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Marizza181 [45]
3 years ago
8

Nonmetals gain electrons under certain conditions to attain a noble-gas electron configuration. how many electrons must be gaine

d by the element c?
Chemistry
1 answer:
r-ruslan [8.4K]3 years ago
8 0
System is said to have achieved noble-gas configuration, when it's valance shell is completely filled.

Atomic number of carbon is 6. Thus, it has 6 electrons.

The electronic configuration of carbon is 1s2 2s2 2p2

Now, the inert gas closest to C is Ne, whose atomic number is 10.

Thus, there are excess of 4 electrons in Ne as compared to C.

Hence, carbon must gain 4 electrons to achieve noble-gas configuration.

Alternatively,  C can also lose 4 electron to achieve noble gas configuration of He.
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Hydrogen, iron, and gallium is the correct answer. Hence, option B is correct.

<h3>What is an electron?</h3>

An electron is a negatively charged subatomic particle.

Let us take another look at the periodic table. The periodic table is composed of metals, metalloids and nonmetals.

Metals lose electrons more easily, that is why they form cations. Metals are found on the left-hand side of the periodic table.

As we move away from the left-hand side of the periodic table, the ability to form cations decreases as metallic character decreases and elements become more nonmetallic.

The noble gas helium rarely loses electrons. Fluorine would rather accept an electron to form a negative ion.

Hence, the elements hydrogen, iron and gallium are metals hence they lose electrons more easily.

Hence, option B is correct.

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The element iridium exists in nature as two isotopes: 191Ir has a mass of 190.9606 u, and 193Ir has a mass of 192.9629 u. The av
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<u>Answer:</u> The percentage abundance of _{77}^{191}\textrm{Ir} and _{77}^{193}\textrm{Ir} isotopes are 37.10% and 62.90% respectively.

<u>Explanation:</u>

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i   .....(1)

Let the fractional abundance of _{77}^{191}\textrm{Ir} isotope be 'x'. So, fractional abundance of _{77}^{193}\textrm{Ir} isotope will be '1 - x'

  • <u>For _{77}^{191}\textrm{Ir} isotope:</u>

Mass of _{77}^{191}\textrm{Ir} isotope = 190.9606 amu

Fractional abundance of _{77}^{191}\textrm{Ir} isotope = x

  • <u>For _{77}^{193}\textrm{Ir} isotope:</u>

Mass of _{77}^{193}\textrm{Ir} isotope = 192.9629 amu

Fractional abundance of _{77}^{193}\textrm{Ir} isotope = 1 - x

Average atomic mass of iridium = 192.22 amu

Putting values in equation 1, we get:

192.22=[(190.9606\times x)+(192.9629\times (1-x))]\\\\x=0.3710

Percentage abundance of _{77}^{191}\textrm{Ir} isotope = 0.3710\times 100=37.10\%

Percentage abundance of _{77}^{193}\textrm{Ir} isotope = (1-0.3710)=0.6290\times 100=62.90\%

Hence, the percentage abundance of _{77}^{191}\textrm{Ir} and _{77}^{193}\textrm{Ir} isotopes are 37.10% and 62.90% respectively.

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