Practice Question and Answer
8Q: The outer electron configuration of Gd (Atomic No : 64) is 1. 4f75d16s2 2. 4f35d56s2 3. 4f85d06s2 4. 4f45d46s2 1443 05b5cc785e4d2b41977750262
5b5cc785e4d2b41977750262- Show AnswerHide Answer
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Answer :
Explanation :
Answer : 1 Sol : 1. 4f75d16s2
Q: The structure of IF7 is 1677 15b5cc785e4d2b4197775025d
5b5cc785e4d2b4197775025d- 1Pentagonal bipyramidtrue
- 2Square pyramidfalse
- 3Trigonal bipyramidfalse
- 4Octahedralfalse
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Answer : 1. "Pentagonal bipyramid"
Explanation :
Answer: A) Pentagonal bipyramid Explanation: Hybridisation of iodine is sp3d3 So, the structure is pentagonal bipyramid.
Q: Ozonolysis of an organic compound gives formaldehyde as one of the products. This confirms the presence of : 1493 05b5cc785e4d2b41977750258
5b5cc785e4d2b41977750258- 1An acetylenic triple bondfalse
- 2Two ethylenic double bondsfalse
- 3A vinyl grouptrue
- 4An isopropyl groupfalse
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Answer : 3. "A vinyl group"
Explanation :
Answer: C) A vinyl group Explanation: C=}CH2→OzonolysisHCHO Vinylic group.
Q: The degree of dissociation (α) of a weak electrolyte, AxBy is related to van't Hoff factor (i) by the expression: 1. α=x+y+1i-1 2.α=x+y-1i-1 3. α=i-1x+y+1 4. α=i-1x+y-1 2155 05b5cc785e4d2b41977750253
5b5cc785e4d2b41977750253- 11 is correctfalse
- 22 is correctfalse
- 33 is correctfalse
- 44 is correcttrue
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Answer : 4. "4 is correct"
Explanation :
Answer: D) 4 is correct Explanation: Van't Hoff factor (i) = Observed Colligative PropertyNormal Colligative Property AxBy1-α→xA+yxα+yB-xyα Total Moles = 1-α+xα+yα = 1-αx+y-1 i=1+αx+y-11 ⇒α=i-1x+y-1
Q: In a face centred cubic lattice, atom A occupies the corner positions and atom B occupies the face centre positions. If one atom of B is missing from one of the face centred points, the formula of the compound is 1) A2B5 2) A3B2 3) A2B3 4) A3B4 1560 05b5cc785e4d2b4197775024e
5b5cc785e4d2b4197775024e- 1Option 1true
- 2Option 2false
- 3Option 3false
- 4Option 4false
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Answer : 1. "Option 1"
Explanation :
Answer: A) Option 1 Explanation: ZA = 88 ; ZB = 52 So formula of compound is AB52 i.e., A2B5.
Q: A gas absorbs a photon of 355 nm and emits at two wavelengths. If one of the emissions is at 680 nm, the other is at : 1263 05b5cc783e4d2b41977750249
5b5cc783e4d2b41977750249- 1518 nmfalse
- 21035 nmfalse
- 3325 nmfalse
- 4743 nmtrue
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Answer : 4. "743 nm"
Explanation :
Answer: D) 743 nm Explanation: 1λ=1λ1+1λ2 1355=1680+1λ2 1λ2=680-355680×355 ⇒λ2=743nm.
Q: The entropy change involved in the isothermal reversible expansion of 2 moles of an ideal gas from a volume of 10 dm3 at 27°C is to a volume of 100 dm3 1417 05b5cc783e4d2b41977750244
5b5cc783e4d2b41977750244- 142.3 J/ mole / Kfalse
- 238.3 J/ mole / Ktrue
- 335.8 J/ mole / Kfalse
- 432.3 J/ mole / Kfalse
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Answer : 2. "38.3 J/ mole / K"
Explanation :
Answer: B) 38.3 J/ mole / K Explanation: ∆s=nRlnv2v1 =2.303 nR logv2v1 = 2.303×2×8.314×log10010 = 38.3 J mol-1k-1
Q: Silver Mirror test is given by which one of the following compounds? 1. Benzophenone 2. Acetaldehyde 3. Acetone 4. Formaldehyde 1554 05b5cc783e4d2b4197775023f
5b5cc783e4d2b4197775023f- 1Only 1false
- 2Only 2false
- 31 and 3false
- 42 and 4true
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Answer : 4. "2 and 4"
Explanation :
Answer: D) 2 and 4 Explanation: Both Formaldehyde and Acetaldehyde will give this test HCHO→[Ag(NH3)2+] Silver mirrorAg + Organic compound CH3-CHO →[Ag(NH3)2+] AgSilver mirror+ Organic compound