Ncert Solutions Chemistry Class 12th

Nucleophilic addition of sodium hydrogen sulphite to aldehyde or ketone is a;

$NaHS{O}_3?N{a}^{+}+HS{O}_3^{?}$                

So; nucleopilic addition of sodium hydrogen sulphite to an aldehyde or a ketone involves proton transfer to form a stable ion.

Addition of hydrogen cyanide;

Final product is cyanohydrin.

BOD is biological oxygen demand which represents the amount of oxygen required to degrade organic matter in water.

Higher the BOD more polluted the water is

So; here water sample with BOD = 3 ppm is the cleanest

Spin only moment ;  = $\mu$ 3.87 BM

$\mu =\sqrt[]{n\left(n+2\right)}B.M$                

3.87 =    $\sqrt[]{n\left(n+2\right)}BM$

So, number of unpaired electrons; n = 3

ln ;

$C{r}^{3+}-4s^{0}3d^{3}n=3$

So; $M^{z+}$ cannot be V³⁺

Impurities present in electrolytic refining of blister Cu, removed as anode mud

= Sb, Se, Te, Ag, Au, Pt

Ans. = 6

Given are the oxide of alkali and alkaline earth metals which are ionic in nature.

Simple oxide are Li₂O, CaO, MgO and K₂O.

Peroxide is Na₂O₂ and superoxide is KO₂.

All simple oxides are diamagnetic as it has no unpaired electron.

Effective number of atom in C.C.P

$\therefore =\frac{1}{8}*8+\frac{1}{2}*6$     

= 4

Number of octahedral void = 4

Number of cations = 4

Number of anion = 4

Formula of compound = A₄B₄

Empirical formula = AB

Ans. x = 1

$\left(\frac{x}{m}\right)=K{P}^{1/n}$           

$\frac{10}{1}=K{\left(100\right)}^{1/n}................\left(1\right)$     

$\frac{15}{1}=K{\left(200\right)}^{1/n}................\left(2\right)$

$\frac{V}{1}=K{\left(300\right)}^{1/n}...............\left(3\right)$

Dividing (2) by (1)

$\frac{15}{10}={\left(\frac{2}{1}\right)}^{1/n}$ = (2)^1/n

$\frac{V}{10}=10^{0.279}$

$V=10*10^{0.279}=10^{1.279}=10^x$

$\therefore x=1.279=127.91*10^{-2}\approx 128*10^{-2}$

Ans. = 128

$C_2{H}_{7}N+\left(2x+\frac{y}{2}\right)CuO\to xC{O}_2+\frac{y}{2}{H}_{2}O+\frac{Z}{2}{N}_2+\left(2x+\frac{y}{2}\right)Cu$

$C_2{H}_7{N}_1+\left(2*2+\frac{7}{2}\right)CuO\to 2C{O}_2+\frac{7}{2}{H}_{2}O+\frac{1}{2}{N}_2+\left(2*2+\frac{7}{2}\right)Cu$              

$\therefore y=7$              

Ans. = 7

Kindly Consider the following Image