Polarization of electrons in acrolein may be written as: 

1. Cδ-H2=CH-Cδ+H=O

2. Cδ-H2=CH-CH=Oδ+

3. Cδ-H2=Cδ+H-CH=O

4. Cδ+H2=CH-CH=Oδ-

Subtopic:  Resonance & Nature of Compounds |
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PCl5 exists but NCl5 does not because:

1. Nitrogen has no vacant d-subshells .

2. NCl5 is unstable .

3. Nitrogen atom is much smaller than p .

4. Nitrogen is highly inert . 

Subtopic:  Acidic, Basic Character & Fajan's Rule |
 87%
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The bond angle between two hybrid orbitals is 105°. The % s character in orbital is-

1. Between 30-31% 2. Between  9- 12%
3. Between 25 -26% 4. Between 22-23%
Subtopic:  Hybridisation |
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Which bond angle would result in the maximum dipole moment for the triatomic molecule XY2 

1. 90°

2. 120°

3. 150°

4. 180°

Subtopic:  Polarity |
 57%
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The strongest hydrogen bond present among the following is-

1. OHS

2. SHO

3. FHF

4. FHO

Subtopic:  van der Waal Force & Hydrogen Bonding |
 78%
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The ionisation energy will be maximum for the process:

(a) BaBa2+

(b) BeBe2+

(c) CsCs+

(d) LiLi+

Subtopic:  Ionization Energy (IE) |
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In which of the process, the bond order increases and magnetic behaviour changes?

(a) N2N2+

(b) C2C2+

(c) NONO+

(d) O2O2+

Subtopic:  M.O.T |
 65%
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In piperidine "N" atom has hybridization:

piperidine

1. sp

2. sp2

3. sp3

4. dsp2

Subtopic:  Hybridisation |
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The coordination geometry around and hybridization of N and B atoms in a 1:1 complex of NH3 and BF3 is:

(1) N: tetrahedral, sp3;  B: tetrahedral, sp3

(2) N: pyramidal, sp3; B: pyramidal, sp3

(3) N: pyramidal, sp3; B: planar, sp2

(4) N: pyramidal, sp3; B: tetrahedral, sp3

Subtopic:  Hybridisation |
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The formation of the oxide ion O2-(g) requries first an exothermic and then an endothermic step as shown below,

O(g)+eO-(g);             H=-142 KJ/mol

O-(g)+eO2-(g);           H=844 KJ/mol

 

 

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