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Umnica [9.8K]
4 years ago
12

Identify the outer electron configurations for the (a) alkali metals, (b) alkaline earth metals, (c) halogens, (d) noble gases.

Chemistry
1 answer:
Doss [256]4 years ago
6 0
a) Alkali metals

=> group 1

=> Li:  1s2 2s  => 1s
     Na: [Ne] 3s => 3s
     K: [Ar] 4s => 4s
     Rb: [Kr] 5s => 5s
     Cs: [Xe] 6s => 6s
     Fr: [Rn] 7s => 7s

=> outer electron configuration is ns, where n is the main energy level: 1, 2, 3, 4, 5, 6,7.
     
b) Alkaline earth metals

=> group 2 => you have to add 1 electron to the alkaly metal of the same row.

=> Be: [He] 2s2 => 2s2
     Mg: [Ne] 3s2 => 3s2
     Ca: [Ar] 4s2 => 4s2
     Sr: [Kr] 5s2 => 5s2
     Ba: [Xe] 6s2 => 6s2
     Ra: [Rn[ 7s2 => 7s2

=>outer electron configuration is n s2, where n is the main energy level: 1, 2, 3, 4, 5, 6, 7

c) halogens

=> group 7

=> F: [He] 2s2 2p5 => 2s2 2p5
     Cl: [Ne] 3s2 3p5 => 3s2 3p5
     Br: [Ar] 3d10 4s2 4p5 => 4s2 4p5
     I: [Kr] 4d10 5s2 5p6 => 5s2 5p5
     At: [Xe] 4f14 5d10 6s2 6p5

=> outer electron configuration is ns2 np5, where n is the main energy level 1, 2, 3, 4, 5, 6, 7

d) Noble gases

=> group 8

I will show only the outer shell which is what is requested

=> He: 1s2
     Ne: ... 2s2 2p6
     Ar: ... 3s2 3p6
     Kr: ... 4s2 4p6
     Xe: ... 5s2 5p6
     Rn: ... 6s2 6p6

=> the outer electron configuration is ns2 np6, except for He for which it is 1s2
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The equilibrium constant, K, for the following reaction is 5.10X10 at 548 K. NH_CH(s) 2 NH3(E) + HC1(2) Calculate the equilibriu
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Answer:

The equilibrium concentration of HCl is 0.01707 M.

Explanation:

Equilibrium constant of the reaction = K_c=5.10\times 10^{-6}

Moles of ammonium chloride = 0.573 mol

Concentration of ammonium chloride = \frac{0.573 mol}{1.00 L}=0.573 M

     NH_4HCl(s)\rightleftharpoons 2 NH_3(g) + HCl(g)

Initial:            0.573     0           0

At eq'm:      (0.573-x)   x           x

We are given:

[NH_4Cl]_{eq}=(0.573-x)

[HCl]_{eq}=x

[NH_3]_{eq}=x

Calculating for 'x'. we get:

The expression of K_{c} for above reaction follows:

K_c=\frac{[HCl][NH_3]}{[NH_4Cl]}

Putting values in above equation, we get:

5.10\times 10^{-6}=\frac{x\times x}{(0.573-x)}

2.9223\times 10^{-6}-5.10x\times 10^{-6}=x^2

x^2-2.9223\times 10^{-6}+(5.10\times 10^{-6})x=0

On solving this quadratic equation we get:

x = 0.01707 M

The equilibrium concentration of HCl is 0.01707 M.

3 0
3 years ago
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