orbital diagrams and electron configurations 1 chem
complete these in your notebook!
for each neutral element and charged ion below:
1. draw an orbital diagram, the orbital lines and arrows
2. write the electron configuration, or “manager’s code”
3. for cations with (+) charge, take away electrons from the outermost shell, or “boarding house floor”.
4. for anions with (-) charge, add electrons to the outermost shell, or “boarding house floor”.
1) p
2) p³⁻
3) k
4) k¹⁺
5) br
6) br¹⁻
7) fe
8) fe²⁺
9) s
10) s²⁻

Question

Accepted Answer

# Explanation:
## Step1: Analyze the problem for P (Phosphorus)
Phosphorus has an atomic number of 15, so it has 15 electrons. The electron configuration follows the Aufbau principle, Pauli exclusion principle, and Hund's rule. The orbital filling order is $1s$, $2s$, $2p$, $3s$, $3p$.
- $1s$ can hold 2 electrons: $1s^2$
- $2s$ can hold 2 electrons: $2s^2$
- $2p$ can hold 6 electrons: $2p^6$
- $3s$ can hold 2 electrons: $3s^2$
- $3p$ has 3 electrons (since $2 + 2+ 6 + 2 + 3=15$): $3p^3$

Orbital diagram:
- $1s$: $\uparrow\downarrow$
- $2s$: $\uparrow\downarrow$
- $2p$: $\uparrow\downarrow$ $\uparrow\downarrow$ $\uparrow\downarrow$
- $3s$: $\uparrow\downarrow$
- $3p$: $\uparrow$ $\uparrow$ $\uparrow$

## Step2: Analyze $P^{3-}$ (Phosphide ion)
A $3-$ charge means we add 3 electrons to P. So total electrons are $15 + 3=18$.
Electron configuration:
- $1s^2$, $2s^2$, $2p^6$, $3s^2$, $3p^6$ (since $3p$ now has $3 + 3 = 6$ electrons)

Orbital diagram for $3p$ will be $\uparrow\downarrow$ $\uparrow\downarrow$ $\uparrow\downarrow$ (filled $3p$ orbitals), others same as P up to $3s$.

## Step3: Analyze K (Potassium)
Atomic number 19, 19 electrons. Orbital filling:
- $1s^2$, $2s^2$, $2p^6$, $3s^2$, $3p^6$, $4s^1$ (since 4s fills before 3d, and $2+2 + 6+2 + 6+1=19$)

Orbital diagram:
- $1s$: $\uparrow\downarrow$
- $2s$: $\uparrow\downarrow$
- $2p$: $\uparrow\downarrow$ $\uparrow\downarrow$ $\uparrow\downarrow$
- $3s$: $\uparrow\downarrow$
- $3p$: $\uparrow\downarrow$ $\uparrow\downarrow$ $\uparrow\downarrow$
- $4s$: $\uparrow$

## Step4: Analyze $K^{2+}$ (Potassium ion with +2 charge)
Lose 2 electrons from the outermost shell (4s first, then 3p? Wait, no: for cations, we remove from outermost shell. K's outermost is 4s. So remove 2 electrons from 4s? Wait, K has electron config $1s^22s^22p^63s^23p^64s^1$. Losing 2 electrons: first lose 1 from 4s (becomes $K^+$: $1s^2...3p^6$), then lose another from 3p? Wait, no, the rule says "take away electrons from the outermost shell, or 'boarding house floor'". The outermost shell for K is n=4 (4s). But 4s has 1 electron. So to lose 2 electrons, we lose 1 from 4s and 1 from 3p? Wait, maybe the problem has a typo, but following the rule: outermost shell is the highest n. For K, n=4 (4s). But 4s has 1 electron. So losing 2 electrons: first 4s (1 electron), then 3p (which is n=3, but maybe the rule is simplified. Alternatively, maybe the problem meant $K^+$ (1+), but for $K^{2+}$, total electrons $19 - 2 = 17$. Electron config: $1s^22s^22p^63s^23p^5$ (since we remove 2 electrons: 1 from 4s, 1 from 3p? Wait, no, correct way: electron removal from highest energy level first. The electron configuration of K is $[Ar]4s^1$ (Ar is $1s^22s^22p^63s^23p^6$). Removing 2 electrons: first remove $4s^1$ (now $4s^0$), then remove one from $3p^6$ (becomes $3p^5$). So config is $1s^22s^22p^63s^23p^5$.

## Step5: Analyze Br (Bromine)
Atomic number 35, 35 electrons. Electron config: $[Ar]3d^{10}4s^24p^5$ (since after $3p^6$, we fill $4s$, then $3d$ (10 electrons), then $4p$ (5 electrons: $2 + 10+ 5=35$ (wait, $1s^22s^22p^63s^23p^64s^23d^{10}4p^5$)

Orbital diagram: $1s$ filled, $2s$ filled, $2p$ filled, $3s$ filled, $3p$ filled, $4s$ $\uparrow\downarrow$, $3d$ (10 electrons: 5 orbitals, each $\uparrow\downarrow$), $4p$: $\uparrow\downarrow$ $\uparrow\downarrow$ $\uparrow$

## Step6: Analyze $Br^-$ (Bromide ion)
$1-$ charge, add 1 electron to Br. Total electrons $35 + 1 = 36$. Electron config: $[Ar]3d^{10}4s^24p^6$ (4p now has $5 + 1 = 6$ electrons, filled).

## Step7: Analyze Fe (Iron)
Atomic number 26, 26 electrons. Electron config: $[Ar]3d^64s^2$ (since $1s^22s^22p^63s^23p^64s^23d^6$)

Orbital diagram: $4s$: $\uparrow\downarrow$, $3d$: $\uparrow$ $\uparrow$ $\uparrow$ $\uparrow$ $\uparrow$ $\uparrow$ (wait, Hund's rule: 3d has 6 electrons, so first 5 orbitals with $\uparrow$, then one $\downarrow$: $\uparrow$ $\uparrow$ $\uparrow$ $\uparrow$ $\uparrow$ $\downarrow$)

## Step8: Analyze $Fe^{2+}$ (Ferrous ion)
Lose 2 electrons, from outermost shell (4s first, since 4s is higher energy than 3d in Fe? Wait, in transition metals, 4s electrons are removed first. So Fe has $4s^23d^6$, losing 2 electrons from 4s: becomes $3d^6$. So electron config: $[Ar]3d^6$ (or $1s^22s^22p^63s^23p^63d^6$)

## Step9: Analyze S (Sulfur)
Atomic number 16, 16 electrons. Electron config: $1s^22s^22p^63s^23p^4$

Orbital diagram: $3p$ has 4 electrons: $\uparrow\downarrow$ $\uparrow$ $\uparrow$

## Step10: Analyze $S^{2-}$ (Sulfide ion)
Add 2 electrons to S, total electrons $16 + 2 = 18$. Electron config: $1s^22s^22p^63s^23p^6$ (3p now has $4 + 2 = 6$ electrons, filled).

# Answer:
The solutions involve drawing orbital diagrams and writing electron configurations as analyzed above for each species (P, $P^{3-}$, K, $K^{2+}$, Br, $Br^-$, Fe, $Fe^{2+}$, S, $S^{2-}$) following the rules of electron filling (Aufbau, Pauli, Hund) and ion charge effects (adding/removing electrons from outermost shell). For example, electron configuration of P is $1s^22s^22p^63s^23p^3$, $P^{3-}$ is $1s^22s^22p^63s^23p^6$, K is $1s^22s^22p^63s^23p^64s^1$, etc., with corresponding orbital diagrams showing electron spins and orbital filling.

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QUESTION IMAGE

Question

Explanation:

Step1: Analyze the problem for P (Phosphorus)

Step2: Analyze \(P^{3-}\) (Phosphide ion)

Step3: Analyze K (Potassium)

Step4: Analyze \(K^{2+}\) (Potassium ion with +2 charge)

Step5: Analyze Br (Bromine)

Step6: Analyze \(Br^-\) (Bromide ion)

Answer: