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Evgen [1.6K]
3 years ago
5

The trend for ionization energy is a general increase from left to right across a period. However, magnesium (Mg) is found to ha

ve a higher first ionization energy value than aluminum (Al). Explain this exception to the general trend in terms of electron arrangements and attraction/repulsion
Chemistry
1 answer:
Advocard [28]3 years ago
5 0

Answer:

Explanation:

The noticeable exception in the trends of ionization energy of Mg and Al can best be explained using the arrangements of the electrons in the sub-levels.

Mg with 12 electrons has an electronic configuration of 2,8,2 = 1S²2S²2P⁶3S²

Al with 13 electrons has an electronic configuration of 2,8,3 = 1S²2S²2P⁶3S²3P¹

Lets take a little peep into what ionization energy is.

Ionization energy is a measure of the readiness of an atom to lose electron. The amount of energy needed to remove an electron is called the ionization energy. In this regard, the first ionization energy is the energy need to remove the valence or the most loosely held electron in an atom.

The magnitude of the ionization energy depends on a number of factors. The most important to our discuss here are:

  • Nuclear charge
  • Atomic radius
  • Sublevel accomodating the electron to be removed
  • Special stability of filled or half-filled sublevels.

The arrangement of electrons in Mg confers a special stability on it because of the filled sublevel.

The maximum number of electrons in each sub-levels are:

s-orbital = 2 electrons

p- orbital = 6 electrons

d - orbital = 10 electrons

f - orbital = 14 electrons

In Mg, 1S²2S²2P⁶3S² , the outermost s-sublevel have 2 electrons and s-orbital can accomodate a maximum of 2 electrons in this subshell. This makes the arrangement stable.

For Al, 1S²2S²2P⁶3S²3P¹,  the outermost p-subshell houses just one electron but it can accommodate a maximum of 6 electrons. This arrangement is unstable.

Due to this electron arrangements, Mg would have a very stable configuration and a greater first ionization energy compared to Al.

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The equation for the reaction between methane and oxygen to yield carbon dioxide and water is confirmed to be balanced per this approach, as shown here: CH 4 + 2O2⟶CO2 + 2H2O

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C 1×1 = 1 1×1 = 1 1 = 1, yes

H 4×1 = 4 2×2 = 4 4 = 4, yes

O 2×2 = 4 (1×2) + (2×1) = 4 4 = 4, yes

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Comparing the number of H and O atoms on either side of this equation confirms its imbalance:

Element Reactants Products Balanced?

H 1×2 = 2 1×2 = 2 2 = 2, yes

O 1×1 = 1 1×2 = 2 1 ≠ 2, no

H2O to H2 O2 would yield balance in the number of atoms, but doing so also changes the reactant’s identity (it’s now hydrogen peroxide and not water). The O atom balance may be achieved by changing the coefficient for H2O to 2.

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2H2O⟶2H2 + O2

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7 0
3 years ago
1.* Write the name and symbol of the ion formed when
wlad13 [49]
1a) anion and S-2
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6 0
3 years ago
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Which statement is TRUE about oxidation-reduction reactions? a. Every oxidation must be accompanied by a reduction. b. There are
Tomtit [17]

Answer:

Every oxidation must be accompanied by a reduction.

Explanation:

Oxidation and reduction are complementary processes. There can be no oxidation without reduction and vice versa. It is actually a given an take affair. A specie looses electrons which must be gained by another specie to complete the process. This explains why the selected option is the correct one.

8 0
3 years ago
What mass of solute is needed to prepare each of the following solutions?
noname [10]

Answer:

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Please see the answers in the picture attached below.

Explanation:

Please see the step-by-step solution in the picture attached below.

Hope this answer can help you. Have a nice day!

5 0
3 years ago
State the oxidation number assigned to each bold element in the formula: NH4+1 a 3 b -3 c -1 d 6
Leya [2.2K]
The fomula is NH4 (1+)


There are only two elements N and H.


As per oxidation state rules, the most electronegative element will have a negative oxidation state and the other element will have a positive oxidation state.


N is more electronative than H, so H will have a positive oxidation state and nitrogen will have a negative oxidation state.


You can also use the rule that states the hydrogen mostly has 1+ oxidation state,except when it is bonded to metals.


In conclusion the oxidation state of H in NH4 (1+) is 1+.


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8 0
4 years ago
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