Question

name: _ teacher: _ date: _ q1 chemistry: u2l15 electron config day 5
prepwork part 1:
for each of the atoms/ions below, write the full electron configuration.

1. *neutral atom of cobalt 2. chlorine ion with a charge of - 1

for each of the atoms/ions below, write the abbreviated notation.

1. neutral atom of iodine 2. *neutral atom of vanadium

for each of the atoms/ions below, write the orbital notation.

1. **oxygen ion with a charge of - 2 2. neutral atom of sodium
2. **write the electron configuration of an ion that is isoelectronic to na. include the ion’s name and charge. (there is more than one possible answer!)
3. **two students are studying an ion that has similar properties to al. they determine that it is isoelectronic to al. they also know based on evidence that this ion is positively charged.

• student a says that the ion must be: p3+
• student b says that the ion must be: si+1

with which student do you agree, and why?

1. bismuth is the main ingredient in pepto - bismol, the popular drug used to cure digestive ailments. in victorian times, before bismuth was well known as a digestive aid, people used a different metal

Explanation:

Step1: Recall electron - configuration rules

Use the Aufbau principle, Pauli - exclusion principle, and Hund's rule.

Step2: Find atomic number of cobalt

Cobalt (Co) has an atomic number of 27. The full electron - configuration is $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{7}$.

Step3: Consider chlorine ion

Chlorine (Cl) has an atomic number of 17. The $Cl^{-}$ ion has 18 electrons. Its full electron - configuration is $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}$.

Step4: Iodine atomic number

Iodine (I) has an atomic number of 53. The abbreviated notation uses the noble - gas core. [Kr] represents the electron configuration of krypton (36 electrons). So, the abbreviated notation for I is $[Kr]5s^{2}4d^{10}5p^{5}$.

Step5: Vanadium atomic number

Vanadium (V) has an atomic number of 23. Using the noble - gas core of argon (18 electrons), the abbreviated notation is $[Ar]4s^{2}3d^{3}$.

Step6: Oxygen ion

Oxygen (O) has an atomic number of 8. The $O^{2 - }$ ion has 10 electrons. The orbital notation:
1s: $\uparrow\downarrow$
2s: $\uparrow\downarrow$
2p: $\uparrow\downarrow$ $\uparrow\downarrow$ $\uparrow\downarrow$

Step7: Sodium atomic number

Sodium (Na) has an atomic number of 11. The orbital notation:
1s: $\uparrow\downarrow$
2s: $\uparrow\downarrow$
2p: $\uparrow\downarrow$ $\uparrow\downarrow$ $\uparrow\downarrow$
3s: $\uparrow$

Step8: Isoelectronic with Na

Sodium (Na) has 11 electrons. A $Mg^{+}$ ion has 11 electrons. Its electron - configuration is $1s^{2}2s^{2}2p^{6}3s^{1}$.

Step9: Analyze Al isoelectronic ion

Aluminum (Al) has 13 electrons. $P^{3 + }$ has 15 - 3=12 electrons. $Si^{+}$ has 14 - 1 = 13 electrons. So, Student B is correct because $Si^{+}$ is isoelectronic to Al.

Answer:

1. Neutral atom of cobalt: $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{7}$
2. Chlorine ion with a charge of - 1: $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}$
3. Neutral atom of iodine (abbreviated notation): $[Kr]5s^{2}4d^{10}5p^{5}$
4. Neutral atom of vanadium (abbreviated notation): $[Ar]4s^{2}3d^{3}$
5. Oxygen ion with a charge of - 2 (orbital notation):

1s: $\uparrow\downarrow$
2s: $\uparrow\downarrow$
2p: $\uparrow\downarrow$ $\uparrow\downarrow$ $\uparrow\downarrow$

1. Neutral atom of sodium (orbital notation):

1s: $\uparrow\downarrow$
2s: $\uparrow\downarrow$
2p: $\uparrow\downarrow$ $\uparrow\downarrow$ $\uparrow\downarrow$
3s: $\uparrow$

1. Ion isoelectronic to Na: $Mg^{+}$, $1s^{2}2s^{2}2p^{6}3s^{1}$
2. With which student do you agree: Student B, because $Si^{+}$ is isoelectronic to Al.