbital.
C. The last electron goes to orbital with $ n = 2 $ and $ l = 1 $.
D. The sum of angular nodes of all the atomic orbitals is 1.
Choose the correct answer from the options given below: