- Achlorobis(ethylenediamine)nitrito $-O-$ cobaltate $(III)$ ion
- Bchlorodiethyldiaminenitrito $-O-$ cobalt $(III)$ ion
- ✓chloronitrito $-O-$ diethyldiamine cobaltate $(III)$ ion
- Dchlorobis(ethylenediamine)nitrito $-O-$ cobalt $(III)$ ion
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Assertion $(A)$ : Haloalkanes react with $\mathrm{KCN}$ to form alkyl cyanides as a main product while with $\mathrm{AgCN}$ form isocyanide as the main product.
Reason $(R)$ : $\mathrm{KCN}$ and $\mathrm{AgCN}$ both are highly ionic compounds.
In the light of the above statement, choose the most appropriate answer from the options given below:
$Fe(OH)_{3(s)} \rightleftharpoons Fe^{3+}_{(aq)} + 3OH^-_{(aq)}$
is decreased by $1/4$ times, then equilibrium concentration of $Fe^{3+}$ will increase by.....$ times$
$Hb{O_2} + {H_3}{O^ + }C{O_2} \rightleftharpoons \left( {{H^ + } - Hb - C{O_2}} \right) + {O_2} + {H_2}O$
Release of $O_2$ is favoured when there is
$(I)$ It is easier to remove $2 \mathrm{p}$ electron than $2 \mathrm{s}$ electron
$(II)$ $2 \mathrm{p}$ electron of $\mathrm{B}$ is more shielded from the nucleus by the inner core of electrons than the $2 s$ electrons of $Be.$
$(III)\; 2 s$ electron has more penetration power than $2 \mathrm{p}$ electron.
$(IV)$ atomic radius of $\mathrm{B}$ is more than $\mathrm{Be}$
(Atomic number $\mathrm{B}=5, \mathrm{Be}=4$ )
The correct statements are
