Class 11th

46. Let a be the first term and r be the common ratio of the G.P.

Then $a_1+a_2+a_3=16$

$a+ar+a{r}^2+16^{}$

$a\left(1+r+r^2\right)=16$ ………. I

Also $a_4+a_5+a_6=128$

$a{r}^3+a{r}^4+a{r}^5=128$

$a{r}^3\left(1+r+r^2\right)=128$ …… II

Dividing II and I we get,

$\frac{a{r}^3\left(1+r+r^2\right)}{a\left(1+r+r^2\right)}=\frac{128}{16}$

r³= 8

r³ = 2³

 r = 2 >1

So, putting r = 2 in I we get,

$a\left(1+2+2^2\right)=16$

$\Rightarrow$ $a\left(1+2+4\right)=16$

$\Rightarrow$ $a\left(7\right)=16$

$\Rightarrow$ $a=\frac{16}{7}$

And $s_x=\frac{a\left(r^n-1\right)}{r-1}$

$\Rightarrow$ $\frac{\frac{16}{7}\left(2^n-1\right)}{2-1}$

$\Rightarrow$ $\frac{16}{7}\left(2^n-1\right)$

The molecules of CO₂ are held together by weak van der Waals forces of attraction which can be easily overcome by collisions of the molecules at room temperature. Consequently, CO₂ is a gas.
On the other hand, silicon atom forms four single covalent bonds with O-atom which are tetrahedrally arranged and form a three-dimensional structure. Thus, SiO₂ is a high melting solid.

Carbon dioxide can be obtained as a solid in the form of dry ice by allowing the liquified CO₂ to expand rapidly.
Unlike ordinary ice it does not melt and hence does not wet the surface on which it is kept. Thus, it is called dry ice.

45. Given, a=3

$r=\frac{3^3}{3}=3$>1

If s_n=120

Then $\frac{a\left(r^n-1\right)}{r-1}=120$

$\Rightarrow$ $\frac{3\left(3^n-1\right)}{3-1}=120$

$\Rightarrow$ $\frac{3\left(3^n-1\right)}{2}=120$

$\Rightarrow$ $3^n-1=120*\frac{2}{3}$

$\Rightarrow$3ⁿ= 80+1

$\Rightarrow$ 3ⁿ= 81

= 3ⁿ = 34

$\therefore r=4$

So, 120 is the sum of first 4 terms of the G.P

The catenation depends on the strength of the element-element bond. As we move down the group 14, the element-element bond energies decrease rapidly, viz. C–C (355 kJ mol–1), Si–Si (222 kJ mol–1), Ge–Ge (167 kJ mol–1) and Sn–Sn (155 kJ mol–1), so the tendency for catenation decreases in the order C > Si > Ge > Sn.

Carbon atoms have the tendency to linkwith one another through covalent bonds to form chains and rings. This property is calledcatenation. This is because C—C bonds are very strong.

44. Let the three terms of the G.P be $\frac{a}{r},a,ar.$

So, $\frac{a}{r}a.ar=1$

= $a^3=1$

= $a=1$

And $\frac{a}{r}+a+ar=\frac{39}{10}$

= $\frac{1}{r}+1+r=\frac{39}{10}$ (as a = 1)

= $\frac{1+r+r^2}{r}=\frac{39}{10}$

= $\left(r^2+r+1\right)*10=39*r$

= $10r^2+10r+10=39r$

= $10r^2+10r-39r+10=0$

= $10r^2-29r+10=0$

= $10r^2-25r-4r+10=0$

= $5r\left(2r-5\right)-2\left(2r-5\right)=10$

$=\left(2x-5\right)\left(5r-2\right)=0$

= $\left(2x-5\right)=0$ or $\left(5r-2\right)=0$

= $r=\frac{5}{2}$ or $r=\frac{2}{5}$

When $a=1,$ $r=\frac{5}{2}$ the terms are,

 $\frac{1}{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.},1,$ $1*\frac{5}{2}=\frac{2}{5},1,\frac{5}{2}$

When $a=1,$ $r=\frac{2}{5}$ the terms are,

$\frac{1}{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$5$}\right.},$ 1, $1*\frac{2}{5}=\frac{5}{2},$ 1, $\frac{2}{5}$

Boron shows anomalous behaviour due to small atomic size, high electronegativity, high ionization energy and absence of d-orbital of B

  (b) SiO₄^4- is the basic unit of all silicates.