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Mazyrski [523]
2 years ago
13

9) Give the set of four quantum numbers that could represent the extra electron added (using the Aufbau principle) to the neutra

l Ne atom.
10) Which 2nd period element has the most negative electron affinity? Why? Bonus questions
11) Which ionization process requires the most energy? Why?
A) Se(g) → Set(g) + e-
B) Set(g) Se2+(g) +
C) Br(g) → Brt(g) + e-
D) Brt(g) Br2+(g) + e-
12) Place the following in order of decreasing metallic character. Explain your reason.
P As K
Chemistry
1 answer:
Pie2 years ago
3 0

Answer:

See explanation

Explanation:

9) If an extra electron is added to the neon atom, then the electronic configuration becomes; 1s2 2s2 2p6 3s1

This last electron has quantum numbers;

n=3, l=0, m=0 and s = +1/2

This is so because the 2s level is already filled so the extra electron must go into the 3s level. The orbital quantum number and the magnetic quantum number for the s orbital is zero.

10) Electron affinity is the energy released when one mole of gaseous atoms accept one mole of gaseous electrons to form one mole of gaseous ions having a negative charge.

In the second period, fluorine has the greatest electron affinity since electron affinity increases across the period. The noble gas, neon has an electron affinity of 0KJ/mol.

11) Ionization energy decreases down the group but increases across the period due to increase in the size of the nuclear charge and decrease in the distance between the nucleus and the outermost electron. Hence, the process; Br+(g) ---->Br2+(g) + e- has the greatest ionization energy. Recall that the second ionization is always higher than the first ionization energy.

12)The order of decreasing metallic character here is K> As> P. Even though As and P belong to the same group, we must note that metallic character increases down the group hence the order written above.

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\boxed{\text{2408 min}}

Explanation:

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\begin{array}{rcl}\ln \dfrac{100}{90} & = & k \times 366\\\\1.054 & = & 366k\\\\k & = & \dfrac{1.054 }{366}\\\\k & = & 2.879 \times 10^{-4} \text{ min}^{-1}\\\end{array}\\\\

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