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Evgen [1.6K]
4 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]4 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 it is 37 by 46 x 806

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3 years ago
A change in the identity of a substance where a new substance is formed
slamgirl [31]
A change that alters the identity of a substance resulting in a new substance or substances with different properties. A change to a substance that occurs without forming a new substance, such as a change in size or state of matter.
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4 years ago
Calculate the [H+] and pH of a 0.0040 M hydrazoic acid solution. Keep in mind that the Ka of hydrazoic acid is 2.20×10−5. Use th
Vera_Pavlovna [14]

Answer:

[H^+]=0.000285

pH=3.55

Explanation:

In this, we can with the <u>ionization equation</u> for the hydrazoic acid (HN_3). So:

HN_3~~H^+~+~N_3^-

Now, due to the Ka constant value, we have to use the whole equilibrium because this <u>is not a strong acid</u>. So, we have to write the <u>Ka expression</u>:

Ka=\frac{[H^+][N_3^-]}{[HN_3]}

For each mol of H^+ produced we will have 1 mol of N_3^-. So, we can use <u>"X" for the unknown</u> values and replace in the Ka equation:

Ka=\frac{X*X}{[HN_3]}

Additionally, we have to keep in mind that HN_3 is a reagent, this means that we will be <u>consumed</u>. We dont know how much acid would be consumed but we can express a<u> subtraction from the initial value</u>, so:

Ka=\frac{X*X}{0.004-X}

Finally, we can put the ka value and <u>solve for "X"</u>:

2.2X10^-^5=\frac{X*X}{0.004-X}

2.2X10^-^5=\frac{X^2}{0.004-X}

X= 0.000285

So, we have a concentration of 0.000285 for H^+. With this in mind, we can calculate the <u>pH value</u>:

pH=-Log[H^+]=-Log[0.000285]=3.55

I hope it helps!

8 0
3 years ago
The rate of decay of a radioactive substances is calculated by
KIM [24]
Answer is highlighted

5 0
2 years ago
Is the temperature of absolute zero measured in celsius A.-373 B.-73 C.-173 D.-273
Andrew [12]

Answer:

Option D is correct = -273 °C

Explanation:

Data given:

Temperature of absolute zero

Absolute Zero in Celsius = ?

Solution:

As we know

internationally the temperature of Absolute zero on kelvin scale  = 0 K

So to convert Kelvin temperature to degree Celsius formula will used

                  T(K) = T(°C) + 273

Rearrange the above equation for °C

                  T(°C) = T(K) - 273 . . . . (1)

Put value in above eq.1

                  T(°C) = 0 - 273

                  T(°C) =  -273

So,

The absolute zero in °C = -273

so option D is correct

3 0
3 years ago
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