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.
Let the matrix $ A = \begin{pmatrix} 1 & 0 & 0 \\1 & 0 & 1 \\0 & 1 & 0 \end{pmatrix} $ satisfy $ A^n = A^{n-2} + A^2 - I $ for $ n \geq 3 $. Then the sum of all the elements of $ A^{50} $ is:
A point particle of charge \( Q \) is located at \( P \) along the axis of an electric dipole 1 at a distance \( r \) as shown in the figure. The point \( P \) is also on the equatorial plane of a second electric dipole 2 at a distance \( r \). The dipoles are made of opposite charge \( q \) separated by a distance \( 2a \). For the charge particle at \( P \) not to experience any net force, which of the following correctly describes the situation?

For the thermal decomposition of \( N_2O_5(g) \) at constant volume, the following table can be formed, for the reaction mentioned below: \[ 2 N_2O_5(g) \rightarrow 2 N_2O_4(g) + O_2(g) \] Given: Rate constant for the reaction is \( 4.606 \times 10^{-2} \text{ s}^{-1} \).
