Question 1

The element Z = 114 has been discovered recently. It will belong to which of the following family/group and electronic configuration?

Accepted Answer

Carbon family, [Rn]5f14 6d10 7s2 7p2. Element Z = 114 belongs to the carbon family (Group 14) as it has 4 valence electrons in the 7p orbital, corresponding to the electronic configuration [Rn]5f14 6d10 7s2 7p2.

Question 2

Identify the incorrect match.

Accepted Answer

Unununnium - Darmstadtium. The incorrect match is D) Unununnium - Darmstadtium. Unununnium corresponds to the element with atomic number 111, which is now named Roentgenium, not Darmstadtium.

Question 3

An atom has electronic configuration 1s2 2s2 2p6 3s2 3p6 3d4 4s2, you will place it in which group?

Accepted Answer

Fifth. The given electronic configuration ends with 3d4 4s2, indicating the element belongs to the d-block. The total number of valence electrons is 6 (4 from 3d and 2 from 4s), placing it in Group 6 (or the fifth group in the modern periodic table).

Question 4

If the atomic number of an element is 33, it will be placed in the periodic table in the:

Accepted Answer

Fifth group. The element with atomic number 33 is arsenic (As), which belongs to Group 15 (the fifth group in the periodic table).

Question 5

The electronic configuration of an element is 1s2 2s2 2p6 3s2 3p3. What is the atomic number of the element, which is just below the above element in the periodic table?

Accepted Answer

34. The given electronic configuration corresponds to atomic number 15 (phosphorus). The element just below phosphorus in the periodic table is sulfur, which has an atomic number of 16. Therefore, the atomic number of the element below sulfur is 34.

Question 6

In the periodic table from left to right in a period, the atomic volume

Accepted Answer

First decrease then increases. As we move from left to right in a period, atomic volume first decreases due to increasing nuclear charge pulling electrons closer, and then increases slightly for noble gases due to their closed-shell configuration.

Question 7

Which electronic configuration of an element has abnormally high difference between second and third ionization energy?

Accepted Answer

1s2, 2s2, 2p6, 3s1. The configuration 1s2, 2s2, 2p6, 3s1 corresponds to an element with a single electron in the 3s orbital. After the removal of the first electron (first ionization energy) and the second electron (second ionization energy), the remaining configuration is 1s2, 2s2, 2p6, which is a stable noble gas configuration (Neon). The third ionization energy will be abnormally high due to the stability of this noble gas configuration.

Question 8

The element, with atomic number 11, will be:

Accepted Answer

alkali. An element with atomic number 11 is sodium, which is in Group 1 (alkali metals) of the periodic table.

Question 9

Na+, Mg2+, Al3+ and Si4+ are isoelectronic. The order of their ionic size is

Accepted Answer

Na+ > Mg2+ > Al3+ > Si4+. Isoelectronic species have the same number of electrons, but their ionic size decreases with increasing nuclear charge due to stronger attraction between the nucleus and electrons. Thus, the order is Na+ > Mg2+ > Al3+ > Si4+.

Question 10

Which of the following sets has strongest tendency to form anions?

Accepted Answer

N, O, F. The elements N, O, and F are highly electronegative and have a strong tendency to gain electrons to form anions, especially fluorine, which is the most electronegative element.

Question 11

Elements of which of the following groups will form anions most readily?

Accepted Answer

Halogens. Halogens (Group 17) have high electron affinity and readily gain electrons to form anions due to their high electronegativity and need to complete their octet.

Question 12

In the periodic table, with the increase in atomic number, the metallic character of an element

Accepted Answer

Decreases in a period and increases in a group. Metallic character decreases across a period due to increasing nuclear charge, which makes it harder to lose electrons. In a group, it increases as atomic size grows, making electron loss easier.

Question 13

The electronic configuration of four elements are given below. Which element does not belong to the same family as others?

Accepted Answer

[Ne]3s23p5. Option C ([Ne]3s23p5) represents a halogen, while the others belong to different families (transition metals or p-block elements). Thus, C does not belong to the same family.

Question 14

A nuclide of an alkaline earth metal undergoes radioactive decay by emission of the α-particle in succession. The group of the periodic table to which the resulting daughter element would belong is:

Accepted Answer

Gr. 6. An alkaline earth metal belongs to Group 2. Emission of an α-particle reduces the atomic number by 2, moving the element to Group 16 in the periodic table.

Question 15

The ions O2−, F−, Na+, Mg2+ and Al3+ are isoelectronic. Their ionic radii show

Accepted Answer

a significant decrease from O2− to Al3+. The ions O2−, F−, Na+, Mg2+, and Al3+ are isoelectronic, meaning they have the same number of electrons. As the nuclear charge increases from O2− to Al3+, the attraction between the nucleus and the electrons increases, leading to a decrease in ionic radii.

Question 16

For the second period elements the correct increasing order of first ionisation enthalpy is:

Accepted Answer

Li < Be < B < C < O < N < F < Ne. The correct order is based on periodic trends and exceptions. Ionization enthalpy generally increases across a period, but Be > B and N > O due to electronic configurations. Option B reflects this correctly.

Question 17

The correct order of atomic radii in group 13 elements is:

Accepted Answer

B < Al < Ga < In < Tl. In group 13, atomic radii generally increase down the group. However, Ga is slightly smaller than Al due to the poor shielding effect of d-electrons in Ga. Thus, the correct order is B < Al < Ga < In < Tl.

Question 18

The species Ar, K+ and Ca2+ contain the same number of electrons. In which order do their radii increase?

Accepted Answer

Ca2+ < Ar < K+. All three species have the same number of electrons (18), but their nuclear charges differ. Higher nuclear charge pulls electrons closer, reducing size. Thus, Ca2+ (20 protons) < Ar (18 protons) < K+ (19 protons) in size.

Question 19

In which of the following options the order of arrangement does not agree with the variation of property indicated against it?

Accepted Answer

I < Br < F < Cl (increasing electron gain enthalpy). The correct order for increasing electron gain enthalpy is F < Cl < Br < I, as fluorine has a lower electron gain enthalpy than chlorine due to its small size and high electron-electron repulsion. Thus, option C is incorrect.

Question 20

The formation of the oxide ion O2−(g), from oxygen atom requires first an exothermic and then an endothermic step as shown below:
O(g) + e− → O−(g); ΔH° = −141 kJ mol−1
O−(g) + e− → O2−(g); ΔH° = +780 kJ mol−1
Thus process of formation of O2− in gas phase is unfavourable even though O2− is

Accepted Answer

Electron repulsion outweighs the stability gained by achieving noble gas configuration. The second electron addition to form O2− is endothermic because the strong electron-electron repulsion in the already negatively charged O− ion outweighs the stability gained by achieving a noble gas configuration.

NEET Chemistry: Classification of Elements and Periodicity questions with solutions

Questions