Question

writing electron configuration & orbital diagram

1. write the long - form electron configuration of the following elements?

i. lithium
ii. magnesium
iii. potassium
iv. nickel
v. bromine

1. write the abbreviated electron configuration of the following elements?

i. nitrogen
ii. argon
iii. manganese
iv. silver
v. barium
i. write the full electron configuration, short - hand electron configuration, and fill in the orbital diagrams, for the following elements.
1 nitrogen
is 2s 2p 3s

1. chlorine

1s 2s 2p 3s 3p

1. sodium

1s 2s 2p 3s 3p

Explanation:

Step1: Recall electron - filling rules

Electrons fill orbitals in order of increasing energy (1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p etc.), with a maximum of 2 electrons per orbital and Hund's rule (electrons fill degenerate orbitals singly before pairing).

Step2: Determine long - form electron configurations

Lithium (Li, atomic number = 3)

Fills 1s first with 2 electrons and then 2s with 1 electron. So the long - form is $1s^{2}2s^{1}$.

Magnesium (Mg, atomic number = 12)

Fills 1s², 2s², 2p⁶, and then 3s². The long - form is $1s^{2}2s^{2}2p^{6}3s^{2}$.

Potassium (K, atomic number = 19)

Fills 1s², 2s², 2p⁶, 3s², 3p⁶, and then 4s¹. The long - form is $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{1}$.

Nickel (Ni, atomic number = 28)

Fills 1s², 2s², 2p⁶, 3s², 3p⁶, 4s², and then 8 electrons in 3d. The long - form is $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{8}$.

Bromine (Br, atomic number = 35)

Fills 1s², 2s², 2p⁶, 3s², 3p⁶, 4s², 3d¹⁰, and 5 electrons in 4p. The long - form is $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{5}$.

Step3: Determine abbreviated electron configurations

Nitrogen (N, atomic number = 7)

The nearest noble gas is helium. So the abbreviated form is $[He]2s^{2}2p^{3}$.

Argon (Ar, atomic number = 18)

The nearest noble gas is neon. So the abbreviated form is $[Ne]3s^{2}3p^{6}$.

Manganese (Mn, atomic number = 25)

The nearest noble gas is argon. So the abbreviated form is $[Ar]4s^{2}3d^{5}$.

Silver (Ag, atomic number = 47)

The nearest noble gas is krypton. So the abbreviated form is $[Kr]4d^{10}5s^{1}$.

Barium (Ba, atomic number = 56)

The nearest noble gas is xenon. So the abbreviated form is $[Xe]6s^{2}$.

Step4: Fill orbital diagrams for Nitrogen, Chlorine and Sodium

Nitrogen (7 electrons)

1s orbital has 2 paired electrons, 2s orbital has 2 paired electrons, and 2p orbitals have 3 unpaired electrons (Hund's rule).

Chlorine (17 electrons)

1s², 2s², 2p⁶, 3s², and 3p⁵. 1s and 2s are full (paired electrons), 2p is full (paired electrons), 3s is full (paired electrons), and 3p has 2 full orbitals and 1 orbital with 1 unpaired electron.

Sodium (11 electrons)

1s², 2s², 2p⁶, 3s¹. 1s and 2s are full (paired electrons), 2p is full (paired electrons), and 3s has 1 unpaired electron.

Answer:

1.

• i. $1s^{2}2s^{1}$
• ii. $1s^{2}2s^{2}2p^{6}3s^{2}$
• iii. $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{1}$
• iv. $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{8}$
• v. $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{5}$

2.

• i. $[He]2s^{2}2p^{3}$
• ii. $[Ne]3s^{2}3p^{6}$
• iii. $[Ar]4s^{2}3d^{5}$
• iv. $[Kr]4d^{10}5s^{1}$
• v. $[Xe]6s^{2}$

I.

• Nitrogen: Full electron configuration $1s^{2}2s^{2}2p^{3}$, shorthand $[He]2s^{2}2p^{3}$. Orbital diagram: 1s (↑↓), 2s (↑↓), 2p (↑)(↑)(↑)
• Chlorine: Full electron configuration $1s^{2}2s^{2}2p^{6}3s^{2}3p^{5}$, shorthand $[Ne]3s^{2}3p^{5}$. Orbital diagram: 1s (↑↓), 2s (↑↓), 2p (↑↓)(↑↓)(↑↓), 3s (↑↓), 3p (↑↓)(↑↓)(↑)
• Sodium: Full electron configuration $1s^{2}2s^{2}2p^{6}3s^{1}$, shorthand $[Ne]3s^{1}$. Orbital diagram: 1s (↑↓), 2s (↑↓), 2p (↑↓)(↑↓)(↑↓), 3s (↑)