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Indian Navy AA (Artificer Apprentice) Inorganic Chemistry — practice questions

Explore 3 free Inorganic Chemistry questions in Indian Navy AA (Artificer Apprentice) Chemistry. Open each question or exercise to read its content, reference answers and available explanation.

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Inorganic Chemistry questions and exercises with answers

Showing questions 1–3 of 3.

Which of the following arrangements is in correct order of increasing electronegativity (Pauling scale)?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, predicting the formula formed by SILICON (Group 14) and BROMINE (Group 17) gives which compound?Read 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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As per your syllabus, the IUPAC OFFICIAL name + symbol for Z = 104 is which?

  1. Kurchatovium, Ku
  2. Dubnium, Db
  3. Rutherfordium, Rf
  4. Seaborgium, Sg
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Correct answer: C. Rutherfordium, Rf

As per your syllabus, Z=104 is Rutherfordium (Rf). Kurchatovium was the Soviet rival name. 105 = Db, 106 = Sg.

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As per your syllabus, predicting the formula formed by SILICON (Group 14) and BROMINE (Group 17) gives which compound?

  1. SiBr₄
  2. Si₂Br
  3. SiBr₂
  4. SiBr
Show answer and solution

Correct answer: A. SiBr₄

As per your syllabus, Si valence = 4, Br valence = 1 ⇒ SiBr₄.

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