The decreasing order of ionic character of the N-H, F-H, C-H, and O-H, is :  

1. N-H > F-H > C-H > O-H
2. F-H > N-H > C-H > O-H
3. O-H > C-H > F-H > N-H
4. F-H > O-H > N-H > C-H

Subtopic:  Ionic Bond |
 82%
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The right order of increase in the ionic character of the molecules,
 
LiF, K2O, N2, SO2, ClF3 is :

1. N< SO< ClF< K2O < LiF

2. N2 > SO2 > ClF> K2O < LiF

3. N> SO> K2O > ClF3 > LiF

4. LiF > K2O < ClF3 > SO2 > N2

Subtopic:  Ionic Bond |
 63%
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The number of (i) sp2 hybridized carbon atoms and (ii) π bonds are present in the following compound are:

1. 7, 5

2. 8, 6

3. 7, 6

4. 8, 5

Subtopic:  Hybridisation |
 73%
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From the perspective of molecular orbital theory, which statement is false?

1. Be2 is not a stable molecule.

2. He2 is not stable but He2+ is expected to exist.

3. Bond strength of N2 is maximum amongst the homonuclear diatomic molecules belonging to the second period.

4. The order of energies of molecular orbitals in N2 molecule is: 

σ2s<σ2s*<σ2pz<(π2pxπ2py)
<(π2px*π2py*)<σ2pz*

Subtopic:  M.O.T |
 75%
From NCERT
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Combination of atoms A and B that forms an anti-bonding molecular orbital is :

1. ΨA2ΨB2

2. ΨA2×ΨB2

3. ΨA+ΨB

4. ΨA-ΨB

Subtopic:  M.O.T |
 74%
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In PO4-3, formal charge on every oxygen atom and P-O bond order respectively are:
1.  -0.75 and 1.25
2.  -0.5 and 2
3.  1 and 1.5
4.  -0.75 and 2

Subtopic:  Octate, Isoelectronic Species, Lewis Structure & Formal Charge |
 73%
From NCERT
AIPMT - 1998
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In a regular octahedral molecule, MX6, the number of X–M–X bonds at 180º is:

1. Two 2. Six
3. Four 4. Three
Subtopic:  Hybridisation |
 70%
From NCERT
AIPMT - 2004
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In NO3 ion, the number of bond pair and lone pair of electrons on nitrogen atom are respectively:

1. 2, 2 2. 3, 1
3. 1, 3 4. 4, 0
Subtopic:  Covalent Bond |
 67%
From NCERT
AIPMT - 2002
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The incorrect statements among the following are-

(a) NaCl being an ionic compound is a good conductor of electricity in the solid-state
(b) In canonical structure, there is a difference in the arrangement of atoms
(c) Hybrid orbitals form stronger bonds than pure orbitals
(d) VSEPR theory can explain the square planar geometry of XeF4
 
1. (a) and (b) only 2. (b) and (c) only
3. (c) and (d) only 4. (b) and (d) only
Subtopic:  Resonance & Nature of Compounds | V.S.E.P.R & V.B.T |
 69%
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A pair in which both species are not likely to exist is:

1. \(H^+_2,He^{2-}_2\) 2. \(H^-_2,He^{2+}_2\)
3. \(H^{2+}_2,He_2\) 4. \(H^-_2,He^{2+}_2\)
Subtopic:  M.O.T |
 82%
From NCERT
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