To solve the problem, we need to identify the element with the electronic configuration [Ar]3d\(^{10}\)4s\(^1\).
1. Understanding the Configuration:
The configuration [Ar] represents the electron configuration of argon, which is 1s\(^2\)2s\(^2\)2p\(^6\)3s\(^2\)3p\(^6\), accounting for 18 electrons. The additional 3d\(^{10}\)4s\(^1\) indicates:
- 3d\(^{10}\): 10 electrons in the 3d subshell.
- 4s\(^1\): 1 electron in the 4s subshell.
Total electrons = 18 (from [Ar]) + 10 (3d) + 1 (4s) = 29 electrons.
2. Determining the Element:
The number of electrons in a neutral atom equals its atomic number. An element with 29 electrons has an atomic number of 29, which corresponds to copper (Cu) on the periodic table.
3. Verifying the Configuration:
Copper (Cu, atomic number 29) has the expected ground state configuration. The noble gas core [Ar] is followed by the filling of the 4s and 3d orbitals. Typically, for transition metals, the 4s orbital fills before 3d. However, copper is an exception:
- Expected: [Ar]4s\(^2\)3d\(^9\)
- Actual: [Ar]4s\(^1\)3d\(^{10}\)
This exception occurs because a fully filled 3d subshell (3d\(^{10}\)) is more stable than a partially filled one, so an electron from 4s moves to 3d, resulting in [Ar]3d\(^{10}\)4s\(^1\), which matches the given configuration.
Final Answer:
The element with the electronic configuration [Ar]3d\(^{10}\)4s\(^1\) is copper (Cu).
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:

| S. No. | Particulars | Amount (in ₹ crore) |
|---|---|---|
| (i) | Operating Surplus | 3,740 |
| (ii) | Increase in unsold stock | 600 |
| (iii) | Sales | 10,625 |
| (iv) | Purchase of raw materials | 2,625 |
| (v) | Consumption of fixed capital | 500 |
| (vi) | Subsidies | 400 |
| (vii) | Indirect taxes | 1,200 |