Calculate the overall complex dissociation equilibrium constant for the [Cu(NH_{3})_{4}]^{2+} ions, given that stability constant (${\mathrm{\beta}}_{4}$) for this complex is 2.1 $\times $ 10^{13}.
8.27 $\times $ 10^{-13}
4.76 $\times $ 10^{-14}
2.39 $\times $ 10^{-7 }
1.83 $\times $ 10^{14}
B.
4.76 $\times $ 10^{-14}
Dissociation constant is the reciprocal of the stability constant $\left(\mathrm{\beta}=\frac{1}{\mathrm{K}}\right)$
Overa;; complex dissociation equilibrium constant,
K = $\frac{1}{{\mathrm{\beta}}_{4}}$
= $\frac{1}{2.1\times {10}^{13}}=4.76\times {10}^{-14}$
For the reaction, H_{2} + I_{2} $\rightleftharpoons $ 2HI, K = 47.6. If the initial number of moles of each reactant and product is 1 mole then at equilibrium
[I_{2}] = [H_{2}] , [I_{2}] > [HI]
[I_{2}] = [H_{2}] , [I_{2}] < [HI]
[I_{2}] < [H_{2}] , [I_{2}] = [HI]
[I_{2}] > [H_{2}] , [I_{2}] = [HI]
B.
[I_{2}] = [H_{2}] , [I_{2}] < [HI]
For the given reaction, K = $\frac{[\mathrm{HI}{]}^{2}}{\left[{\mathrm{H}}_{2}\right]\left[{\mathrm{I}}_{2}\right]}$
As 1 mole of H_{2} reacts with 1 mole of I_{2}, even at equilibrium, [H_{2}] = [I_{2}]
Hence, K = $\frac{[\mathrm{HI}{]}^{2}}{[{\mathrm{I}}_{2}{]}^{2}}$ or $\sqrt{\mathrm{K}}=\frac{\left[\mathrm{HI}\right]}{\left[{\mathrm{I}}_{2}\right]}=\sqrt{47.6}$
i.e. [HI] > [I_{2}]
K_{a} for HCN is 5 x 10^{-10 }at 25°C. For maintaining a constant pH = 9, the volume of 5 M KCN solution required to be added to 10 mL of 2 M HCN solution is
4 mL
2.5 mL
2 mL
6.4 mL
C.
2 mL
pH = pK_{a} + log $\frac{\left[\mathrm{Salt}\right]}{\left[\mathrm{Acid}\right]}$
= pK_{a} + log $\frac{\left[\mathrm{KCN}\right]}{\left[\mathrm{HCN}\right]}$ ...(i)
Let the volume of KCN solution required be V mL
$\therefore $ [KCN] = $\frac{5\times \mathrm{V}}{\mathrm{V}+10}$ and [HCN] = $\frac{10\times 2}{\mathrm{V}+10}$
Now from equation (i),
pH = -log (5 $\times $ 10^{-10}) + log $\left[\frac{5\times \mathrm{V}}{\mathrm{V}+10}/\frac{10\times 2}{\mathrm{V}+10}\right]$
9 = -log (5 $\times $ 10^{-10}) + log $\frac{\mathrm{V}}{4}$
On solving, V = 1.99 $\approx $2mL.
The AsF_{5} molecule is trigonal bipyramidal. The hybrid orbitals used by the As atoms for bonding are
d_{x2-y2}, d_{z2}, s, p_{x}, p_{y}
d_{xy}, s, p_{x}, p_{y}, p_{z}
d_{x2-y2}, s, p_{x}, p_{y}
s, p_{x}, p_{y}, p_{z}, d_{z2}
D.
s, p_{x}, p_{y}, p_{z}, d_{z2}
AsF_{5} has sp^{3}d hybridisation. In sp^{3}d hybridisation, the d_{z2} orbital is used along with the 's' and three 'p' orbitals to form three equatorial bonds and two equally strong axial bonds for a trigonal bipyramid.
Ionisation potential of hydrogen atom is 13.6 eV. Hydrogen atom in ground state is excited by monochromatic light of energy 12.1 eV. The spectral lines emitted by hydrogen according to Bohr's theory will be
one
two
three
four
C.
three
Since ionisation potential of hydrogen atom is 13.6 eV.
$\therefore $E_{1} = -13.6 eV
Now, E_{n} - E_{1} = $\frac{-13.6}{{\mathrm{n}}^{2}}$-(-13.6) = 12.1
$\frac{-13.6}{{\mathrm{n}}^{2}}$+13.6 = 12.1
Therefore, n = 3
After absorbing 12.1 eV, the electron of H-atom is excited to 3^{rd} shell.
Thus, possible transitions are 3 i.e., 3 $\to $ 2, 2 $\to $ 1 and 3 $\to $ 1.
Which of the following electronic configurations has maximum energy?
A.
Among all the given option, option a has maximum energy as the two electrons are in excited state.
The order of increasing energy of the sub-atomic orbitals is s < p < d < f. The energy in excited state is more than that in the ground state.
Which of the following oxoacids of phosphorus is a reducing agent and a monobasic acid as well?
H_{4}P_{2}O_{5}
HPO_{3}
H_{3}PO_{3}
H_{3}PO_{2}
D.
H_{3}PO_{2}
Acids which contain P-H bonds have strong reducing properties. Hypophosphorus acid (H3PO_{2}) is a good reducing agent as it contains two P-H bonds. It has one P-OH bond, so it is monobasic.
The density of a gas A is thrice that of a gas B at the same temperature. The molecular weight of gas B is twice that of A. What will be the ratio of the pressures acting on B and A?
$\frac{1}{4}$
$\frac{7}{8}$
$\frac{2}{5}$
$\frac{1}{6}$
D.
$\frac{1}{6}$
$\frac{d}{p}=\frac{M}{RT}$
Let density of gas B be d.
$\therefore $ density of gas A = 3d
And let molecular weight of A be M.
$\therefore $ molecular weight of B = 2M
Since, R is a gas constant and T is same for both gases,
p_{A} = $\frac{{\mathrm{d}}_{\mathrm{A}}\mathrm{RT}}{{\mathrm{M}}_{\mathrm{A}}}$ and p_{B }= $\frac{{\mathrm{d}}_{\mathrm{B}}\mathrm{RT}}{{\mathrm{M}}_{\mathrm{B}}}$
$\frac{{\mathrm{p}}_{\mathrm{B}}}{{\mathrm{p}}_{\mathrm{A}}}=\frac{{\mathrm{d}}_{\mathrm{B}}}{{\mathrm{d}}_{\mathrm{A}}}\times \frac{{\mathrm{M}}_{\mathrm{A}}}{{\mathrm{M}}_{\mathrm{B}}}=\frac{\mathrm{d}}{3\mathrm{d}}\times \frac{\mathrm{M}}{2\mathrm{M}}=\frac{1}{6}$
Therefore, $\frac{1}{6}$ is the ration of the pressures acting on B and A.
Hydride of boron occurs as B_{2}H_{6} but B_{2}Cl_{6} does not exist. This is because
p$\mathrm{\pi}$ - d$\mathrm{\pi}$ back bonding is possible in B_{2}H_{6} but not in B_{2}Cl_{6}
boron and hydrogen have almost equal values of electronegativity
boron and chlorine have almost equal atomic sizes
small hydrogen atoms can easily fit in between boron atoms but large chlorine atoms do not.
D.
small hydrogen atoms can easily fit in between boron atoms but large chlorine atoms do not.
Hydride of boron occurs as B_{2}H_{6} but B_{2}Cl_{6} does not exist because small hydrogen atoms can easily fir in between boron atoms but large atoms such as chlorine do not fit easily.
The temperature dependence of a reaction is represented by the Arrhenius equation:
ln k = -$\frac{{\mathrm{E}}_{\mathrm{a}}}{\mathrm{RT}}+\mathrm{ln}\mathrm{A}$
Which among the following is wrong conclusion about the given plot?
Intercept of the line= ln A
Slope = -$\frac{{\mathrm{E}}_{\mathrm{a}}}{\mathrm{RT}}$
Reaction with high activation energy is more temperature sensitive than that of low activation energy (E_{a}).
Slope = - $\frac{{\mathrm{E}}_{\mathrm{a}}}{\mathrm{R}}$
B.
Slope = -$\frac{{\mathrm{E}}_{\mathrm{a}}}{\mathrm{RT}}$
In the above given figure, option b is incorrect. The slope of the figure is
Slope = -$\frac{{\mathrm{E}}_{\mathrm{a}}}{\mathrm{R}}$
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