Although +3 oxidation state is most common in lanthanoids, cerium still shows +4 oxidation state because:
After losing one more electron, it acquires $4 f^{14}$ electronic configuration.
Its nearest inert gas is Radon.
Its atomic number is 61 .
After losing one more electron, it acquires $4 f^0$ electronic configuration.
$$ \text { In the following reaction sequence, } \mathrm{X} \text { and } \mathrm{Z} \text { respectively are : } $$


$$ \mathrm{X}=\mathrm{POCl}_3 ; \mathrm{Z}=\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2-\mathrm{Br} $$

$$ \mathrm{X}=\mathrm{H}_3 \mathrm{PO}_3 ; \mathrm{Z}=\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2-\mathrm{Br} $$
$$ \text { Match List I with List II : } $$
| List I (Complex/ion) |
List II (Shape/geometry) |
||
|---|---|---|---|
| A. | $$ \left[\mathrm{Pt}\left(\mathrm{Cl}_2\right)\left(\mathrm{NH}_3\right)_2\right] $$ |
I. | Octahedral |
| B. | $$ \left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right] \mathrm{Cl}_3 $$ |
II. | Trigonal bipyramidal |
| C. | $$ \left[\mathrm{NiCl}_4\right]^{2-} $$ |
III. | Square planar |
| D. | $$ \left[\mathrm{Fe}(\mathrm{CO})_5\right] $$ |
IV. | Tetrahedral |
A-III, B-IV, C-I, D-II
A-III, B-I, C-IV, D-II
A-IV, B-I, C-III, D-II
A-I, B-III, C-IV, D-II
The functional group that can be identified through phthalein dye test is :
Aldehyde
Phenolic
Carboxylic acid
Alcohol
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