Class 12th

K₂Cr₂O₇ + H₂O₂ + H₂SO₄->  + K₂SO₄ + H₂O compound 'X' is ->CrO₅

Oxidation state of chromium = +6.

A. Vitamin B₁₂– Co

B.  Wilkinson catalyst –Rh ( [Rh (PPh₃)₃Cl])

C.  Ziegler-Natta catalyst –Ti (TiCl₄ +Al (C₂H₅)₃)

D.  Haemoglobin – Fe

Ionisation enthalpy increases in a period. Z dominates over screening effect (s) in a period as Z_eff. increases.

Na+ C + N + S ®NaSCN

Fe³⁺ + SCN^– $\stackrel{}{\to }$ $\underset{Bloodred}{{\left[Fe\left(SCN\right)\right]}^{2+}}$

Potassium permanganate in alkaline medium oxidise lodide to lodate.

$2Mn{O}_4^{-}+H_{2}O+I^{\Theta }\stackrel{}{\to }2Mn{O}_2+2O{H}^{\Theta }+I{O}_3^{\Theta }(A)$            

Compound A is  . Therefore, oxidation state of 1is + 5.

From structure it is clear [CO₂ (CO)₂] has bridging carbonyl ligand.

In presence of light allylic substitution occur.

In presence of CCl₄, addition reaction will occur.

KMnO₄ decomposes upon heating at 513 K and forms K₂MnO₄ and MnO₂.

2KMnO₄  K₂MnO₄ + O₂ + MnO₂

  $\overrightarrow{a}+5\overrightarrow{b}$  is collinear with $\overrightarrow{c}$  

⇒   $\overrightarrow{a}+5\overrightarrow{b}$ = $\overrightarrow{c}$           …(1)

$\overrightarrow{b}+6\overrightarrow{c}$ is collinear with $\overrightarrow{a}$  

⇒   $\overrightarrow{b}+6\overrightarrow{c}=\mu \overrightarrow{a}$               …(2)

From (1) and (2)

  $\overrightarrow{b}+6\overrightarrow{c}=\mu \left(\lambda \overrightarrow{c}-5\overrightarrow{b}\right)$          

-> $\left(1+5\mu \right)\overrightarrow{b}+\left(6-\lambda \mu \right)\overrightarrow{c}=0$

$?\overrightarrow{b}$ and $\overrightarrow{c}$  are non-collinear

-> 1 + 5m = 0 $\mu =\frac{-1}{5}$  and 6 – lm = 0 Þ lm = 6

-> l = – 30

Now,

$\overrightarrow{b}=6\overrightarrow{c}=\frac{-1}{5}\overrightarrow{a}$

$5\overrightarrow{b}+30\overrightarrow{c}=-\overrightarrow{a}$

$\begin{array}{cc}\hfill \overrightarrow{a}+5\overrightarrow{b}+30\overrightarrow{c}=0\hfill & \hfill \overrightarrow{a}+\alpha \overrightarrow{b}+\beta \overrightarrow{c}=0\hfill \end{array}]$

On comparing

α = 5, β = 30 ⇒ α + β = 35