Question

For the electrons of an oxygen atom, which of the following statements is correct?
A) The effective nuclear charge ${Z_{eff}}$ for an electron in $2s$ orbital is the same as ${Z_{eff}}$ for an electron in an $1p$ orbital.
B) An electron in the $2s$ orbital has the same energy as an electron in the $2p$ orbital.
C) The effective nuclear charge ${Z_{eff}}$ for an electron in $1s$ orbital is the same as ${Z_{eff}}$ for an electron in an $2s$ orbital.
D) The two electrons present in the $2s$ orbital have spin quantum numbers ${m_s}$ but of opposite sign.

Answer 1

We know Pauli’s Exclusion principle:
An atom or molecule no two electrons can have the same four quantum numbers. The condition at which the second electron can enter an orbital is if the spin of that electron is in the opposite direction of the electron which is already present in the orbit.

Complete step by step answer:
Let us discuss the option A.
We know that s-orbital has a high probability density near the nucleus, unlike p-orbitals that have a node at the nucleus. The electrons in the s-orbital are more tightly and closely bound to the nucleus and thus it experiences high effective nuclear charge. Hence ${Z_{eff}}$ for an electron in an $2s$ orbital is greater than that in $2p$ orbital. Therefore, the option A is incorrect.
We know that,
The principal quantum number is denoted by n. The value of shell ‘K’ has been given n=1, the ‘L’ shell has been given the value n=2. The secondary quantum number l divides the shells up into smaller groups of subshells and orbitals.
Energy of an orbital is governed by $n + l$ value where n=principal quantum number and l=azimuthal quantum number.
For $2s$ orbital $n + l = 2 + 0 = 2$ and for $2p$ it is $2 + 1 = 3$ , and since $n + l$ value is lower for $2s$ orbital then, its energy will be lower than that of $2p$ -orbital. Therefore, the option B is incorrect.
If the distance of the outer electrons from the nucleus increases the effective nuclear charge decreases. As $\;2s$ orbital is larger in size than $1s$ orbital. The electrons of 2s will experience lower ${Z_{eff}}$ than $1s$ electrons. Therefore, the option C is incorrect.
According to Pauli's exclusion principle: The two electrons present in any orbital have spin quantum numbers with an opposite sign ${m_s}{\text{ }} = + {\text{ }}1\;/2 and \; - 1/2$.

So, the correct answer is Option D .

Note:
Now we discuss how principal quantum number, azimuthal quantum number, magnetic quantum number and spin quantum number differ from each other.
The principal energy level of an electron refers to the orbital in which the electron is located relative to the atom’s nucleus.
The azimuthal quantum number denotes the subshell (orbital) to which the electron belongs and also determines the shape of the orbital and the energy associated with the angular momentum of the electron.
The magnetic quantum number determines the preferred orientations of orbitals in space.
The spin quantum number’ is related to electron spin.