Class 11th

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.

(a) Producer gas is a mixture of CO and N₂ in the ratio of 2 :1

(d) CO₂ is regarded as anhydride of carbonic acid.

H₂CO₃ ————> H₂O + CO₂

(d) is an incorrect statement because non-metal oxides are acidic or neutral whereas metal oxides are basic in nature.