Respuesta :
Answer:
d. Ar, because of its higher effective nuclear charge
For the secon part see explanation below.
Explanation:
The first ionization energy is the energy required to remove an electron from the atom from its outermost shell. It depends on the nuclear charge, distance from the nucleus and the screening of other electrons in the inner shells of the atom.
Comparing Cl and Ar we see that being both elements of the third period, the Ar atom has one more proton than Cl and therefore the electron feels more nuclear charge making the first ionization of Ar greater than Cl.
a) False, electronegativity relates to attraction for an electron and not to the first ionization.
b) False, again electron affinity is not first ionization, it is defined as the energy released when the atom captures an added electron.
c) False,athough it is true that Ar has a complete octet, the higher first ionization is affected by nuclear charge. The screening of electrons in the n= 1 and 2 shells is almost the same so what is important is that the electrons in the n= 3 shell feel more nuclear charge.
d) True for all the reasons given previously : the higher effective nuclear charge in Ar.
For the second part, we have to make an inventory of the bonds being broken and formed:
ΔHºrxn = H broken - H formed, where H is the bond energy
H2 C = CH_2 + H-Br ⇒ CH_3CH_2Br
ΔHºrxn = ( 1 C=C + 4 C-H + 1 H-Br) - ( 1 C-C + 5 C-H + 1 C-Br)
ΔHºrxn (kJ) = (614 + 4(413) + 363) - ( 347 + 5 (413) + 276)
ΔHºrxn (kJ) = 2629 - 2688 = -59 kJ
This value is not in the choices due to mistaken bond energy values from the tables.
Answer:
1. Ar, because of its higher effective nuclear charge.
2. ∆Hrxn = -200 KJ/mol
Explanation:
The size of the atoms of chemical elements can be measured from their atomic radius which is also affected by the effective nuclear charge.
Recall that elements in a particular period have the same number of electron shells. Also, along a given period, atomic radius decreases due to an increase in the effective (positive) nuclear charge. This is because as the atomic (proton) number increases along that period, the charge on the nucleus also increases. With more protons in the nucleus the overall attraction between the positively charged nucleus and the negatively charged surrounding electrons increases, so the electrons are pulled closer to the nucleus thereby leading to a decrease in the atomic size.
So, along a given period atomic size decreases due to an increase in the effective nuclear charge.
The first ionization energy is the minimum energy (in kilojoules) needed to strip one mole of electrons from one mole of a gaseous atom of an element to form one mole of a gaseous unipositive ion.
Along a particular period, ionization energy increases due to an increase in the effective nuclear charge and a decrease in atomic radius. This is because, the smaller the atom the more stable it is and the more difficult it will be to remove an electron.
For the second question,
The enthalpy change of a reaction is the difference in the bond dissociation energies of the reactants and products. Bonds are broken in reactant molecules and formed in product molecules. Bond breaking energies are usually intrinsic ( endothermic, +be ∆H ) while bond forming energies are usually extrinsic ( exothermic, -ve ∆H ).
So,
∆Hrxn = n∆H(reactants/bonds broken) - m∆H(products/bonds formed)
Where n and m = stoichiometric coefficients of the products and reactants respectively from the balanced chemical equation.
First, draw the correct Lewis structures of the compounds.
Next, identify all the bonds broken and formed.
Then, from the bond dissociation energies ( usually given or looked up in texts ), sum up the bond breaking energies and the bond forming energies and subtract the bond forming energies from the bond breaking energies.
Considering this equation:
H_2C = CH_2 + H-Br rightarrow CH_3CH_2Br
The equation is balanced.
Bonds broken (number of bonds ):
I. C=C (1)
II. H-Br (1)
III. C-H (4)
Bonds formed:
I. C-C (1)
II. C-H (5)
III. C-Br (1)
∆Hrxn = [ ( 1 x C=C ) + ( 4 x C-H ) + ( 1 x H-Br ) ] – [ ( 1 x C-C ) + ( 5 x C-H ) + ( 1 x C-Br ) ]
∆Hrxn = [ ( 1 x 614 ) + ( 4 x 413 ) + ( 1 x 141 ) ] – [ ( 1 x 348 ) + ( 5x 413 ) + ( 1 x 194 ) ]
∆Hrxn = [ ( 614+1652+141) ] – [ ( 348 + 2065 + 194 ) ]
∆Hrxn = 2407 – 2607
∆Hrxn = -200KJ/mol
