Block D and F Elements

Wt of Cl^- in 100 ml = 1.8 * 10^-3 gm

Mol. of Cl^- in 100 ml = $\frac{1.8*10^{-3}}{35.5}=0.0507*10^{-3}mole$  

$\therefore \left[C{l}^{-}\right]=\frac{0.0507*10^{-3}*1000}{100}=5.07*10^{-4}M$   

i.e. 0.507 milli mole in one lit required in one hr.

Coagulation value = (millimole/lit) required in one hr = 0.507

= 1 (the nearest integer)

ln $Cul,C{u}^{+}has3d^{10}4s^0,\left[n=0and\mu =0BM\right]$

ln   $\left[Cu{\left(N{H}_3\right)}_4\right]C{l}_2,C{u}^{++}has3d^{9}4s^0\left[n=1and\mu =1.73BM\right]$

$ln{O}_2^{-}\&{O_2}^{+}$ only one electron is unpaired in anti bonding MO,

Hence [n = 1 and $\mu$ = 1.73 BM]

Kl acts as a reducing agent for $C{u}^{2+}\left[2C{u}^{++}+4Kl\to C{u}_2{l}_2+l_2+4K^{+}\right]$

$M{n}^{+2}\to groupIV\left(M{n}^{+2},C{o}^{+2},Z{n}^{+2},N{i}^{2+}\right)$

$A{s}^{+3}\to groupIIB\left(A{s}^{+3},A{s}^{+5},S{b}^{+3},S{b}^{+5},S{n}^{+2},S{n}^{+4}\right)$

$\begin{array}{l}C{u}^{+2}\to groupIIA\left(C{u}^{+2},P{b}^{+2},H{g}^{+2},C{d}^{+2},B{i}^{+3}\right)\\ A{\mathcal{l}}^{+3}\to groupIII\left(F{e}^{+3},A{\mathcal{l}}^{+3},C{r}^{+3}\right)\end{array}$

Here A is CH₃ - CH₂ – CH₂ – COO – CH₂ – CH₃

The given sequence of reaction is:

$C{H}_3-C{H}_2-C{H}_2-C{H}_2-OH\underset{}{\overset{Oxidation}{\to }}C{H}_3-C{H}_2-C{H}_2-COOH$

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.

La⁺² (z = 57) = 5d¹

Ce⁺² (z = 58) = 4f²

Nd⁺² (z = 60) = 4f⁴

Yb⁺² (z = 70) = 4f¹⁴

Yb⁺² has no unpaired electrons thus diamagnetic in nature.

Non-linear and symmetrical Cr-O-Cr bond due to resonance.

One water molecule is associated with hydrogen bond.

Ans. = 1