MCQ
The most common oxidation state of an element is $-2.$ The number of electrons present in its outermost shell is
- A$4$
- B$2$
- ✓$6$
- D$8$
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$A.6{H_2}{O_{\left( s \right)}} \rightleftharpoons A.2{H_2}{O_{(s)}} + \mathop {4{H_2}{O_{(g)}}};{{K_P} = 1.6 \times {{10}^{ - 11}}} $
$B.12{H_2}{O_{\left( s \right)}} \rightleftharpoons B.7{H_2}{O_{(s)}} + \mathop {5{H_2}{O_{(g)}}};{{K_P} = 2.43 \times {{10}^{ - 13}}} $
$B.10{H_2}{O_{\left( s \right)}} \rightleftharpoons {C_{(s)}} + \mathop {10{H_2}{O_{(g)}}};{{K_P} = {{10}^{ - 30}}} $
Aqueous tension of $H_2O$ at $0\,^oC$ is given as $0.76\ torr$
$(I)$ The most effective drying agent will be $C_{(s)}$. Out of $C_{(s)}$ , $B. 7H_2O_{(s)}$ & $A. 2H_2O_{(s)}$
$(II)$ At $0\,^oC$, $A.6H_2O_{(s)}$ & $B.12H_2O_{(s)}$ will
$(III)$ If $R.H$ is less than $100\%$ in a chamber at $0\,^oC$, then none of the substance can act as deliquescent
$A.$ For 1 s orbital, the probability density is maximum at the nucleus.
$B.$ For $2 s$ orbital, the probability density first increases to maximum and then decreases sharply to zero.
$C.$ Boundary surface diagrams of the orbitals encloses a region of $100 \%$ probability of finding the electron.
$D.$ $p$ and d-orbitals have $1$ and $2$ angular nodes respectively.
$E.$ Probability density of p-orbital is zero at the nucleus.
$I$. Valence bond theory cannot explain the color exhibited by transition metal complexes
$II$. Valence bond theory can predict quantitatively the magnetic properties of transition metal complexes
$III$. Valence bond theory cannot distinguish ligands as weak and strong field ones
(Round off to the Nearest Integer).
[Use : $\sqrt{3}=1.73, h =6.63 \times 10^{-34} Js$ $m _{ e }=9.1 \times 10^{-31} kg ; c =3.0 \times 10^{8} ms ^{-1}$ $\left.1 eV =1.6 \times 10^{-19} J \right]$