Chemistry

Ba has outer electronic configuration 6s².

CaC₂O₄ is insoluble in water.

Compound of Li are covalent so soluble in organic solvent.

Na forms strong monoacidic base.

So; A is H₃PO₂ & B is CH₃ – CH₂ – Cl

Sulphide ion (s^2-) form ores commonly with Pb & Ag as PbS and Ag₂S

3-Ethylhex-3-ene will not show stereo isomerism

Methylhex-1-ene will show steroisomerism (optical isomerism)

Since it has chiral carbon.

3, 4-Dimethylhex-3-ene will show stereoisomerism (Geometrical isomerism)

4-Methylhex-1-ene will show stereoisomerism (optical) since it has one chiral carbon

$2NaB{H}_4+l_2\to 2Nal+H_2+B_2{H}_6$

Diborane is produced when NaBH₄ is reacted with I₂

Both boron atoms are sp³ hybridised & molecule is non-planar. Diborane as two bridged 3 centre-2-electron bonds.

Compound A is phenol since phenol gives dark green colour with FeCl₃.

(A) $N{i}^{2+}-4s^{0}3d^8$ - Paramagnetic & coloured

$M{n}^{7+}-4s^{0}3d^0$       - Diamagnetic & colourless

$H{g}^{2+}-6s^{0}4f^{14}5d^{10}$ - Diamagnetic & colourless

(B)  $C{u}^{+}-4s^{0}3d^{10}$    - Diamagnetic & colourless

$Z{n}^{2+}-4s^{0}4d^{10}$          - Diamagnetic & colourless

$M{n}^{4+}-4s^{0}3d^3$    - paramagnetic & coloured

(C)  $S{c}^{3+}-4s^{0}3d^0$

$\begin{array}{l}{V}^{5+}-4s^{0}3d^0\\ T{i}^{4+}-4s^{0}3d^0\end{array}$  

All are diamagnetic & colourless

(D)  $C{u}^{2+}-4s^{0}3d^9$  

$\begin{array}{l}C{r}^{3+}-4s^{0}3d^3\\ S{c}^{+}-4s^{1}3d^1\end{array}$

All are paramagnetic & coloured

All d- & f block paramagnetic cations are coloured also.

  $C{H}_3-C{H}_2-C{H}_2-C-N{H}_2\to \pi -\mathcal{l}pconjugation$  

              $C{H}_3-C{H}_2-CH=CH-C{H}_2-N{H}_2\to$ No conjugation

          $C{H}_3-C{H}_2-O-CH=C{H}_2\to \pi -\mathcal{l}p$     conjugation