p block Elements Class 11 Important Questions with Answers

Ans. Electronic configuration of $\mathrm{p}$ -block elements is $n s^{2} n p^{1-6}$

Ans. The element having highest melting point is carbon (diamond).

Ans. Amorphous and Crystalline.

Ans. Tincal is an ore of boron. Its formula is $\mathrm{Na}_{2} \mathrm{B}_{4} \mathrm{O}_{10} \cdot 10 \mathrm{H}_{2} \mathrm{O}$

Ans. Boron has very small size and has very high sum of three ionisation enthalpies $\left(I E+I E_{2}+I E_{3}\right)$ therefore it cannot lose its three electrons to form $B^{3+}$ ions.

Ans. $2 \mathrm{BCl}_{3}(\mathrm{g})+3 \mathrm{H}_{2}(\mathrm{g}) \frac{\mathrm{sun}}{_{1270 \mathrm{K}}} 2 \mathrm{B}(\mathrm{s})+6 \mathrm{HCl}(\mathrm{g})$

Ans. $B F_{3}$ is weaker lewis acid than $B C l_{3}$ because of more effecttive back bonding in case of $F$ due to smaller size than $C l$

Ans. It is because in boric acid, boron does not have its octet complete. It accepts $O H^{-}$ from water.

Ans. Boron does not have vacant d-orbitals as second shell is the outer most shell.

Ans. $B(O H)_{3}$ is a weak monobasic acid. It acts as a lewis acid by accepting electrons from a hydroxy ion in water. $B(O H)_{3}+2 H O H \longrightarrow\left[B(O H)_{4}\right]^{-}+H_{3} O^{+}$

Ans. Sodium borohydride is formed $2 \mathrm{NaH}+\mathrm{B}_{2} \mathrm{H}_{6} \stackrel{\text { heat }}{\longrightarrow} 2 \mathrm{NaBH}_{4}$

Ans. Pyrex glass is a glass which is heat resistant. It can with stand high temperature.

Ans. Boron can be obtained by reduction of Boron oxide with electropositive metal like Mg. $B_{2} O_{3}(s)+3 M g(g) \longrightarrow 2 B(s)+3 M g O(s)$

Ans. Boric acid is $H_{3} B O_{3}$ Oxidation state of boron $=+3$ Boric acid is tribasic.

Ans. Boron does not have vacant $d$ -orbitals in its valence shell. Therefore, it cannot expand its octet. Therefore, maximum covalency of boron cannot exceed and it does not form $B F_{6}^{3-}$ ion.

Ans. Boron halides are electron deficient, therefore, they can form addition compounds with electron rich amines $e . g .$ $\left(\mathrm{CH}_{3}\right)_{3} \mathrm{N} \longrightarrow \mathrm{BF}_{3}$

Ans. When boric acid is heated at $370 K .$ metaboric acid $\left(H B O_{2}\right)$ is formed. On heating strongly $B_{2} O_{3}$ is formed. (i) $\mathrm{H}_{3} \mathrm{BO}_{3} \stackrel{370 \mathrm{K}}{\longrightarrow} \mathrm{HBO}_{2}+\mathrm{H}_{2} \mathrm{O}$ (ii) $2 H B O_{2} \stackrel{\text { heat }}{\longrightarrow} B_{2} O_{3}+H_{2} O$

Ans. $\mathrm{Na}_{2} \mathrm{B}_{4} \mathrm{O}_{7} \cdot 10 \mathrm{H}_{2} \mathrm{O}+2 \mathrm{HCl}+5 \mathrm{H}_{2} \mathrm{O} \longrightarrow 2 \mathrm{NaCl}$ $+4 H_{2} B Q_{3}+12 H_{3} Q$

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Ans. $=B-C l$ bond has dipole moment due to difference in electronegativity of $B$ and chlorine atom. $B C l_{3}$ on the contrary has a zero dipole moment because dipole moment is a vector quantity. The net resultant of the three bonds is zero due to its symmetrical planer triangular structure.

Ans. - In $B F_{3}$ the hybridisation of boron is $s p^{2}$ and hence the molecule has a planar triangular shape. In $B H_{4}^{-}$ the hybridisation of boron is $s p^{3}$ and hence it has a tetrahedral shape.

Ans. Electron deficient compounds are those which have incomplete octet. $B C l_{3}$ is an electron deficient compound as it has sextet of electrons. On the contrary $\mathrm{SiCl}_{4}$ is not an electron deficient compound. However, due to the presence of vacant $d$ -orbitals in its valence shell it can extend its co-ordination number and form species like $S i F_{3}^{-}, S i F_{6}^{2-},$ etc.

Ans. - The boron family exhibits two oxidation states of $+3$ and $+1$ in their compounds on moving down the group from boron to thallium, the stability of lower oxidation state increases $i . e .,+1$ oxidation state becomes more prominent due to " inert pair effect?

Ans. $B C l_{3}$ has more stability than $T I C l_{3} .$ It is because boron due to its small size can exhibit only one oxidation state of $+3$ in its compounds. On the contrary. Thallium exhibits two oxidation states of $+3$ and $+1$ in its compounds. The lower oxidation state of $+1$ is more stable because of inert pair effect. To add further the increase in the magnitude of nuclear charge is not compensated by the increase in size as a result of which the $n s$ -electrons do not participate in bonding.

Ans. It is obtained by reacting $A I C I_{3}$ with $L i H$ in the presence of dry ether Uses : It is a very good reducing agent used in many organic synthesis.

Ans. $\mathrm{GaCl}_{3}+\mathrm{NaOH} \longrightarrow \mathrm{Ga}(\mathrm{OH})_{3}+3 \mathrm{NaCl}$ $\mathrm{Ga}(\mathrm{OH})_{3}+\mathrm{NaOH} \longrightarrow \underset{\text { Souble Complex }}{\mathrm{NaGaO}_{2}}+2 \mathrm{H}_{2} \mathrm{O}$

Ans. Uses of boron compounds : (i) The main use of borax and boric acid is in the preparation of heat resistant borosilicate glass such as pyrex glass. (ii) Borax is used as flux for soldering of metals and procelain enamels. (iii) Dilute aqueous solution of boric acid is used as mild antiseptic. (iv) Boron rods are used as control rods in nuclear reactors.

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Ans. (i) $\quad 2 B+3 C l_{2} \stackrel{\text { heat }}{\longrightarrow} 2 B C l_{3}$ (ii) $\quad 4 B+3 O_{2} \stackrel{\text { heat }}{\longrightarrow} 2 B_{2} O_{3}$ (iii) $\quad 2 B+N_{2} \stackrel{\text { heat }}{\longrightarrow} 2 B N$

Ans. $B-F$ bond length in $B F_{3}$ is 130 pm because boron in is hybridised and has a planar triangular structure. As a result of deficiency of electrons on the boron atom the excessive electron density on the fluorine atoms (due to its small size and strong interelectronic repulsions) is pushed on to the boron atom $i . e .$ back donation or back bonding occurs. This introduces a partially double bond character shortening the bond. However, in the boron atom is hybridised with bond length of $143 \mathrm{pm}$

Ans. Both $B F_{3}$ and $B B r_{3}$ are electron deficient molecules due to presence of only six electrons around boron atom. Therefore, these molecules act as Lewis acids due to their tendency to accept a pair of electrons. In , the 2 p-orbital of adjoining $F$ atom containing lone pair of electrons form effective bonding with empty $2 \mathrm{p}$ -orbital of boron atom. As a result, the lone pair of $F$ is donated to $B$ atom and hence the electron-deficiency ofboron decreases. In contrast in, the size of $B r$ atom is much bigger than the $F$ atom, hence donation of lone pair of electrons of $B r$ to $B$ does not occur to any significant extent. As a result, the electron-deficiency of $B$ is much higher in $\mathrm{BBr}_{3}$ than that in $\mathrm{BF}_{3},$ and hence $\mathrm{BBr}_{3}$ is a stronger Lewis acid than $\mathrm{BF}_{3}$

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Ans. - Boric acid is $H_{3} B O_{3} .$ It can also be represented by $B(O H)_{3} .$ It has layer type structure in which $B O_{3}$ units are linked to one another through $H$ -bonds. The $H$ -atom constitutes covalent bond with one unit and $H$ -bond with other unit. Boric acid is a weak monobasic acid, it does not act as proton donor but as Lewis acid. $\mathrm{B}(\mathrm{OH})_{3}+\mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{B}(\mathrm{OH})_{4}^{-}+\mathrm{H}^{+}$ Preparation (i) By the action of $\mathrm{SO}_{2}$ on boiling suspension of colemanite in water. \[ \mathrm{Ca}_{2} \mathrm{B}_{6} \mathrm{O}_{11}+2 \mathrm{SO}_{2}+9 \mathrm{H}_{2} \mathrm{O} \longrightarrow 2 \mathrm{CaSO}_{3}+6 \mathrm{H}_{3} \mathrm{BO}_{3} \] (ii) By acidifying aqueous solution of borax with mineral acids. \[ \mathrm{Na}_{2} \mathrm{B}_{4} \mathrm{O}_{7}+2 \mathrm{HCl}+5 \mathrm{H}_{2} \mathrm{O} \longrightarrow 2 \mathrm{NaCl}+4 \mathrm{H}_{3} \mathrm{BO}_{3} \] Effect of heat. Heating orthoboric acid above $370 K$ leads to partial removal of water to yield metaboric acid, $\left(\mathrm{HBO}_{2}\right) ;$ further heating yields.