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JEE Advanced Periodic Classification of Elements — practice questions

Explore 19 free Periodic Classification of Elements questions in JEE Advanced Science. Open each question or exercise to read its content, reference answers and available explanation.

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Periodic Classification of Elements questions and exercises with answers

Showing questions 1–19 of 19.

Which of the following arrangements is in correct order of increasing electronegativity (Pauling scale)?Read the answer and worked solution →The element with electronic configuration $1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^3$ belongs to which group and period?Read the answer and worked solution →The radius of the $n$-th Bohr orbit of a hydrogen-like atom of atomic number $Z$ varies as:Read the answer and worked solution →The minimum energy required to remove an electron from the ground state of a hydrogen atom is:Read the answer and worked solution →The kinetic energy of an electron in the $n$-th Bohr orbit of a hydrogen atom is related to its total energy $E_n$ by:Read the answer and worked solution →Which spectral series of hydrogen lies entirely in the ultraviolet region?Read the answer and worked solution →The hydrogen spectral lines that lie in the visible region of the electromagnetic spectrum belong to the:Read the answer and worked solution →The major shortcoming of the Rutherford atomic model was that it could not explain:Read the answer and worked solution →For a hydrogen-like ion of atomic number $Z$, the ground-state energy is:Read the answer and worked solution →An electron in a hydrogen atom undergoes a transition from $n = 3$ to $n = 2$. The energy of the emitted photon is approximately:Read the answer and worked solution →As per your syllabus, the MODERN periodic law (post-Moseley 1913) states that properties of elements are a periodic function of which quantity?Read the answer and worked solution →As per your syllabus, the IUPAC system numbers the groups in the modern table from which range, replacing the older IA…VIIA / VIII / IB…VIIB / 0?Read the answer and worked solution →As per your syllabus, in IUPAC systematic nomenclature for Z > 100, what are the digit roots for 1, 2 and 0 respectively?Read the answer and worked solution →As per your syllabus, the IUPAC systematic name + symbol for the element with atomic number 120 are which?Read the answer and worked solution →As per your syllabus, the IUPAC OFFICIAL name + symbol for Z = 104 is which?Read the answer and worked solution →As per your syllabus, the FIRST IONISATION ENTHALPY of an element X represents which process?Read the answer and worked solution →As per your syllabus, the ELECTRON GAIN ENTHALPY ΔₑgH represents enthalpy change for which reaction?Read the answer and worked solution →As per your syllabus, predicting the formula formed by SILICON (Group 14) and BROMINE (Group 17) gives which compound?Read the answer and worked solution →JEEMAIN · Chemistry · Spot the Error — Chemistry BasicsRead the answer and worked solution →

Worked practice questions

Which of the following arrangements is in correct order of increasing electronegativity (Pauling scale)?

  1. F < O < N < C
  2. N < C < O < F
  3. C < N < O < F
  4. O < F < N < C
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Correct answer: C. C < N < O < F

Electronegativity increases across a period and decreases down a group. The Pauling-scale values for period 2 nonmetals: - C: 2.55 - N: 3.04 - O: 3.44 - F: 3.98 (highest of any element) So the correct ascending order is **C < N < O < F**. Option A reverses everything. Option C breaks the trend by putting F before O. Memory aid: "FONClBrISCHP" is the electronegativity ladder from highest to lowest among common nonmetals (F first, P last among this set).

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The element with electronic configuration $1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^3$ belongs to which group and period?

  1. Group 13, Period 3
  2. Group 15, Period 4
  3. Group 2, Period 4
  4. Group 17, Period 4
Show answer and solution

Correct answer: B. Group 15, Period 4

**Period** equals the highest principal quantum number, which is 4 here. So Period 4. **Group**: the outermost shell ($n=4$) has $4s^2 4p^3$. For p-block elements, group number = 10 + (number of $s$ valence + number of $p$ valence) = 10 + (2 + 3) = **15**. This configuration matches **arsenic (As)**, atomic number 33. Check: As is below P and N in group 15 (nitrogen family), period 4. ✓

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The radius of the $n$-th Bohr orbit of a hydrogen-like atom of atomic number $Z$ varies as:

  1. $r \propto n / Z^2$
  2. $r \propto Z / n^2$
  3. $r \propto n^2 / Z$
  4. $r \propto Z n^2$
Show answer and solution

Correct answer: C. $r \propto n^2 / Z$

From Bohr's derivation, $r_n = \dfrac{n^2 h^2}{4\pi^2 m k Z e^2}$, hence $r_n \propto n^2 / Z$. **Therefore higher orbits are larger ($n^2$) and higher-$Z$ atoms hold their electrons in tighter orbits ($1/Z$).**

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