Which statement best describes how electrons fill orbitals in the periodic table? Electrons fill orbitals in order of their incr
2 answers:
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Answer:
<u>The molecular formula of the gas sample =
</u>
<u>Lewis structure is shown in the image below.</u>
<u>The geometry around each carbon atom is linear.</u>
Explanation:
Given that:
Temperature = 294.0 K
V = 10.0 mL = 0.01 L ( 1 mL = 0.001 L )
Pressure = 1.10 atm
Using ideal gas equation as:
where,
P is the pressure
V is the volume
n is the number of moles
T is the temperature
R is Gas constant having value = 0.0821 L atm/ K mol
Applying the equation as:
1.10 atm × 0.01 L = n ×0.0821 L atm/ K mol × 294.0 K
⇒n = 0.0004557 mol
Given, mass = 0.04328 g
The formula for the calculation of moles is shown below:
Thus,
<u>Molar mass of the gas sample = 94.9747 g/mol</u>
Given that:-
% of C = 25.305
Molar mass of C = 12.0107 g/mol
% moles of C = 25.305 / 12.0107 = 2.1069
% of Cl = 74.695
Molar mass of Cl = 35.453 g/mol
% moles of Cl = 74.695 / 35.453 = 2.1069
Taking the simplest ratio for C and Cl as:
2.1069 : 2.1069 = 1 : 1
<u>The empirical formula is = </u>
Molecular formulas is the actual number of atoms of each element in the compound while empirical formulas is the simplest or reduced ratio of the elements in the compound.
Thus,
Molecular mass = n × Empirical mass
Where, n is any positive number from 1, 2, 3...
Mass from the Empirical formula = 12 + 35.5 = 47.5 g/mol
Molar mass = 94.9747 g/mol
So,
Molecular mass = n × Empirical mass
94.9747 = n × 47.5
⇒ n = 2
<u>The molecular formula of the gas sample =
</u>
Also,
Valence electrons of carbon = 4
Valence electrons of Chlorine = 7
The total number of the valence electrons = 4*2 + 7*2 = 22
The Lewis structure is drawn in such a way that the octet of each atom in the molecule is complete. So,
The Lewis structure is shown in the image below.
According to the theory, the atoms will form a geometry in such a way that there is minimum repulsion and maximum stability. The carbon atoms are sp hybridized.
<u>So, it is of linear shape.</u>
