Question

1. how many unpaired electrons are there in the following atoms? (consider the last orbital diagram.)

draw the last orbital diagram here
a. nitrogen atom =

b. iodine atom =

c. nickel atom =

d. barium atom =

Explanation:

Part (a): Nitrogen Atom

Step 1: Determine Electron Configuration

Nitrogen has an atomic number of 7, so its electron configuration is $1s^2 2s^2 2p^3$. The last orbital is the 2p orbital.

Step 2: Apply Hund's Rule

Hund's rule states that electrons fill degenerate orbitals (orbitals of the same energy) singly first, with parallel spins. The 2p subshell has 3 orbitals, and we have 3 electrons to fill them. So each 2p orbital gets one electron, resulting in 3 unpaired electrons.

Step 1: Determine Electron Configuration

Iodine has an atomic number of 53. Its electron configuration is $[Kr] 4d^{10} 5s^2 5p^5$. The last orbital is the 5p orbital.

Step 2: Analyze 5p Subshell

The 5p subshell has 3 orbitals. We have 5 electrons in the 5p subshell. First, we fill each of the 3 orbitals with one electron (3 electrons), then we pair electrons in the orbitals. So we have $5 - 3 = 2$ paired electrons and $3 - 2 = 1$ unpaired electron (since two orbitals will have paired electrons and one will have a single unpaired electron).

Step 1: Determine Electron Configuration

Nickel has an atomic number of 28. Its electron configuration is $[Ar] 3d^8 4s^2$. The last orbital (valence orbital, and the d - orbital is the one with unpaired electrons here) is the 3d orbital.

Step 2: Analyze 3d Subshell

The 3d subshell has 5 orbitals. We have 8 electrons in the 3d subshell. Using Hund's rule, we first fill each of the 5 orbitals with one electron (5 electrons), then we pair electrons in the orbitals. So we have $8 - 5 = 3$ paired electrons? Wait, no. Wait, 5 orbitals, first 5 electrons are unpaired (one in each orbital), then the next 3 electrons will pair with 3 of the 5 orbitals. So the number of unpaired electrons is $5 - 3 = 2$? Wait, no. Wait, 3d^8: the 3d orbitals are 5 in number. Let's fill them:

Orbital 1: $\uparrow \downarrow$

Orbital 2: $\uparrow \downarrow$

Orbital 3: $\uparrow \downarrow$

Orbital 4: $\uparrow$

Orbital 5: $\uparrow$

Wait, no, Hund's rule: electrons fill singly first. So for 8 electrons in 5 orbitals:

First, fill 5 orbitals with one electron each: 5 electrons (all unpaired). Then we have $8 - 5 = 3$ electrons left. These 3 electrons will pair with 3 of the 5 orbitals. So the number of unpaired electrons is $5 - 3 = 2$? Wait, no, when we add the 3 electrons, we pair them with 3 of the 5 singly - occupied orbitals. So now, 3 orbitals have $\uparrow \downarrow$ (paired) and 2 orbitals have $\uparrow$ (unpaired). So the number of unpaired electrons is 2. Wait, but actually, the correct electron configuration for Ni is $[Ar] 3d^8 4s^2$, and the 3d subshell has 8 electrons. The 3d orbitals:

The five 3d orbitals: we fill them as per Hund's rule. The first 5 electrons go into each orbital with parallel spins (unpaired). Then the next 3 electrons will pair with 3 of the 5 orbitals. So the number of unpaired electrons is $5 - 3 = 2$? Wait, no, 5 orbitals, 8 electrons:

Orbital 1: $\uparrow \downarrow$

Orbital 2: $\uparrow \downarrow$

Orbital 3: $\uparrow \downarrow$

Orbital 4: $\uparrow$

Orbital 5: $\uparrow$

Yes, so there are 2 unpaired electrons.

Answer:

3 (Orbital diagram for 2p: $\uparrow \; \uparrow \; \uparrow$ in three separate boxes representing the 2p orbitals)

Part (b): Iodine Atom