Step 1: Recall the electron configuration for oxygen
Oxygen (O) has an atomic number of 8, meaning it has 8 electrons.
Step 2: Distribute the electrons in orbitals
The electron configuration follows the Aufbau principle, which fills the lowest energy orbitals first:
- The first shell can hold up to 2 electrons, so the \( 1s \) orbital is filled first: \( 1s^2 \).
- The second shell can hold up to 8 electrons, so the \( 2s \) orbital is filled next: \( 2s^2 \).
- After that, the \( 2p \) orbital starts filling.
Oxygen has 8 electrons in total, so the next 4 electrons will go into the \( 2p \) orbital: \( 2p^4 \).
Therefore, the electron configuration of oxygen is: \[ 1s^2 2s^2 2p^4 \]
Answer:
Therefore, the correct electron configuration for an oxygen atom is \( 1s^2 2s^2 2p^4 \). So, the correct answer is option (1).
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)