Rubidium (Rb) has the electron configuration: [Kr]5s1.
For the valence electron in the 5s orbital: - Principal quantum number, n = 5. - Azimuthal quantum number, l = 0 (since it is an s-orbital).
Magnetic quantum number, m = 0 (as m can range from –l to +l, and l = 0 allows only m = 0).
Spin quantum number, s = \(+\frac{1}{2}\) (or \(-\frac{1}{2}\) as it can have either spin).
Thus, the correct set of quantum numbers is (5, 0, 0, \(+\frac{1}{2}\)).
So, the correct answer is: 5, 0, 0, \(+\frac{1}{2}\)
Let \( y = f(x) \) be the solution of the differential equation
\[ \frac{dy}{dx} + 3y \tan^2 x + 3y = \sec^2 x \]
such that \( f(0) = \frac{e^3}{3} + 1 \), then \( f\left( \frac{\pi}{4} \right) \) is equal to:
Find the IUPAC name of the compound.
If \( \lim_{x \to 0} \left( \frac{\tan x}{x} \right)^{\frac{1}{x^2}} = p \), then \( 96 \ln p \) is: 32