Nitrogen exists as a diatomic molecule and phosphorus as P4 because on moving
1. | up a group, the tendency to form pπ−pπ bonds decreases. |
2. | down a group, the tendency to form pπ−pπ bonds decreases. |
3. | down a group, the tendency to form dπ−dπ bonds decreases. |
4. | up a group, the tendency to form dπ−dπ bonds decreases. |
Oxygen forms hydrogen bonds while chlorine does not, because :
1. | Chlorine has a bigger size and as a result, a higher electron density per unit volume. |
2. | Oxygen has a smaller size and as a result, a lower electron density per unit volume. |
3. | Oxygen has a smaller size and as a result, a higher electron density per unit volume. |
4. | Oxygen has a bigger size and as a result, a higher electron density per unit volume. |
The incorrect statement regarding diamond and graphite is-
1. | In diamond each carbon is sp2 hybridised and in graphite, each carbon is sp3 hybridised. |
2. | Diamond is an insulator |
3. | Graphite has a planar geometry. |
4. | Both (2) and (3) |
Statement I: | Acid strength increases in the order given as HF < HCl < HBr < HI. |
Statement II: | As the size of the elements F, Cl, Br, I increases down the group, the bond strength of HF, HCI, HBr, and HI decreases and so the acid strength increases. |
1. | Statement I is correct and Statement II is incorrect |
2. | Statement I is incorrect and Statement II is correct. |
3. | Both Statement I and Statement II are correct. |
4. | Both Statement I and Statement II are incorrect. |
Boric acid is polymeric due to-
1. Its acidic nature.
2. The presence of hydrogen bonds.
3. Its monobasic nature.
4. Its geometry.
Match the compounds of Xe in Column I with the molecular structure in Column II.
Column-I | Column-II | ||
(a) | XeF2 | (i) | Square planar |
(b) | XeF4 | (ii) | Linear |
(c) | XeO3 | (iii) | Square pyramidal |
(d) | XeOF4 | (iv) | Pyramidal |
(a) | (b) | (c) | (d) | |
1. | (ii) | (i) | (iii) | (iv) |
2. | (ii) | (iv) | (iii) | (i) |
3. | (ii) | (iii) | (i) | (iv) |
4. | (ii) | (i) | (iv) | (iii) |
Assertion (A): | Both rhombic and monoclinic sulphur exist as S8 but oxygen exists as O2. |
Reason (R): | Oxygen forms pπ - pπ multiple bonds due to its small size and small bond length but pπ - pπ bonding is not possible in sulphur. |
1. | Both (A) and (R) are True and (R) is the correct explanation of (A). |
2. | Both (A) and (R) are True but (R) is not the correct explanation of (A). |
3. | (A) is True but (R) is False. |
4. | Both (A) and (R) are False. |
The condition that does not exist in the Haber process
for the manufacturing of ammonia is:
1. Pressure (around 200 ×105 Pa)
2. Temperature (700 K)
3. Catalyst such as iron oxide.
4. Presence of inert gases.
The disproportionation reaction of is as follows:
X and Y in the above reaction are:
1. H3PO2, HPO2
2. H3PO4, PH3
3. H2PO6, PO3
4. None of the above.
The reason behind carbon showing catenation property but lead does not is:
1. Due to the smaller size of C than that of Pb
2. Due to the smaller size of Pb than that of C
3. Due to the smaller ionization energy of C than that of Pb
4. Due to the inert pair effect