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andreyandreev [35.5K]
3 years ago
8

Without consulting Figure 28 or Table 22, determine whether each of the following electron configurations is an inert gas,a halo

gen, an alkali metal, an alkaline earth metal. or a transition metal. Justify your choices.(a) 1522522p63$23p5(b)1$22$22p63$23p63d74$2(c) 1522522p63523p63d104524p6(d) 1522522p63$23p6451(e) 1522522p63§23p63d104§24p64¢3552(f) 1522522p6352
Chemistry
1 answer:
Anna [14]3 years ago
7 0

This is an incomplete question, here is a complete question.

Determine whether each of the following electron configurations is an inert gas, a halogen, an alkali metal, an alkaline earth metal, or a transition metal. Justify your choices.

(a) 1s^22s^22p^63s^23p^5

(b) 1s^22s^22p^63s^23p^63d^74s^2

(c) 1s^22s^22p^63s^23p^63d^{10}4s^24p^6

(d) 1s^22s^22p^63s^23p^64s^1

(e) 1s^22s^22p^63s^23p^63d^24s^2

(f) 1s^22s^22p^63s^2

Answer :

(a) 1s^22s^22p^63s^23p^5   → Halogen

(b) 1s^22s^22p^63s^23p^63d^74s^2    → Transition metal

(c) 1s^22s^22p^63s^23p^63d^{10}4s^24p^6   → Transition metal

(d) 1s^22s^22p^63s^23p^64s^1   → Halogen metal

(e) 1s^22s^22p^63s^23p^63d^24s^2     → Transition metal

(f) 1s^22s^22p^63s^2   → Alkaline earth metal

Explanation :

Inert gas : These are the gases which lie in group 18.

Their general electronic configuration is: ns^2np^6 where n is the outermost shell.

Halogen : These are the elements which lie in group 17.

Their general electronic configuration is: ns^2np^5 where n is the outermost shell.

An alkali metal : These are the elements which lie in group 1.

Their general electronic configuration is: ns^1 where n is the outermost shell.

An alkaline earth metal : These are the elements which lie in group 2.

Their general electronic configuration is: ns^2 where n is the outermost shell.

Transition elements : They are the elements which lie between 's' and 'p' block elements. These are the elements which lie in group 3 to 12. The valence electrons of these elements enter d-orbital.

Their general electronic configuration is:  where n is the outermost shell.

(a) 1s^22s^22p^63s^23p^5

The element having this electronic configuration belongs to the halogen family.

(b) 1s^22s^22p^63s^23p^63d^74s^2

The element having this electronic configuration belongs to the transition family.

(c) 1s^22s^22p^63s^23p^63d^{10}4s^24p^6

The element having this electronic configuration belongs to the transition family.

(d) 1s^22s^22p^63s^23p^64s^1

The element having this electronic configuration belongs to the halogen family.

(e) 1s^22s^22p^63s^23p^63d^24s^2

The element having this electronic configuration belongs to the transition family.

(f) 1s^22s^22p^63s^2

The element having this electronic configuration belongs to the an alkaline earth metal.

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<u>Answer:</u> The volume of chlorine gas produced in the reaction is 2.06 L.

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To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Given mass of potassium permanganate = 6.23 g

Molar mass of potassium permanganate = 158.034 g/mol

Putting values in above equation, we get:

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To calculate the moles of hydrochloric acid, we use the equation:

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Molarity of HCl = 6.00 M

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Putting values in above equation, we get:

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By Stoichiometry of the reaction:

16 moles of hydrochloric acid reacts with 2 moles of potassium permanganate.

So, 0.27 moles of hydrochloric acid will react with = \frac{2}{16}\times 0.27=0.033moles of potassium permanganate.

As, given amount of potassium permanganate is more than the required amount. So, it is considered as an excess reagent.

Thus, hydrochloric acid is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

16 moles of hydrochloric acid reacts with 5 moles of chlorine gas.

So, 0.27 moles of hydrochloric acid will react with = \frac{5}{16}\times 0.27=0.0843moles of chlorine gas.

  • To calculate the volume of gas, we use the equation given by ideal gas equation:

PV=nRT

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P = pressure of the gas = 1.05 atm

V = Volume of gas = ? L

n = Number of moles = 0.0843 mol

R = Gas constant = 0.0820\text{ L atm }mol^{-1}K^{-1}

T = temperature of the gas = 40^oC=[40+273]K=313K

Putting values in above equation, we get:

1.05atm\times V=0.0843\times 0.0820\text{ L atm }mol^{-1}K^{-1}\times 313K\\\\V=2.06L

Hence, the volume of chlorine gas produced in the reaction is 2.06 L.

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