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38
Subjective

Match the species given in Column I with the hybridisation given in Column II.

Column I Column II
A. Boron in [B(OH$$_4$$)]$$^-$$ 1. sp$$^2$$
B. Aluminium in $\left[\mathrm{Al}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+}$ 2. sp$$^3$$
C. Boron in B$$_2$$H$$_6$$ 3. sp$$^3$$d$$^2$$
D. Carbon in buckminster fullerene
E. Silicon in SiO$$_4^{-4}$$
F. Germanium in $\left[\mathrm{GeCl}_6\right]^{2-}$

Explanation

A. $\rightarrow(2)$

B. $\rightarrow$ (3)

C. $\rightarrow(2)$

D. $\rightarrow$ (1)

E. $\rightarrow(2)$

F. $\rightarrow(3)$

A. Boron in $\left[\mathrm{B}(\mathrm{OH})_4\right]^{-} s p^3$ hybridised.

B. Aluminium in $\left[\mathrm{Al}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+} s p^3 d^2$ hybridised.

C. Boron in $\mathrm{B}_2 \mathrm{H}_6 s p^3$ hybridised.

D. Carbon in Buckminsterfullerene $s p^2$ hybridised.

E. Silicon in $\mathrm{SiO}_4^{4-} s p^3$ hybridised.

F. Germanium in $\left[\mathrm{GeCl}_6\right]^{2-} s p^3 d^2$ hybridised.

39
MCQ (Single Correct Answer)

Assertion (A) If aluminium atoms replace a few silicon atoms in three dimensional network of silicon dioxide, the overall structure acquires a negative charge.

Reason (R) Aluminium is trivalent while silicon is tetravalent.

A
Both $A$ and $R$ are correct and $R$ is the correct explanation of $A$
B
Both $A$ and $R$ are correct but $R$ is not the correct explanation of $A$
C
Both $A$ and $R$ are not correct
D
$A$ is not correct but $R$ is correct
40
MCQ (Single Correct Answer)

Assertion (A) Silicones are water repelling in nature.

Reason (R) Silicones are organosilicon polymers, which have $\left(-R_2 \mathrm{SiO}-\right)$ as repeating unit.

A
$A$ and $R$ both are correct and $R$ is the correct explanation of $A$
B
Both $A$ and $R$ are correct but $R$ is not the correct explanation of $A$
C
A and R both are correct
D
$A$ is not true but $R$ is true
41
Subjective

Describe the general trends in the following properties of the elements in groups 13 and 14.

(a) Atomic size

(b) Ionisation enthalpy

(c) Metallic character

(d) 0xidation states

(e) Nature of halides

Explanation

For Group 13

(a) Atomic Size On moving down the group for each successive member, one extra shell of electrons is added and therefore, atomic radius is expected to increase. However, a deviation can be seen.

Atomic radius of Ga is less than that of Al due to presence of additional 10 d - electrons, which offer poor screening effect to the outer electron.

(b) Ionisation Enthalpy The ionisation enthalpy values as expected from general trends do not decrease smoothly down the group. The decrease from B to Al is associated with increase in size.

The observed discontinued between Al and Ga and between In and TI due to low screening effect of $d$ and $f$-electrons which compensates increased nuclear charge.

(c) Metallic or Electropositive Character Boron is a semi-metal (metalloid) due to very high ionisation enthalpy. All others are metals and metallic character first increases from B to Al as size increases. From Al to TI decrease due to poor shielding of $d$ - and $f$-electrons.

(d) Oxidation States As we move down the group, the stability of +3 oxidation state decreases while that of +1 oxidation state progressively increases. In other words, the order of stability of +1 oxidation state increase in the order. $\mathrm{Al}<\mathrm{Ga}<\ln <\mathrm{TI}$. Infact, in Ga , In and TI , both +1 and +3 oxidation states are observed.

(e) Nature of Halides These elements react with halogens to form trihalids (except $\mathrm{TIl}_3$ )

$$2 E(s)+3 X_2(g) \longrightarrow 2 \mathrm{EX}_3(s) \quad[X=\mathrm{F}, \mathrm{Cl}, \mathrm{Br}, \mathrm{I}]$$

Boron in halides are electron deficient molecules and behave as Lewis acids. The Lewis character decreases in the order: $\mathrm{BI}_3>\mathrm{BBr}_3>\mathrm{BCl}_3>\mathrm{BF}_3$

For Group 14

(a) Atomic Size There is considerable increase in covalent radius from C to Si thereafter from Si to Pb as small increase in radius is observed. This is due to the presence of completely filled $d$ and $f$-orbitals in heavier member.

(b) Ionisation Enthalpy The first ionisation enthalpy of group 14 members is higher than the corresponding members of group 13. The influence of inner core electrons is visible here. In general the ionisation enthalpy decreases down the group. Small decrease in $\Delta$; H from Si to Ge to Sn and slight increase in $\Delta ; \mathrm{H}$ from Sn to Pb is the consequence of poor shielding effect of intervening $d$ and $f$-orbitals and increase in size of the atom.

(c) Metallic Character Metallic character increases down the group C (non-metal) $\mathrm{Si}, \mathrm{Ge}$ (metalloid) $\mathrm{Sn}, \mathrm{Pb}$ (metals).

(d) Oxidation States The group 14 elements have four electrons in the outermost shell. The common oxidation states exhibited by these elements are +4 and +2 . Carbon also exhibits negative oxidation states. Since, the sum of the first four ionisation enthalpies is very high, compounds in +4 oxidation states are generally covalent in nature. In heavier members the tendency to show +2 oxidation state increases in the $\mathrm{Ge}<\mathrm{Sn}<\mathrm{Pd}$ due to inert pair effect.

(e) Nature of Halides These elements can form halides of formula $M X_2$ and $M X_4$ (where, $X=F, C l, B r, I)$. Except carbon, all other members react directly with halogen under suitable condition to make halides. Most of $M X_4$ are covalent with $s p^3$ hybridisation and tetrahedral in structure. Exceptions are $\mathrm{SnF}_4$ and $\mathrm{PbF}_4$ which are ionic in nature. Heavier members Ge to Pb are able to make halides of formula $M X_2$. Stability of halides increases down the group.

42
Subjective

Account for the following observations.

(a) $\mathrm{AlCl}_3$ is a Lewis acid.

(b) Though fluorine is more electronegative than chlorine yet $\mathrm{BF}_3$ is a weaker Lewis acid than $\mathrm{CI}_3$.

(c) $\mathrm{PbO}_2$ is a stronger oxidising agent than $\mathrm{SnO}_2$.

(d) The +1 oxidation state of thallium is more stable than its +3 state.

Explanation

(a) In $\mathrm{AlCl}_3, \mathrm{Al}$ has only six electrons in its valence shell. It is an electron deficient species. Therefore, it acts as a Lewis acid (electron acceptor).

(b) In $\mathrm{BF}_3$ boron has a vacant $2 p$-orbital and fluorine has one $2 p$-completely filled unutilised orbital. Both of these orbitals belong to same energy level therefore, they can overlap effectively and form $p \pi-p \pi$ bond. This type of bond formation is known as back bonding.

While back bonding is not possible in $\mathrm{BCl}_3$, because there is no effective overlapping between the $2 p$-orbital of boron and $3 p$-orbital of chlorine. Therefore, electron deficiency of B is higher in $\mathrm{BCl}_3$ than that of $\mathrm{BF}_3$. That's why $\mathrm{BF}_3$ is a weaker Lewis acid than $\mathrm{BCl}_3$.

(c) In $\mathrm{PbO}_2$ and $\mathrm{SnO}_2$, both lead and tin are present in +4 oxidation state. But due to stronger inert pair effect, $\mathrm{Pb}^{2+}$ ion is more stable than $\mathrm{Sn}^{2+}$ ion. In other words, $\mathrm{Pb}^{4+}$ ions i.e., $\mathrm{PbO}_4$ is more easily reduced to $\mathrm{Pb}^{2+}$ ions than $\mathrm{Sn}^{4+}$ ions reduced to $\mathrm{Sn}^{2+}$ ions. Thus, $\mathrm{PbO}_2$ acts as a stronger oxidising agent than $\mathrm{SnO}_2$.

(d) $\mathrm{TI}^{+}$is more stable than $\mathrm{TI}^{3+}$ because of inert pair effect.