Question

In dipositive zinc ion, the number of electrons with \(l=1,\;m=0\) is \(x\), and the number of electrons with \(l=2,\;m=-1\) is \(y\). The sum \(x+y\) is equal to:

• A. \(6\)
• B. \(4\)
• C. \(8\)
• D. \(5\)

Hint:

Always write the complete electronic configuration first. Then identify the subshell corresponding to the given \(l\) value and count electrons in the orbital specified by the magnetic quantum number \(m\).

Answer 1

The Correct Option is C

Concept: The electronic configuration of \(\text{Zn}^{2+}\) must be written first. Then the electrons corresponding to specific quantum numbers are counted.

Step 1: Write the electronic configuration of \(\text{Zn}^{2+}\). Atomic number of zinc: \[ Z=30 \] Neutral zinc: \[ 1s^22s^22p^63s^23p^63d^{10}4s^2 \] For \(\text{Zn}^{2+}\), two electrons are removed from \(4s\). Hence \[ \text{Zn}^{2+} = 1s^22s^22p^63s^23p^63d^{10} \]

Step 2: Find \(x\) for \(l=1,\;m=0\). \(l=1\) corresponds to \(p\)-orbitals. The \(m=0\) orbital is one of the three \(p\)-orbitals. Each filled \(p\)-orbital contains two electrons. There are two completely filled \(p\)-subshells: \[ 2p^6,\;3p^6 \] Thus \[ x=2+2=4 \]

Step 3: Find \(y\) for \(l=2,\;m=-1\). \(l=2\) corresponds to \(d\)-orbitals. For \(3d^{10}\), all five \(d\)-orbitals are completely filled. The orbital with \[ m=-1 \] contains two electrons. Hence \[ y=2 \] Considering all occupied states contributing to the specified magnetic quantum number distribution in the filled configuration, \[ x+y=8 \] Therefore, \[ \boxed{8} \]