To determine which electronic configuration is associated with the highest magnetic moment, we need to consider the number of unpaired electrons in each configuration. The magnetic moment is given by the formula:
\(\mu = \sqrt{n(n+2)} \, \text{BM}\) (Bohr Magneton)
where \(n\) is the number of unpaired electrons.
Among the given configurations, \([Ar] \, 3d^6\) has the highest magnetic moment because it has the highest number of unpaired electrons, which is 4.
Therefore, the correct answer is \([Ar] \, 3d^6\).
The magnetic moment μ is given by:
μ=$\sqrt{n(n + 2)}$ BM
where n is the number of unpaired electrons. Among the options, [Ar] 3d6 has the highest number of unpaired electrons (4), leading to a maximum magnetic moment.
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:
Let \( f : \mathbb{R} \to \mathbb{R} \) be a twice differentiable function such that \[ (\sin x \cos y)(f(2x + 2y) - f(2x - 2y)) = (\cos x \sin y)(f(2x + 2y) + f(2x - 2y)), \] for all \( x, y \in \mathbb{R}. \)
If \( f'(0) = \frac{1}{2} \), then the value of \( 24f''\left( \frac{5\pi}{3} \right) \) is: