The question asks which set of elements all possess a \( d^{10} \) electronic configuration. Let's evaluate each element in the given options by determining their electronic configurations and identifying those with a completely filled \( d \)-subshell.
On evaluating the individual electronic configurations for each of the elements provided in this option, all of them possess a \( d^{10} \) electronic configuration. Therefore, the correct answer consists of elements \({}^{29}\text{Cu}, {}^{30}\text{Zn}, {}^{48}\text{Cd}, {}^{47}\text{Ag}\).
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/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:
