- 1(current)
Q. No.Which of the following explains why boron is unable to form the BF₆³⁻ ion?
1. Non-availability of s orbitals
2. Non-availability of d orbitals
3. Non-availability of f orbitals
4. None of the above
The product formed by the reaction of diborane and ammonia in the presence of heat is:
1.
2.
3.
4.
The geometry of a complex species can be understood from the knowledge of the type of hybridization of orbitals of the central atom. The hybridization of orbitals of a central atom in and the geometry of the complex are respectively
| 1. | sp3, tetrahedral | 2. | sp3, square planar |
| 3. | sp3d2, octahedral | 4. | dsp2, square planar |
\(H_{3}BO_{3}\xrightarrow[>370 K]{\Delta } A \xrightarrow[]{\Delta }B\)
1. \(A = HBO_2 , ~B= B_2O_3\)
2. \(A = BH_3 , ~~B= B_2O_3\)
3. \(A =B_2O_3, ~B = HBO_2\)
4. \(A =BH_3, ~B = HBO_2\)
A compound X, of boron, reacts with NH3 on heating to give another compound Y which is called inorganic benzene. The compound X can be prepared by treating BF3 with lithium aluminum hydride. The compounds X and Y are represented by the formulas.
1. B2H6, B3N3H6
2. B2O3, B3N3H6
3. BF3, B3N3H6
4. B3N3H6, B2H6
| List-I (Compounds) |
List-II (Molecular formula) |
||
| (a) | Borax | (i) | NaBO2+B2O3 |
| (b) | Kernite | (ii) | Na2B4O7.4H2O |
| (c) | Orthoboric acid | (iii) | H3BO3 |
| (d) | Borax bead | (iv) | Na2B4O7.10H2O |
Choose the correct answer from the options given below :
| (a) | (b) | (c) | (d) | |
| 1. | (iv) | (ii) | (iii) | (i) |
| 2. | (ii) | (iv) | (iii) | (i) |
| 3. | (iii) | (i) | (iv) | (ii) |
| 4. | (i) | (iii) | (iv) | (ii) |
Boric acid is an acid because its molecule
1. Contains replaceable H+ ion
2. Gives up a proton
3. Accepts OH- from water releasing proton
4. Combines with proton from water molecule
Which of the following is true regarding the structure of diborane?
| 1. | All hydrogen atoms lie in one plane and boron atoms lie in a plane perpendicular to this plane. |
| 2. | 2 boron atoms and 4 terminal hydrogen atoms lie in the same plane and 2 bridging hydrogen atoms lie in the perpendicular plane. |
| 3. | 4 bridging hydrogen atoms and boron atoms lie in one plane and two terminal hydrogen atoms lie in a plane perpendicular to this plane. |
| 4. | All the atoms are in the same plane. |
\(BCl_3 \) accepts a lone pair of electrons from \(NH_3 \) to form y. The geometry of y is different from that of \(BCl_3 \). Select the correct option for y and change in geometry from \(BCl_3 \) to y :
1. \(y = BCl_3.NH_3 ~; \) planar to tetrahedral.
2. \(y = BCl_3.NH_3 ~; \) planar to square planar.
3. \(y = B(NH_3)Cl ~Cl_2 ~; \) planar to tetrahedral.
4. \(y = B(NH_2Cl_3) ~HCl ~; \) planar to square planar.
- 1(current)
Q. No.
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