Chemistry NCERT Exemplar Solutions Class 11th Chapter Eleven

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B

The four terminal hydrogen atoms and two boron atoms are all in the same plane.

There are two bridging hydrogen atoms above and below this plane.

The four terminal B-H bonds are regular two-centre-two-electron bonds, whereas the two bridge (B-H-B) bonds are unique and can be described as three-centre-two-electron bonds, as shown in figure:

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C

Me₃SiCl is a soleuless liquid That has a major application in the formation of silicones. On its addition in the process, it blocks the end and controls the chain length of the polymer.

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B

In group 14, as we move down from C to Sn, the size increases and electronegativity decreases. This results in the decrease in M-M bond energy. Hence the catenation property decreases.

Bond energy is the amount of energy required to fragment the atoms combined in a molecule are born into separate atoms. Carbon has maximum bond energy as compared to its other group elements. Therefore, shows maximum catenation property.

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C

Boric acid is Lewis acid, having six electrons in its valence shell. It combines with water, accepts electrons

from OH⁻of water molecule and complete it octet to 8 and releases H⁺

Reaction: B (OH)₃ + OH − H→ [B (OH)₄]⁻ + H⁺

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A

Number of items/molecules bonded to the central atom is termed as coordination number. In the given MF₆³⁻

Coordination number of metal is six and boron can have topmost coordination number of four as it consists of s and p orbital only and lacks d-orbital. Therefore, boron cannot be compared in the form of MF₆³⁻.

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A

The B₂O₃ reacts with water to form boric acid hence it is an acidic oxide. As we move down the group electronegativity decreases and the tendency to donate electrons increases therefore basic character increases or acidic character decreases. So, from the given options, the most acidic oxide is B₂O₃.

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A

Central atom in [B (OH)₄ ]⁻ no lone pair. Electronic conf In ground state: 1s² 2s² 2px¹ 2py⁰ 2pz⁰

In excited state, One Electron from 2s shifts to 2py  orbital and the configuration becomes:

1s² 2s² 2p¹ x² p¹ y² pz⁰ 

Now, one s and three p orbitals combined to give sp³ hybridisation and tetrahedral shape.

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A

Lewis acids are the species in which the state is not complete and ready to accept electrons. Because Al is surrounded by 6 electrons in AlCl₃ and all three Cl atoms are surrounded by 8 electrons, AlCl₃ is an electron acceptor. It is a covalent compound.