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Ksju [112]
3 years ago
12

Predict the ground-state electron configuration of the following ions. Write your answers in abbreviated form, that is, beginnin

g with a noble gas in brackets.a). Cr2+b.)Cu2+c.)Co3+
Chemistry
1 answer:
abruzzese [7]3 years ago
6 0

These are three questions and three complete answers

Answer:

a) Cr²⁺: [Ar] 4s² 3d²

b) Cu²⁺: [Ar] 4s² 3d⁷

c) Co³⁺: [Ar] 4s² 3d⁴

Explanation:

<u>a) Cr²⁺</u>

  • Z = 24
  • Number of protons: 24
  • Number of elecrons of the neutral atom: 24
  • Charge of the ions: + 2
  • Number of electrons of the ion: 24 - charge = 24 - 2 = 22.
  • Electron configuration:

       Fill the orbitals in increasing order of energy. Using Aufbau's rules the order is: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ .....

       Hence, for 22 electrons you get:

       1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d²

  • Abbreviated notation: since the last complete level is the number 3s² 3p⁶, you use the noble gas of the period 3, which is Ar, and the configuration is:

       

        [Ar] 4s² 3d²

<u>b) Cu²⁺</u>

  • Z = 29
  • Number of protons: 29
  • Number of elecrons of the neutral atom: 29
  • Charge of the ion: + 2
  • Number of electrons of the ion: 29 - charge = 29 - 2 = 27.
  • Electron configuration:

       Fill the orbitals in increasing order of energy. Using Aufbau's rules the order is: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ .....

       Hence, for 27 electrons you get:

       1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d⁷

  • Abbreviated notation: since the last complete level is the number 3s² 3p⁶, you use the noble gas of the period 3, which is Ar, and the configuration is:

       

        [Ar] 4s² 3d⁷

<u>c) Co³⁺</u>

  • Z = 27
  • Number of protons: 27
  • Number of elecrons of the neutral atom: 27
  • Charge of the ion: + 3
  • Number of electrons of the ion: 27 - charge = 27 - 3 = 24.
  • Electron configuration:

       Fill the orbitals in increasing order of energy. Using Aufbau's rules the order is: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ .....

       Hence, for 24 electrons you get:

       1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d⁴

  • Abbreviated notation: since the last complete level is the number 3s² 3p⁶, you use the noble gas of the period 3, which is Ar, and the configuration is:

       

        [Ar] 4s² 3d⁴

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What is the electron configuration of rubidium?
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3 years ago
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Ad libitum [116K]

Answer:

[Kr] 5s²

Explanation:

From the question given above, the following data were obtained:

Atomic number of strontium (Sr) = 38

Electronic configuration =?

Next, we shall determine the electronic configuration of the noble gas element before strontium (Sr).

The noble gas element before strontium (Sr) is krypton (Kr). Thus, the electronic configuration of krypton (Kr) is given below:

Kr (36) => 1s² 2s²2p⁶ 3s²3p⁶ 3d¹⁰ 4s²4p⁶

Finally, we shall determine the electronic configuration of strontium (Sr). This can be obtained as follow:

Kr (36) => 1s² 2s²2p⁶ 3s²3p⁶ 3d¹⁰ 4s²4p⁶

Sr (38) =>?

Sr (38) => 1s² 2s²2p⁶ 3s²3p⁶ 3d¹⁰ 4s²4p⁶ 5s²

But

Kr (36) => 1s² 2s²2p⁶ 3s²3p⁶ 3d¹⁰ 4s²4p⁶

Therefore,

Sr (38) => [Kr] 5s²

7 0
3 years ago
Which property of gases best explains the ability of air bags to cushion the force of impact during a car accident?
Citrus2011 [14]
Gases are comressible
7 0
3 years ago
The Ka of a monoprotic weak acid is 0.00732. What is the percent ionization of a 0.123 M solution of this acid?
MaRussiya [10]

Answer:

21.60% is the percent ionization of a 0.123 M solution of this acid.

Explanation:

The equilibrium reaction for dissociation of weak acidis,

HA\rightleftharpoons A^-+H^+

initially conc.         c                       0         0

At eqm.         c(1-\alpha )                  c\alpha     c\alpha  

Concentration of the weak acid (c) = 0.123 M

Acid dissociation constant = k_a=0.00732

Degree of ionization of weak acid = \alpha

An expression of dissociation constant is given as:

k_a=\frac{(c\alpha)(c\alpha)}{c(1-\alpha )}=\frac{c\times (\alpha )^2}{(1-\alpha )}

0.00732=\frac{0.123 M\times (\alpha )^2}{(1-\alpha )}

\alpha =0.2160

Percent ionization of weak acid:

\% ionization=\frac{c\alpha }{c}\times 100

\frac{0.123 M\times 0.2160}{0.123 M}\times 100=21.60\%

5 0
3 years ago
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