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.

Subtopic:  Group 15 - Preparation, Properties & Uses |
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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.

Subtopic:  Group 16 - Preparation, Properties & Uses |
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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)

Subtopic:   Structure & Properties of Allotropes of C |
 77%
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In the light of the below statements, choose the correct answer from the options given below:
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.
 
Subtopic:  Group 17 - Preparation, Properties & Uses |
 81%
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NEET - 2021
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Boric acid is polymeric due to-

1. Its acidic nature.

2. The presence of hydrogen bonds.

3. Its monobasic nature.

4. Its geometry.

Subtopic:  Compounds of Boron- Preparations, Properties & Uses |
 72%
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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)
Subtopic:  V.S.E.P.R & V.B.T |
 78%
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NEET - 2020
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Select the correct option based on statements below:
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.
Subtopic:  Group 16 - Preparation, Properties & Uses |
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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.

Subtopic:  Group 15 - Preparation, Properties & Uses |
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The disproportionation reaction of H3PO3 is as follows:

H3PO3  X + Y

 X and Y in the above reaction are:

1. H3PO2, HPO2  

2. H3PO4, PH3

3. H2PO6, PO3

4. None of the above.

Subtopic:  Group 15 - Preparation, Properties & Uses |
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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

Subtopic:  Carbon Family - Preparations, Properties & Uses |
 73%
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