Questions: Chapter 7 Assessment Question 5 of 25 (4 points) Question Attempt: 1 of 1 The following orbital diagram corresponds to the element uparrow downarrow / 1s uparrow downarrow / 2s uparrow downarrow uparrow downarrow / 2p uparrow downarrow / uparrow downarrow / 3s uparrow downarrow / 3p uparrow uparrow P Si S Se Cl

Solution Steps

The given orbital diagram is: \[ \frac{\uparrow \downarrow}{1s} \frac{\uparrow \downarrow}{2s} \frac{\uparrow \downarrow \uparrow \downarrow \uparrow \downarrow}{2p} \frac{\uparrow \downarrow}{3s} \frac{\uparrow \downarrow \uparrow \downarrow \uparrow}{3p} \]

To identify the element, we need to count the total number of electrons in the orbital diagram:

• \(1s\) orbital: 2 electrons
• \(2s\) orbital: 2 electrons
• \(2p\) orbitals: 6 electrons
• \(3s\) orbital: 2 electrons
• \(3p\) orbitals: 5 electrons

Total number of electrons = \(2 + 2 + 6 + 2 + 5 = 17\)

The element with 17 electrons is chlorine (Cl).

Final Answer