Elements such as Cu, Zn, Ag, and Cd exhibit a \(d^{10}\) electronic configuration:
- \([ \text{Cu} ] = [ \text{Ar} ] 3d^{10} 4s^1\),
- \([ \text{Zn} ] = [ \text{Ar} ] 3d^{10} 4s^2\),
- \([ \text{Ag} ] = [ \text{Kr} ] 4d^{10} 5s^1\),
- \([ \text{Cd} ] = [ \text{Kr} ] 4d^{10} 5s^2\).
The Correct answer is: \( {}^{29}\text{Cu}, {}^{30}\text{Zn}, {}^{48}\text{Cd}, {}^{47}\text{Ag} \)
Which of the following is the correct electronic configuration for \( \text{Oxygen (O)} \)?
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:
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)
Let a line passing through the point $ (4,1,0) $ intersect the line $ L_1: \frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4} $ at the point $ A(\alpha, \beta, \gamma) $ and the line $ L_2: x - 6 = y = -z + 4 $ at the point $ B(a, b, c) $. Then $ \begin{vmatrix} 1 & 0 & 1 \\ \alpha & \beta & \gamma \\ a & b & c \end{vmatrix} \text{ is equal to} $
Resonance in X$_2$Y can be represented as
The enthalpy of formation of X$_2$Y is 80 kJ mol$^{-1}$, and the magnitude of resonance energy of X$_2$Y is: