Periodic Properties and Periodic Table ‌–‌ ‌JEE‌ ‌Advanced‌ ‌Previous‌ ‌Year‌ ‌Questions with Solutions

Periodic Properties and Periodic Table JEE‌ ‌Advanced‌ ‌Previous‌ ‌Year‌ ‌Questions‌ ‌with‌ ‌Solutions‌ can be accessed here. JEE aspirants will find a set of questions that have appeared in the previous years of JEE Advanced. They will find detailed solutions which will further help them to understand the right answers and gain better concept clarity. As students go through the questions and solutions it will also allow them to revise the important topics before the examination.

In any case, the chapter ‘Periodic Properties and Periodic Table’ is one of the most important chapters for JEE Advanced. Few questions are usually asked from this chapter. With the previous year questions related to this chapter, aspirants will further gain a clear idea of the types of questions, difficulty level and prepare accordingly. Periodic Properties and Periodic Table JEE‌ ‌Advanced‌ ‌Previous‌ ‌Year‌ ‌Questions‌ ‌with‌ ‌Solutions‌ are available in the form of a PDF and can be downloaded freely.

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JEE Advanced Previous Year Questions on Periodic Properties and Periodic Table

Question 1. The option(s) with only amphoteric oxides is(are)

A. Cr₂O₃, BeO, SnO, SnO₂

B. Cr₂O₃, CrO, SnO, PbO

C. NO, B₂O₃, PbO, SnO₂

D. ZnO, Al₂O₃, PbO, PbO₂

Solution: (A and D)

Amphoteric oxides are

Cr₂O₃, BeO, SnO, SnO₂, ZnO, Al₂O₃, PbO, PbO₂

NO is a neutral oxide

CrO is a basic oxide

B₂O₃ is an acidic oxide

Question 2. The statement that is not correct for periodic classification of elements is.

A. The properties of elements are the periodic functions of their atomic numbers

B. Nonmetallic elements are less in number than metallic elements.

C. For transition elements, the 3d-orbitals are filled with electrons after 3p-orbitals and before 4s-orbitals.

D. The first ionisation enthalpies of elements generally increases with an increase in atomic number as we go along a period.

Solution: (C)

In cases of transition elements (or any elements), the order of filling of electrons in the various orbital is 3p < 4s <3d, Thus 3d orbital is filled when 4s orbital gets completely filled (Aufbau Principle).

According to Aufbau Principle, the ground state of an atom or ion e^_s fill atomic orbitals of the lowest available energy levels before occupying higher levels.

Question 3. The 1st, 2nd, and the 3rd ionization enthalpies, 𝐼₁, 𝐼₂, and 𝐼₃, of four atoms with atomic numbers n, 𝑛 + 1, 𝑛 + 2, and 𝑛 + 3, where 𝑛 < 10, are tabulated below. What is the value of n?

Solution: (9)

According to the tabulated data,

Element with Atomic number (n + 2), should be an alkali metal

As we see, the first ionization enthalpy (I₁) is very less but the second ionization enthalpy (I₂) is very large.

Hence, the atomic number can be = 11

That is, (n + 2) = 11

n = 9

Note: ‘n’ can’t be ‘1’.

Question 4. Among the following,

O, Cl, F, N, P, Sn, Tl, Na, Ti

The number of elements showing only one non – zero oxidation state is.

Solution: (2)

Only F and Na show only one non – zero oxidation state. Thus, the number of elements showing only one non-zero oxidation state is 2.

Na exhibits +1 and F exhibits only -1 oxidation state are the two elements.

O = O^–, O^2-, O²⁺

Cl = -1 to + 7

N = -3 to +5

P = -3 to + 5

Sn = +2, +4

Tl = +1, +3 (rare but does exist)

Ti = +2, +3, +4.

Question 5. Which of the following has the maximum number of unpaired electrons?

A. Mg²⁺

B. Fe²⁺

C. Ti³⁺

D. V³⁺

Solution: (B)

Mg²⁺ = 1s² 2s² 2p⁶ = no unpaired electron

Ti³⁺ = 1s² 2s² 2p⁶ 3p⁶ 3d¹ = one unpaired electron

V³⁺ = 1s² 2s² 2p⁶ 3s² 3p⁶ 3d² = two unpaired electron

Fe²⁺ = 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁶ = four unpaired electron

Question 6. The correct order of radii is

A. N < Be < B

B. F^– < O^2- < N^3-

C. Na < Li < K

D. Fe³⁺ < Fe²⁺ < Fe⁴⁺

Solution: (B)

The correct order of radii is F^– < O^2- < N^3-. These all are isoelectronic species. They all have 10 electrons. As the negative charge increases, the atomic size increases. Also, the size of isoelectronic decreases with an increase in atomic number.

Question 7. The electronegativity of the following elements increases in the order

A. C, N, Si, P

B. N, Si, C, P

C. Si, P, C, N

D. P, Si, N, C

Solution: (C)

In general, electronegativity increases across a period from left to right and decreases down the group.

Si < P < C < N is the correct order of electronegativity.

Question 8. Atomic radii of fluorine and neon in Angstrom units are respectively given by:

A. 0.72, 1.60

B. 1.60, 1.60

C. 0.72, 0.72

D. None of these values

Solution: (A)

The atomic radii of fluorine and neon in the Angstrom unit are 0.72 and 1.60. The atomic radii generally decrease across the period due to an increase in the effective nuclear charge with the increase in the atomic number. The outer electrons go to the same valance shell, hence, it results in the increased attraction of electrons to the nucleus.

The non-bonded radii of noble gases are very large due to completely filled orbitals. It is larger than the precedent elements of the same period due to completely filled orbitals and interelectronic repulsion.

Question 9. Identify the correct order of acidic strength of CO₂, CuO, CaO, H₂O.

A. CaO < CuO < H₂O < CO₂

B. H₂O < CuO < CaO < CO₂

C. CaO < H₂O < CuO < CO₂

D. H₂O < CO₂ < CaO < CuO

Solution: (A)

CO₂ is an acidic oxide, H₂O is neutral, CaO is a strong base and CuO is a weak base. Therefore, the order of acid strength is: CaO < CuO < H₂O < CO₂

Question 10. Amongst the following elements (whose electronic configurations are given below), the one having the highest ionisation energy is.

A. [Ne] 3s² 3p¹

B. [Ne] 3s² 3p³

C. [Ne] 3s² 3p²

D. [Ar] 3d¹⁰ 4s² 4s² 4p³

Solution: (B)

s- Block Elements – Video Lesson

p- Block Elements – Video Lesson

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