has higher boiling point than because.
(1) is a smaller molecule and hence more closely packed
(2) The bond angle in is more than in and hence the former molecules are more tightly packed
(3) Of intermolecular hydrogen bonding in liquid
(4) The latent heat of vaporisation is higher for than for
Match the compounds given in Column I with the hybridization and shape given in Column II and mark the correct option.
Column I | Column II |
A. XeF6 | 1. Distorted octahedral |
B. XeO3 | 2. Square planar |
C. XeOF4 | 3. Pyramidal |
D. XeF4 | 4. Square pyramidal |
A | B | C | D | |
1. | 1 | 2 | 4 | 3 |
2. | 4 | 3 | 1 | 2 |
3. | 4 | 1 | 2 | 3 |
4. | 1 | 3 | 4 | 2 |
A molecule that has a complete octet is:
1. BeCl2(dimer)
2. BeH2(dimer)
3. BeH2(s)
4. BeCl2(s)
The maximum number of 90° angles between bond pair-bond pair of electrons is observed in:
1. sp3d2-hybridization.
2. sp3d-hybridization.
3. dsp2-hybridization.
4. dsp3-hybridization.
The molecule/s that will have non zero dipole moments is /are :
1. 1,2-dichloro benzene.
2. Cis-3-hexene.
3. Trans-2-pentene.
4. All of the above.
Which molecule is least likely to form hydrogen bonds?
1. | \(\mathrm{NH}_3\) | 2. | \(\mathrm{H_2NOH}\) |
3. | \(\mathrm{HF}\) | 4. | \(\mathrm{CH_3F}\) |
A molecule among the following with non-zero dipole moment is:
1.
2.
3.
4.
Amongst the one with the highest boiling point is:
1. H2O because of H-bonding.
2. H2Te because of higher molecular weight.
3. H2S because of H-bonding.
4. H2Se because of lower molecular weight.
Arrange the given oxides in decreasing order of their basic character:
1. Na2O > MgO > Al2O3 > CuO
2. MgO > Al2O3 > CuO > Na2O
3. Al2O3 > MgO > CuO > Na2O
4. CuO > Na2O > MgO > Al2O3
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. | \(H^+_2,He^{2+}_2\) |