Periodic Table and Structure Of Atom- Combined Test

According to Pauli's exclusion principle, no two electrons in an atom can have the same set of four quantum numbers. One of these quantum numbers is the spin quantum number, which can have two possible values: +1/2 or -1/2. This means that in any subshell, the spin quantum number of two electrons must be different. Therefore, the correct answer is -1/2 , +1/2.

The quantum numbers describe the energy levels, orbital shapes, orientations, and spin states of electrons in an atom. For an electron in the 4f orbital, the principal quantum number (n) is 4, indicating the energy level. The azimuthal quantum number (l) is 3, representing the orbital shape. The magnetic quantum number (m) is +1, indicating the orientation of the orbital in space. The spin quantum number (s) is +1/2, representing the spin state of the electron. Therefore, the set of quantum numbers n=4, l=3, m=+1, s=+1/2 is correct for an electron in the 4f orbital.

The correct answer is atomic numbers. According to the periodic law of elements, the variation in properties of elements is related to their atomic numbers. Atomic number represents the number of protons in an atom's nucleus, which determines the element's identity. As the atomic number increases, the properties of elements change in a periodic manner, leading to the organization of the periodic table. Atomic mass, nuclear masses, and neutron-proton number ratios are not directly related to the variation in properties of elements.

The similarity between the characteristics of Li and Mg is referred to as a diagonal relation. This term is used to describe elements that are located diagonally from each other in the periodic table and share similar properties. Li and Mg are both located in Group 2 of the periodic table and have similar characteristics such as being lightweight, having low melting and boiling points, and forming ionic compounds with similar properties. This diagonal relationship is due to the similarities in the electronic configurations of these elements.

The correct answer is "Extra stability of half filled 'p' orbital". Nitrogen has a half-filled p orbital in its ground state, which provides it with extra stability. This stability makes it more difficult to remove an electron from nitrogen compared to oxygen, which does not have a half-filled p orbital. The increased attraction of electrons towards the nucleus and the small size of N2 are not the main reasons for the higher ionization energy of nitrogen.

The electron gain enthalpy is the energy released when an atom gains an electron. It is generally observed that the electron gain enthalpy becomes less negative as we move down a group in the periodic table. This is because the atomic radius increases down the group, making it easier for the atom to accommodate an additional electron.

In this case, we are comparing the electron gain enthalpy of F, Cl, Br, and I. Since Cl has a smaller atomic number than F, Br, and I, it has a smaller atomic radius. Therefore, it has a higher electron gain enthalpy. Among F, Br, and I, F has the highest electron gain enthalpy because it has the smallest atomic radius.

So, the correct order of electron gain enthalpy with a negative sign is Cl > F > Br > I.

Fe3+ and Mn2+ are isoelectronic because they have the same number of electrons. Fe3+ has lost 3 electrons to become positively charged, while Mn2+ has lost 2 electrons to become positively charged. Therefore, both ions have the same electron configuration and are isoelectronic.

The first ionization enthalpy refers to the energy required to remove one electron from an atom in its gaseous state. As we move across a period from left to right, the atomic radius decreases, resulting in stronger attractions between the nucleus and the outermost electron. This makes it more difficult to remove an electron, thus increasing the ionization enthalpy. Among the given options, Ba has the largest atomic radius and therefore the weakest attraction to its outermost electron, making it the easiest to remove. Ca has a smaller atomic radius than Ba but larger than Se, S, and Ar, hence it has a higher ionization enthalpy than Ba but lower than the remaining elements. Se has a smaller atomic radius than Ca but larger than S and Ar, resulting in a higher ionization enthalpy than Ca but lower than S and Ar. S has a smaller atomic radius than Se but larger than Ar, giving it a higher ionization enthalpy than Se but lower than Ar. Ar has the smallest atomic radius among the given elements, resulting in the strongest attraction to its outermost electron and the highest ionization enthalpy. Therefore, the correct order of increasing first ionization enthalpy is Ba < Ca < Se < S < Ar.

The correct order of 1st ionization energy in alkali metals is Li > Na > K > Rb. This means that it requires the most energy to remove an electron from a lithium atom, followed by sodium, potassium, and finally rubidium. This order can be explained by the trend in atomic radius and effective nuclear charge. As you move down the group, the atomic radius increases, resulting in a weaker attraction between the nucleus and the outermost electron. Additionally, the effective nuclear charge decreases due to the shielding effect of inner electrons. These factors make it easier to remove an electron, leading to a decrease in ionization energy.

The quantum numbers n, l, m, and s represent different properties of an electron in an atom. The principal quantum number (n) represents the energy level or shell of the electron. The angular momentum quantum number (l) represents the shape of the orbital. The magnetic quantum number (m) represents the orientation of the orbital in space. The spin quantum number (s) represents the spin of the electron.

In this case, the set of quantum numbers n=3, l=2, m=1, s=+1/2 represents the highest energy of an atom because the principal quantum number (n=3) indicates that the electron is in the third energy level, which is higher than the other options. Additionally, the angular momentum quantum number (l=2) indicates that the electron is in a d orbital, which has higher energy than s or p orbitals. The combination of these quantum numbers suggests a higher energy state for the electron.

The collection of iso-electronic species refers to a group of ions that have the same number of electrons. In this case, K+ has lost one electron, Ca2+ has lost two electrons, Sc3+ has lost three electrons, and Cl- has gained one electron. Therefore, all four ions have the same number of electrons, making them iso-electronic.

Electronegativity is a measure of an atom's ability to attract electrons in a chemical bond. It is a relative scale and does not have a specific unit of measurement. Electronegativity values are determined based on a comparison between different elements, rather than being measured in a specific unit. In contrast, electron affinity, ionization energy, and excitation potential are all physical properties that can be measured and have specific units associated with them.

In a multi-electron atom, the energy of an orbital is determined by the values of the principal quantum number (n), azimuthal quantum number (l), and magnetic quantum number (m). The orbitals with the same energy are those that have the same values of n, l, and m.

Option D has n=3, l=2, and m=1. Option E has n=3, l=2, and m=0. Since both options have the same values for n and l, they will have the same energy in the absence of magnetic and electric fields.

Elements with similar chemical properties have the same number of electrons in their outer shell. The outer shell is the valence shell, which determines the element's reactivity and ability to form bonds with other elements. Elements with the same number of electrons in their outer shell have similar chemical behaviors because they have similar valence electron configurations. This similarity allows them to exhibit similar chemical reactions and form similar compounds.

The maximum number of electrons present in an energy level can be calculated using the formula 2n^2, where n is the principal quantum number. For n=5, the maximum number of electrons would be 2(5^2) = 50.

The electronic configuration 1s2 2s2 2p6 3s1 corresponds to the atom with the lowest ionisation enthalpy because it has the highest number of valence electrons (in this case, 1) in the outermost energy level. The ionisation enthalpy is the energy required to remove an electron from an atom, and atoms with more valence electrons have a lower ionisation enthalpy because the outermost electron is less strongly attracted to the nucleus and therefore easier to remove.

The ionic radii of ions are determined by the number of electrons and the effective nuclear charge. In this case, all the ions are iso-electronic, meaning they have the same number of electrons. However, the effective nuclear charge increases as you move across a period in the periodic table, resulting in a decrease in ionic radius. Therefore, the larger ions will be towards the left side of the given options. Among the given options, Ca2+ has the largest ionic radius because it has the least effective nuclear charge. K+ has the next largest ionic radius, followed by Cl- and S2- with the smallest ionic radius due to their higher effective nuclear charges.

The correct answer states that in alkali metals, the reactivity increases but in the halogens it decreases with an increase in atomic number down the group. This is because alkali metals have a single valence electron that is easily lost, leading to increased reactivity as the atomic number increases. On the other hand, halogens have a high electron affinity and tend to gain electrons, resulting in decreased reactivity as the atomic number increases. Therefore, this statement accurately describes the periodic trends of chemical reactivity in alkali metals and halogens.

An atom with electronic configuration 2, 8, 7 refers to an atom with 17 electrons. This corresponds to the element chlorine (Cl) in the periodic table. Chlorine is a non-metal element and does not form diatomic molecules. It forms a diatomic molecule called chlorine gas (Cl2). Its valency is 1, meaning it tends to gain one electron to achieve a stable electron configuration. Chlorine forms acidic oxides, not basic oxides, as it tends to combine with oxygen to form compounds like chlorine dioxide (ClO2) or chlorine trioxide (ClO3).

The correct answer is O2- > F- > Na+ > Mg2+ > Al3+. This is because as we move across a period in the periodic table, the atomic radius decreases due to the increasing effective nuclear charge. However, when an atom gains electrons to form an anion, the added electrons repel each other, causing the electron cloud to expand and the ionic radius to increase. Therefore, O2- has the largest ionic radius as it has gained two electrons, followed by F- with one added electron. Na+ has lost one electron, making it smaller than F-. Mg2+ has lost two electrons, making it smaller than Na+. Lastly, Al3+ has lost three electrons, making it the smallest ion in the sequence.

The assertion states that Na, K, and Rb form Dobereiner's triad. However, this is incorrect as Dobereiner's triad consists of elements with similar chemical properties and atomic weights. Na, K, and Rb do not fulfill this criteria. The reason states that the atomic weight of K is twice the atomic weight of Rb, which is also incorrect. Therefore, both the assertion and reason are false.

Chromium (Cr) may be expected to have the highest second ionization enthalpy because it has the highest atomic number among the given elements. As the atomic number increases, the number of protons in the nucleus increases, leading to a stronger attraction between the nucleus and the electrons. This stronger attraction requires more energy to remove an electron, resulting in a higher ionization enthalpy.

O2- is an oxygen ion with a charge of -2. Oxygen normally has 6 electrons, but since it has a charge of -2, it gains 2 extra electrons, resulting in a total of 8 electrons. In order to determine the number of unpaired electrons, we need to look at the electron configuration. The electron configuration of oxygen is 1s2 2s2 2p4. In O2-, the two extra electrons will occupy the 2p orbital, filling it completely. Since all the electrons are paired, there are no unpaired electrons in O2-, so the answer is 0.

The correct answer is B < C < N < O because the trend for first ionization energy is typically to increase from left to right across a period. However, in this arrangement, the order is reversed, with B having a higher first ionization energy than C, N, and O. This indicates that there may be an error in the arrangement or a different property is being indicated.

The given statement "Zeff increases down the group" is false. Zeff, also known as effective nuclear charge, refers to the net positive charge experienced by an electron in an atom. It is determined by the number of protons in the nucleus and the shielding effect of inner electrons. As we move down a group in the periodic table, the number of protons and inner electrons both increase, resulting in an increase in screening effect and a decrease in Zeff. Therefore, Zeff decreases down the group, making the statement false.

Fe+3 has the maximum number of unpaired electrons. When Fe loses three electrons to become Fe+3, it loses them from its 3d orbitals. Fe has 6 electrons in its 3d orbitals, and when it loses three electrons, it leaves behind 3 unpaired electrons. Fe, Fe+2, and Fe+4 do not have any unpaired electrons.

The lanthanide contraction refers to the decrease in both atomic and M3+ radii. This phenomenon occurs because as you move across the lanthanide series, the increasing positive charge in the nucleus attracts the outermost electrons more strongly, causing the electron cloud to contract. This contraction affects both the atomic radii, which is the size of the atom as a whole, and the M3+ radii, which is the size of the ion with a +3 charge. Therefore, the correct answer is atomic as well as M3+ radii.

The assertion states that Na, K, and Rb form Dobereiner's triad. This means that these three elements have similar chemical properties and can be grouped together. The reason given is that the atomic weight of K is twice the atomic weight of Rb. However, this is incorrect as the atomic weight of K is actually less than twice the atomic weight of Rb. Therefore, the reason is false, but the assertion is true. Hence, the correct answer is C.

Zinc and Copper have similarity in their group number because they both belong to Group 11 of the periodic table. Group number refers to the vertical columns in the periodic table which indicate the number of valence electrons an element has. Zinc and Copper both have one valence electron, making them part of the same group.

Hund's rule states that electrons will occupy empty orbitals of the same energy level before pairing up. In the case of a 2p orbital of a Nitrogen atom, there are three orbitals (2px, 2py, 2pz) that are degenerate, meaning they have the same energy. According to Hund's rule, each orbital must be singly occupied before any of them can start pairing up. Therefore, if Hund's rule is violated, it means that there are no electrons present in the 2p orbital, resulting in a minimum of 0 electrons.