Start learning →

Indian Navy AA (Artificer Apprentice) Classification of Elements and Periodicity — practice questions

Explore 56 free Classification of Elements and Periodicity questions in Indian Navy AA (Artificer Apprentice) Chemistry. Open each question or exercise to read its content, reference answers and available explanation.

Practise Classification of Elements and Periodicity with Quill →

Classification of Elements and Periodicity questions and exercises with answers

Showing questions 1–30 of 56.

Which of the following arrangements is in correct order of increasing electronegativity (Pauling scale)?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 bond order of the oxygen molecule O₂ (by molecular orbital theory) isRead the answer and worked solution →CLASS10 · Chemistry · Periodic Table Trends — Gap FillRead the answer and worked solution →Sodium (atomic number 11) belongs to which group and period?Read the answer and worked solution →Which of the following elements will form an acidic oxide?Read the answer and worked solution →Order the atomic models as they were proposedRead the answer and worked solution →Match each scientist to their modelRead the answer and worked solution →Match each model to what it could NOT explainRead the answer and worked solution →True or false: Rutherford's modelRead the answer and worked solution →True or false: Bohr's modelRead the answer and worked solution →Complete the reason Bohr's model failsRead the answer and worked solution →Which model does this belong to?Read the answer and worked solution →Match each quantum number to what it is calledRead the answer and worked solution →True or false: the quantum mechanical modelRead the answer and worked solution →Order the structure: from atom down to electronRead the answer and worked solution →Complete the (n + l) ruleRead the answer and worked solution →True or false: orbital energiesRead the answer and worked solution →Match each rule to what it forbids or requiresRead the answer and worked solution →Tap every statement Hund's rule requiresRead the answer and worked solution →The fundamental particlesRead the answer and worked solution →Assertion and reasonRead the answer and worked solution →Two are true, one is notRead the answer and worked solution →Three of these are wrong. Find the one that holds.Read the answer and worked solution →Which principle is being described?Read the answer and worked solution →Flip and pair: symbol with meaningRead the answer and worked solution →Spot the one that does not belongRead the answer and worked solution →Tap every statement true of the Schrodinger modelRead the answer and worked solution →Reading the models backRead the answer and worked solution →Assertion and reason: filling orbitalsRead 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
Show answer and solution

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).

Open this question →

The hydrogen spectral lines that lie in the visible region of the electromagnetic spectrum belong to the:

  1. Lyman series
  2. Paschen series
  3. Pfund series
  4. Balmer series
Show answer and solution

Correct answer: D. Balmer series

Balmer series corresponds to transitions terminating in $n_1 = 2$. The first four lines ($n_2 = 3, 4, 5, 6$ to $n_1 = 2$) lie at approximately $656\,\text{nm}$ (H$_\alpha$, red), $486\,\text{nm}$ (H$_\beta$, blue-green), $434\,\text{nm}$ (H$_\gamma$, blue) and $410\,\text{nm}$ (H$_\delta$, violet), all in the visible range.

Open this question →

The bond order of the oxygen molecule O₂ (by molecular orbital theory) is

  1. 2
  2. 1
  3. 2.5
  4. 3
Show answer and solution

Correct answer: A. 2

MO config gives bond order = (bonding − antibonding)/2 = (8−4)/2 = 2, consistent with the O=O double bond.

Open this question →

How to practise

  1. Choose a Classification of Elements and Periodicity question and try to work out your answer.
  2. Open the question to check the correct answer and read its worked solution.
  3. Continue in the app for bite-sized lessons, explanations and review with Quill. You can start without creating an account.

More Chemistry topics

Chemical EquilibriumAlcohols Phenols EthersAlkenesOrganic ChemistryPolymersAtoms and MoleculesClassification of ElementsS Block Elements