JEE Main Basic Concepts & Atomic Structure: Atomic Structure – Chapter Test

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Difficulty: 9 medium · 11 hard

Atomic Structure MCQ questions with answers

The first 3 of 20 questions. Unlock the series for the rest, each with a solution.

1. Q1. The total number of orbitals present in the shell with principal quantum number n = 4 is:

   मुख्य क्वांटम संख्या n = 4 वाले कोश में उपस्थित कक्षकों की कुल संख्या है:

   • A. 32
   • B. 16
   • C. 9
   • D. 8
   Solution

   The total number of orbitals in a shell is n². For n = 4 this is 4² = 16 orbitals (one 4s, three 4p, five 4d and seven 4f). The value 32 is the maximum number of electrons (2n²).

   किसी कोश में कक्षकों की कुल संख्या n² होती है। n = 4 के लिए यह 4² = 16 कक्षक है (एक 4s, तीन 4p, पाँच 4d और सात 4f)। मान 32 इलेक्ट्रॉनों की अधिकतम संख्या (2n²) है।

2. Q2. The correct ground-state electronic configuration of chromium (Z = 24) is:

   क्रोमियम (Z = 24) का सही मूल-अवस्था इलेक्ट्रॉनिक विन्यास है:

   • A. [Ar]3d⁵4s²
   • B. [Ar]3d⁴4s²
   • C. [Ar]3d⁵4s¹
   • D. [Ar]3d⁶
   Solution

   Chromium adopts [Ar]3d⁵4s¹ rather than [Ar]3d⁴4s², because the half-filled 3d subshell together with the half-filled 4s gives extra stability due to symmetrical distribution and exchange energy.

   क्रोमियम [Ar]3d⁴4s² के बजाय [Ar]3d⁵4s¹ ग्रहण करता है, क्योंकि अर्ध-पूरित 3d उपकोश के साथ अर्ध-पूरित 4s सममित वितरण और विनिमय ऊर्जा के कारण अतिरिक्त स्थायित्व प्रदान करते हैं।

3. Q3. In the emission spectrum of atomic hydrogen, which of the following transitions produces the photon of highest frequency?

   परमाण्वीय हाइड्रोजन के उत्सर्जन स्पेक्ट्रम में, निम्नलिखित में से कौन-सा संक्रमण उच्चतम आवृत्ति का फोटॉन उत्पन्न करता है?

   • A. n = 2 → n = 1
   • B. n = 5 → n = 4
   • C. n = 4 → n = 3
   • D. n = 3 → n = 2
   Solution

   The energy gap between consecutive levels decreases rapidly with n, so ΔE for 2 → 1 (0.75 × 13.6 eV) is far larger than for 3 → 2, 4 → 3 or 5 → 4, giving the highest-frequency photon.

   क्रमागत ऊर्जा स्तरों के बीच ऊर्जा अंतराल n बढ़ने पर तेज़ी से घटता है, अतः 2 → 1 के लिए ΔE (0.75 × 13.6 eV) 3 → 2, 4 → 3 या 5 → 4 की तुलना में कहीं अधिक है, जिससे उच्चतम आवृत्ति का फोटॉन प्राप्त होता है।

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