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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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A chemistry graduate student is given of a chlorous acid solution. Chlorous acid is a weak acid with . What mass of should the s
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Answer:

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Explanation:

<em> Is given 250mL of a 1.60M chlorous acid HClO2 solution. Ka is 1.110x10⁻². What mass of NaClO₂ should the student dissolve in the HClO2 solution to turn it into a buffer with pH =1.45? </em>

It is possible to answer this question using Henderson-Hasselbalch equation:

pH = pKa + log₁₀ [A⁻] / [HA]

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You can change the concentration of the substance if you write the moles of the substances:

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Replacing in H-H expression, as the pH you want is 1.45:

1.45 = 1.9547 + log₁₀ [Moles NaClO₂] / [0.40 moles HClO₂]

-0.5047 = log₁₀ [Moles NaClO₂] / [0.40 moles HClO₂]

<em>0.3128 = </em>[Moles NaClO₂] / [0.40 moles HClO₂]

0.1251 = Moles NaClO₂

As molar mass of NaClO₂ is 90.44g/mol, mass of 0.1251 moles of NaClO₂ is:

0.1251 moles NaClO₂ ₓ (90.44g / mol) =

<h3>11.31g NaClO₂</h3>
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<span>Then convert to kJ/mol


</span>
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3 years ago
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