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10 Questions MCQ Test - Test: Periodic Table

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Test: Periodic Table - Question 1

Which of the following does NOT depend on the attraction of the bonding pair towards the nucleus?

Detailed Solution for Test: Periodic Table - Question 1

Electronegativity is the tendency for an atom to attract a bonding pair of electrons. Electronegativity unlike other periodic trends is measured within a bond.

Effective nuclear charge has a direct effect on each of the periodic trends. The greater the effective nuclear charge the greater the electronegativity.

Effective nuclear charge depends on the number of protons in the nucleus, the distance from the nucleus, and the amount of shielding by inner shell electrons. The number of protons increases Ze ff. while the distance and amount of shielding decreases Ze ff.

The repulsion by the electrons in the same valence shell is not measured by Ze ff and the attraction of the bonding pair does not depend on that repulsion.

Test: Periodic Table - Question 2

Which of the following statements accurately describes the different classifications of elements?

Detailed Solution for Test: Periodic Table - Question 2

The common classifications of elements are the alkali metals (Group I), alkali earth metals (Group II), pnictogens (Group V), chalcogens (Group VI), halogens (Group VII), and noble gases (group VIII).

As for alkali metals, they have distinctive flame colors because of their easily excited electron in the s shell. Not all alkali metals even have d electrons.

As for alkali earth metals, they react with water to form hydrogen gas and the respective alkaline hydroxides in aqueous solution. Metal oxides became hydroxides in water.

All the halogens do react with hydrogen gas to form hydrogen halides, i.e. HF, HCl, HBr, HI, HAt. Not all of them form strong acids since HF is considered a weak acid because of the high electronegativity of fluorine. HAt would be the strongest acid but it readily decomposes back to hydrogen gas and astatine, and astatine has a short half-life.

The transition metals include the designation of noble metals, which comprises ruthenium, rhodium, palladium, silver, osmium, iridium, platinum, and gold. The noble metals resist corrosion because their d orbitals are fully-filled.

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Test: Periodic Table - Question 3

Which of the following statements most accurately describes the characteristics of the transition metals?

Detailed Solution for Test: Periodic Table - Question 3

Transition metals are the ones which form one or more stable ions which have incompletely filled d orbitals. The filling of the d orbitals imparts particular properties to the transition metals.

Transition metals are characterized by the formation of many paramagnetic compounds due to the presence of unpaired 
d electrons. Diamagnetic compounds have paired electrons and do not possess any magnetic capabilities.

Transition metals also are characterized by many oxidation states due to the low reactivity of the unpaired d electrons. High reactivity would limit the number of oxidation states.

One difference between transition metals and main group metals is that transition metals can form coordination complexes, while main group metals tend to form binary compounds like CaCl₂.

The transition metals have little variability in their ionization energies and electronegativities due to their similar valence electron shells, which is the s shell. Even though, for instance, 3d fills after 4s, when forming ions, we remove from the highest energy orbital or the s orbital.

Test: Periodic Table - Question 4

Which of the following inequalities accurately describes the relationship between the two ionic species in terms of size?

Detailed Solution for Test: Periodic Table - Question 4

In comparing ionic radii, there are two guiding rules. First, in comparing isoelectronic species, anions are always bigger than the cations such that A3- > B2- > C- > noble gas > X+ > Y2 + > Z3

Secondly, in comparing elements in a group, the ionic radii always increase going down the group or with each extra energy level such that l- > Br- > Cl- > F-

For land Ca2+ using the isoelectronic trend, we know that the iodide ion is larger than the barium cation. Since calcium and barium are in the same group, it can be deduced that calcium is smaller than barium. I⁻  must be larger than Ca2+

For At- and Li+ start superscript, plus, end superscript, using the isoelectronic trend, we know that the astatine ion is larger than the cesium cation. Since cesium and lithium are in the same group, it can be deduced that lithium is smaller than cesium. At- must be larger than Li+

For Se2- and p3-, since they are neither isoelectronic or in the same group, it cannot be stated that Se2- > p3-. In fact Se2- < p3- with values of 198 pm and 212 pm.

As for As3- and Y3+, they are both isoelectronic. Anions are larger than corresponding isoelectronic cations, so it can be said with certainty that has As3- a larger ionic radius than  Y3+.

Test: Periodic Table - Question 5

Which of the following statements about atomic radii accurately describes the data below?

Detailed Solution for Test: Periodic Table - Question 5

Atomic radius as a general trend decreases as we move across the period due to the increased effective nuclear charge. The outer valence electrons are not effective at shielding each other from the feel and pull of the nucleus, while the inner shell provides a shielding effect on the valence electrons.

The peaks correspond to the alkali metals because they have the smallest effective nuclear charge and the electron is not held very tightly towards the nucleus.

The minimum peak corresponds to the noble gases because they have the greatest effective nuclear charge. There may be some discrepancies because noble gas does not form bonds like the others, so its covalent radius is not easily determined.

The elements right after the d block, namely Ga, Ge, As, Se, and Br, experience a phenomenon called d-block contraction, which is due to the poor shielding of the nuclear charge by the electrons in the d orbitals. The outer valence electrons are more strongly attracted by the nucleus causing the observed decrease in atomic radius.

The transition metals have relatively the same atomic size because their valence shell essentially remains the same, which is the s orbital. The d electrons instead provide a shielding effect on the s orbital which allows the atomic size to remain the same.

Test: Periodic Table - Question 6

Which of the following statements accurately describes the first ionization energy data shown in the table?

Detailed Solution for Test: Periodic Table - Question 6

Ionization energy is the energy needed to remove an electron from a neutral atom in the gaseous state and is considered endothermic.

As a general trend, ionization energy increases moving across a period on the periodic table, but the data seems to indicate there are dips along the way.

The local peaks labelled in black correspond to all the noble gases plus Hg or mercury, and the local minima correspond to the alkali metals.

The plateaus in the data correspond to the transition metals and the inner transition metals. Between K and Kr, there is one set of transition metals, and another between Rb and Xe.

The exceptions to the trend or the small dips correspond to half-filled and fully-filled orbitals. Between Li and Ne, there are two dips which correspond to B and O.

What comes prior to each is an element with half-filled or fully-filled orbitals. Be comes before B, and because of the stability of the fully-filled s orbital, it takes more energy to rip off an electron from the neutral atom. The same applies to N with half-filled p orbitals.

Test: Periodic Table - Question 7

Based on the values for the first through fifth ionization energies in the table, what is the most likely identity of Element A and B?

Detailed Solution for Test: Periodic Table - Question 7

Ionization energies increase moving across a period, as a rule of thumb, since we are removing an electron first from a neutral atom then from a cation. To reiterate, second ionization energies are always larger than first ionization energies and are roughly double.

In analyzing the data for element A, look for the biggest gap. One way to approach this is to look at the factor between ionization energies.

The second ionization energy is almost 10 times as great as the first. As for the other ionization energies, the factor is only somewhere between 1 and 2.

The significance of the high amount of energy for IE2, is to due to the stability of the noble gas configuration. One electron was removed to reach IE2, so element A must be in group I.

As for element B, the doubling of IEto get IE2 is to be expected. The large gap does not occur until reaching IE5 whereby IE4 increases by a factor of almost 4.

Test: Periodic Table - Question 8

Which of the following statements accurately describes the electron affinity data in the chart below?

Detailed Solution for Test: Periodic Table - Question 8

The minima correspond to the fully-filled s and p orbitals and some of the half-filled p and d orbitals. The stability of such configurations makes adding an electron energetically unfavorable where some elements have electron affinities (EA) that are endothermic.

There is no clear trend moving down a group due to increasing electron shells. Analyzing the data for the halides, for instance, which are the peaks, will reveal that fluorine has a lower EA than chlorine, which has the highest EA out of the group.

The maxima correspond to the halides since by acquiring one more electron they achieve a complete octet or the noble gas configuration.

The values are smaller for the nonmetals in Period II, i.e. O and F, because the repulsion between the electron being added and the electrons already present is greatest. Upon adding an electron to the neutral atom, the repulsion is more greatly felt due to their small size, and hence less energy is released.

Test: Periodic Table - Question 9

Which of the following statements describes a correct step in the formation of an ionic bond between sodium and chlorine?

Detailed Solution for Test: Periodic Table - Question 9

In order for the formation of an ionic bond to be favorable, the energy released must be greater than the energy consumed. Essentially, the cation and anion separately are formed and then brought together for bond formation.

Ionization energy is endothermic, so it would require energy. No energy is recovered from the ionization energy of sodium.

Electron affinity is exothermic, so it would release and provide energy for bond formation.

Once the ions are formed, they have to be brought together for bond formation. When the two charges are brought within picometers of each other, their electron shells which are both negatively charged will exert a certain repulsion on each other, and this repulsion will require energy to overcome.

However, the Pauli repulsion energy is small compared to the energy released in the loss in electrostatic or Coulombic potential when the opposite charges are brought together.

Test: Periodic Table - Question 10

Which of the following statements does NOT accurately describe the atomic trends?

Detailed Solution for Test: Periodic Table - Question 10

Effective nuclear charge increases moving across and decreases moving down.

Ionic radius decreases starting from the +1 to +3, sometimes +4, ion. Progressing to either the -4 or -3 ion, there is an increase in ionic radius since the anions are larger than the cations. Progressing from the -3 to the -1 ion, the radius decreases again.

Electronegativity increases moving across and decreases moving down.

The first ionization energy increases moving across and decreases moving down. For the second ionization energy, in period 2, lithium has the highest 2nd ionization energy because the electron is being removed from the noble gas configuration.

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